ADAPTOR = Allow Difference Aplication TO Re stability and ilustration technic from Lord for Matthew 8 : 23 - 27 AMNIMARJESLOW AL DO FOUR DO AL ONE LJBUSAF thankyume orbit

  
                              ADAPTOR  =  Allow Difference Aplication TO Re stability
                               and ilustration technic from Lord for   Matthew 8 : 23 - 27


       


Adaptor is a plane / electronic equipment that serves to change the power from the AC power voltage (back and forth) into a DC voltage (direction) a large / high voltage in accordance with the needs of electronic equipment used.   

Adapter Parts
1. Input Voltage Parts: the part that receives the voltage from the voltage source. The components inside the input voltage:• Plug (St)• Switch (S)1. On-Off switch: switch on and off the input voltage plane2. Voltage Switch / Voltage Transmitter (Transformer): lower the voltage from 220V to 110V / move the voltage from 110V to 220V to adjust the available PLN power supply.• LED (Indicator Light): checks whether or not the utility voltage is present• Fuses: as a safety, to avoid damage to aircraft due to overcurrent voltage.
2. Voltage Reduction Section: the part that lowers the AC power voltage is too large (220V-110V) into AC power voltage according to the needs of the equipment used. Components that lower the voltage:• Voltage transformer (Step-Down Transformer). The output voltage chosen by the transformer is fixed.• Rotate Switch: To convert the fixed output voltage to a high voltage as needed. 

3. Voltage Rectifier Parts: the part that converts AC mains voltage to DC voltage. However, it is still a pulse of DC voltage / pulse. The types:• Half-wave Rectifier• Full-wave Rectifier with CT Transformer• Full-wave Rectifier with Bridge
4. Filter Section (Filter): the part that flatten the DC pulse / voltage pulse so that more stable and flat. Filter types:• Filter Condensator (C)• Condensator-Lens-Condensator (CLC) Filter• Condensator-Resistor-Condensator (CRC) Filter• Filter Condensator-Resistor-Condensator (CRC) with transistors
5. Output Voltage (Output Voltage): parts that are directly related to the electronic equipment used. The components:• Crocodile cap / jack• LED (pilot lamp): the advantages: LED output function to know the out / absence of power supply voltage even though there is no electric current. Meanwhile, the LED input only serves as an indication of the entry / absence of electrical voltage.  



illustration matching situational  :


MATTHEW 8: 23-27

23 And he got into the boat, and his disciples followed him.

24 And suddenly the wind stormed in the lake, so that the boat was taken with a wave, but Jesus was asleep.

25 Then His disciples came and woke Him, saying, "Lord, help us perish."

26 He said to them, "Why are you afraid, you who lack faith?" And he arose and rebuked the wind and the lake, and the lake became very calm.

27 And the men were amazed, saying, "What kind of man is this, that even the winds and lakes obey him?" (Mark 4: 35-41; Luke 8: 22-25; Acts 27: 22-34; )   

LORD JESUS TO RE STABILITY WAVE UNTIL TO BE  QUIET

      


                                                                 X  .  I    
                                                    Adapter resource set   


                             


 adaptor to transformer  ; 

Travel Power Adapters: How to Choose

Are you preparing to travel internationally and want to take items that require electricity? In most cases, you’ll need only an adapter plug; in some cases, you'll need a voltage converter or transformer, too. 

   

Adapter Plugs

If you’re traveling outside of North America, you’ll most likely need an adapter plug. All over the world, there are different types of electrical wall outlets. Unless your destination country has the same outlet configuration as your home country, you’ll need an adapter. Adapter plugs do not convert electricity, they simply allow your device’s plug to fit into the foreign outlet.
Here are some common plug types:

The chart below lists suitable adapter plug models for the most common travel destinations.

Adapter Plug Quick Guide

Common Destinations	Outlet Type
Japan, Taiwan, Central America, Caribbean, South America	A,B
Europe, Middle East, Israel, some Asian countries, some African countries	C,E,F
United Kingdom, Ireland, Hong Kong, Singapore, Malaysia, some African countries	G
China, Australia, New Zealand, Fiji	I

For a complete list of countries and their power supply and outlet types, see our companion REI Expert Advice article, Voltage and Outlets by Country.

Electricity Converters and Transformers

If you’re traveling with certain devices, such as older hair dryers and irons, you may also need a voltage converter or transformer.
The world runs on two types of electricity: 110/125V or 220/240V. North American devices run on 110/125V electricity while the majority of the world runs on 220/240V. Converters and transformers change the voltage of electricity to match the voltage of your device.

How to Determine if You Need a Transformer or Converter

The label on your device will help determine if a voltage converter or transformer is necessary. This label may be: a) affixed directly to the back of the device; b) on the AC transformer box of the power supply lead; or c) molded into the plastic on the plug. It is often in very small print.
The INPUT line contains the key information—whether the voltage (V) is single, dual or multi.
Single-voltage items have a small voltage range (such as 100–120V). These small ranges are designed to accommodate voltage fluctuations only and will not accommodate a 220V power supply. Single-voltage devices include older appliances, such as hair dryers and irons.
Dual-voltage devices use a slash to separate the 2 voltages. Example: 120V/240V. Common dual-voltage devices include newer hair dryers, electric shavers and toothbrushes, irons, coffee makers and tea kettles. These do not require a transformer or converter.
Multi-voltage items use a dash to indicate the range of voltages. Example: 100–240V. Common multi-voltage devices include laptops, e-readers, tablets, smartphones, cell phones, MP3 players, cameras and battery chargers. These do not require a transformer or converter.

Converter or Transformer?

Single-voltage electrical devices (ones that use heating elements or mechanical motors) can use a converter or a transformer.
Single-voltage electronic devices (ones that use chips, circuits or electronic motors) require a transformer.
Good news: Many converters operate as both a converter for high-watt electrical devices and a transformer for low-watt electronic devices.
Watts (W) is the amount of power a device uses. Low watts range up to 25W or 50W, depending on the converter. This would be typical of small personal electronics. Electrical heating units will require a “high” setting as they may consume 1000W to 2000W.
Make sure to check the product specifications on your devices and make certain that your converter is rated for the specified power.

Device Conversion Chart

If your device is rated for a single voltage (such as 110V), and this is different than the power supply at your destination (such as 220V), you will need a voltage converter or transformer.

Device and Type of Voltage (INPUT)	Power Supply in Destination Country	Converter Needed?	Transformer Needed?
Electrical, single: 110, 115, 120, 125V	220, 230, 240V	Yes, or a transformer	Yes, or a converter
Electronic, single: 110, 115, 120, 125V	220, 230, 240V	No	Yes

                                                                           X  .  II 
                                                                 AC ADAPTOR   

An AC adapter, AC/DC adapter, or AC/DC converter is a type of external power supply, often enclosed in a case similar to an AC plug. Other common names include plug pack, plug-in adapter, adapter block, domestic mains adapter, line power adapter, wall wart, power brick, and power adapter. Adapters for battery-powered equipment may be described as chargers or rechargers (see also battery charger). AC adapters are used with electrical devices that require power but do not contain internal components to derive the required voltage and power from mains power. The internal circuitry of an external power supply is very similar to the design that would be used for a built-in or internal supply.
External power supplies are used both with equipment with no other source of power and with battery-powered equipment, where the supply, when plugged in, can sometimes charge the battery in addition to powering the equipment.
Use of an external power supply allows portability of equipment powered either by mains or battery without the added bulk of internal power components, and makes it unnecessary to produce equipment for use only with a specified power source; the same device can be powered from 120 VAC or 230 VAC mains, vehicle or aircraft battery by using a different adapter.
     
A "wall wart" type AC adapter for a household game console and Internal adapter circuitry 

Modes of operation

An AC adapter disassembled to reveal a simple, unregulated linear DC supply circuit: a transformer, four diodes in a bridge rectifier, and an electrolytic capacitor to smooth the waveform

Originally, most AC/DC adapters were linear power supplies, containing a transformer to convert the mains electricity voltage to a lower voltage, a rectifier to convert it to pulsating DC, and a filter to smooth the pulsating waveform to DC, with residual ripple variations small enough to leave the powered device unaffected. Size and weight of the device was largely determined by the transformer, which in turn was determined by the power output and mains frequency. Ratings over a few watts made the devices too large and heavy to be physically supported by a wall outlet. The output voltage of these adapters varied with load; for equipment requiring a more stable voltage, linear voltage regulator circuitry was added. Losses in the transformer and the linear regulator were considerable; efficiency was relatively low, and significant power dissipated as heat even when not driving a load.
In the early twenty-first century, switched-mode power supplies (SMPSs) became almost ubiquitous for this purpose. Mains voltage is rectified to a high direct voltage driving a switching circuit, which contains a transformer operating at a high frequency and outputs direct current at the desired voltage. The high-frequency ripple is more easily filtered out than mains-frequency. The high frequency allows the transformer to be small, which reduces its losses; and the switching regulator can be much more efficient than a linear regulator. The result is a much more efficient, smaller, and lighter device. Safety is ensured, as in the older linear circuit, because there is still a transformer which electrically isolates the output from the mains.
A linear circuit must be designed for a specific, narrow range of input voltages (e.g., 220–240 VAC) and must use a transformer appropriate for the frequency (usually 50 or 60 Hz), but a switched-mode supply can work efficiently over a very wide range of voltages and frequencies; a single 100–240 VAC unit will handle almost any mains supply in the world.
However, unless very carefully designed and using suitable components, switching adapters are more likely to fail than the older type, due in part to complex circuitry and the use of semiconductors. Unless designed well, these adapters may be easily damaged by overloads, even transient ones, which can come from lightning, brief mains overvoltage (sometimes caused by an incandescent light on the same power circuit failing), component degradation, etc. A very common mode of failure is due to the use of electrolytic capacitors whose equivalent series resistance (ESR) increases with age; switching regulators are very sensitive to high ESR (the older linear circuit also used electrolytic capacitors, but the effect of degradation is much less dramatic). Well-designed circuits pay attention to the ESR, ripple current rating, pulse operation, and temperature rating of capacitors.

Advantages

External AC adapters are widely used to power small or portable electronic devices. The advantages include:

• Safety – External power adapters can free product designers from worrying about some safety issues. Much of this style of equipment uses only voltages low enough not to be a safety hazard internally, although the power supply must out of necessity use dangerous mains voltage. If an external power supply is used (usually via a power connector, often of coaxial type), the equipment need not be designed with concern for hazardous voltages inside the enclosure. This is particularly relevant for equipment with lightweight cases which may break and expose internal electrical parts.
• Heat reduction – Heat reduces reliability and longevity of electronic components, and can cause sensitive circuits to become inaccurate or malfunction. A separate power supply removes a source of heat from the apparatus.
• Electrical noise reduction – Because radiated electrical noise falls off with the square of the distance, it is to the manufacturer's advantage to convert potentially noisy AC line power or automotive power to "clean", filtered DC in an external adapter, at a safe distance from noise-sensitive circuitry.
• Weight and size reduction – Removing power components and the mains connection plug from equipment powered by rechargeable batteries reduces the weight and size which must be carried.
• Ease of replacement – Power supplies are more prone to failure than other circuitry due to their exposure to power spikes and their internal generation of waste heat. External power supplies can be replaced quickly by a user without the need to have the powered device repaired.

AC adapter supporting four different AC plug systems

• Configuration versatility – Externally powered electronic products can be used with different power sources as needed (e.g. 120VAC, 240VAC, 12VDC, or external battery pack), for convenient use in the field, or when traveling.
• Simplified product inventory, distribution, and certification – An electronic product that is sold and used internationally must be powered from a wide range of power sources, and must meet product safety regulations in many jurisdictions, usually requiring expensive certification by national or regional safety agencies such as Underwriters Laboratories or Technischer Überwachungsverein. A single version of a device may be used in many markets, with the different power requirements met by different external power supplies, so that only one version of the device need be manufactured, stocked, and tested. If the design of the device is modified over time (a frequent occurrence), the power supply design itself need not be retested (and vice versa).
• Constant voltage is produced by a specific type of adapter used for computers and laptops. These types of adapters are commonly known as eliminators.

Problems

"Power brick" in-line configuration, with detachable AC cord

Problems with this type of power supply include, but are not limited to:

• Size – Power supplies which plug into the mains directly without using a plug on a cable (true wall warts) are bulkier than bare plugs; sometimes they are too large to plug into power sockets with restricted space, or into adjacent sockets on power strips (due to the fact that they can block other plugs also).
• Weight – Some AC adapters can be heavy, exerting excess weight on the power socket (this depends on the socket design of the country in question). Some external power supplies are "power bricks" (also known as "line lumps") having a short AC cord so they can lie on the floor, thus relieving strain, at the expense of clutter. Other wall-hanging types are made long and thin, minimizing the leverage of their weight vector that pulls the plug out, but exacerbating the size problem. The weight for equipment that must be carried (e.g. for traveling) is not a disadvantage of external supplies, as the alternative is an equally heavy internal supply; but in many cases a single universal supply could replace several proprietary ones.
• Inefficiency – Some idling power is wasted as the power supply is left running when the equipment power switch is off or the equipment is disconnected from the power supply. In recent years, it has become common for equipment with internal supplies to share this problem due to the use of a "soft" power switch.
• Confusion – External power supplies are often generic and not clearly marked to identify the equipment they are designed to power. It is very easy to separate power supply and equipment, and can be difficult to re-match the many devices with their power supplies.
• Compatibility problems – There is no standardization of connectors; the same connector is often used for different voltages, and for both DC supplies and AC-to-AC transformers. Incompatible voltage or polarity may be present on physically interchangeable connectors. This easily leads to using the wrong power supply, which can destroy equipment.

A survey of consumers showed widespread dissatisfaction with the cost, inconvenience, and wastefulness of the profusion of power adapters used by electronic devices. Science fiction author and satirist Douglas Adams wrote an essay bemoaning the profusion and confusion of power adapters, and calling for more standardization.

Dangerous and unreliable adapters

Manufacturers of equipment supplied with AC adapters often supply replacements at high prices; this has encouraged the manufacture of compatible third-party aftermarket replacements, which may be of satisfactory quality and performance at significantly lower prices. However, some adapters, usually at very low prices, and sometimes with unknown brands or sometimes fraudulently marked with the name of a reputable manufacturer, have various deficiencies which can cause inadequate performance (e.g. poor regulation and ripple, maximum power capacity lower than specified, hot running), unreliability (e.g. overheating to temperatures exceeding component ratings), and electrical or fire danger to users (e.g. insulation which frays with wear, lack of fuse).
Spurious marks of conformity to standards may be present; in one case it was reported that "Chinese manufacturers were submitting well-engineered electrical products to obtain conformity testing reports, but then removing non-essential components in production to reduce costs". A test of 27 chargers found that all the eight legitimately branded with a reputable name met safety standards, but none of those unbranded or with minor names did, despite bearing the mark of conformity.

Efficiency (historical)

Millions of still-usable AC power adapters are thrown out annually, because of poor or unknown compatibility with new equipment.

The issue of inefficiency of some power supplies has become well known, with U.S. president George W. Bush referring in 2001 to such devices as "Energy Vampires". Legislation is being enacted in the EU and a number of U.S. states, to reduce the level of energy wasted by some of these devices. Such initiatives include standby power and the One Watt Initiative.
But others have argued that these inefficient devices are low powered, e.g., devices that are used for small battery chargers, so even if they have a low efficiency, the amount of energy they waste is less than 1% of household consumption of electric energy.
Considering the total efficiency of power supplies for small electronic equipment, the older mains-frequency linear transformer-based power supply was found in a 2002 report to have efficiencies from 20–75%, and have considerable energy loss even when powered up but not supplying power. Switched-mode power supplies (SMPSs) are much more efficient; a good design can be 80–90% efficient, and is also much smaller and lighter. In 2002 most external plug-in "wall wart" power adapters commonly used for low-power consumer electronics devices were of linear design, as well as supplies built into some equipment.
External supplies are usually left plugged in even when not in use, and consume from a few watts to 35 watts of power in that state. The report concluded that about 32 billion kilowatt-hours (kWh) per year, about 1% of total electrical energy consumption, could be saved in the United States by replacing all linear power supplies (average efficiency 40–50%) with advanced switching designs (efficiency 80–90%), by replacing older switching supplies (efficiencies of less than 70%) with advanced designs (efficiency of at least 80%), and by reducing standby consumption of supplies to not more than 1 watt.
Since the report was published, SMPSs have indeed replaced linear supplies to a great extent, even in wall warts. The 2002 report estimated that 6% of electrical energy used in the U.S. "flows through" power supplies (not counting only the wall warts). The website where the report was published said in 2010 that despite the spread of SMPSs, "today's power supplies consume at least 2% of all U.S. electricity production. More efficient power supply designs could cut that usage in half".
Since wasted electrical energy is released as heat, an inefficient power supply is hot to the touch, as is one that wastes power without an electrical load. This waste heat is itself a problem in warm weather, since it may require additional air conditioning to prevent overheating, and even to remove the unwanted heat from large supplies.

Reuse

AC adapters are often reused on other appliances, but there are five parameters which all must suit the appliance:

• Voltage
• Current capacity
• Polarity (not relevant with AC)
• Voltage regulation (or stabilization)
• Connector type

Universal power adapters

 

A six-way connector on a "universal" DC power supply, consisting of a four-way X connector and two separate individual connectors (one is the nine-volt battery connector). The X-connector here provides 3.5 and 2.5 mm phone plugs and two sizes of coaxial power connector

External power adapters can fail, and can become separated from the product they are intended to power. Consequently, there is a market for replacement adapters. The replacement must match input and output voltages, match or exceed current capability, and be fitted with a matching connector. Many electrical products are poorly labeled with information concerning the power supply they require, so it is prudent to record the specifications of the original power supply in advance, to ease replacement if the original is later lost. Careful labeling of power adapters can also reduce the likelihood of a mixup which could cause equipment damage.
Some "universal" replacement power supplies allow output voltage and polarity to be switched to match a range of equipment. With the advent of switch-mode supplies, adapters which can work with any voltage from 110 VAC to 240 VAC became widely available; previously either 100–120 VAC or 200–240 VAC versions were used. Adaptors which can also be used with motor vehicle and aircraft power (see EmPower) are available.
Four-way X connectors or six-way star connectors, also known as spider connectors, with multiple plug sizes and types are common on generic power supplies. Other replacement power supplies have arrangements for changing the power connector, with four to nine different alternatives available when purchased in a set. RadioShack sells universal AC adapters of various capacities, branded as "Enercell Adaptaplug", and fitted with two-pin female sockets compatible with their Adaptaplug connector lineup. This allows many different configurations of AC adapters to be put together, without requiring soldering. Philmore and other competing brands offer similar AC adapters with interchangeable connectors.
The label on a power supply may not be a reliable guide to the actual voltage it supplies under varying conditions. Many low-cost power supplies are "unregulated", in that their voltage can change considerably with load. If they are lightly loaded, they may put out much more than the nominal "name plate" voltage, which could damage the load. If they are heavily loaded, the output voltage may droop appreciably, in some cases well below the nominal label voltage even within the nominal rated current, causing the equipment being supplied to malfunction or be damaged. Supplies with linear (as against switched) regulators are heavy, bulky, and expensive.
Modern switched-mode power supplies (SMPSs) are smaller, lighter, and more efficient. They put out a much more constant voltage than unregulated supplies as the input voltage and the load current vary. When introduced, their prices were high, but by the early 21st century the prices of switch-mode components had dropped to a degree which allowed even cheap supplies to use this technology, saving the cost of a larger and heavier mains-frequency transformer.

Auto-sensing adapters

Some universal adapters automatically set their output voltage and maximum current according to which of a range of interchangeable tips is fitted; tips are available to fit and supply appropriate power to many notebook computers and mobile devices. Different tips may use the same connector, but automatically supply different power; it is essential to use the right tip for the apparatus being powered, but no switch needs to be set correctly by the user. The advent of switch-mode power supplies has allowed adapters to work from any AC mains supply from 100 to 240V with an appropriate plug; operation from standard 12V DC vehicle and aircraft supplies can also be supported. With the appropriate adapter, accessories, and tips, a variety of equipment can be powered from almost any source of power.
A "Green Plug" system has been proposed, based on USB technology, by which the consuming device would tell the external power supply what kind of power is needed.

Use of USB

 

Common sizes of USB AC adapters

The USB connector (and voltage) has emerged as a de facto standard in low-power AC adapters for many portable devices. In addition to serial digital data exchange, the USB standard also provides 5 VDC power, up to 500 mA (900 mA over USB 3.0). Numerous accessory gadgets ("USB decorations") were designed to connect to USB only for DC power and not for data interchange. The USB Implementers Forum in March, 2007 released the USB Battery Charging Specification which defines, "...limits as well as detection, control and reporting mechanisms to permit devices to draw current in excess of the USB 2.0 specification for charging ...".Electric fans, lamps, alarms, coffee warmers, battery chargers, and even toys have been designed to tap power from a USB connector. Plug-in adapters equipped with USB receptacles are widely available to convert 120 VAC or 240 VAC power or 12 VDC automotive power to 5 VDC USB power (see photo at right).

Standards

The trend towards more-compact electronic devices has driven a shift towards the micro-USB and mini-USB connectors, which are electrically compatible in function to the original USB connector but physically smaller. In 2009, the International Telecommunication Union (ITU) announced support of the Open Mobile Terminal Platform's (OMTP) "Common Charging and Local Data Connectivity" standard. The ITU published Recommendation ITU-T L.1000, "Universal power adapter and charger solution for mobile terminals and other hand-held ICT devices", which specifies a charger similar in most respects to that of the GSMA/OMTP proposal and to the European Common External Power Supply. The ITU recommendation was expanded and updated in June, 2011. The hope is to markedly reduce the profusion of non-interchangeable power adapters.
The European Union defined a Common External Power Supply for "hand-held data-enabled mobile phones" (smartphones) sold from 2010, intended to replace the many incompatible proprietary power supplies and eliminate waste by reducing the total number of supplies manufactured. Conformant supplies deliver 5 VDC via a micro-USB connector, with preferred input voltage handled ranging from 90 to 264 VAC.
In 2012, a USB Power Delivery Specification was proposed to standardize delivery of up to 100 watts, suitable for devices such as laptop computers that usually depend on proprietary adapters.
Larry Page, a founder of Google, proposed a 12 V and up to 15 A standard for almost all equipment requiring an external converter, with new buildings fitted with 12 VDC wiring, making external AC-to-DC adapter circuitry unnecessary.
                                                                          X  .  IIII

       

                Basic AC-DC Power Supplies

                                             Discrete Semiconductor Devices and Circuits

                                                    

An adapter is a physical device that allows one hardware or electronic interface to be adapted (accommodated without loss of function) to another hardware or electronic interface. In a computer, an adapter is often built into a card that can be inserted into a slot on the computer's motherboard. The card adapts information that is exchanged between the computer's microprocessor and the devices that the card supports.    

Question 1

Don’t just sit there! Build something!!

Learning to mathematically analyze circuits requires much study and practice. Typically, students practice by working through lots of sample problems and checking their answers against those provided by the textbook or the instructor. While this is good, there is a much better way.
You will learn much more by actually building and analyzing real circuits, letting your test equipment provide the “answers” instead of a book or another person. For successful circuit-building exercises, follow these steps:

1. Carefully measure and record all component values prior to circuit construction, choosing resistor values high enough to make damage to any active components unlikely.
2. Draw the schematic diagram for the circuit to be analyzed.
3. Carefully build this circuit on a breadboard or other convenient medium.
4. Check the accuracy of the circuit’s construction, following each wire to each connection point, and verifying these elements one-by-one on the diagram.
5. Mathematically analyze the circuit, solving for all voltage and current values.
6. Carefully measure all voltages and currents, to verify the accuracy of your analysis.
7. If there are any substantial errors (greater than a few percent), carefully check your circuit’s construction against the diagram, then carefully re-calculate the values and re-measure.

When students are first learning about semiconductor devices, and are most likely to damage them by making improper connections in their circuits, I recommend they experiment with large, high-wattage components (1N4001 rectifying diodes, TO-220 or TO-3 case power transistors, etc.), and using dry-cell battery power sources rather than a benchtop power supply. This decreases the likelihood of component damage.
As usual, avoid very high and very low resistor values, to avoid measurement errors caused by meter “loading” (on the high end) and to avoid transistor burnout (on the low end). I recommend resistors between 1 kΩ and 100 kΩ.
One way you can save time and reduce the possibility of error is to begin with a very simple circuit and incrementally add components to increase its complexity after each analysis, rather than building a whole new circuit for each practice problem. Another time-saving technique is to re-use the same components in a variety of different circuit configurations. This way, you won’t have to measure any component’s value more than once.

Let the electrons themselves give you the answers to your own “practice problems”!

Notes:
It has been my experience that students require much practice with circuit analysis to become proficient. To this end, instructors usually provide their students with lots of practice problems to work through, and provide answers for students to check their work against. While this approach makes students proficient in circuit theory, it fails to fully educate them.
Students don’t just need mathematical practice. They also need real, hands-on practice building circuits and using test equipment. So, I suggest the following alternative approach: students should build their own “practice problems” with real components, and try to mathematically predict the various voltage and current values. This way, the mathematical theory “comes alive,” and students gain practical proficiency they wouldn’t gain merely by solving equations.
Another reason for following this method of practice is to teach students scientific method: the process of testing a hypothesis (in this case, mathematical predictions) by performing a real experiment. Students will also develop real troubleshooting skills as they occasionally make circuit construction errors.
Spend a few moments of time with your class to review some of the “rules” for building circuits before they begin. Discuss these issues with your students in the same Socratic manner you would normally discuss the worksheet questions, rather than simply telling them what they should and should not do. I never cease to be amazed at how poorly students grasp instructions when presented in a typical lecture (instructor monologue) format!
A note to those instructors who may complain about the “wasted” time required to have students build real circuits instead of just mathematically analyzing theoretical circuits:
What is the purpose of students taking your course?
If your students will be working with real circuits, then they should learn on real circuits whenever possible. If your goal is to educate theoretical physicists, then stick with abstract analysis, by all means! But most of us plan for our students to do something in the real world with the education we give them. The “wasted” time spent building real circuits will pay huge dividends when it comes time for them to apply their knowledge to practical problems.
Furthermore, having students build their own practice problems teaches them how to perform primary research, thus empowering them to continue their electrical/electronics education autonomously.
In most sciences, realistic experiments are much more difficult and expensive to set up than electrical circuits. Nuclear physics, biology, geology, and chemistry professors would just love to be able to have their students apply advanced mathematics to real experiments posing no safety hazard and costing less than a textbook. They can’t, but you can. Exploit the convenience inherent to your science, and get those students of yours practicing their math on lots of real circuits!

Question 2

A technician builds a simple half-wave rectifier circuit for a project, but is surprised to find that the diode keeps failing:



This comes as a surprise because the diode has a repetitive peak reverse voltage rating of 50 volts, which the technician knows is greater than the peak voltage output by the step-down transformer. However, the technician has overlooked something very important in this circuit design. Explain what the problem is, and how to solve it.

The diode’s peak inverse voltage (“PIV”) rating is insufficient. It needs to be about 85 volts or greater in order to withstand the demands of this circuit.
Follow-up question: suggest a part number for a diode capable of withstanding the reverse voltage generated by this circuit, and able to handle at least 1 amp of continuous current.

Notes:
If students experience difficulty calculating the necessary PIV rating for this circuit’s diode, ask them to analyze the peak output from the transformer’s secondary winding for each half-cycle of the AC waveform, noting the voltage drops across all circuit components. Once a full-cycle voltage analysis is performed for all circuit components, the necessary diode rating should become obvious.
Though it may not be obvious at first reading, this question may actually serve as a lead-in for discussing voltage multiplier circuits. The fact that the diode experiences a reverse voltage twice that of the peak AC voltage is something we may exploit!
Another reliability factor most students won’t recognize in this circuit is the “inrush” current experienced by the diode every time the circuit is powered up and the capacitor recharges. Certainly, the diode was not properly rated for the reverse voltage it was being subjected to, but this might not be the only form of abuse! If time permits, discuss this possibility as well.

Question 3

Diodes and capacitors may be interconnected to form a type of circuit that boosts voltage in the process of rectification. This type of circuit is generally known as a voltage multiplier. Shown here are a few different voltage multiplier circuits:



Determine the degree of voltage multiplication (double, triple, etc.) provided by each circuit.

Notes:
Ask your students how they analyzed each of these voltage multiplier circuits, and to explain their technique(s) to the rest of the class during discussion.

Question 4

Suppose a technician measures the voltage output by an AC-DC power supply circuit:



The waveform shown by the oscilloscope is mostly DC, with just a little bit of AC “ripple” voltage appearing as a ripple pattern on what would otherwise be a straight, horizontal line. This is quite normal for the output of an AC-DC power supply.
Suppose we wished to take a closer view of this “ripple” voltage. We want to make the ripples more pronounced on the screen, so that we may better discern their shape. Unfortunately, though, when we decrease the number of volts per division on the “vertical” control knob to magnify the vertical amplification of the oscilloscope, the pattern completely disappears from the screen!
Explain what the problem is, and how we might correct it so as to be able to magnify the ripple voltage waveform without having it disappear off the oscilloscope screen.

The problem is that the vertical axis input is DC-coupled.
Follow-up question: predict the frequency of the ripple voltage in this power supply circuit.

Notes:
As usual, what I’m looking for in an answer here is an explanation for what is happening. If a student simply tells you, “the vertical input is DC-coupled,” press them for more detail. What does it mean for the input to be “DC-coupled,” and why does this cause the line to disappear from the screen when we increase the vertical sensitivity? What alternative do we have to “DC coupling” on an oscilloscope?
One nice thing about oscilloscopes is that they cannot be damaged by “pegging” the display, as can analog multimeters. The same concept applies, though, and is useful in explaining to students why waveforms disappear from the screen when the vertical sensitivity is too great.

Question 5

AC-DC power supply circuits are one of the most common circuit configurations in electronic systems. Though designs may vary, the task of converting AC power to DC power is vital in the functioning of a great many electronic devices.
Why is this? What is it about this kind of circuit that makes it such a necessary part of many electronic systems?

Most electric power distribution systems are AC, yet most electronic circuits function on DC power.

Notes:
One factor not mentioned in the answer is circuit operating voltage. How do the operating voltages of a typical AC power system and a typical electronic circuit (radio, alarm clock, computer) compare? Ask your students what purpose a power supply has with regard to voltage.
Ask your students if the word “supply” is truly appropriate for this type of circuit. Does it really supply energy, or does it just convert energy from one form into another?

Question 6

Although not a popular design, some power supply circuits are transformerless. Direct rectification of AC line power is a viable option in some applications:



However, this form of AC-to-DC power conversion has some significant limits. Explain why most power supply circuits utilize a transformer instead of directly rectifying the line power as this circuit does.

Transformers provide voltage/current ratio transformation, and also electrical isolation between the AC line circuit and the DC circuit. The issue of isolation is a safety concern, as neither of the output conductors in a non-isolated (direct) rectifier circuit is at the same potential as either of the line conductors.
Follow-up question: explain in detail how the issue of non-isolation could create a safety hazard if this rectifier circuit were energized by an earth-grounded AC line circuit.

Notes:
Many old television sets used such transformer less rectifier circuits to save money, but this meant the metal circuit chassis inside the plastic cover was energized rather than being at ground potential! Very dangerous for technicians to work on.

Question 7

An essential part of an AC-DC power supply circuit is the filter, used to separate the residual AC (called the “ripple” voltage) from the DC voltage prior to output. Here are two simple AC-DC power supply circuits, one without a filter and one with:



Draw the respective output voltage waveforms of these two power supply circuits (V_unfiltered versus V_filtered). Also identify the type of filter circuit needed for the task (low pass, high pass, band pass, or band stop), and explain why that type of filter circuit is needed.

A low pass filter is the kind needed to filter “ripple voltage” from the power supply output.

Notes:
Many years ago, when I was first learning about power supplies, I tried to power an automotive radio with voltage from a battery charger. The battery charger was a simple power supply suitable for charging 12-volt automotive batteries, I reasoned, so what harm would there be in using it to power an automotive radio? After a few moments of LOUD humming from the radio speaker, my rhetorical question was answered by a puff of smoke from the radio, then silence.
Part of the problem was the output voltage of the battery charger, but a large part of the problem was the fact that the charger’s output was unfiltered as well. For the same reasons my radio did not function properly on unfiltered, rectified AC, many electronic circuits will not function on it either.

Question 8

Observe the following two waveforms, as represented on an oscilloscope display measuring output voltage of a filtered power supply:



If both of these waveforms were measured on the same power supply circuit, at different times, determine which waveform was measured during a period of heavier “loading” (a “heavier” load being defined as a load drawing greater current).

The left-hand waveform was measured during a period of heavier loading.

Notes:
Ask your students what the term “loading” means in this context. Some of them may not comprehend the term accurately, and so it is good to review just to make sure.
More importantly, discuss with your students why the ripple is more severe under conditions of heavy loading. What, exactly, is happening in the circuit to produce this kind of waveform? If it is necessary for us to maintain a low amount of ripple under this heavy loading, what must we change in the power supply circuit?

Question 9

What does it mean if a power supply has a DC output with 5% ripple?

This means the peak-to-peak ripple voltage is equal to 5% of the DC (average) voltage.

Notes:
The purpose of this question is to get students to look up the formula for calculating ripple voltage percentage. Notice how I did not simply ask them to regurgitate a formula; rather, I presented a realistic figure for them to interpret. When at all possible, try to format your questions in this sort of practical context!

Question 10

What parameters determine the frequency of a power supply’s ripple voltage?

For linear power supplies (those designs having a transformer-rectifier-filter topology), the parameters determining ripple frequency are line frequency and rectification pulses.

Notes:
Note that I did not simply say the ripple frequency is equal to line frequency for half-wave rectification and double the line frequency for full-wave. Such an answer is misleading, since it completely ignores poly phase AC rectification!

Question 11

Suppose a power supply is energized by an AC source of 119 V RMS. The transformer step-down ratio is 8:1, it uses a full-wave bridge rectifier circuit with silicon diodes, and the filter is nothing but a single electrolytic capacitor. Calculate the unloaded DC output voltage for this supply (assume 0.7 volts drop across each diode). Also, write an equation solving for DC output voltage (V_out), given all these parameters.

V_out = 19.6 volts

Vout = Vin r 0.707 − 2 Vf

Where,
V_out = DC output voltage, in volts
V_in = AC input voltage, in volts RMS
r = Transformer step-down ratio
V_f = Forward voltage drop of each diode, in volts
Follow-up question: algebraically manipulate this equation to solve for V_in.

Notes:
It is important for students to understand where this equation comes from. Ask your students to explain, step by step, the process of calculating output voltage for a simple power supply circuit. It is helpful in this process to calculate the voltage at each “stage” of the power supply (transformer primary, transformer secondary, etc.), as though we were building the circuit one component at a time.
Incidentally, the method of building a project (such as a power supply) in a step-by-step fashion rather than all at once, saves a lot of time and effort when things go wrong. The same “step-by-step” strategy works well for mathematical analysis, and other problem-solving tasks as well: try to analyze the circuit one “block” at a time instead of the whole thing at once.

Question 12

What does it mean if a power supply exhibits 2% voltage regulation?

This means the difference between no-load output voltage and full-load output voltage is 2% of the full-load output voltage.

Notes:
The purpose of this question is to get students to look up the formula for calculating voltage regulation percentage. Notice how I did not simply ask them to regurgitate a formula; rather, I presented a realistic figure for them to interpret. When at all possible, try to format your questions in this sort of practical context!

Question 13

What will be the consequence of one diode failing open in the bridge rectifier of a single-phase power supply?

The (unfiltered) output voltage will be half-wave, not full-wave.

Notes:
A question such as this is best discussed while viewing the schematic diagram for a bridge rectifier. I recommend projecting an image of a bridge rectifier circuit on a whiteboard, then having students use dry-erase markers to “mark up” the schematic with arrows for current, voltage drop indications, etc. This way, mistakes can be corrected, or alternate cycles erased and re-drawn, without having to erase and re-draw the schematic diagram itself.

Question 14

Suppose you suspected a failed-open diode in this power supply circuit. Describe how you could detect its presence without using an oscilloscope:



Incidentally, the “low voltage AC power supply” is nothing more than a step-down transformer with a center-tapped secondary winding.

“Remove all diodes from the circuit and test them individually” is not an acceptable answer to this question. Think of a way that they could be checked while in-circuit (ideally, without having to shut off power to the circuit).

Notes:
A common tendency for students is to troubleshoot using the “shotgun approach,” which is to remove each component one-by-one and test it. This is a very time-intensive and inefficient method of troubleshooting. Instead, students need to develop diagnostic procedures not requiring removal of components from the circuit. At the very least, there should be some way we can narrow the range of possibilities using in-circuit tests prior to removing components.

Question 15

A student learns that a rectifier circuit is often followed by a low-pass filter circuit in an AC-DC power supply to reduce “ripple” voltage on the output. Looking over his notes from AC theory, the student proceeds to build this power supply circuit complete with a low-pass filter at the output:



While this design will work, there are better filter configurations for this application. Describe the limitations of the circuit shown, and explain how some of the other filters would do a better job.

The resistor R tends to limit the output current, resulting in less-than-optimal voltage regulation (the output voltage “sagging” under load). Better filter configurations include all forms of LC ripple filters, including the popular “pi” (π) filter.
Follow-up question: in some applications - especially where very large filter capacitors are used - it is a good idea to place a series resistor before the capacitor. Such a resistor is typically rated at a low value so as to not cause excessive output voltage “sag” under load, but its resistance does serve a practical purpose. Explain what this purpose might be.

Notes:
Challenge your students with this question: is this the right kind of filter circuit (low pass, high pass, band pass, band stop) to be using, anyway? This question presents a good opportunity to review basic filter theory.
The follow-up question asks students to think carefully about the possible positive benefits of having a series resistor before the capacitor as shown in the student’s original design. If your students are experiencing difficulty understanding why a resistor would ever be necessary, jog their memories with this formula:

i = C dv dt

Question 16

Identify the voltages that are supposed to appear between the listed test points:

1. V_TP1−TP2 =
2. V_TP1−TP3 =
3. V_TP2−TP3 =
4. V_TP4−TP5 =
5. V_TP5−TP6 =
6. V_TP7−TP8 =
7. V_TP9−TP10 =

Assume that the power transformer has a step-down ratio of 9.5:1.

1. V_TP1−TP2 = 120 volts AC
2. V_TP1−TP3 = 120 volts AC
3. V_TP2−TP3 = 0 volts
4. V_TP4−TP5 = 12.63 volts AC
5. V_TP5−TP6 = 12.63 volts AC
6. V_TP7−TP8 = 16.47 volts DC
7. V_TP9−TP10 = 16.47 volts DC

Notes:
Before one can troubleshoot a malfunctioning circuit, one must know what voltages and currents are supposed to be in various portions of the circuit. This question, therefore, is a prelude to further troubleshooting questions.

Question 17

A technician is troubleshooting a power supply circuit with no DC output voltage. The output voltage is supposed to be 15 volts DC:



The technician begins making voltage measurements between some of the test points (TP) on the circuit board. What follows is a sequential record of his measurements:

1. V_TP9−TP10 = 0 volts DC
2. V_TP8−TP7 = 0 volts DC
3. V_TP8−TP5 = 0 volts DC
4. V_TP6−TP7 = 0 volts DC
5. V_TP4−TP5 = 0 volts AC
6. V_TP1−TP3 = 0 volts AC
7. V_TP1−TP2 = 116 volts AC

Based on these measurements, what do you suspect has failed in this supply circuit? Explain your answer. Also, critique this technician’s troubleshooting technique and make your own suggestions for a more efficient pattern of steps.

The fuse is blown open.
Follow-up question: with regard to the troubleshooting technique, this technician seems to have started from one end of the circuit and moved incrementally toward the other, checking voltage at almost every point in between. Can you think of a more efficient strategy than to start at one end and work slowly toward the other?

Notes:
Troubleshooting scenarios are always good for stimulating class discussion. Be sure to spend plenty of time in class with your students developing efficient and logical diagnostic procedures, as this will assist them greatly in their careers.

Question 18

A technician is troubleshooting a power supply circuit with no DC output voltage. The output voltage is supposed to be 15 volts DC:



The technician begins making voltage measurements between some of the test points (TP) on the circuit board. What follows is a sequential record of her measurements:

1. V_TP1−TP2 = 118 volts AC
2. V_TP3−TP2 = 0 volts AC
3. V_TP1−TP3 = 118 volts AC
4. V_TP4−TP5 = 0.5 volts AC
5. V_TP7−TP8 = 1.1 volts DC
6. V_TP9−TP10 = 1.1 volts DC

Based on these measurements, what do you suspect has failed in this supply circuit? Explain your answer. Also, critique this technician’s troubleshooting technique and make your own suggestions for a more efficient pattern of steps.

The transformer has an open winding.
Follow-up question #1: with regard to the troubleshooting technique, this technician seems to have started from one end of the circuit and moved incrementally toward the other, checking voltage at almost every point in between. Can you think of a more efficient strategy than to start at one end and work slowly toward the other?
Challenge question: based on the voltage measurements taken, which do you think is the more likely failure, an open primary winding or an open secondary winding?
Follow-up question #2: how could you test the two windings of the transformer for a possible open fault? In other words, is there another type of measurement that could verify our hypothesis of a failed winding?

Notes:
Troubleshooting scenarios are always good for stimulating class discussion. Be sure to spend plenty of time in class with your students developing efficient and logical diagnostic procedures, as this will assist them greatly in their careers.
Students may be puzzled by the presence of DC voltage between TP7 and TP8, and also between TP9 and TP10 (1.1 volts), given that there is less than that amount of AC voltage at the rectifier’s input. However, this is a common phenomenon with electrolytic capacitors, to “recover” a small voltage after having been discharged.

Question 19

A technician is troubleshooting a power supply circuit with no DC output voltage. The output voltage is supposed to be 15 volts DC:



The technician begins making voltage measurements between some of the test points (TP) on the circuit board. What follows is a sequential record of his measurements:

1. V_TP9−TP10 = 0 volts DC
2. V_TP1−TP2 = 117 volts AC
3. V_TP1−TP3 = 117 volts AC
4. V_TP5−TP6 = 0 volts AC
5. V_TP7−TP8 = 0.1 volts DC
6. V_TP5−TP4 = 12 volts AC
7. V_TP7−TP6 = 0 volts DC

Based on these measurements, what do you suspect has failed in this supply circuit? Explain your answer. Also, critique this technician’s troubleshooting technique and make your own suggestions for a more efficient pattern of steps.

There is an “open” fault between TP4 and TP6.
Follow-up question: with regard to the troubleshooting technique, this technician seems to have started from one end of the circuit and moved incrementally toward the other, checking voltage at almost every point in between. Can you think of a more efficient strategy than to start at one end and work slowly toward the other?

Notes:
Troubleshooting scenarios are always good for stimulating class discussion. Be sure to spend plenty of time in class with your students developing efficient and logical diagnostic procedures, as this will assist them greatly in their careers.

Question 20

AC-DC power supplies are a cause of harmonic currents in AC power systems, especially large AC-DC power supplies used in motor control circuits and other high-power controls. In this example, I show the waveforms for output voltage and input current for an unloaded AC-DC power supply with a step-down transformer, full-wave rectifier, and capacitive filter circuit (the unfiltered DC voltage waveform is shown as a dashed line for reference):



As you can see, the input current waveform lags the voltage waveform by 90^o, because when the power supply is unloaded, the only input current is the magnetizing current of the transformer’s primary winding.
With increased loading, the output ripple voltage becomes more pronounced. This also changes the input current waveform significantly, making it non-sinusoidal. Trace the shape of the input current waveform, given the output voltage waveform and magnetizing current waveform (dotted line) shown here:



The non-filtered DC output waveform is still shown as a dotted line, for reference purposes.

Challenge question: does the input current waveform shown here contain even-numbered harmonics (i.e. 120 Hz, 240 Hz, 360 Hz)?

Notes:
In a filtered DC power supply, the only time current is drawn from the rectifier is when the filter capacitor charges. Thus, the only time you see input current above and beyond the magnetizing current waveform is when the capacitor voltage requires charging.
Note that although the (sinusoidal) magnetizing current waveform is 90^o out of phase with the voltage waveform, the input current transients are precisely in-phase with the current transients on the transformer’s secondary winding. This reviews an important principle of transformers: that whatever primary current is the result of secondary winding load is in-phase with that secondary load current. In this regard, a transformer does not act as a reactive device, but a direct power-coupling device.
Note also that after the initial surge (rising pulse edge) of current, the input current waveform follows a different curve from the voltage waveform, because i = C[dv/dt] for a capacitor.
In case you haven’t guessed by now, there is a lot of stuff happening in this circuit! I would consider this question to be ädvanced” for most introductory-level courses, and may be skipped at your discretion.

Question 21

Power supplies are sometimes equipped with EMI/RFI filters on their inputs, to prevent high-frequency “noise” voltage created within the power supply circuit from getting back to the power source where it might interfere with other powered equipment. This is especially useful for “switching” power supply circuits, where transistors are used to switch power on and off very rapidly in the voltage transformation and regulation process:



Determine what type of filter circuit this is (LP, HP, BP, or BS), and also determine the inductive and capacitive reactances of its components at 60 Hz, if the inductors are 100 μH each and the capacitors are 0.022 μF each.

X_L = 0.0377 Ω (each)
X_C = 120.6 kΩ (each)

Notes:
Ask your students how they determined the identity of this filter. Are they strictly memorizing filter configurations, or do they have a technique for determining what type of filter circuit it is based on basic electrical principles (reactance of components to different frequencies)? Remind them that rote memorization is a very poor form of learning!

Question 22

Complete this schematic diagram, turning it into a split (or dual power supply, with three output terminals: +V, Ground, and -V:

Examples of “split” or “dual” power supply schematic diagrams abound in textbooks. I’ll let you do the research here and present your answer(s) during class discussion!

Notes:
Students need not provide details of voltage regulation, but merely show how AC from a center-tapped transformer winding may be rectified into two distinct DC outputs with a common “ground” connection.

Question 23

Predict how all component voltages and currents in this circuit will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

Any one diode fails open:

Transformer secondary winding fails open:

Inductor L₁ fails open:

Capacitor C₁ fails shorted:

For each of these conditions, explain why the resulting effects will occur.

Any one diode fails open: Half-wave rectification rather than full-wave, less DC voltage across load, more ripple (AC) voltage across load.

Transformer secondary winding fails open: no voltage or current on secondary side of circuit after C₁ discharges through load, little current through primary winding.

Inductor L₁ fails open: no voltage across load, no current through load, no current through rest of secondary-side components, little current through primary winding.

Capacitor C₁ fails shorted: increased current through both transformer windings, increased current through diodes, increased current through inductor, little voltage across or current through load, capacitor and all diodes will likely get hot.

Notes:
The purpose of this question is to approach the domain of circuit troubleshooting from a perspective of knowing what the fault is, rather than only knowing what the symptoms are. Although this is not necessarily a realistic perspective, it helps students build the foundational knowledge necessary to diagnose a faulted circuit from empirical data. Questions such as this should be followed (eventually) by other questions asking students to identify likely faults based on measurements.

Question 24

Suppose this power supply circuit was working fine for several years, then one day failed to output any DC voltage at all:



When you open the case of this power supply, you immediately notice the strong odor of burnt components. From this information, determine some likely component faults and explain your reasoning.

Shorted capacitor, open transformer winding (as a result of overloading), shorted diode(s) resulting in blown fuse.

Notes:
Troubleshooting scenarios are always good for stimulating class discussion. Be sure to spend plenty of time in class with your students developing efficient and logical diagnostic procedures, as this will assist them greatly in their careers.
Remind your students that test instrument readings are not the only viable source of diagnostic data! Burnt electronic components usually produce a strong and easily-recognized odor, always indicative of overheating. It is important to keep in mind that often the burnt component is not the original source of trouble, but may be a casualty of some other component fault.

Question 25

The ripple frequency of a half-wave rectifier circuit powered by 60 Hz AC is measured to be 60 Hz. The ripple frequency of a full-wave rectifier circuit powered by the exact same 60 Hz AC line voltage is measured to be 120 Hz. Explain why the ripple frequency of the full-wave rectifier is twice that of the half-wave rectifier.

There are double the number of pulses in the full-wave rectifier’s output, meaning the wave-shape repeats itself twice as often.

Notes:
I have heard students concoct very interesting (and wrong) explanations for why the two frequencies differ. One common misconception is that is has something to do with the transformer, as though a transformer had the ability to step frequency up and down just as easily as voltage or current! If students are not understanding why there is a frequency difference, you might want to help them out by asking two students to come up to the front of the class and draw two waveforms: half-wave and full-wave, along with their original AC (unrectified) waveforms.

Question 26

Calculate the approximate DC output voltage of this power supply when it is not loaded:

V_out ≈ 21.4 volts

Notes:
Ask your students to explain how they solved for this output voltage, step by step.

Question 27

Calculate the approximate DC output voltage of this power supply when it is not loaded:

V_out ≈ 11.6 volts

Notes:
Ask your students to explain how they solved for this output voltage, step by step.

Question 28

A simple AC-DC power supply circuit outputs about 6.1 volts DC without a filter capacitor connected, and about 9.3 volts DC with a filter capacitor connected:

Explain why this is. How can the addition of nothing but a capacitor have such a great effect on the amount of DC voltage output by the circuit?

The filter capacitor captures the peak voltage level of each pulse from the rectifier circuit, holding that peak level during the time between pulses.

Notes:
Many new students find this phenomenon paradoxical, especially when they see a DC output voltage greater than the AC (RMS) output voltage of the transformer’s secondary winding. Case in point: being able to build a 30 volt DC power supply using a transformer with a secondary voltage rating of only 24 volts. The purpose of this question is to get students to face this paradox if they have not recognized and resolved it already.

Question 29

A technician is troubleshooting a power supply circuit that outputs substantially less DC voltage than it should. The output voltage is supposed to be 15 volts DC, but instead it is actually outputting less than 8 volts DC:



The technician measures approximately 18 volts AC (RMS) across the secondary winding of the transformer. Based on this voltage measurement and the knowledge that there is reduced DC output voltage, identify two possible faults that could account for the problem and all measured values in this circuit, and also identify two circuit elements that could not possibly be to blame (i.e. two things that you know must be functioning properly, no matter what else may be faulted). The circuit elements you identify as either possibly faulted or properly functioning can be wires, traces, and connections as well as components. Be as specific as you can in your answers, identifying both the circuit element and the type of fault.

Circuit elements that are possibly faulted

1.

2.

Circuit elements that must be functioning properly

1.

2.

I’ll let you and your classmates figure out some possibilities here!
 

Notes:
Troubleshooting scenarios are always good for stimulating class discussion. Be sure to spend plenty of time in class with your students developing efficient and logical diagnostic procedures, as this will assist them greatly in their careers.

I’ll let you and your classmates figure out some possibilities here!

Notes:
Troubleshooting scenarios are always good for stimulating class discussion. Be sure to spend plenty of time in class with your students developing efficient and logical diagnostic procedures, as this will assist them greatly in their careers.

 

Question 30

A technician is troubleshooting a power supply circuit with no DC output voltage. The output voltage is supposed to be 15 volts DC, but instead it is actually outputting nothing at all (zero volts): 



The technician measures 120 volts AC between test points TP1 and TP3. Based on this voltage measurement and the knowledge that there is zero DC output voltage, identify two possible faults that could account for the problem and all measured values in this circuit, and also identify two circuit elements that could not possibly be to blame (i.e. two things that you know must be functioning properly, no matter what else may be faulted). The circuit elements you identify as either possibly faulted or properly functioning can be wires, traces, and connections as well as components. Be as specific as you can in your answers, identifying both the circuit element and the type of fault.

Circuit elements that are possibly faulted

1.

2.

Circuit elements that must be functioning properly

1.

2.

I’ll let you and your classmates figure out some possibilities here!I’ll let you and your classmates figure out some possibilities here!

Notes:
Troubleshooting scenarios are always good for stimulating class discussion. Be sure to spend plenty of time in class with your students developing efficient and logical diagnostic procedures, as this will assist them greatly in their careers.

Notes:
Troubleshooting scenarios are always good for stimulating class discussion. Be sure to spend plenty of time in class with your students developing efficient and logical diagnostic procedures, as this will assist them greatly in their careers.

Question 31

A common sub-circuit in power supplies of all kinds is an EMI/RFI filter. This LC network is all but “transparent” to the 50 or 60 Hz power line frequency, so that the transformer sees full line voltage at all times:



If this EMI/RFI filter does nothing to or with the line power, what purpose does it serve?  

An EMI/RFI filter has no purpose in the process of converting AC power into DC power. It does, however, help prevent the power supply circuitry from interfering with other equipment energized by the same AC line power, by filtering out unwanted high-frequency noise generated within the power supply by diode switching.
Follow-up question: calculate the total inductive reactance posed by the two inductors to 60 Hz power if the inductance of each is 100 μH.

Notes:
These filters are very common in switch-mode power supplies, but they are not out of place in linear (“brute-force”) power supply circuits such as this one. Mention to your students how the abatement of electromagnetic and radio frequency interference is a high priority in all kinds of electronic device design.    



                                                          X  .  IIIII
 

             Calculus for Electric Circuits

Mathematics for Electronics

---

Question 1

∫f(x) dx Calculus alert!

Calculus is a branch of mathematics that originated with scientific questions concerning rates of change. The easiest rates of change for most people to understand are those dealing with time. For example, a student watching their savings account dwindle over time as they pay for tuition and other expenses is very concerned with rates of change (dollars per year being spent).
In calculus, we have a special word to describe rates of change: derivative. One of the notations used to express a derivative (rate of change) appears as a fraction. For example, if the variable S represents the amount of money in the student’s savings account and t represents time, the rate of change of dollars over time would be written like this:

dS dt

The following set of figures puts actual numbers to this hypothetical scenario:

Date: November 20

Saving account balance (S) = $12,527.33

Rate of spending ([dS/dt]) = -5,749.01 per year

List some of the equations you have seen in your study of electronics containing derivatives, and explain how rate of change relates to the real-life phenomena described by those equations.

Voltage and current for a capacitor:

i = C dv dt

Voltage and current for an inductor:

v = L di dt

Electromagnetic induction:

v = N d φ dt

I leave it to you to describe how the rate-of-change over time of one variable relates to the other variables in each of the scenarios described by these equations.
Follow-up question: why is the derivative quantity in the student’s savings account example expressed as a negative number? What would a positive [dS/dt] represent in real life?
Challenge question: describe actual circuits you could build to demonstrate each of these equations, so that others could see what it means for one variable’s rate-of-change over time to affect another variable.

Notes:
The purpose of this question is to introduce the concept of the derivative to students in ways that are familiar to them. Hopefully the opening scenario of a dwindling savings account is something they can relate to!
A very important aspect of this question is the discussion it will engender between you and your students regarding the relationship between rates of change in the three equations given in the answer. It is very important to your students’ comprehension of this concept to be able to verbally describe how the derivative works in each of these formulae. You may want to have them phrase their responses in realistic terms, as if they were describing how to set up an illustrative experiment for a classroom demonstration.

Question 2

∫f(x) dx Calculus alert!

According to the “Ohm’s Law” formula for a capacitor, capacitor current is proportional to the time-derivative of capacitor voltage:

i = C dv dt

Another way of saying this is to state that the capacitors differentiate voltage with respect to time, and express this time-derivative of voltage as a current.
Suppose we had an oscilloscope capable of directly measuring current, or at least a current-to-voltage converter that we could attach to one of the probe inputs to allow direct measurement of current on one channel. With such an instrument set-up, we could directly plot capacitor voltage and capacitor current together on the same display:



For each of the following voltage waveforms (channel B), plot the corresponding capacitor current waveform (channel A) as it would appear on the oscilloscope screen:









Note: the amplitude of your current plots is arbitrary. What I’m interested in here is the shape of each current waveform!

Follow-up question: what electronic device could perform the function of a “current-to-voltage converter” so we could use an oscilloscope to measure capacitor current? Be as specific as you can in your answer.

Notes:
Here, I ask students to relate the instantaneous rate-of-change of the voltage waveform to the instantaneous amplitude of the current waveform. Just a conceptual exercise in derivatives.

Question 3

∫f(x) dx Calculus alert!

Potentiometers are very useful devices in the field of robotics, because they allow us to represent the position of a machine part in terms of a voltage. In this particular case, a potentiometer mechanically linked to the joint of a robotic arm represents that arm’s angular position by outputting a corresponding voltage signal:



As the robotic arm rotates up and down, the potentiometer wire moves along the resistive strip inside, producing a voltage directly proportional to the arm’s position. A voltmeter connected between the potentiometer wiper and ground will then indicate arm position. A computer with an analog input port connected to the same points will be able to measure, record, and (if also connected to the arm’s motor drive circuits) control the arm’s position.
If we connect the potentiometer’s output to a differentiator circuit, we will obtain another signal representing something else about the robotic arm’s action. What physical variable does the differentiator output signal represent?

The differentiator circuit’s output signal represents the angular velocity of the robotic arm, according to the following equation:

v = dx dt

Where,
v = velocity
x = position
t = time
Follow-up question: what type of signal will we obtain if we differentiate the position signal twice (i.e. connect the output of the first differentiator circuit to the input of a second differentiator circuit)?

Notes:
This question asks students to relate the concept of time-differentiation to physical motion, as well as giving them a very practical example of how a passive differentiator circuit could be used. In reality, one must be very careful to use differentiator circuits for real-world signals because differentiators tend to amplify high-frequency noise. Since real-world signals are often “noisy,” this leads to a lot of noise in the differentiated signals.

Question 4

∫f(x) dx Calculus alert!

One of the fundamental principles of calculus is a process called integration. This principle is important to understand because it is manifested in the behavior of capacitance. Thankfully, there are more familiar physical systems which also manifest the process of integration, making it easier to comprehend.
If we introduce a constant flow of water into a cylindrical tank with water, the water level inside that tank will rise at a constant rate over time:



In calculus terms, we would say that the tank integrates water flow into water height. That is, one quantity (flow) dictates the rate-of-change over time of another quantity (height).
Like the water tank, electrical capacitance also exhibits the phenomenon of integration with respect to time. Which electrical quantity (voltage or current) dictates the rate-of-change over time of which other quantity (voltage or current) in a capacitance? Or, to re-phrase the question, which quantity (voltage or current), when maintained at a constant value, results in which other quantity (current or voltage) steadily ramping either up or down over time?

In a capacitance, voltage is the time-integral of current. That is, the applied current “through” the capacitor dictates the rate-of-change of voltage across the capacitor over time.
Challenge question: can you think of a way we could exploit the similarity of capacitive voltage/current integration to simulate the behavior of a water tank’s filling, or any other physical process described by the same mathematical relationship?

Notes:
The concept of integration doesn’t have to be overwhelmingly complex. Electrical phenomena such as capacitance and inductance may serve as excellent contexts in which students may explore and comprehend the abstract principles of calculus. The amount of time you choose to devote to a discussion of this question will depend on how mathematically adept your students are.
Hopefully, the challenge question will stir your students’ imaginations, as they realize the usefulness of electrical components as analogues for other types of physical systems.

Question 5

∫f(x) dx Calculus alert!

One of the fundamental principles of calculus is a process called integration. This principle is important to understand because it is manifested in the behavior of inductance. Thankfully, there are more familiar physical systems which also manifest the process of integration, making it easier to comprehend.
If we introduce a constant flow of water into a cylindrical tank with water, the water level inside that tank will rise at a constant rate over time:



In calculus terms, we would say that the tank integrates water flow into water height. That is, one quantity (flow) dictates the rate-of-change over time of another quantity (height).
Like the water tank, electrical inductance also exhibits the phenomenon of integration with respect to time. Which electrical quantity (voltage or current) dictates the rate-of-change over time of which other quantity (voltage or current) in an inductance? Or, to re-phrase the question, which quantity (voltage or current), when maintained at a constant value, results in which other quantity (current or voltage) steadily ramping either up or down over time?

In an inductance, current is the time-integral of voltage. That is, the applied voltage across the inductor dictates the rate-of-change of current through the inductor over time.
Challenge question: can you think of a way we could exploit the similarity of inductive voltage/current integration to simulate the behavior of a water tank’s filling, or any other physical process described by the same mathematical relationship?

Notes:
The concept of integration doesn’t have to be overwhelmingly complex. Electrical phenomena such as capacitance and inductance may serve as excellent contexts in which students may explore and comprehend the abstract principles of calculus. The amount of time you choose to devote to a discussion of this question will depend on how mathematically adept your students are.

Question 6

∫f(x) dx Calculus alert!

Both the input and the output of this circuit are square waves, although the output waveform is slightly distorted and also has much less amplitude:



You recognize one of the RC networks as a passive integrator, and the other as a passive differentiator. What does the likeness of the output waveform compared to the input waveform indicate to you about differentiation and integration as functions applied to waveforms?

Differentiation and integration are mathematically inverse functions of one another. With regard to waveshape, either function is reversible by subsequently applying the other function.
Follow-up question: this circuit will not work as shown if both R values are the same, and both C values are the same as well. Explain why, and also describe what value(s) would have to be different to allow the original square-waveshape to be recovered at the final output terminals.

Notes:
That integration and differentiation are inverse functions will probably be obvious already to your more mathematically inclined students. To others, it may be a revelation.
If time permits, you might want to elaborate on the limits of this complementarity. As anyone with calculus background knows, integration introduces an arbitrary constant of integration. So, if the integrator stage follows the differentiator stage, there may be a DC bias added to the output that is not present in the input (or visa-versa!).

⌠ ⌡ d dx [ f(x) ] dx = f(x) + C

In a circuit such as this where integration precedes differentiation, ideally there is no DC bias (constant) loss:

d dx   ⌠ ⌡ f(x) dx   = f(x)

However, since these are actually first-order “lag” and “lead” networks rather than true integration and differentiation stages, respectively, a DC bias applied to the input will not be faithfully reproduced on the output. Whereas a true integrator would take a DC bias input and produce an output with a linearly ramping bias, a passive integrator will assume an output bias equal to the input bias.


f Therefore, the subsequent differentiation stage, perfect or not, has no slope to differentiate, and thus there will be no DC bias on the output.
Incidentally, the following values work well for a demonstration circuit:

---

Footnotes:

If this is not apparent to you, I suggest performing Superposition analysis on a passive integrator (consider AC, then consider DC separately), and verify that V_DC(out) = V_DC(in). A passive differentiator circuit would have to possess an infinite time constant (τ = ∞) in order to generate this ramping output bias
!

Question 7

∫f(x) dx Calculus alert!

Determine what the response will be to a constant DC voltage applied at the input of these (ideal) circuits:

Notes:
Ask your students to frame their answers in a practical context, such as speed and distance for a moving object (where speed is the time-derivative of distance and distance is the time-integral of speed).

Question 8

∫f(x) dx Calculus alert!

In calculus, differentiation is the inverse operation of something else called integration. That is to say, differentiation “un-does” integration to arrive back at the original function (or signal). To illustrate this electronically, we may connect a differentiator circuit to the output of an integrator circuit and (ideally) get the exact same signal out that we put in:



Based on what you know about differentiation and differentiator circuits, what must the signal look like in between the integrator and differentiator circuits to produce a final square-wave output? In other words, if we were to connect an oscilloscope in between these two circuits, what sort of signal would it show us?

Follow-up question: what do the schematic diagrams of passive integrator and differentiator circuits look like? How are they similar to one another and how do they differ?

Notes:
This question introduces students to the concept of integration, following their prior familiarity with differentiation. Since they should already be familiar with other examples of inverse mathematical functions (arcfunctions in trigonometry, logs and powers, squares and roots, etc.), this should not be too much of a stretch. The fact that we may show them the cancellation of integration with differentiation should be proof enough.
In case you wish to demonstrate this principle “live” in the classroom, I suggest you bring a signal generator and oscilloscope to the class, and build the following circuit on a breadboard:



The output is not a perfect square wave, given the loading effects of the differentiator circuit on the integrator circuit, and also the imperfections of each operation (being passive rather than active integrator and differentiator circuits). However, the wave-shapes are clear enough to illustrate the basic concept.

Question 9

∫f(x) dx Calculus alert!

Plot the relationships between voltage and current for resistors of three different values (1 Ω, 2 Ω, and 3 Ω), all on the same graph:



What pattern do you see represented by your three plots? What relationship is there between the amount of resistance and the nature of the voltage/current function as it appears on the graph?
Advanced question: in calculus, the instantaneous rate-of-change of an (x,y) function is expressed through the use of the derivative notation: [dy/dx]. How would the derivative for each of these three plots be properly expressed using calculus notation? Explain how the derivatives of these functions relate to real electrical quantities.

The greater the resistance, the steeper the slope of the plotted line.
Advanced answer: the proper way to express the derivative of each of these plots is [dv/di]. The derivative of a linear function is a constant, and in each of these three cases that constant equals the resistor resistance in ohms. So, we could say that for simple resistor circuits, the instantaneous rate-of-change for a voltage/current function is the resistance of the circuit.

Notes:
Students need to become comfortable with graphs, and creating their own simple graphs is an excellent way to develop this understanding. A graphical representation of the Ohm’s Law function allows students another “view” of the concept, allowing them to more easily understand more advanced concepts such as negative resistance.
If students have access to either a graphing calculator or computer software capable of drawing 2-dimensional graphs, encourage them to plot the functions using these technological resources.
I have found it a good habit to “sneak” mathematical concepts into physical science courses whenever possible. For so many people, math is an abstract and confusing subject, which may be understood only in the context of real-life application. The studies of electricity and electronics are rich in mathematical context, so exploit it whenever possible! Your students will greatly benefit.

Question 10

∫f(x) dx Calculus alert!

Ohm’s Law tells us that the amount of current through a fixed resistance may be calculated as such:

I = E R

We could also express this relationship in terms of conductance rather than resistance, knowing that G = ¹/_R:

I = EG

However, the relationship between current and voltage for a fixed capacitance is quite different. The “Ohm’s Law” formula for a capacitor is as such:

i = C de dt

What significance is there in the use of lower-case variables for current (i) and voltage (e)? Also, what does the expression [de/dt] mean? Note: in case you think that the d’s are variables, and should cancel out in this fraction, think again: this is no ordinary quotient! The d letters represent a calculus concept known as a differential, and a quotient of two d terms is called a derivative.

Lower-case variables represent instantaneous values, as opposed to average values. The expression [de/dt], which may also be written as [dv/dt], represents the instantaneous rate of change of voltage over time.
Follow-up question: manipulate this equation to solve for the other two variables ([de/dt] = … ; C = …).

Notes:
I have found that the topics of capacitance and inductance are excellent contexts in which to introduce fundamental principles of calculus to students. The time you spend discussing this question and questions like it will vary according to your students’ mathematical abilities.
Even if your students are not ready to explore calculus, it is still a good idea to discuss how the relationship between current and voltage for a capacitance involves time. This is a radical departure from the time-independent nature of resistors, and of Ohm’s Law!

Question 11

∫f(x) dx Calculus alert!

Capacitors store energy in the form of an electric field. We may calculate the energy stored in a capacitance by integrating the product of capacitor voltage and capacitor current (P = IV) over time, since we know that power is the rate at which work (W) is done, and the amount of work done to a capacitor taking it from zero voltage to some non-zero amount of voltage constitutes energy stored (U):

P = dW dt

dW = P dt

U = W = ⌠ ⌡ P dt

Find a way to substitute capacitance (C) and voltage (V) into the integrand so you may integrate to find an equation describing the amount of energy stored in a capacitor for any given capacitance and voltage values.

U = 1 2 CV2

Notes:
The integration required to obtain the answer is commonly found in calculus-based physics textbooks, and is an easy (power rule) integration.

Question 12

∫f(x) dx Calculus alert!

Integrator circuits may be understood in terms of their response to DC input signals: if an integrator receives a steady, unchanging DC input voltage signal, it will output a voltage that changes with a steady rate over time. The rate of the changing output voltage is directly proportional to the magnitude of the input voltage:



A symbolic way of expressing this input/output relationship is by using the concept of the derivative in calculus (a rate of change of one variable compared to another). For an integrator circuit, the rate of output voltage change over time is proportional to the input voltage:

dVout dt ∝ Vin

A more sophisticated way of saying this is, “The time-derivative of output voltage is proportional to the input voltage in an integrator circuit.” However, in calculus there is a special symbol used to express this same relationship in reverse terms: expressing the output voltage as a function of the input. For an integrator circuit, this special symbol is called the integration symbol, and it looks like an elongated letter “S”:

Vout ∝ ⌠ ⌡ T 0 Vin dt

Here, we would say that output voltage is proportional to the time-integral of the input voltage, accumulated over a period of time from time=0 to some point in time we call T.
“This is all very interesting,” you say, “but what does this have to do with anything in real life?” Well, there are actually a great deal of applications where physical quantities are related to each other by time-derivatives and time-integrals. Take this water tank, for example:



One of these variables (either height H or flow F, I’m not saying yet!) is the time-integral of the other, just as V_out is the time-integral of V_in in an integrator circuit. What this means is that we could electrically measure one of these two variables in the water tank system (either height or flow) so that it becomes represented as a voltage, then send that voltage signal to an integrator and have the output of the integrator derive the other variable in the system without having to measure it!
Your task is to determine which variable in the water tank scenario would have to be measured so we could electronically predict the other variable using an integrator circuit.

Flow (F) is the variable we would have to measure, and that the integrator circuit would time-integrate into a height prediction.

Notes:
Your more alert students will note that the output voltage for a simple integrator circuit is of inverse polarity with respect to the input voltage, so the graphs should really look like this:



I have chosen to express all variables as positive quantities in order to avoid any unnecessary confusion as students attempt to grasp the concept of time integration.

Question 13

∫f(x) dx Calculus alert!

We know that the output of an integrator circuit is proportional to the time-integral of the input voltage:

Vout ∝ ⌠ ⌡ T 0 Vin dt

But how do we turn this proportionality into an exact equality, so that it accounts for the values of R and C? Although the answer to this question is easy enough to simply look up in an electronics reference book, it would be great to actually derive the exact equation from your knowledge of electronic component behaviors! Here are a couple of hints:

I = V R i = C dv dt

Vout = − 1 RC ⌠ ⌡ T 0 Vin dt

Follow-up question: why is there a negative sign in the equation?

Notes:
The two “hint” equations given at the end of the question beg for algebraic substitution, but students must be careful which variable(s) to substitute! Both equations contain an I, and both equations also contain a V. The answer to that question can only be found by looking at the schematic diagram: do the resistor and capacitor share the same current, the same voltage, or both?

Question 14

∫f(x) dx Calculus alert!

If an object moves in a straight line, such as an automobile traveling down a straight road, there are three common measurements we may apply to it: position (x), velocity (v), and acceleration (a). Position, of course, is nothing more than a measure of how far the object has traveled from its starting point. Velocity is a measure of how fast its position is changing over time. Acceleration is a measure of how fast the velocity is changing over time.
These three measurements are excellent illustrations of calculus in action. Whenever we speak of “rates of change,” we are really referring to what mathematicians call derivatives. Thus, when we say that velocity (v) is a measure of how fast the object’s position (x) is changing over time, what we are really saying is that velocity is the “time-derivative” of position. Symbolically, we would express this using the following notation:

v = dx dt

Likewise, if acceleration (a) is a measure of how fast the object’s velocity (v) is changing over time, we could use the same notation and say that acceleration is the time-derivative of velocity:

a = dv dt

Since it took two differentiations to get from position to acceleration, we could also say that acceleration is the second time-derivative of position:

a = d2x dt2

“What has this got to do with electronics,” you ask? Quite a bit! Suppose we were to measure the velocity of an automobile using a tachogenerator sensor connected to one of the wheels: the faster the wheel turns, the more DC voltage is output by the generator, so that voltage becomes a direct representation of velocity. Now we send this voltage signal to the input of a differentiator circuit, which performs the time-differentiation function on that signal. What would the output of this differentiator circuit then represent with respect to the automobile, position or acceleration? What practical use do you see for such a circuit?
Now suppose we send the same tachogenerator voltage signal (representing the automobile’s velocity) to the input of an integrator circuit, which performs the time-integration function on that signal (which is the mathematical inverse of differentiation, just as multiplication is the mathematical inverse of division). What would the output of this integrator then represent with respect to the automobile, position or acceleration? What practical use do you see for such a circuit?

The differentiator’s output signal would be proportional to the automobile’s acceleration, while the integrator’s output signal would be proportional to the automobile’s position.

a ∝ dv dt Output of differentiator

x ∝ ⌠ ⌡ T 0 v dt Output of integrator

Follow-up question: draw the schematic diagrams for these two circuits (differentiator and integrator).

Notes:
The calculus relationships between position, velocity, and acceleration are fantastic examples of how time-differentiation and time-integration works, primarily because everyone has first-hand, tangible experience with all three. Everyone inherently understands the relationship between distance, velocity, and time, because everyone has had to travel somewhere at some point in their lives. Whenever you as an instructor can help bridge difficult conceptual leaps by appeal to common experience, do so!

Question 15

∫f(x) dx Calculus alert!

A familiar context in which to apply and understand basic principles of calculus is the motion of an object, in terms of position (x), velocity (v), and acceleration (a). We know that velocity is the time-derivative of position (v = [dx/dt]) and that acceleration is the time-derivative of velocity (a = [dv/dt]). Another way of saying this is that velocity is the rate of position change over time, and that acceleration is the rate of velocity change over time.
It is easy to construct circuits which input a voltage signal and output either the time-derivative or the time-integral (the opposite of the derivative) of that input signal. We call these circuits “differentiators” and “integrators,” respectively.



Integrator and differentiator circuits are highly useful for motion signal processing, because they allow us to take voltage signals from motion sensors and convert them into signals representing other motion variables. For each of the following cases, determine whether we would need to use an integrator circuit or a differentiator circuit to convert the first type of motion signal into the second:

Converting velocity signal to position signal: (integrator or differentiator?)

Converting acceleration signal to velocity signal: (integrator or differentiator?)

Converting position signal to velocity signal: (integrator or differentiator?)

Converting velocity signal to acceleration signal: (integrator or differentiator?)

Converting acceleration signal to position signal: (integrator or differentiator?)

Also, draw the schematic diagrams for these two different circuits.

Converting velocity signal to position signal: (integrator)

Converting acceleration signal to velocity signal: (integrator)

Converting position signal to velocity signal: (differentiator)

Converting velocity signal to acceleration signal: (differentiator)

Converting acceleration signal to position signal: (two integrators!)

I’ll let you figure out the schematic diagrams on your own!

Notes:
The purpose of this question is to have students apply the concepts of time-integration and time-differentiation to the variables associated with moving objects. I like to use the context of moving objects to teach basic calculus concepts because of its everyday familiarity: anyone who has ever driven a car knows what position, velocity, and acceleration are, and the differences between them.
One way I like to think of these three variables is as a verbal sequence:



Arranged as shown, differentiation is the process of stepping to the right (measuring the rate of change of the previous variable). Integration, then, is simply the process of stepping to the left.
Ask your students to come to the front of the class and draw their integrator and differentiator circuits. Then, ask the whole class to think of some scenarios where these circuits would be used in the same manner suggested by the question: motion signal processing. Having them explain how their schematic-drawn circuits would work in such scenarios will do much to strengthen their grasp on the concept of practical integration and differentiation.

Question 16

∫f(x) dx Calculus alert!

We know that the output of a differentiator circuit is proportional to the time-derivative of the input voltage:

Vout ∝ dVin dt

But how do we turn this proportionality into an exact equality, so that it accounts for the values of R and C? Although the answer to this question is easy enough to simply look up in an electronics reference book, it would be great to actually derive the exact equation from your knowledge of electronic component behaviors! Here are a couple of hints:

I = V R i = C dv dt

Vout = − RC   dVin dt  

Follow-up question: why is there a negative sign in the equation?

Notes:
The two “hint” equations given at the end of the question beg for algebraic substitution, but students must be careful which variable(s) to substitute! Both equations contain an I, and both equations also contain a V. The answer to that question can only be found by looking at the schematic diagram: do the resistor and capacitor share the same current, the same voltage, or both?

Question 17

∫f(x) dx Calculus alert!

You are part of a team building a rocket to carry research instruments into the high atmosphere. One of the variables needed by the on-board flight-control computer is velocity, so it can throttle engine power and achieve maximum fuel efficiency. The problem is, none of the electronic sensors on board the rocket has the ability to directly measure velocity. What is available is an altimeter, which infers the rocket’s altitude (it position away from ground) by measuring ambient air pressure; and also an accelerometer, which infers acceleration (rate-of-change of velocity) by measuring the inertial force exerted by a small mass.
The lack of a “speedometer” for the rocket may have been an engineering design oversight, but it is still your responsibility as a development technician to figure out a workable solution to the dilemma. How do you propose we obtain the electronic velocity measurement the rocket’s flight-control computer needs?

One possible solution is to use an electronic integrator circuit to derive a velocity measurement from the accelerometer’s signal. However, this is not the only possible solution!

Notes:
This question simply puts students’ comprehension of basic calculus concepts (and their implementation in electronic circuitry) to a practical test.

Question 18

∫f(x) dx Calculus alert!

A Rogowski Coil is essentially an air-core current transformer that may be used to measure DC currents as well as AC currents. Like all current transformers, it measures the current going through whatever conductor(s) it encircles.
Normally transformers are considered AC-only devices, because electromagnetic induction requires a changing magnetic field ([(d φ)/dt]) to induce voltage in a conductor. The same is true for a Rogowski coil: it produces a voltage only when there is a change in the measured current. However, we may measure any current (DC or AC) using a Rogowski coil if its output signal feeds into an integrator circuit as shown:



Connected as such, the output of the integrator circuit will be a direct representation of the amount of current going through the wire.
Explain why an integrator circuit is necessary to condition the Rogowski coil’s output so that output voltage truly represents conductor current.

The coil produces a voltage proportional to the conductor current’s rate of change over time (v_coil = M [di/dt]). The integrator circuit produces an output voltage changing at a rate proportional to the input voltage magnitude ([(dv_out)/dt] ∝ v_in). Substituting algebraically:

dvout dt = M di dt

Review question: Rogowski coils are rated in terms of their mutual inductance (M). Define what “mutual inductance” is, and why this is an appropriate parameter to specify for a Rogowski coil.
Follow-up question: the operation of a Rogowski coil (and the integrator circuit) is probably easiest to comprehend if one imagines the measured current starting at 0 amps and linearly increasing over time. Qualitatively explain what the coil’s output would be in this scenario and then what the integrator’s output would be.
Challenge question: the integrator circuit shown here is an “active” integrator rather than a “passive” integrator. That is, it contains an amplifier (an “active” device). We could use a passive integrator circuit instead to condition the output signal of the Rogowski coil, but only if the measured current is purely AC. A passive integrator circuit would be insufficient for the task if we tried to measure a DC current - only an active integrator would be adequate to measure DC. Explain why.

Notes:
This question provides a great opportunity to review Faraday’s Law of electromagnetic induction, and also to apply simple calculus concepts to a practical problem. The coil’s natural function is to differentiate the current going through the conductor, producing an output voltage proportional to the current’s rate of change over time (v_out ∝ [(di_in)/dt]). The integrator’s function is just the opposite. Discuss with your students how the integrator circuit “undoes” the natural calculus operation inherent to the coil (differentiation).
The subject of Rogowski coils also provides a great opportunity to review what mutual inductance is. Usually introduced at the beginning of lectures on transformers and quickly forgotten, the principle of mutual inductance is at the heart of every Rogowski coil: the coefficient relating instantaneous current change through one conductor to the voltage induced in an adjacent conductor (magnetically linked).

v2 = M di1 dt

Unlike the iron-core current transformers (CT’s) widely used for AC power system current measurement, Rogowski coils are inherently linear. Being air-core devices, they lack the potential for saturation, hysteresis, and other nonlinearities which may corrupt the measured current signal. This makes Rogowski coils well-suited for high frequency (even RF!) current measurements, as well as measurements of current where there is a strong DC bias current in the conductor. By the way, this DC bias current may be “nulled” simply by re-setting the integrator after the initial DC power-up!
If time permits, this would be an excellent point of departure to other realms of physics, where op-amp signal conditioning circuits can be used to “undo” the calculus functions inherent to certain physical measurements (acceleration vs. velocity vs. position, for example).

Question 19

∫f(x) dx Calculus alert!

A Rogowski coil has a mutual inductance rating of 5 μH. Calculate the size of the resistor necessary in the integrator circuit to give the integrator output a 1:1 scaling with the measured current, given a capacitor size of 4.7 nF:



That is, size the resistor such that a current through the conductor changing at a rate of 1 amp per second will generate an integrator output voltage changing at a rate of 1 volt per second.

R = 1.064 kΩ

Notes:
This question not only tests students’ comprehension of the Rogowski coil and its associated calculus (differentiating the power conductor current, as well as the need to integrate its output voltage signal), but it also tests students’ quantitative comprehension of integrator circuit operation and problem-solving technique. Besides, it gives some practical context to integrator circuits!

Question 20

∫f(x) dx Calculus alert!

The chain rule of calculus states that:

dx dy dy dz = dx dz

Similarly, the following mathematical principle is also true:

dx dy = dx dz dy dz

It is very easy to build an opamp circuit that differentiates a voltage signal with respect to time, such that an input of x produces an output of [dx/dt], but there is no simple circuit that will output the differential of one input signal with respect to a second input signal.
However, this does not mean that the task is impossible. Draw a block diagram for a circuit that calculates [dy/dx], given the input voltages x and y. Hint: this circuit will make use of differentiators.
Challenge question: draw a full opamp circuit to perform this function!

Notes:
Differentiator circuits are very useful devices for making “live” calculations of time-derivatives for variables represented in voltage form. Explain to your students, for example, that the physical measurement of velocity, when differentiated with respect to time, is acceleration. Thus, a differentiator circuit connected to a tachogenerator measuring the speed of something provides a voltage output representing acceleration.
Being able to differentiate one signal in terms of another, although equally useful in physics, is not so easy to accomplish with opamps. A question such as this one highlights a practical use of calculus (the “chain rule”), where the differentiator circuit’s natural function is exploited to achieve a more advanced function.

Question 21

∫f(x) dx Calculus alert!

Ohm’s Law tells us that the amount of voltage dropped by a fixed resistance may be calculated as such:

E = IR

However, the relationship between voltage and current for a fixed inductance is quite different. The “Ohm’s Law” formula for an inductor is as such:

e = L di dt

What significance is there in the use of lower-case variables for current (i) and voltage (e)? Also, what does the expression [di/dt] mean? Note: in case you think that the d’s are variables, and should cancel out in this fraction, think again: this is no ordinary quotient! The d letters represent a calculus concept known as a differential, and a quotient of two d terms is called a derivative.

Lower-case variables represent instantaneous values, as opposed to average values. The expression [di/dt] represents the instantaneous rate of change of current over time.
Follow-up question: manipulate this equation to solve for the other two variables ([di/dt] = … ; L = …).

Notes:
I have found that the topics of capacitance and inductance are excellent contexts in which to introduce fundamental principles of calculus to students. The time you spend discussing this question and questions like it will vary according to your students’ mathematical abilities.
Even if your students are not ready to explore calculus, it is still a good idea to discuss how the relationship between current and voltage for an inductance involves time. This is a radical departure from the time-independent nature of resistors, and of Ohm’s Law!

Question 22

∫f(x) dx Calculus alert!

Digital logic circuits, which comprise the inner workings of computers, are essentially nothing more than arrays of switches made from semiconductor components called transistors. As switches, these circuits have but two states: on and off, which represent the binary states of 1 and 0, respectively.
The faster these switch circuits are able to change state, the faster the computer can perform arithmetic and do all the other tasks computers do. To this end, computer engineers keep pushing the limits of transistor circuit design to achieve faster and faster switching rates.
This race for speed causes problems for the power supply circuitry of computers, though, because of the current “surges” (technically called transients) created in the conductors carrying power from the supply to the logic circuits. The faster these logic circuits change state, the greater the [di/dt] rates-of-change exist in the conductors carrying current to power them. Significant voltage drops can occur along the length of these conductors due to their parasitic inductance:



Suppose a logic gate circuit creates transient currents of 175 amps per nanosecond (175 A/ns) when switching from the “off” state to the “on” state. If the total inductance of the power supply conductors is 10 picohenrys (9.5 pH), and the power supply voltage is 5 volts DC, how much voltage remains at the power terminals of the logic gate during one of these “surges”?

Voltage remaining at logic gate terminals during current transient = 3.338 V

Notes:
Students will likely marvel at the [di/dt] rate of 175 amps per nanosecond, which equates to 175 billion amps per second. Not only is this figure realistic, though, it is also low by some estimates (see IEEE Spectrum
magazine, July 2003, Volume 40, Number 7, in the article “Putting Passives In Their Place”). Some of your students may be very skeptical of this figure, not willing to believe that ä computer power supply is capable of outputting 175 billion amps?!”
This last statement represents a very common error students commit, and it is based on a fundamental misunderstanding of [di/dt]. “175 billion amps per second” is not the same thing as “175 billion amps”. The latter is an absolute measure, while the former is a rate of change over time. It is the difference between saying “1500 miles per hour” and “1500 miles”. Just because a bullet travels at 1500 miles per hour does not mean it will travel 1500 miles! And just because a power supply is incapable of outputting 175 billion amps does not mean it cannot output a current that changes at a rate of 175 billion amps per second!

Question 23

∫f(x) dx Calculus alert!

Inductors store energy in the form of a magnetic field. We may calculate the energy stored in an inductance by integrating the product of inductor voltage and inductor current (P = IV) over time, since we know that power is the rate at which work (W) is done, and the amount of work done to an inductor taking it from zero current to some non-zero amount of current constitutes energy stored (U):

P = dW dt

dW = P dt

U = W = ⌠ ⌡ P dt

Find a way to substitute inductance (L) and current (I) into the integrand so you may integrate to find an equation describing the amount of energy stored in an inductor for any given inductance and current values.

U = 1 2 LI2

Notes:
The integration required to obtain the answer is commonly found in calculus-based physics textbooks, and is an easy (power rule) integration.

Question 24

∫f(x) dx Calculus alert!

Define what “derivative” means when applied to the graph of a function. For instance, examine this graph:



Label all the points where the derivative of the function ([dy/dx]) is positive, where it is negative, and where it is equal to zero.

The graphical interpretation of “derivative” means the slope of the function at any given point.

Notes:
Usually students find the concept of the derivative easiest to understand in graphical form: being the slope of the graph. This is true whether or not the independent variable is time (an important point given that most “intuitive” examples of the derivative are time-based!).

Question 25

∫f(x) dx Calculus alert!

Shown here is the graph for the function y = x²:



Sketch an approximate plot for the derivative of this function.

Challenge question: derivatives of power functions are easy to determine if you know the procedure. In this case, the derivative of the function y = x² is [dy/dx] = 2x. Examine the following functions and their derivatives to see if you can recognize the “rule” we follow:

y = x³ [dy/dx] = 3x²

y = x⁴ [dy/dx] = 4x³

y = 2x⁴ [dy/dx] = 8x³

y = 10x⁵ [dy/dx] = 50x⁴

y = 2x³ + 5x² − 7x [dy/dx] = 6x² + 10x − 7

y = 5x³ − 2x − 16 [dy/dx] = 15x² − 2

y = 4x⁷ − 6x³ + 9x + 1 [dy/dx] = 28x⁶ − 18x² + 9

Notes:
Usually students find the concept of the derivative easiest to understand in graphical form: being the slope of the graph. This is true whether or not the independent variable is time (an important point given that most “intuitive” examples of the derivative are time-based!).
Even if your students are not yet familiar with the power rule for calculating derivatives, they should be able to tell that [dy/dx] is zero when x = 0, positive when x > 0, and negative when x < 0.

Question 26

∫f(x) dx Calculus alert!

Calculus is widely (and falsely!) believed to be too complicated for the average person to understand. Yet, anyone who has ever driven a car has an intuitive grasp of calculus’ most basic concepts: differentiation and integration. These two complementary operations may be seen at work on the instrument panel of every automobile:



On this one instrument, two measurements are given: speed in miles per hour, and distance traveled in miles. In areas where metric units are used, the units would be kilometers per hour and kilometers, respectively. Regardless of units, the two variables of speed and distance are related to each other over time by the calculus operations of integration and differentiation. My question for you is which operation goes which way?
We know that speed is the rate of change of distance over time. This much is apparent simply by examining the units (miles per hour indicates a rate of change over time). Of these two variables, speed and distance, which is the derivative of the other, and which is the integral of the other? Also, determine what happens to the value of each one as the other maintains a constant (non-zero) value.

Speed is the derivative of distance; distance is the integral of speed.
If the speed holds steady at some non-zero value, the distance will accumulate at a steady rate. If the distance holds steady, the speed indication will be zero because the car is at rest.

Notes:
The goal of this question is to get students thinking in terms of derivative and integral every time they look at their car’s speedometer/odometer, and ultimately to grasp the nature of these two calculus operations in terms they are already familiar with.

Question 27

∫f(x) dx Calculus alert!

Define what “integral” means when applied to the graph of a function. For instance, examine this graph:



Sketch an approximate plot for the integral of this function.

The graphical interpretation of “integral” means the area accumulated underneath the function for a given domain.

Notes:
Usually students find the concept of the integral a bit harder to grasp than the concept of the derivative, even when interpreted in graphical form. One way to help them make this “leap” is to remind them that integration and differentiation are inverse functions, then ask them to analyze the answer “backwards” (looking at the red integral plot and seeing how the blue function is the derivative of the red function). The thought process is analogous to explaining logarithms to students for the very first time: when we take the logarithm of a number, we are figuring out what power we would have to raise the base to get that number (e.g. log1000 = 3 ; 10³ = 1000). When we determine the integral of a function, we are figuring out what other function, when differentiated, would result in the given function. This is the essence of what we mean by inverse functions, and it is an important concept in algebra, trigonometry, and calculus alike.

Question 28

∫f(x) dx Calculus alert!

A forward-biased PN semiconductor junction does not possess a “resistance” in the same manner as a resistor or a length of wire. Any attempt at applying Ohm’s Law to a diode, then, is doomed from the start.
This is not to say that we cannot assign a dynamic value of resistance to a PN junction, though. The fundamental definition of resistance comes from Ohm’s Law, and it is expressed in derivative form as such:

R = dV dI

The fundamental equation relating current and voltage together for a PN junction is Shockley’s diode equation:

I = IS (e[qV/NkT] − 1)

At room temperature (approximately 21 degrees C, or 294 degrees K), the thermal voltage of a PN junction is about 25 millivolts. Substituting 1 for the non-ideality coefficient, we may simply the diode equation as such:

I = IS (e[V/0.025] − 1) or I = IS (e40 V − 1)

Differentiate this equation with respect to V, so as to determine [dI/dV], and then reciprocate to find a mathematical definition for dynamic resistance ([dV/dI]) of a PN junction. Hints: saturation current (I_S) is a very small constant for most diodes, and the final equation should express dynamic resistance in terms of thermal voltage (25 mV) and diode current (I).

r ≈ 25 mV I

Notes:
The result of this derivation is important in the analysis of certain transistor amplifiers, where the dynamic resistance of the base-emitter PN junction is significant to bias and gain approximations. I show the solution steps for you here because it is a neat application of differentiation (and substitution) to solve a real-world problem:

I = IS (e40 V − 1)

dI dV = IS (40e40 V − 0)

dI dV = 40 IS e40 V

Now, we manipulate the original equation to obtain a definition for I_S e^{40 V} in terms of current, for the sake of substitution:

I = IS (e40 V − 1)

I = IS e40 V − IS

I + IS = IS e40 V

Substituting this expression into the derivative:

dI dV = 40 (I + IS)

Reciprocating to get voltage over current (the proper form for resistance):

dV dI = 0.025 I + IS

Now we may get rid of the saturation current term, because it is negligibly small:

dV dI ≈ 0.025 I

r ≈ 25 mV I

The constant of 25 millivolts is not set in stone, by any means. Its value varies with temperature, and is sometimes given as 26 millivolts or even 30 millivolts.

Question 29

∫f(x) dx Calculus alert!

Just as addition is the inverse operation of subtraction, and multiplication is the inverse operation of division, a calculus concept known as integration is the inverse function of differentiation. Symbolically, integration is represented by a long “S”-shaped symbol called the integrand:

If x = dy dt ( x is the derivative of y with respect to t)

Then y = ⌠ ⌡ x dt ( y is the integral of x with respect to t)

To be truthful, there is a bit more to this reciprocal relationship than what is shown above, but the basic idea you need to grasp is that integration “un-does” differentiation, and visa-versa. Derivatives are a bit easier for most people to understand, so these are generally presented before integrals in calculus courses. One common application of derivatives is in the relationship between position, velocity, and acceleration of a moving object. Velocity is nothing more than rate-of-change of position over time, and acceleration is nothing more than rate-of-change of velocity over time:

v = dx dt Velocity (v ) is the time−derivative of position (x )

a = dv dt Acceleration (a ) is the time−derivative of velocity (v )

Illustrating this in such a way that shows differentiation as a process:



Given that you know integration is the inverse-function of differentiation, show how position, velocity, and acceleration are related by integration. Show this both in symbolic (proper mathematical) form as well as in an illustration similar to that shown above.

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Footnotes:

It is perfectly accurate to say that differentiation undoes integration, so that [d/dt] ∫x dt = x, but to say that integration undoes differentiation is not entirely true because indefinite integration always leaves a constant C that may very well be non-zero, so that ∫[dx/dt] dt = x + C rather than simply being x.

x = ⌠ ⌡ v dt Position (x ) is the time−integral of velocity (v )

v = ⌠ ⌡ a dt Velocity (v ) is the time−integral of acceleration (a )

Challenge question: explain why the following equations are more accurate than those shown in the answer.

x = ⌠ ⌡ v dt + x0 Position (x ) is the time−integral of velocity (v )

v = ⌠ ⌡ a dt + v0 Velocity (v ) is the time−integral of acceleration (a )

Where,
x₀ is the initial position (at time = 0)
v₀ is the initial velocity (at time = 0)

Notes:
The purpose of this question is to introduce the integral as an inverse-operation to the derivative. Introducing the integral in this manner (rather than in its historical origin as an accumulation of parts) builds on what students already know about derivatives, and prepares them to see integrator circuits as counterparts to differentiator circuits rather than as unrelated entities.

Question 30

∫f(x) dx Calculus alert!

Calculus is a branch of mathematics that originated with scientific questions concerning rates of change. The easiest rates of change for most people to understand are those dealing with time. For example, a student watching their savings account dwindle over time as they pay for tuition and other expenses is very concerned with rates of change (dollars per day being spent).
The “derivative” is how rates of change are symbolically expressed in mathematical equations. For example, if the variable S represents the amount of money in the student’s savings account and t represents time, the rate of change of dollars over time (the time-derivative of the student’s account balance) would be written as [dS/dt]. The process of calculating this rate of change from a record of the account balance over time, or from an equation describing the balance over time, is called differentiation.
Suppose, though, that instead of the bank providing the student with a statement every month showing the account balance on different dates, the bank were to provide the student with a statement every month showing the rates of change of the balance over time, in dollars per day, calculated at the end of each day:



Explain how the Isaac Newton Credit Union calculates the derivative ([dS/dt]) from the regular account balance numbers (S in the Humongous Savings & Loan statement), and then explain how the student who banks at Isaac Newton Credit Union could figure out how much money is in their account at any given time.
Hint: the process of calculating a variable’s value from rates of change is called integration in calculus. It is the opposite (inverse) function of differentiation.   
The Isaac Newton Credit Union differentiates S by dividing the difference between consecutive balances by the number of days between those balance figures. Differentiation is fundamentally a process of division.
To integrate the [dS/dt] values shown on the Credit Union’s statement so as to arrive at values for S, we must either repeatedly add or subtract the days’ rate-of-change figures, beginning with a starting balance. Thus, integration is fundamentally a process of multiplication.
Follow-up question: explain why a starting balance is absolutely necessary for the student banking at Isaac Newton Credit Union to know in order for them to determine their account balance at any time. Why would it be impossible for them to figure out how much money was in their account if the only information they possessed was the [dS/dt] figures?

Notes:
The purpose of this question is to introduce the concept of the integral to students in a way that is familiar to them. Hopefully the opening scenario of a dwindling savings account is something they can relate to!
Some students may ask why the differential notation [dS/dt] is used rather than the difference notation [(∆S)/(∆t)] in this example, since the rates of change are always calculated by subtraction of two data points (thus implying a ∆). Given that the function here is piecewise and not continuous, one could argue that it is not differentiable at the points of interest. My purpose in using differential notation is to familiarize students with the concept of the derivative in the context of something they can easily relate to, even if the particular details of the application suggest a more correct notation. 


                                                                  X  .  IIIIII   

  

                             JFET Amplifiers

Discrete Semiconductor Devices and Circuits

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Question 1

Don’t just sit there! Build something!!

Learning to mathematically analyze circuits requires much study and practice. Typically, students practice by working through lots of sample problems and checking their answers against those provided by the textbook or the instructor. While this is good, there is a much better way.
You will learn much more by actually building and analyzing real circuits, letting your test equipment provide the “answers” instead of a book or another person. For successful circuit-building exercises, follow these steps:

1. Carefully measure and record all component values prior to circuit construction, choosing resistor values high enough to make damage to any active components unlikely.
2. Draw the schematic diagram for the circuit to be analyzed.
3. Carefully build this circuit on a breadboard or other convenient medium.
4. Check the accuracy of the circuit’s construction, following each wire to each connection point, and verifying these elements one-by-one on the diagram.
5. Mathematically analyze the circuit, solving for all voltage and current values.
6. Carefully measure all voltages and currents, to verify the accuracy of your analysis.
7. If there are any substantial errors (greater than a few percent), carefully check your circuit’s construction against the diagram, then carefully re-calculate the values and re-measure.

When students are first learning about semiconductor devices, and are most likely to damage them by making improper connections in their circuits, I recommend they experiment with large, high-wattage components (1N4001 rectifying diodes, TO-220 or TO-3 case power transistors, etc.), and using dry-cell battery power sources rather than a benchtop power supply. This decreases the likelihood of component damage.
As usual, avoid very high and very low resistor values, to avoid measurement errors caused by meter “loading” (on the high end) and to avoid transistor burnout (on the low end). I recommend resistors between 1 kΩ and 100 kΩ.
One way you can save time and reduce the possibility of error is to begin with a very simple circuit and incrementally add components to increase its complexity after each analysis, rather than building a whole new circuit for each practice problem. Another time-saving technique is to re-use the same components in a variety of different circuit configurations. This way, you won’t have to measure any component’s value more than once.

Let the electrons themselves give you the answers to your own “practice problems”!

Notes:
It has been my experience that students require much practice with circuit analysis to become proficient. To this end, instructors usually provide their students with lots of practice problems to work through, and provide answers for students to check their work against. While this approach makes students proficient in circuit theory, it fails to fully educate them.
Students don’t just need mathematical practice. They also need real, hands-on practice building circuits and using test equipment. So, I suggest the following alternative approach: students should build their own “practice problems” with real components, and try to mathematically predict the various voltage and current values. This way, the mathematical theory “comes alive,” and students gain practical proficiency they wouldn’t gain merely by solving equations.
Another reason for following this method of practice is to teach students scientific method: the process of testing a hypothesis (in this case, mathematical predictions) by performing a real experiment. Students will also develop real troubleshooting skills as they occasionally make circuit construction errors.
Spend a few moments of time with your class to review some of the “rules” for building circuits before they begin. Discuss these issues with your students in the same Socratic manner you would normally discuss the worksheet questions, rather than simply telling them what they should and should not do. I never cease to be amazed at how poorly students grasp instructions when presented in a typical lecture (instructor monologue) format!
A note to those instructors who may complain about the “wasted” time required to have students build real circuits instead of just mathematically analyzing theoretical circuits:
What is the purpose of students taking your course?
If your students will be working with real circuits, then they should learn on real circuits whenever possible. If your goal is to educate theoretical physicists, then stick with abstract analysis, by all means! But most of us plan for our students to do something in the real world with the education we give them. The “wasted” time spent building real circuits will pay huge dividends when it comes time for them to apply their knowledge to practical problems.
Furthermore, having students build their own practice problems teaches them how to perform primary research, thus empowering them to continue their electrical/electronics education autonomously.
In most sciences, realistic experiments are much more difficult and expensive to set up than electrical circuits. Nuclear physics, biology, geology, and chemistry professors would just love to be able to have their students apply advanced mathematics to real experiments posing no safety hazard and costing less than a textbook. They can’t, but you can. Exploit the convenience inherent to your science, and get those students of yours practicing their math on lots of real circuits!

Question 2

This relaxation oscillator circuit uses a resistor-capacitor combination (R₁ - C₁) to establish the time delay between output pulses:



The voltage measured between TP1 and ground looks like this on the oscilloscope display:



A slightly different version of this circuit adds a JFET to the capacitor’s charge current path:



Now, the voltage at TP1 looks like this:



What function does the JFET perform in this circuit, based on your analysis of the new TP1 signal waveform? The straight-line charging voltage pattern shown on the second oscilloscope display indicates what the JFET is doing in this circuit.
Hint: you don’t need to know anything about the function of the unijunction transistor (at the circuit’s output) other than it acts as an on/off switch to periodically discharge the capacitor when the TP1 voltage reaches a certain threshold level.
Challenge question: write a formula predicting the slope of the ramping voltage waveform measured at TP1.

The JFET in this circuit functions as a constant current regulator.
Answer to challenge question: Slope = [dv/dt] = [(I_D)/C]

Notes:
Ask your students how they would know to relate “constant current” to the peculiar charging action of this capacitor. Ask them to explain this mathematically.
Then, ask them to explain exactly how the JFET works to regulate charging current.
Note: the schematic diagram for this circuit was derived from one found on page 958 of John Markus’
Guidebook of Electronic Circuits, first edition. Apparently, the design originated from a Motorola publication on using unijunction transistors (“Unijunction Transistor Timers and Oscillators,” AN-294, 1972).

Question 3

A student builds this transistor amplifier circuit on a solderless “breadboard”:



The purpose of the potentiometer is to provide an adjustable DC bias voltage for the transistor, so it may be operated in Class-A mode. After some adjustment of this potentiometer, the student is able to obtain good amplification from the transistor (signal generators and oscilloscopes have been omitted from the illustration for simplicity).
Later, the student accidently adjusts the power supply voltage to a level beyond the JFET’s rating, destroying the transistor. Re-setting the power supply voltage back where the student began the experiment and replacing the transistor, the student discovers that the biasing potentiometer must be re-adjusted to achieve good Class-A operation.
Intrigued by this discovery, the student decides to replace this transistor with a third (of the same part number, of course), just to see if the biasing potentiometer needs to be adjusted again for good Class-A operation. It does.
Explain why this is so. Why must the gate biasing potentiometer be re-adjusted every time the transistor is replaced, even if the replacement transistor(s) are of the exact same type?

This amplifier circuit uses gate bias, which is a notoriously unstable method of biasing a JFET amplifier circuit.

Notes:
Ask your students to explain exactly what it is that causes the Q point of this amplifier circuit to change with each new transistor. Is it something in the transistor itself, or in some other part of the circuit?
Given the instability of gate biasing, should this method be used in mass-produced amplifier circuits? Ask your students to elaborate on why or why not.

Question 4

The simple JFET amplifier circuit shown here (built with surface-mount components) employs a biasing technique known as self-biasing:



Self-biasing provides much greater Q-point stability than gate-biasing. Draw a schematic diagram of this circuit, and then explain how self-biasing works.

Self-biasing uses the negative feedback created by a source resistor to establish a “natural” Q-point for the amplifier circuit, rather than having to supply an external voltage as is done with gate biasing.

Notes:
The concept of negative feedback is extremely important in electronic circuits, but it is not easily grasped by all. Self-biasing of JFET transistors is a relatively easy-to-understand application of negative feedback, so be sure to take advantage of this opportunity to explore the concept with your students.
Ask your students to explain why Q-point stability is a desirable feature for mass-produced amplifier circuits, as well as circuits subject to component-level repair.

Question 5

The voltage gain for a “bypassed” common-emitter BJT amplifier circuit is as follows:



Common-source JFET amplifier circuits are very similar:



One of the problems with “bypassed” amplifier configurations such as the common-emitter and common-source is voltage gain variability. It is difficult to keep the voltage gain stable in either type of amplifier, due to changing factors within the transistors themselves which cannot be tightly controlled (r′_e and g_m, respectively). One solution to this dilemma is to “swamp” those uncontrollable factors by not bypassing the emitter (or source) resistor. The result is greater A_V stability at the expense of A_V magnitude:



Write the voltage gain equations for both “swamped” BJT and JFET amplifier configurations, and explain why they are similar to each other.

AV ≈ RC RE Common−emitter BJT amplifier

AV ≈ RD RS Common−source JFET amplifier

I’ll let you explain why these two voltage gain approximations share the same form. Hint: it has something to do with the magnitudes of the currents through each transistor terminal!
Follow-up question: explain mathematically why the emitter/source resistances succeed in “swamping” r′_e and g_m, respectively, in these more precise formulae. You should provide typical values for r′_e and g_m as part of your argument:

AV = RC RE + r′e Common−emitter BJT amplifier

AV = RD RS + 1 gm Common−source JFET amplifier

Notes:
Swamping is a common engineering practice, and one that students would do well to understand. It is unfortunate that parameters such as dynamic emitter resistance (r′_e) and transconductance (g_m) are so variable, but this does not have to be the end of the story. To be able to work around practical limitations such as these is the essence of engineering practice, in my opinion.

Question 6

The circuit shown here is a precision DC voltmeter:



Explain why this circuit design requires the use of a field-effect transistor, and not a bipolar junction transistor (BJT).
Also, answer the following questions about the circuit:

Explain, step by step, how an increasing input voltage between the test probes causes the meter movement to deflect further.

If the most sensitive range of this voltmeter is 0.1 volts (full-scale), calculate the other range values, and label them on the schematic next to their respective switch positions.

What type of JFET configuration is this (common-gate, common-source, or common-drain)?

What purpose does the capacitor serve in this circuit?

What detrimental effect would result from installing a capacitor that was too large?

Estimate a reasonable value for the capacitor’s capacitance.

Explain the functions of the “Zero” and “Span” calibration potentiometers.

The voltage ranges for this meter are as follows:

0.1 volts

0.2 volts

1.0 volts

2.0 volts

10 volts

20 volts

The JFET is being used in the common drain configuration. A reasonable value for the capacitor would be 0.01 μF.

Notes:
This relatively simple DC voltage amplifier circuit provides a wealth of educational value, both for understanding the function of the JFET, and also for review on past electrical/electronics concepts.
Note: John Markus’
Guidebook of Electronic Circuits, first edition, page 469, provided the inspiration for this circuit.

Question 7

This is a schematic of an RF amplifier using a JFET as the active element:



What configuration of JFET amplifier is this (common drain, common gate, or common source)? Also, explain the purpose of the two iron-core inductors in this circuit. Hint: inductors L₁ and L₂ are often referred to as RF chokes.

This is a common-gate amplifier. The iron-core inductors block (“choke”) the high-frequency AC signals from getting to the DC power supply.

Notes:
Be sure to ask your students why it would not be good for the RF signals to find their way to the DC power supply. There is more than one possible answer to this question!
This schematic was derived from an evaluation amplifier schematic shown in an
ON Semiconductor J308/J309/J310 transistor datasheet.

Question 8

Calculate the approximate input impedance of this JFET amplifier circuit:



Explain why it is easier to calculate the Z_in of a JFET circuit like this than it is to calculate the Z_in of a similar bipolar transistor amplifier circuit. Also, explain how calculation of this amplifier’s output impedance compares with that of a similar BJT amplifier circuit - same approach or different approach?

Z_in = 89.2 kΩ

Notes:
Ask your students to explain why input impedance is an important factor in amplifier design. Why should we care how much input impedance an amplifier has?
Also, ask your students to explain why such high-value bias resistors (150 kΩ and 220 kΩ) would probably not be practical in a BJT amplifier circuit.

Question 9

Define what a common-source transistor amplifier circuit is. What distinguishes this amplifier configuration from the other single-FET amplifier configurations, namely common-drain and common-gate? What configuration of BJT amplifier circuit does the common-source FET circuit most resemble in form and behavior?
Also, describe the typical voltage gains of this amplifier configuration, and whether it is inverting or non-inverting.

The common-source amplifier configuration is defined by having the input and output signals referenced to the gate and drain terminals (respectively), with the source terminal of the transistor typically having a low AC impedance to ground and thus being “common” to one pole of both the input and output voltages.
The common-source amplifier configuration most resembles the common-emitter BJT amplifier configuration in both form and behavior.
Common-source amplifiers are characterized by moderate voltage gains, and an inverting phase relationship between input and output.

Notes:
The answers to the question may be easily found in any fundamental electronics text, but it is important to ensure students know why these characteristics are such. I always like to tell my students, “Memory will fail you, so you need to build an understanding of why things are, not just what things are.”
One exercise you might have your students do is come up to the board in front of the room and draw an example of this circuit, then everyone may refer to the drawn image when discussing the circuit’s characteristics.

Question 10

Define what a common-gate transistor amplifier circuit is. What distinguishes this amplifier configuration from the other single-FET amplifier configurations, namely common-drain and common-source? What configuration of BJT amplifier circuit does the common-gate FET circuit most resemble in form and behavior?
Also, describe the typical voltage gains of this amplifier configuration, and whether it is inverting or non-inverting.

The common-gate amplifier configuration is defined by having the input and output signals referenced to the source and drain terminals (respectively), with the gate terminal of the transistor typically having a low AC impedance to ground and thus being “common” to one pole of both the input and output voltages.
The common-gate amplifier configuration most resembles the common-base BJT amplifier configuration in both form and behavior.
Common-gate amplifiers are characterized by moderate voltage gains, and a non-inverting phase relationship between input and output.

Notes:
The answers to the question may be easily found in any fundamental electronics text, but it is important to ensure students know why these characteristics are such. I always like to tell my students, “Memory will fail you, so you need to build an understanding of why things are, not just what things are.”
One exercise you might have your students do is come up to the board in front of the room and draw an example of this circuit, then everyone may refer to the drawn image when discussing the circuit’s characteristics.

Question 11

Define what a common-drain transistor amplifier circuit is. What distinguishes this amplifier configuration from the other single-FET amplifier configurations, namely common-source and common-gate? What configuration of BJT amplifier circuit does the common-drain FET circuit most resemble in form and behavior?
Also, describe the typical voltage gains of this amplifier configuration, and whether it is inverting or non-inverting.

The common-drain amplifier configuration is defined by having the input and output signals referenced to the gate and source terminals (respectively), with the drain terminal of the transistor typically having a low AC impedance to ground and thus being “common” to one pole of both the input and output voltages.
The common-drain amplifier configuration most resembles the common-collector BJT amplifier configuration in both form and behavior.
Common-drain amplifiers are characterized by low voltage gains (less than unity), and a non-inverting phase relationship between input and output.

Notes:
The answers to the question may be easily found in any fundamental electronics text, but it is important to ensure students know why these characteristics are such. I always like to tell my students, “Memory will fail you, so you need to build an understanding of why things are, not just what things are.”
One exercise you might have your students do is come up to the board in front of the room and draw an example of this circuit, then everyone may refer to the drawn image when discussing the circuit’s characteristics.

Question 12

Determine whether this amplifier circuit is inverting or non-inverting (i.e. the phase shift between input and output waveforms):



Be sure to explain, step by step, how you were able to determine the phase relationship between input and output in this circuit. Also identify the type of amplifier each transistor represents (common-???).

Non-inverting. The JFET is connected as a common-source, while the BJT is connected as a common-emitter.

Notes:
There are several other questions you could ask about this amplifier circuit. For example:

How is the Q-point bias established for the JFET?

How is the Q-point bias established for the BJT?

What purpose does the potentiometer serve?

Is there another possible location for the potentiometer that would perform the same function?

Note: the schematic diagram for this circuit was derived from one found on page 36 of John Markus’
Guidebook of Electronic Circuits, first edition. Apparently, the design originated from a Motorola publication on using field effect transistors (“Tips on using FET’s,” HMA-33, 1971).

Question 13

It is a well-known fact that temperature affects the operating parameters of bipolar junction transistors. This is why grounded-emitter circuits (with no emitter feedback resistor) are not practical as stand-alone amplifier circuits.
Does temperature affect junction field-effect transistors in the same way, or to the same extent? Design an experiment to determine the answer to this question.

Did you really think I would tell you the answer to this question? Build the circuit(s) and discover the answer for yourselves!

Notes:
The purpose of this question is to get students thinking in an experimental mode. It is very important that students learn to set up and run their own experiments, so they will be able to verify (or perhaps discover!) electronic principles after they have graduated from school. There will be times when the answers they seek are not to be found in a book, and they will have to “let the electrons teach them” what they need to know.
Remind your students that proper scientific experiments include both experimental and control subjects, so that results are based upon a comparison of measurements.

Question 14

Identify what type of amplifier circuit this is, and also what would happen to the output voltage if V_in2 were to become more positive:

This is a differential amplifier circuit. If V_in2 were to become more positive, V_out would become more negative.

Notes:
Students should be able to relate this circuit to its bipolar transistor counterpart. Ask them to explain what advantages or disadvantages this circuit holds over a bipolar differential amplifier circuit.

Question 15

The following circuit is a “multi-coupler” for audio signals: one audio signal source (such as a microphone) is distributed to three different outputs:



Suppose an audio signal is getting through from the input to outputs 2 and 3, but not through to output 1. Identify possible failures in the circuit that could cause this. Be as specific as you can, and identify how you would confirm each type of failure using a multimeter. 


X  .  IIIIIII 

Digital Communication

Digital Circuits

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Question 1

Don’t just sit there! Build something!!

Learning to analyze digital circuits requires much study and practice. Typically, students practice by working through lots of sample problems and checking their answers against those provided by the textbook or the instructor. While this is good, there is a much better way.
You will learn much more by actually building and analyzing real circuits, letting your test equipment provide the “answers” instead of a book or another person. For successful circuit-building exercises, follow these steps:

1. Draw the schematic diagram for the digital circuit to be analyzed.
2. Carefully build this circuit on a breadboard or other convenient medium.
3. Check the accuracy of the circuit’s construction, following each wire to each connection point, and verifying these elements one-by-one on the diagram.
4. Analyze the circuit, determining all output logic states for given input conditions.
5. Carefully measure those logic states, to verify the accuracy of your analysis.
6. If there are any errors, carefully check your circuit’s construction against the diagram, then carefully re-analyze the circuit and re-measure.

Always be sure that the power supply voltage levels are within specification for the logic circuits you plan to use. If TTL, the power supply must be a 5-volt regulated supply, adjusted to a value as close to 5.0 volts DC as possible.
One way you can save time and reduce the possibility of error is to begin with a very simple circuit and incrementally add components to increase its complexity after each analysis, rather than building a whole new circuit for each practice problem. Another time-saving technique is to re-use the same components in a variety of different circuit configurations. This way, you won’t have to measure any component’s value more than once.

Let the electrons themselves give you the answers to your own “practice problems”!

Notes:
It has been my experience that students require much practice with circuit analysis to become proficient. To this end, instructors usually provide their students with lots of practice problems to work through, and provide answers for students to check their work against. While this approach makes students proficient in circuit theory, it fails to fully educate them.
Students don’t just need mathematical practice. They also need real, hands-on practice building circuits and using test equipment. So, I suggest the following alternative approach: students should build their own “practice problems” with real components, and try to predict the various logic states. This way, the digital theory “comes alive,” and students gain practical proficiency they wouldn’t gain merely by solving Boolean equations or simplifying Karnaugh maps.
Another reason for following this method of practice is to teach students scientific method: the process of testing a hypothesis (in this case, logic state predictions) by performing a real experiment. Students will also develop real troubleshooting skills as they occasionally make circuit construction errors.
Spend a few moments of time with your class to review some of the “rules” for building circuits before they begin. Discuss these issues with your students in the same Socratic manner you would normally discuss the worksheet questions, rather than simply telling them what they should and should not do. I never cease to be amazed at how poorly students grasp instructions when presented in a typical lecture (instructor monologue) format!
I highly recommend CMOS logic circuitry for at-home experiments, where students may not have access to a 5-volt regulated power supply. Modern CMOS circuitry is far more rugged with regard to static discharge than the first CMOS circuits, so fears of students harming these devices by not having a “proper” laboratory set up at home are largely unfounded.
A note to those instructors who may complain about the “wasted” time required to have students build real circuits instead of just mathematically analyzing theoretical circuits:
What is the purpose of students taking your course?
If your students will be working with real circuits, then they should learn on real circuits whenever possible. If your goal is to educate theoretical physicists, then stick with abstract analysis, by all means! But most of us plan for our students to do something in the real world with the education we give them. The “wasted” time spent building real circuits will pay huge dividends when it comes time for them to apply their knowledge to practical problems.
Furthermore, having students build their own practice problems teaches them how to perform primary research, thus empowering them to continue their electrical/electronics education autonomously.
In most sciences, realistic experiments are much more difficult and expensive to set up than electrical circuits. Nuclear physics, biology, geology, and chemistry professors would just love to be able to have their students apply advanced mathematics to real experiments posing no safety hazard and costing less than a textbook. They can’t, but you can. Exploit the convenience inherent to your science, and get those students of yours practicing their math on lots of real circuits!

Question 2

Explain the difference between serial digital data and parallel digital data.

Serial data is transmitted along one line, one bit at a time; parallel data is transmitted all at once.

Notes:
Ask your students of they have ever heard of “serial” and “parallel” ports on personal computers. If time permits, have them examine the two types of ports on the back of a PC, contrasting the number of pins used for each connector.

Question 3

The following schematic diagram shows two four-bit universal shift registers used to communicate data serially over a coaxial cable of unspecified length:



Specify what logic states would have to be input at the PL, CE, and Clk terminals of each shift register, and at what times, to successfully load four bits of parallel data, shift them serially over the coaxial data cable, and then hold them at the outputs (Q) of the receiving shift register.

I won’t give you all the details here, but I will get you started with a few steps:

De-activate the clock enable (CE) inputs of both shift registers.

Apply the four desired bits (logic levels) to the D₀ through D₃ inputs of the left-hand shift register.

Briefly activate the parallel load (PL) input of the left-hand shift register.

Activate the clock enable (CE) inputs of both shift registers simultaneously for four clock pulses.

etc.

etc. . . .

Notes:
This question asks students to think their way through the operation of two coupled shift registers to accomplish the task of parallel-to-serial-to-parallel data conversion. Not only is this a good review of shift register operation, but it shows some (not all!) of what happens during the seemingly simple procedure of sending four bits of data in serial form over a cable.
A challenging detail to figure out in this scheme is how to keep both shift registers synchronized so that one receives the serial data bits at the same time the other sends them. There is more than one way to do this, of course, but the easiest would be to connect the two clock inputs together through another cable conductor.

Question 4

Personal computers and peripheral devices provide a rich source of examples for both serial and parallel data transmission. Identify some common examples of both serial and parallel data transmission networks (and standards) at work in a common personal computer. Examples may include communication between computers, between computers and peripheral devices (printers, scanners, cameras, special cards), or between fundamental components of the computer (CPU, disk drive, monitor, etc.).

Examples of serial data communication include the 9-pin and/or 25 pin “serial” connectors for RS-232C communication, Ethernet communication, USB ports, and most “mice.” Examples of parallel data communication include 25-pin “parallel” connectors to printer and scanner devices, and cables between motherboard and disk drives (legacy IDE technology).

Notes:
As computer-savvy as most young students are, questions such as these tend to evoke ready responses and strong interest. You may find that little effort is required on your part to introduce these technologies to your students, as they may be more familiar with certain areas and features of it than you!

Question 5

A ubiquitous example of serial data communication is the cable linking a keyboard to a personal computer: for every key switch pressed, an ASCII character is transmitted to the computer. An interesting characteristic of this particular communication protocol is the random rate at which the ASCII characters are sent. Because the characters are generated at the rate the computer user happens to type, the rate is completely unpredictable. Consequently, this form of serial data communication is known as asynchronous.
Compare and contrast this against synchronous serial data communication, giving an example of a synchronous data communications standard.

One widespread synchronous data communications standard is SONET, used in long-distance data communication applications. I’ll let you do the research to compare and contrast synchronous against asynchronous.
Challenge question: the data sent between computers along serial-format networks such as RS-232C and Ethernet is “clocked” by precise oscillators at both the transmitting and receiving ends, yet is not considered “synchronous,” even if each byte of data is sent at regular (non-random) intervals. Explain why.

Notes:
At first, it seems as though any communication between digital devices occurring at a pre-determined frequency (bps) and rate (characters per second) would be synchronous, because everything is happening on fixed intervals. However, the precision inherent to a true synchronous communications network is more rigorous than this. Let your students elaborate on what they have found through their research.

Question 6

An important integrated circuit (IC) used in digital data communication is a UART. Describe what this acronym stands for, and explain the purpose of this circuit.

“UART” stands for Universal Asynchronous Receiver Transmitter, and its job is to act as an interface between two parallel-data devices, managing communications in serial format along a communication line of some sort.
Follow-up question: give an example of a UART IC available for purchase today.

Notes:
When students research what a UART is, they will invariably stumble upon terms such as parity, start bit, and stop bit. If they are not yet familiar with the details of asynchronous data communications, this could lead to some enlightening discoveries. Be sure to discuss these terms and details with your students if they bring them up in class, because it means they will be very receptive to your instruction (having been “primed” for learning by wanting to know).

Question 7

Shown here are three different telegraph circuits. Determine which of these could be classified as simplex, full-duplex, and half-duplex, in terms of serial data transmission:

Simplex: one-way communication

Half-duplex: two-way communication, one way at a time.

Full-duplex: two-way communication, both ways simultaneously.

Follow-up question: trace all currents in these circuits using conventional flow, and then electron flow.

Notes:
I could have just asked for definitions here, but relating these concepts to real circuits, however simple, carries with it more educational benefit. It is also important to show students that the basic concept of digital communication is really no more complex than the old telegraph, just faster.

Question 8

An early form of digital, serial data communication was Morse code. Explain what “Morse code” is (or was), and how it compares to more modern codes such as ASCII.

Morse code was a simple convention used to represent alphanumeric characters for telegraph data transmission. At first, human operators served the task of parallel-to-serial-to-parallel data converters, but then machines were built to do this automatically.

Notes:
An interesting feature of Morse code likely not recognized by your students is inherent compression. Because some Morse characters are shorter than others (fewer pulses) as opposed to ASCII where all characters are the same length, messages sent in Morse tend to require fewer bits than messages sent in ASCII.

Question 9

An important performance parameter of digital communications networks is the number of bits per second (bps) of data it can handle. Unfortunately, a different term called baud is often used interchangeably with bps. Define what “baud” is, and how it differs from “bits per second.”

“Baud” technically refers to the number of logic level transitions (low-to-high or high-to-low) per second on a network, while “bps” actually refers to the number of data bits transmitted per second. For a specific application where the two terms significantly differ, research a method of data modulation known as Manchester encoding.

Notes:
While some may argue the difference to be academic, I believe that precision of language and precision of thinking are closely related. The person who does not recognize the difference between “baud” and “bps” most likely knows little about how digital information is encoded for serial transmission. Of course, this is the ultimate issue - understanding how digital data is transmitted. So while we’re at it, we might as well address a common mis-use of language and gain a deeper understanding of how things work, right?

Question 10

In Claude Shannon’s famous 1948 paper entitled A Mathematical Theory of Communication, he opens with the following statement:

“The recent development of various methods of modulation such as PCM and PPM which exchange bandwidth for signal-to-noise ratio has intensified the interest in a general theory of communication.”

Explain what Shannon was referring to when he said, “exchange bandwidth for signal-to-noise ratio”. In many cases, the superior signal-to-noise ratio of digital communication over analog communication is the primary reason justifying the much greater complexity of digital communications equipment. Also, elaborate on how bandwidth becomes sacrificed in order to achieve relatively noiseless signal transmission.  
Digital signals are highly resistant to corruption from noise, because they are composed of discrete (“high” and “low”) states rather than continuously-variable quantities as analog signals are. However, in order to communicate any significant measure of digital information in serial form, many pulses are needed. This requires a high-bandwidth data path to be comparable in speed to analog.
 

Notes:
Shannon’s language is perhaps a bit above the norm for technician-level education, but it nevertheless captures an important quality of digital communication: that the noise immunity enjoyed by digital communication comes at a price: high bandwidth. Without a high-bandwidth medium in which to exchange digital information, communication is either slow or completely impractical.

 


X  .  IIIIIII 

Design Project: Pulse-Width Modulation (PWM) Signal Generator

Analog Integrated Circuits

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Question 1

What shape of voltage waveform would you expect to measure (using an oscilloscope) across capacitor C₁? How does this waveform interact with the DC reference voltage at the wiper of R_pot2 to produce a pulse-width modulated square wave output?

The waveform will be a “sawtooth” shape. When compared against the DC reference voltage at R_pot2‘s wiper by the LM339 comparator IC, the result is a square wave of varying duty cycle.

Notes:
Many students will initially be puzzled by the operating principle of this circuit. The best way I have found to answer their questions, I have found, is with a multi-trace oscilloscope (preferably one that can show three traces simultaneously). Connect one channel to the top of C₁, the next channel to R_pot2‘s wiper, and the third to the comparator’s output terminal. A picture, as they say, is worth a thousand words.

Question 2

Which direction would you have to move the wiper on potentiometer R_pot2 to increase the duty cycle of the output waveform? Explain your answer.

Moving the wiper up (as drawn in the schematic) produces greater duty cycles.

Notes:
Notice that I did not even hint at why this is. Ask your students to fully explain their answers!

Question 3

The design recommendation for this circuit is to make resistors R₂ and R₃ both equal to the resistance of the potentiometer R_pot2. This way, the full mechanical range of the potentiometer will be useful for adjusting duty cycle: fully turning it one way will just produce a 0% duty cycle, while fully turning it the other way will just produce a 100% duty cycle.
Explain why those resistor values need to be equal to achieve this optimum usage of pot range. Hint: it has something to do with the internal workings of the 555 timer IC.

The sawtooth signal seen at capacitor C₁ (with reference to ground) oscillates between ¹/₃ and ²/₃ supply voltage.

Notes:
This question forces students to explore what the 555 is really doing, and to recognize the comparator’s function in generating PWM square waves.

Question 4

It is important to not make resistor R₁ too small in value. Explain why, and what might happen if it were.

In the 555’s discharge cycle, resistor R₁ will drop (almost) the full power supply voltage when R_pot1 is set for minimum resistance. Not only could this overheat R₁, but it could also damage the discharge transistor within the 555 timer.

Notes:
In order to really understand why insufficient resistance at R₁ would be a bad thing, students must understand the astable operating cycle of the 555 timer. This question provides a very practical context in which to explore and/or review it!

Question 5

For those of you who are used to working with regular opamps, the presence of resistor R₄ may be a mystery. It is necessary because of a special limitation of the type LM339 comparator IC. Research and explain what this limitation is.

The LM339 comparator is only able to sink current at its output. Therefore, R₄ acts as a pullup resistor.
Follow-up question: is this (overall) circuit capable of sourcing current to a load? Explain why or why not.

Notes:
This characteristic of the LM339 caused problems for me the first few times I tried to use it in my designs. Despite this limitation, though, the LM339 is very aptly suited for this application with its fast response and very wide power supply voltage limits.

Question 6

The following modification may be made to the circuit to provide it with additional output current capability:



Here, six CMOS inverters (IC part number 4049) are ganged together in parallel to supply significantly more sourcing and sinking current capability than the LM339-with-pullup-resistor could on its own. Since CMOS logic gates are inherently on-or-off devices, they have no trouble handling the square wave output of the LM339.
Two questions here: first, why all the resistors at the outputs of the inverter gates? Why not just connect all the inverter outputs directly in parallel? Secondly, what value of resistor would you suggest at the output of each gate (R₅ through R₁₀)?  

The resistors prevent the (unlikely) occurrence of a short-circuit between two or more inverter gates, if one gate happens to fail high or low, or if some gates are significantly slower than others (and thus will “fight” with the faster gates during each transition).
Challenge question: adding inverter gates to the output of this circuit has an interesting effect on the duty cycle control. The potentiometer will now act backward from the way it was before (decreasing duty cycle when it formerly increased duty cycle, and visa-versa). Explain why.

Notes:
This question may or may not be suitable for your students, depending on whether or not they have studied logic gates yet. If they have not yet studied digital circuits, it may be wise to skip this question!
Note that I did not answer the question of resistor sizes. Discuss this with your students, and let them determine how these resistors should be sized. Let them have access to datasheets for the 4049 inverter IC, and ask them what parameter(s) would be the most important in this decision.
If no one notices, point out how the power supply wires are drawn for the six-gate (hex) inverter. This is a common way to show power wiring for multiple gates (or opamps, or any other sort of IC where complex elements are duplicated).