Power Supply notes

7/28/2019 Power Supply notes

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Basic Power Supply Rectification Tutorial

by Lewis Loflin

Many devices, in particular electronics, must use DC or direct current. A diode is a

solid-state device that conducts in one direction only. When the anode (A) is positiveand the cathode (K) is negative (though the load) current (I'm assuming electron flow

from negative to positive) will flow through the load, through the diode and back to

the power supply.

Thus current will flow only of the positive half-cycle (0 to 180 degrees) and the diode

will shut-off during the negative half-cycle from 180 degrees to 360 degrees. The

period of a sine wave from 0 degrees to 360 degrees equals 1/F. In the case of 60

Hertz it's 1/60 = 16.7 mSec.

What is power? Voltage (in volts) is the "push" and the current (in Amperes) is what

is being pushed. (Electric charges) Power is voltage times current. Power is measuredin watts. So one amp at one volt equals one watt. (I'm not going into all of Ohm's Law

here. See your text.) We must have voltage and current together to get power, so an

open switch, broken wire, or a shut-off diode delivers no power.

In the case above, we get very poor power transfer with the diode off during the

negative half-cycle and the positive half-cycle changing constantly between zero volts

and peak. Note that Vmax is peak.


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Figure 2

Let's say the AC in is 12.6 volts RMS. To get peak we multiply 12.6 by 1.414, whichequals about 17.8 volts. But the average (or measured) voltage DC is peak times .3185

equals about 5.67 volts. This is what is called pulsating DC. Pure DC, such as from a

12 volt auto battery, has none of the "ripple" and will be a real 12 volts.

Put a DC voltmeter across the load above in figure 1, one will read about 5.66 volts.

Switch the meter to AC, one will still read a voltage of some value. This is normal as

one is reading the "ripple" riding the unfiltered raw D.C. Connect the same AC

voltmeter across a clean DC source such as a car battery, one will read zero volts AC.

In figure 2 we inserted a capacitor across the load. The capacitor charges during thepositive half-cycle, then discharges through the load during the negative half-cycle

when we have no output. The amount of ripple is dependant on the resistance of the

load and the size of the capacitor.

A larger capacitor produces less ripple or a higher resistance load (drawing less

current thus less time for the capacitor to discharge) will reduce the level of ripple

because the capacitor has less time to discharge. With no load at all, just the capacitor

and the rectifier, the capacitor will charge to peak.

A word of caution. If constructing these circuits observe capacitor polarity anddiode polarity. The voltage ratings of the capacitors should exceed the expected

peak voltage by 50%. Also note the current ratings of the transformers and

diodes.


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Figure 3

Full-wave rectification

Full-wave rectification converts both polarities of the input waveform to DC (direct

current), and is more efficient. However, in a circuit with a non-center tapped

transformer, four diodes are required instead of the one needed for half-wave

rectification. This is due to each output polarity requiring two rectifiers each. Four

rectifiers arranged this way are called a diode bridge or bridge rectifier.

Note that in this example the arrows show conventional current flow, not electron

flow I use with my students. This causes endless confusion for students as the

military, etc. use electron flow in their training material while semiconductor classes

use conventional current. Just be aware of this as one follows this material. Electron

flow is from negative to positive, conventional (or charge) flow is from positive to

negative.

In figure 3 D1 and D2 conduct during the positive half-cycle while D3 and D4

conduct during the negative half-cycle. Power delivered here is twice that of half-

wave rectification because we are using both half-cycles. Using 12 volts AC again, wehave 12.6 X 1.414 or 17 volts peak. (17.8 volts) But now to get the average we

multiply by peak (17.8 volts) by 0.637 which equals 10.83 volts, double that of half-

wave.

In addition we can use a smaller filter capacitor to clean out the ripple than we used

with half-wave rectification. We have also doubled the frequency from 60 Hertz to


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120 Hertz. It should be noted that when this circuit is constructed the voltage on the

meter will be about one volt low. This is due to a 0.6 volt drop across the diodes,

meter calibration due to frequency change (from 60 Hz to 120 Hz), and calculation

errors.

About me Bristol VA/TN E-Mail Hobby Electronics Arduino Microcontroller

Figure 4 typical bridge rectifiers.

Figure 5
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Figure 5 above illustrates another method to obtain full-wave rectification. In this case

we use a center-tapped transformer and two diodes. In using the center-tap (C) as a

common, the voltage A and B is 180 degrees out of phase. When A is positive, D1

will be forward biased and conduct, while B will be negative thus reverse-biasing D2,

while is non-conductive. On the negative half cycle in relation to A when D1 doesn't

conduct, D2 will conduct.

It should be noted the output voltage will be cut by half. If we use a 25.2 volt, three

amp transformer, the output voltage will be 12.6 volts. There is some controversy on

output current. We are dealing RMS amps and have to factor in transformer

impedance. (Z) During each half-cycle in this configuration current flow in one-half

the total windings. Based on resistance of the wire, Z, etc. the current can be 1.2 to 1.8

times the rated current. I'd urge caution on these claims and wouldn't go above 1.4.

All of the previous rules for peak, output voltage, etc. still hold true.

Figure 1

Tricks and Tips for the LM78XX Series Voltage Regulators

by Lewis Loflin

In this section we will explore fixed, regulated power supplies. We will make use of

the 78XX and 79XX series of voltage regulators. They are made by several

manufacturers, most are readily available, and are inexpensive. InBasic Power

Supply Rectification Tutorialwe already discussed the process from AC in to

filtering. Later we will examine adjustable, regulated power supplies.
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Figure 2

The LM78XX series of three terminal positive regulators are available in the TO-220

package. Each type employs internal current limiting, thermal shut down and safe

operating area protection, making it essentially indestructible. If adequate heat sinking

is provided, they can deliver over 1A output current. These devices can be used withexternal components to obtain adjustable voltages and currents. Available output

voltages: 5, 6, 8, 9, 10, 12, 15, 18, and 24V. Figure 2 shows the electrical connection

for the LM78XX series.

Figure 3

The LM79XX series of 3-terminal regulators is available with fixed output voltages of

-5V, -12V, and -15V. These devices need only a compensation capacitor (1 uF solid

tantalum or 25F aluminum electrolytic) at the output. The LM79XX series is

packaged in the TO-220 power package and is capable of supplying 1.5A of output

current with proper heat sinking. Like the LM78XX series they hey employ internalcurrent limiting safe area protection and thermal shutdown for protection against

virtually all overload conditions. Figure 3 shows the electrical connections on the

LM79XX series and how they differ from the LM78XX series.


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Figure 4

Figure 4 shows a basic 5 volt general purpose power supply. Any of the other positive

regulators will work the same way as long as one observes proper input voltage levelsand component ratings.

Figure 5

In figure 5 we have added a NPN pass transistor such as a 2N3055 to boost output

current to several amps. Diode D1 was added to compensate for the voltage drop

across the base-emitter junction of Q1.


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Figure 6

In figure 6 we have added a 5.6 volt zener diode for D1. By using zeners we can

produce any number of odd voltage requirements. Q1 works the same as in figure 5 or

could be left out and a 5 volt zener used if current requirement is under 1 amp.

Figure 7

Figure 7 illustrates a regulated bi-polar power supply for use with OP-AMP circuits.


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Building a Parallel Voltage Summer

Building fomula:Parallel voltage summer= V-to-I converters + current summer+ I-to-V converter.

Fig. 1. Parallel voltage summer = V-to-I converters + current summer + I-to-V converter.

Contents

[hide]

1 A problem: the common ground

2 Deriving the simplest current summer from KCL

3 Building a parallel voltage summer

4 Exploring the circuit

5 Do we really need an I-to-V converter?

6 Applications

o 6.1 Op-amp inverting voltage summer

o 6.2 Digital-to-analog converter

o 6.3 Audio mixer

7 Parallel versus series summer

8 Further reading
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[edit]A problem: the common ground

Fig. 2. Where to ground the circuit?

Kirchhoff's Voltage Lawhas already given us an idea how to create the simplest series voltage summer.

Unfortunately, it had a lot of disadvantages; maybe, the most crucial of them was the problem of

thecommon ground.

Do you remember?First, we grounded the common point between the input source V IN1 and the load; as

a result, the sources VIN2 and VIN3became flying.Then, we tried to ground the common point between the

input sources VIN1 and VIN2; now, the source VIN3 and the load became flying.Finally, we were forced to

use only two grounded sources and a differential load.

Only, Kirchhoff has formulated another law - Kirchhoff's Current Law (KCL). Maybe, it can help us to

create a perfect voltage summer having no problems with the common ground? Let's try this speculation!
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[edit]Deriving the simplest current summer from KCL

Fig. 3. In order to sum currents, we have just to connect in parallel the input current sources to the load.

First, we can apply directly KCL to create a current summer. For this purpose, we have just to connect in

parallel the input current sources Ii (let's for concreteness assume again that we have three sources) to

thecurrent loadLI (Fig. 3). Let's first assume that it is a perfect current loadhaving zero resistance (e.g.,

just a piece of wire). The output current IOUT flowing through the load is the sum of the input currents:IOUT = IIN1 + IIN2+ IIN3.

We can ask ourselves again, "Only, what is actually the summer here?", "Where is it?" The input current

sources and the load are external components; so, the rest (the bare node or the junction point) serves

here as a summer!

The simplest current summer is just a node.

Wonderful! We have another "ideal" device - the current summer. It is apparent device because there is

not actually a device:)!
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[edit]Building a parallel voltage summer

Fig. 4. We can build a parallel voltage summer, if we connect V-toI converters before the inputs and I-to-V converter after

the output of a current summer.

Only, we want to sum input voltages; so, we have to convert them into currents. For this purpose, we

connectvoltage-to-current convertersbetween the input voltage sources and the inputs of the current

summer (Fig. 4). Also, we need a voltage output; so, we have to connect the opposite current-to-voltageconverterat the current summer output. In this way, we have assembled a composed voltage summing

circuit:

Parallel voltage summer= V-to-I converters + current summer+ I-to-V converter

Let's compare the two viewpoints: From the classical viewpoint, the parallel voltage summer consists of a

few resistors; from our viewpoint, it contains voltage-to-current converters, a current summer and a

current-to-voltage converter. The use of applying such a system approach is that we see the function of

the resistors in the circuit; we see the forest for the tree!
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[edit]Exploring the circuit

Fig. 5. Visualizing the circuit operation by voltage bars and current loops.

Let's now see how the circuit operates. Attractive voltage bars will help us to visualize the invisible

voltages and voltage drops (Fig. 5); current loops will show us where the currents flow (remember: every

current returns where it has begun flowing).

The input voltage sources VIN produce voltages, the resistors RIN convert them into currents, the junctionpoint sums the currents and finally, the resistor R converts back the current sum into an output voltage

VOUT.

Only, the output voltage introduces an error because it subtracts from the input voltages. Now, the

effective voltages VR1, VR2 and VR3 create the currents instead the whole input voltages VIN1, VIN2 and VIN3:

IIN1 = (VIN1 - VOUT)/R1 = VR1/R1



How do we remove the error?
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[edit]Do we really need an I-to-V converter?

Fig. 6. The circuit continues working fine even, if we remove the I-to-V converter.

Maybe, you have already noted the contradictory role of the resistor R.

In order not to disturb the input sources (seeabove), we want the resistance R to be as small as possible

(preferably R = 0). As a result, the input voltage-to-current converters become each other independent;

only, the output voltage decreases:( We will use this technique later when we build an active parallel

voltage summer.

But why do not we increase the resistance R up to infinity (i.e., just to remove the resistor R)? In this

case, the input voltage-to-current converters become each other absolutely dependent but we obtain a

maximal output voltage. We can use this solution when the load has an infinite internal resistance (e.g.,

when we have buffered the passive summer by a non-inverting amplifier).

Only, there is a sound reason to connect the "unnecessary" resistor R - it can add the input coefficients

up to 1 (see moreexplanations). We will use this technique below to build a simple digital-to-analog

converter.

[edit]Applications

The parallel voltage summer exists in many analog circuits considered in electronics books; only, authors

do not discern and do not pay attention to it. As a result, it is presented rather implicitly than manifestly. In
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this section, we will do our best to show its presence in various electronic circuits. This famous circuit

deserves our attention.

[edit]Op-amp inverting voltage summer

Fig. 7. We can build an op-amp inverting summer by connecting a passive parallel voltage summer and an op-amp.

Maybe, the most important application of the passive summer is building an op-amp active summer. Only,

what is the idea behind such an op-amp inverting summing circuit (Fig. 7)?

From the classical viewpoint, an op-amp inverting summer consists of input resistors Ri, a negative

feedback resistor R and an op-amp. Only, thinking of the active circuit in this way we can't discern the

basic idea behind it; we do not see the forest for the tree.

From our fresh viewpoint, an op-amp inverting summer consists of a passive parallel voltage summer and

an op-amp:

Op-amp inverting summer = parallel voltage summer + op-amp

More precisely speaking, an n-inputop-amp inverting summer needs an (n+1)-input passive parallel

voltage summer. For example, in order to build a 3-input op-amp inverting summer (our case) we have to

add an additional fourth input IN4. A properly supplied op-amp serves as an additional input voltage

source that adjusts its output voltage VOUT (the "input" voltage VIN4) so that to zero the passive summer's

output voltage VA (the virtual ground). As a result, the op-amp's output voltage represents the sum of the

input voltages VIN. Let's repeat again:

We (i.e., the op-amp) introduce(s) an additional "compensating" input and make its voltage equal to the

sum of the rest "true" input voltages. Then, we abandon the "genuine" summer's output voltage VA (now,

it serves only as an indication of an equilibrium) and begin using the compensating voltage VIN4 as an

output.
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[edit]Digital-to-analog converter

Fig. 8. A parallel voltage summer with binary weighted input coefficients can serve as a DAC.

What is a digital-to-analog converter (DAC)? It is just a circuit that materializes an abstract digit (most

frequently, converting it into a voltage). A parallel voltage summer having binary-weighted inputs can do

this work.

In this example (Fig. 8), a 3-bit digital device drives the DAC. Here we suppose that the high voltage

levels of the three digital outputs are relatively equal; this voltage serves as a reference VREF. The output

voltage is:

VOUT = 0.1 x b0 x VREF + 0.2 x b1 x VREF + 0.4 x b2 x VREF

Forexample, you can connect such a simple DAC to the parallel computer port (such as a printer port).

The resistor R is absolutely necessary in this application; it adds the sum of the input coefficients up to 1

(according toDaisy's theorem). You can use the niceBrandy's formulato calculate the resistances.

[edit]Audio mixer
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Fig. 9. A parallel voltage summer can mix audio signals.

Another popular application of the parallel summing circuit isaudio mixingof analog signals from a few

voltage sources.

Note again that the parallel summer not just sums the signals; it also attenuates them. In this case, this

feature is useful.

Summer and subtractor opamp circuitsQuestion 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 nswers" 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.

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.
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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.

Avoid using the model 741 op-amp, unless you want to challenge your circuit design skills. There are more versatile op-amp models

commonly available for the beginner. I recommend the LM324 for DC and low-frequency AC circuits, and the TL082 for AC projects involving

audio or higher frequencies.

As usual, avoid very high and very low resistor values, to avoid measurement errors caused by meter "loading". I recommend resistor values

between 1 k and 100 k.

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.

Reveal Answer

Question 2:

The simple resistor network shown here is known as apassive averager. Describe what the word "passive" means in this context, and write

an equation describing the output voltage (Vd) in terms of the input voltages (Va, Vb, and Vc):

Hint: there is a network theorem that directly applies to this form of circuit, and it is known as Millman's Theorem. Research this theorem and

use it to generate your equation!

Reveal Answer

Question 3:

Add an op-amp circuit to the output of this passive averager network to produce a summercircuit: an operational circuit generating an output

voltage equal to the sum of the four input voltages. Then, write an equation describing the whole circuit's function.
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Reveal Answer

Question 4:

Write a mathematical equation for this op-amp circuit, assuming all resistor values are equal:

What is this circuit typically called?

Reveal Answer

Question 5:

This opamp circuit is known as a difference amplifier, sometimes called a subtractor. Assuming that all resistor values are equal in the circuit,

write an equation expressing the output (y) as a function of the two input voltages (a and b):

Reveal Answer

Question 6:

Determine all current magnitudes and directions, as well as voltage drops, in this circuit:
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Reveal Answer

Question 7:

Determine all current magnitudes and directions, as well as voltage drops, in this circuit:

Reveal Answer

Question 8:

Determine the amount of current from point A to point B in this circuit:


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Reveal Answer

Question 9:

Determine the amount of current from point A to point B in this circuit, and also the output voltage of the operational amplifier:

Reveal Answer

Question 10:

Identify some of the distinguishing characteristics of inverting and noninverting summer circuits. How may you identify which is which, and

how may you determine the proper resistor values to make each one work as it should?

Reveal Answer

Question 11:

Complete the table of values for this opamp circuit, calculating the output voltage for each combination of input voltages shown:

V1 V2 Vout

0 V 0 V


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+1 V 0 V

0 V +1 V

+2 V +1.5 V

+3.4 V +1.2 V

-2 V +4 V

+5 V +5 V

-3 V -3 V

What pattern do you notice in the data? What mathematical relationship is there between the two input voltages and the output voltage?

Reveal Answer

Question 12:

How does the operation of this difference amplifier circuit compare with the resistor values given (2R = twice the resistance of R), versus its

operation with all resistor values equal?

Describe what approach or technique you used to derive your answer, and also explain how your conclusion for this circuit might be

generalized for all difference amplifier circuits.

Reveal Answer

Question 13:


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If a weak voltage signal is conveyed from a source to an amplifier, the amplifier may detect more than just the desired signal. Along with the

desired signal, external electronic "noise" may be coupled to the transmission wire from AC sources such as power line conductors, radio

waves, and other electromagnetic interference sources. Note the two waveshapes, representing voltages along the transmission wire

measured with reference to earth ground:

Shielding of the transmission wire is always a good idea in electrically noisy environments, but there is a more elegant solution than simply

trying to shield interference from getting to the wire. Instead of using a single-ended amplifier to receive the signal, we can transmit the

signal along two wires and use adifference amplifier at the receiving end. Note the four waveforms shown, representing voltages at those

points measured with reference to earth ground:

If the two wires are run parallel to each other the whole distance, so as to be exposed to the exact same noise sources along that distance,

the noise voltage at the end of the bottom wire will be the same noise voltage as that superimposed on the signal at the end of the top wire.

Explain how the difference amplifier is able to restore the original (clean) signal voltage from the two noise-ridden voltages seen at its inputs

with respect to ground, and also how the phrase common-mode voltage applies to this scenario.

Reveal Answer

Question 14:

Singers who wish to practice singing to popular music find that the following vocal eliminatorcircuit is useful:

The circuit works on the principle that vocal tracks are usually recorded through a single microphone at the recording studio, and thus are
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represented equally on each channel of a stereo sound system. This circuit effectively eliminates the vocal track from the song, leaving only

the music to be heard through the headphone or speaker.

Explain how the operational amplifiers accomplish this task of vocal track elimination. What role does each opamp play in this circuit?

Reveal Answer

Question 15:

The following circuit is known as an instrumentation amplifier:

Suppose a DC voltage were to be applied to the noninverting input terminal, +1 volt at V in(+), and the inverting input terminal grounded.

Complete the following table showing the output voltage of this circuit for different values of m:

m Vout

1

2

3

4

5


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6

Reveal Answer

Question 16:

Find the datasheet for a real instrumentation amplifier (packaged as a single integrated circuit) and bring it to class for d iscussion with your

classmates. Analyze and discuss the inner workings of the circuit, and some of its performance parameters. If you do not know where to

begin looking, try researching the Analog Devices model AD623, either in a reference book or on the internet.

Reveal Answer

Question 17:

The following circuit is a type of difference amplifier, similar in behavior to the instrumentation amplifier, but only using two operational

amplifiers instead of three:

Complete the table of values for this opamp circuit, calculating the output voltage for each combination of input voltages shown. From the

calculated values of output voltage, determine which input of this circuit is inverting, and which is noninverting, and also how much

differential voltage gain this circuit has. Express these conclusions in the form of an equation.

V1 V2 Vout

0 V 0 V

+1 V 0 V

0 V +1 V

+2 V +1.5 V


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+3.4 V +1.2 V

-2 V +4 V

+5 V +5 V

-3 V -3 V

Reveal Answer

Question 18:

An important parameter of any differential amplifier - bare opamps and difference amplifiers made from opamps alike - is common-mode

rejection, or CMR. Explain what this parameter means, how the following circuit tests this parameter, and why it is important to us:

Reveal Answer

Question 19:

Explain what common-mode rejection ratio means for a differential amplifier, and give a formula for calculating it.

Reveal Answer

Question 20:

Predict how the operation of this passive averager network will be affected as a result of the following faults. Consider each fault

independently (i.e. one at a time, no multiple faults):


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Resistor R1 fails open:

Solder bridge (short) across resistor R1:





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

Reveal Answer

Question 21:

Predict how the operation of this summer circuit will be affected as a result of the following faults. Consider each fault independently (i.e. one

at a time, no multiple faults):


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Reveal Answer

Question 22:

Predict how the operation of this summer circuit will be affected as a result of the following faults. Consider each fault independently (i.e. one

at a time, no multiple faults):





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Reveal Answer

Question 23:

Predict how the operation of this difference amplifier circuit will be affected as a result of the following faults. Consider each fault

independently (i.e. one at a time, no multiple faults):







Reveal Answer

Question 24:

The instrumentation amplifieris a popular circuit configuration for analog signal conditioning in a wide variety of electronic measurement


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applications. One of the reasons it is so popular is that its differential gain may be set by changing the value of a single resistor, the value of

which is represented in this schematic by a multiplier constant named m:

There is an equation describing the differential gain of an instrumentation amplifier, but it is easy enough to research so I'll leave that detail

up to you. What I'd like you to do here is algebraically derive that equation based on what you know of inverting and noninverting operational

amplifier circuits.

Suppose we apply +1 volt to the noninverting input and ground the inverting input, giving a differential input voltage of 1 volt. Whatevervoltage appears at the output of the instrumentation amplifier circuit, then, directly represents the voltage gain:

A hint for constructing an algebraic explanation for the circuit's output voltage is to view the two "buffer" opamps separately, as inverting and

noninverting amplifiers:


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Note which configuration (inverting or noninverting) each of these circuits resemble, develop transfer functions for each (Output = Input),

then combine the two equations in a manner representing what the subtractor circuit will do. Your final result should be the gain equation for

an instrumentation amplifier in terms of m.

Reveal Answer

Question 25:

Calculate the voltage gain of the following opamp circuit with the potentiometer turned fully up, precisely mid-position, and fully down:

AV (pot fully up) =

AV (pot mid-position) =

AV (pot fully down) =

Reveal Answer


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Voltage doubler

From Wikipedia, the free encyclopedia

A voltage doubleris an electronic circuit which charges capacitors from the input voltage and switches these

charges in such a way that, in the ideal case, exactly twice the voltage is produced at the output as at its input.

The simplest of these circuits are a form ofrectifierwhich take an AC voltage as input and output a doubled DC

voltage. The switching elements are simple diodes and they are driven to switch state merely by the alternating

voltage of the input. DC to DC voltage doublers cannot switch in this way and require a driving circuit to control

the switching. They frequently also require a switching element that can be controlled directly, such as

atransistor, rather than relying on the voltage across the switch as in the simple AC to DC case.

Voltage doublers are a variety ofvoltage multipliercircuit. Many (but not all) voltage doubler circuits can be

viewed as a single stage of a higher order multiplier: cascading identical stages together achieves a greater

voltage multiplication.

Contents

[hide]

1 Voltage doubling rectifiers

o 1.1 Villard circuit

o 1.2 Greinacher circuit

o 1.3 Bridge circuit

2 Switched capacitor circuits

o 2.1 Dickson charge pump

o 2.2 Cross-coupled switched capacitors

3 See also

4 References

5 Bibliography

[edit]Voltage doubling rectifiers

[edit]Villard circuit
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Figure 1. Villard circuit

The Villard circuit consists simply of a capacitor and a diode. While it has the great benefit of simplicity, its

output has very poorripplecharacteristics. Essentially, the circuit is a diodeclampcircuit. The capacitor is

charged on the negative half cycles to the peak AC voltage (Vpk

). The output is the superposition of the input

AC waveform and the steady DC of the capacitor. The effect of the circuit is to shift the DC value of the

waveform. The negative peaks of the AC waveform are "clamped" to 0 V (actually VF, the small forward bias

voltage of the diode) by the diode, therefore the positive peaks of the output waveform are 2 Vpk. The peak-to-

peak ripple is an enormous 2Vpk and cannot besmoothedunless the circuit is effectively turned into one of the

more sophisticated forms.[1]

This is the circuit (with diode reversed) used to supply the negative high voltage for

the magnetron in a microwave oven.

[edit]Greinacher circuit

Figure 2. Greinacher circuit

The Greinachervoltage doubler is a significant improvement over the Villard circuit for a small cost in

additional components. The ripple is much reduced, nominally zero under open-circuit load conditions, but

when current is being drawn depends on the resistance of the load and the value of the capacitors used. Thecircuit works by following a Villard cell stage with what is in essence apeak detectororenvelope

detectorstage. The peak detector cell has the effect of removing most of the ripple while preserving the peak

voltage in the output.

Figure 3. Voltage quadrupler two Greinacher cells of opposite polarities

This circuit was first invented byHeinrich Greinacherin 1913 (published 1914[2]) to provide the 200300 V he

needed for his newly inventedionometer, the 110 V AC supplied by theZurichpower stations of the time being

insufficient.[3]

He later (1920) extended this idea into a cascade of multipliers.[4]

This cascade of Greinacher
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cells is often inaccurately referred to as a Villard cascade. It is also called aCockcroftWalton multiplierafter

theparticle acceleratormachine built byJohn CockcroftandErnest Walton, who independently rediscovered

the circuit in 1932.[5]

The concept in this topology can be extended to a voltage quadrupler circuit by using two Greinacher cells of

opposite polarities driven from the same AC source. The output is taken across the two individual outputs. As

with a bridge circuit, it is impossible to simultaneously ground the input and output of this circuit.[6]

[edit]Bridge circuit

Figure 4. Bridge (Delon) voltage doubler

The Delon circuit uses abridge topologyfor voltage doubling. This form of circuit was, at one time, commonly

found incathode ray tubetelevision sets where it was used to provide ane.h.t.voltage supply. Generating

voltages in excess of 5 kV with atransformerhas safety issues in terms of domestic equipment and in any case

is uneconomic. However, black and white television sets required an e.h.t. of 10 kV and colour sets even more.

Voltage doublers were used to either double the voltage on an e.h.t winding on the mains transformer or were

applied to the waveform on the lineflyback coils.[7]

The circuit consists of two half-wave peak detectors, functioning in exactly the same way as the peak detector

cell in the Greinacher circuit. Each of the two peak detector cells operates on opposite half-cycles of the

incoming waveform. Since their outputs are in series, the output is twice the peak input voltage.

[edit]Switched capacitor circuits

Figure 5. Switched capacitor voltage doubler achieved by simply switching charged capacitors from parallel to series
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It is possible to use the simple diode-capacitor circuits described above to double the voltage of a DC source

by preceding the voltage doubler with achopper circuit. In effect, this converts the DC to AC before application

to the voltage doubler.[8]

More efficient circuits can be built by driving the switching devices from an external

clock so that both functions, the chopping and multiplying, are achieved simultaneously. Such circuits are

known asswitched capacitorcircuits. This approach is especially useful in low-voltage battery-powered

applications where integrated circuits require a voltage supply greater than the battery can deliver. Frequently,

a clock signal is readily available on board the integrated circuit and little or no additional circuitry is needed to

generate it.[9]

Conceptually, perhaps the simplest switched capacitor configuration is that shown schematically in figure 5.

Here two capacitors are simultaneously charged to the same voltage in parallel. The supply is then switched off

and the capacitors are switched into series. The output is taken from across the two capacitors in series

resulting in an output double the supply voltage. There are many different switching devices that could be used

in such a circuit, but in integrated circuitsMOSFETdevices are frequently employed.[10]

Figure 6. Charge-pump voltage doubler schematic

Another basic concept is thecharge pump, a version of which is shown schematically in figure 6. The charge

pump capacitor, CP, is first charged to the input voltage. It is then switched to charging the output capacitor, C O,

in series with the input voltage resulting in CO eventually being charged to twice the input voltage. It may take

several cycles before the charge pump succeeds in fully charging CO but after steady state has been reached it

is only necessary for CP to pump a small amount of charge equivalent to that being supplied to the load from

CO. While CO is disconnected from the charge pump it partially discharges into the load resulting inrippleon the

output voltage. This ripple is smaller for higher clock frequencies since the discharge time is shorter, and is also

easier to filter. Alternatively, the capacitors can be made smaller for a given ripple specification. The practical

maximum clock frequency in integrated circuits is typically in the hundreds of kHz.[11]

[edit]Dickson charge pump
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Figure 7. Dickson charge-pump voltage-doubler

The Dickson charge pump, orDickson multiplier, consists of a cascade of diode/capacitor cells with the bottom

plate of each capacitor driven by aclock pulsetrain. The circuit is a modification of the Cockcroft-Walton

multiplier but takes a DC input with the clock trains providing the switching signal instead of the an AC input.

The Dickson multiplier normally requires that alternate cells are driven from clock pulses of opposite phase.

However, since a voltage doubler, shown in figure 7, requires only one stage of multiplication only one clock

signal is required.[12]

The Dickson multiplier is frequently employed in integrated circuits where the supply voltage (from a battery for

instance) is lower than that required by the circuitry. It is advantageous in integrated circuit manufacture that all

the semiconductor components are of basically the same type.MOSFETsare commonly the standard logic

block in many integrated circuits. For this reason the diodes are often replaced by this type of transistor, but

wired to function as a diode - an arrangement called a diode-wired MOSFET. Figure 8 shows a Dickson voltage

doubler using diode-wired n-channel enhancement type MOSFETs.[13]

Figure 8. Dickson voltage doubler using diode-wired MOSFETs

There are manyvariations and improvementsto the basic Dickson charge pump. Many of these are concerned

with reducing the effect of the transistor drain-source voltage. This can be very significant if the input voltage is

small, such as a low-voltage battery. With ideal switching elements the output is an integral multiple of the input

(two for a doubler) but with a single-cell battery as the input source and MOSFET switches the output will be far

less than this value since much of the voltage will be dropped across the transistors. For a circuit using discrete

components theSchottky diodewould be a better choice of switching element for its extremely low voltage

drop in the on state. However, integrated circuit designers prefer to use the easily available MOSFET and

compensate for its inadequacies with increased circuit complexity.[14]

As an example, analkaline batterycell has a nominal voltage of 1.5 V. A voltage doubler using ideal switching

elements with zero voltage drop will output double this, namely 3.0 V. However the drain-source voltage drop of

a diode-wired MOSFET when it is in the on state must be at least the gate threshold voltage which might

typically be 0.9 V.[15]

This voltage "doubler" will only succeed in raising the output voltage by about 0.6 V to 2.1
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V. If the drop across the final smoothing transistor is also taken into account the circuit may not be able to

increase the voltage at all without using multiple stages. A typical Schottky diode, on the other hand, might

have an on state voltage of 0.3 V.[16]

A doubler using this Schottky diode will result in a voltage of 2.7 V at the

output.[17]

[edit]Cross-coupled switched capacitors

Figure 9. Cross-coupled switched-capacitor voltage doubler

Cross-coupled switched capacitor circuits come into their own for very low input voltages. Wireless battery

driven equipment (pagers, bluetooth devices and the like) may require a single-cell battery to continue to

supply power when it has discharged to under a volt.[18]

When clock is low transistor Q2 is turned off. At the same time clock is high turning on transistor

Q1 resulting in capacitor C1 being charged toVin. When goes high the top plate of C1 is pushed up to

twice Vin. At the same time switch S1 closes so this voltage appears at the output. At the same time Q2 is turned

on allowing C2 to charge. On the next half cycle the roles will be reversed: will be low, will be high,

S1 will open and S2will close. Thus, the output is supplied with 2Vin alternately from each side of the circuit.[19]

The loss is low in this circuit because there are no diode-wired MOSFETs and their associated threshold

voltage problems. The circuit also has the advantage that the ripple frequency is doubled because there are

effectively two voltage doublers both supplying the output from out of phase clocks. The primary disadvantage

of this circuit is that stray capacitances are much more significant than with the Dickson multiplier and account

for the larger part of the losses in this circuit.[20]

[edit]See also

Voltage multiplierFrom Wikipedia, the free encyclopedia

(Redirected fromVoltage-multiplier circuit)
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Villard cascade voltage multiplier.

A voltage multiplieris anelectrical circuitthat converts AC electrical power from a lowervoltageto a higher

DC voltage, typically using a network ofcapacitorsanddiodes.

Voltage multipliers can be used to generate a few volts for electronic appliances, to millions of volts forpurposes such as high-energy physics experiments and lightning safety testing. The most common type of

voltage multiplier is the half-wave series multiplier, also called the Villard cascade (but actually invented

byHeinrich Greinacher).
http://en.wikipedia.org/wiki/Electrical_circuithttp://en.wikipedia.org/wiki/Electrical_circuithttp://en.wikipedia.org/wiki/Electrical_circuithttp://en.wikipedia.org/wiki/Voltagehttp://en.wikipedia.org/wiki/Voltagehttp://en.wikipedia.org/wiki/Voltagehttp://en.wikipedia.org/wiki/Capacitorhttp://en.wikipedia.org/wiki/Capacitorhttp://en.wikipedia.org/wiki/Capacitorhttp://en.wikipedia.org/wiki/Diodehttp://en.wikipedia.org/wiki/Diodehttp://en.wikipedia.org/wiki/Diodehttp://en.wikipedia.org/wiki/Heinrich_Greinacherhttp://en.wikipedia.org/wiki/Heinrich_Greinacherhttp://en.wikipedia.org/wiki/Heinrich_Greinacherhttp://en.wikipedia.org/wiki/File:Voltage_Multiplier_diagram.PNGhttp://en.wikipedia.org/wiki/File:Voltage_Multiplier_diagram.PNGhttp://en.wikipedia.org/wiki/File:Voltage_Multiplier_diagram.PNGhttp://en.wikipedia.org/wiki/File:Voltage_Multiplier_diagram.PNGhttp://en.wikipedia.org/wiki/Heinrich_Greinacherhttp://en.wikipedia.org/wiki/Diodehttp://en.wikipedia.org/wiki/Capacitorhttp://en.wikipedia.org/wiki/Voltagehttp://en.wikipedia.org/wiki/Electrical_circuit


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Contents

[hide]

1 Operation

2 Voltage doubler and tripler

3 Breakdown voltage

4 Other circuit topologies

o 4.1 Dickson charge pump

4.1.1 Modification for RF power

o 4.2 Cross-coupled switched capacitor

5 Applications

6 See also

7 Notes

8 Bibliography

9 External links

[edit]Operation

Assuming that the peak voltage of the AC source is +Us, and that the C values are sufficiently high to allow,

when charged, that a current flows with no significant change in voltage, then the (simplified) working of the

cascade is as follows:

1. negative peak (Us): The C1 capacitor is charged through diode D1 to UsV(potential

differencebetween left and right plate of the capacitor is Us)

2. positive peak (+Us): the potential of C1 adds with that of the source, thus charging C2 to 2Us through D2

3. negative peak: potential of C1 drops to 0 V thus allowing C3 to be charged through D3 to 2Us.

4. positive peak: potential of C1 rises to 2Us (analogously to step 2), also charging C4 to 2Us. The output

voltage (the sum of voltages under C2 and C4) raises till 4Us.
http://en.wikipedia.org/wiki/Voltage-multiplier_circuithttp://en.wikipedia.org/wiki/Voltage-multiplier_circuithttp://en.wikipedia.org/wiki/Voltage-multiplier_circuithttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Operationhttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Operationhttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Voltage_doubler_and_triplerhttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Voltage_doubler_and_triplerhttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Breakdown_voltagehttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Breakdown_voltagehttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Other_circuit_topologieshttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Other_circuit_topologieshttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Dickson_charge_pumphttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Dickson_charge_pumphttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Modification_for_RF_powerhttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Modification_for_RF_powerhttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Cross-coupled_switched_capacitorhttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Cross-coupled_switched_capacitorhttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Applicationshttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Applicationshttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#See_alsohttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#See_alsohttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Noteshttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Noteshttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Bibliographyhttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#Bibliographyhttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#External_linkshttp://en.wikipedia.org/wiki/Voltage-multiplier_circuit#External_linkshttp://en.wikipedia.org/w/index.php?title=Voltage_multiplier&action=edit&section=1http://en.wikipedia.org/w/index.php?title=Voltage_multiplier&action=edit&section=1http://en.wikipedia.org/w/index.php?title=Voltage_multiplier&action=edit&section=1http://en.wikipedia.org/wiki/Volthttp://en.wikipedia.org/wiki/Volthttp://en.wikipedia.org/wiki/Volthttp://en.wikipedia.org/wiki/Potential_differencehttp://en.wikipedia.org/wiki/Poten

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