EET222 Worksheet #2: Operational Amplfiers and Negative Feedback Help for this worksheet may be found in Chapters 16, 17, and 18 of the textbook. This is not the only place to find help. Don’t be afraid to explore. Educational Objectives • Understand the ideal characteristics of an opamp • Understand how to read a data sheet to determine important non-ideal characteristics of an opamp • Explain the relationship between gain and bandwidth • Explain why negative feedback is important and how it works. • Perform circuit analysis on simple linear integrated circuits.

• • • •

If you want additional practice you should try these problems from the book Inverting and NonInverting circuits: Ch16 #7-11 Non-ideal opamp characteristics: Ch16: #1-6 Gain Bandwidth Product: CH17: #17-21 Analyzing linear opamp circuits: Ch18 #1-39

1

Questions Question 1 Write the transfer function (input/output equation) for an operational amplifier with an open-loop voltage gain of 100,000. In other words, write an equation describing the output voltage of this op-amp (Vout ) for any combination of input voltages (Vin(+) and Vin(−) ):

Vin(+) Vin(-)

+ Vout

−

Kuphaldt file 00925 Question 2 An op-amp has +3 volts applied to the inverting input and +3.002 volts applied to the noninverting input. Its open-loop voltage gain is 220,000. Calculate the output voltage as predicted by the following formula: Vout = AV Vin(+) − Vin(−)




How much differential voltage (input) is necessary to drive the output of the op-amp to a voltage of -4.5 volts? Kuphaldt file 00926 Question 3 What does the phrase open-loop voltage gain mean with reference to an operational amplifier? For a typical opamp, this gain figure is extremely high. Why is it important that the open-loop voltage gain be high when using an opamp as a comparator? Kuphaldt file 00873

2

Question 4 A helpful model for understanding opamp function is one where the output of an opamp is thought of as being the wiper of a potentiometer, the wiper position automatically adjusted according to the difference in voltage measured between the two inputs: Positive power supply "rail"

+V

Vin(+) + V -

Vout

Voltmeter

Vin(-)

-V Negative power supply "rail"

To elaborate further, imagine an extremely sensitive, analog, zero-center voltmeter inside the opamp, where the moving-coil mechanism of the voltmeter mechanically drives the potentiometer wiper. The wiper’s position would then be proportional to both the magnitude and polarity of the difference in voltage between the two input terminals. Realistically, building such a voltmeter/potentiometer mechanism with the same sensitivity and dynamic performance as a solid-state opamp circuit would be impossible, but the point here is to model the opamp in terms of components that we are already very familiar with, not to suggest an alternative construction for real opamps. Describe how this model helps to explain the output voltage limits of an opamp, and also where the opamp sources or sinks load current from. Kuphaldt file 02290

3

Question 5 Just as certain assumptions are often made for bipolar transistors in order to simplify their analysis in circuits (an ideal BJT has negligible base current, IC = IE , constant β, etc.), we often make assumptions about operational amplifiers so we may more easily analyze their behavior in closed-loop circuits. Identify some of these ideal opamp assumptions as they relate to the following parameters: • • • • • • • •

Magnitude of input terminal currents: Input impedance: Output impedance: Input voltage range: Output voltage range: Differential voltage (between input terminals) with negative feedback: Output current: Voltage Gain of the opamp:

Additional Discussion

• • • • • •

What are the actual values for a LM741 opamp? Magnitude of input terminal currents: Input impedance: Input voltage range: Output voltage range: Output current: Voltage Gain of the opamp: Based on the realistic values, are the ideal values realistic? Kuphaldt file 02749c

Question 6 What does it mean if an operational amplifier has the ability to ”swing its output rail to rail”? Why is this an important feature to us? Kuphaldt file 00844

4

Question 7 Ideally, when the two input terminals of an op-amp are shorted together (creating a condition of zero differential voltage), and those two inputs are connected directly to ground (creating a condition of zero common-mode voltage), what should this op-amp’s output voltage be?

Vout = ???

-15 V +

−

+15 V

In reality, the output voltage of an op-amp under these conditions is not the same as what would be ideally predicted. Identify the fundamental problem in real op-amps, and also identify the best solution. Kuphaldt file 00847 Question 8 Draw the schematic symbol for a differential opamp. Label all 5 of its terminals. Explain the function of each terminal. file ch18003

5

Question 9 In this circuit, an op-amp turns on an LED if the proper input voltage conditions are met:

+V +V + +V

−

+V Power supply

Trace the complete path of current powering the LED. Where, exactly, does the LED get its power from? Additional Discussion In order to make the LED turn on, what is the relationship between the voltage at the + and − terminals. Kuphaldt file 00801c

6

Question 10 Determine the output voltage polarity of this op-amp (with reference to ground), given the following input conditions:

+V

+V

+ −

+ ???

−

-V

-V

+V

+V

+ −

+ ???

−

-V

-V

+V

+V

+ −

???

???

+ ???

−

-V

???

-V

Kuphaldt file 00803 Question 11 A very important concept in electronics is that of negative feedback. This is an extremely important concept to grasp, as a great many electronic systems exploit this principle for their operation and cannot be properly understood without a comprehension of it. However important negative feedback might be, it is not the easiest concept to understand. In fact, it is quite a conceptual leap for some. The following is a list of examples – some electronic, some not – exhibiting negative feedback: 7

• • • • • • • •

A voltage regulating circuit An auto-pilot system for an aircraft or boat A thermostatic temperature control system (”thermostat”) Emitter resistor in a BJT amplifier circuit Lenz’s Law demonstration (magnetic damping of a moving object) Body temperature of a mammal Natural regulation of prices in a free market economy (Adam Smith’s ”invisible hand”) A scientist learning about the behavior of a natural system through experimentation. For each case, answer the following questions:

• What variable is being stabilized by negative feedback? • How is the feedback taking place (step by step)? • What would the system’s response be like if negative feedback were not present? Kuphaldt file 02253 Question 12 We know that an opamp connected to a voltage divider with a voltage division ratio of 12 will have an overall voltage gain of 2, and that the same circuit with a voltage division ratio of 32 will have an overall voltage gain of 1.5, or 23 :

+

Vin

+

Vin Vout = 2 (Vin)

−

Vout =

−

Voltage divider circuit

3 (Vin) 2

Voltage divider circuit

1 V 2 out

2 V 3 out

There is definitely a mathematical pattern at work in these noninverting opamp circuits: the overall voltage gain of the circuit is the mathematical inverse of the feedback network’s voltage gain. Building on this concept, what do you think would be the overall function of the following opamp circuits?

Vin

+

Vin Vout = ???

− x y

+ Vout = ???

− x

Feedback network

y y = x2

y=x+4 Kuphaldt file 02464

8

Feedback network

Question 13 Define ”Gain-Bandwidth Product” (GBW) as the term applies to operational amplifiers. Kuphaldt file 02593 Question 14 A very important parameter of operational amplifier performance is slew rate. Describe what ”slew rate” is, and why it is important for us to consider in choosing an op-amp for a particular application. Kuphaldt file 00846 Question 15 GBW and slew rate both cause frequency limitations on the opamp circuit. GBW causes the output to attenuate. Attenuation is when the output is a smaller amplitude than expected. The output shape is still the shape of the input waveform. The op-amp is able to switch fast enough, it just doesn’t have the gain at this frequency to make the output the appropriate size. Slewing causes distortion. Distortion is when the output signal does not look like a scaled (bigger or smaller) copy of the input signal, ie a triangle wave output with a Sine wave input is distortion. On Sine waves it is really hard to see the initial distortion. We typically don’t see the distortion until it is large enough that either the wave shape is different or we get attenuation. Slewing is a problem caused because the output driver is not able switch fast enough. The gain is there, but the output is unable to keep up with the input. • If a clock pulse is used as the input to an opamp circuit. Why would you want the opamp to a be LM318 instead of a LM741? • Why does reducing the frequency of the input sine wave reduce the slewing problem? • What does it mean if increasing the frequency caused a smaller output waveform? The output waveform still looks like a SINE wave. file ch18001 Question 16 Shown here are two different voltage amplifier circuits with the same voltage gain. Which of them has greater input impedance, and why? Try to give as specific an answer for each circuit’s input impedance as possible.

Inverting amplifier circuit 5 kΩ Vin

Noninverting amplifier circuit

10 kΩ

5 kΩ

−

5 kΩ −

Vout

Vout

+

Vin

+

Kuphaldt file 02709 Question 17 What is common-mode voltage, and how should a differential amplifier (ideally) respond to it? Kuphaldt file 00939

9

Question 18 When analyzing opamp circuits we make the assumption that the output load doesn’t matter. When could this assumption cause problems? Assume a 10Ω resistor is connected between the output of my opamp and gnd. The circuit is supposed to produce a 1Vpp sine wave. If my opamp is a LM741 will I have any problems? What will my problems look like? Is this a large or small load? file ch18002 Question 19 Many op-amp circuits require a dual or split power supply, consisting of three power terminals: +V, -V, and Ground. Draw the necessary connections between the 6-volt batteries in this schematic diagram to provide +12 V, -12 V, and Ground to this op-amp:

6 volts each +12 V + − Load -12 V

Kuphaldt file 00880

10

Question 20 Calculate the output voltage of this op-amp circuit (using negative feedback):

+ Vout

−

5 kΩ

27 kΩ

1.5 V

Additional Discussion • Calculate the DC voltage gain • Calculate the Bandwidth of this circuit, assume GBW=1MHz • The midpoint of the voltage divider (connecting to the inverting input of the op-amp) is often called a virtual ground in a circuit like this. Explain why. Kuphaldt file 00932c Question 21 Calculate the necessary resistor value (R1 ) in this circuit to give it a voltage gain of 15:

Vin

R1

−

+

22 kΩ

Vout Kuphaldt file 02729

11

Question 22 Determine both the input and output voltage in this circuit:

+ Vout

− 5 kΩ

12 kΩ

Vin I = 2 mA Kuphaldt file 02732 Question 23 Calculate the voltage gain for each stage of this amplifier circuit (both as a ratio and in units of decibels), then calculate the overall voltage gain:

Stage 1 10 kΩ Vin

Stage 2

15 kΩ

3.3 kΩ

−

5.1 kΩ − Vout

+

+

Kuphaldt file 02470 Question 24 How much current will go through the load in this op-amp circuit?

+ Vout − Load 2.2 kΩ 4V

What impact will a change in load resistance have on the operation of this circuit? What will change, if anything, supposing the load resistance were to increase? Kuphaldt file 02694

12

Question 25 Calculate the necessary resistor value (R1 ) in this circuit to give it a voltage gain of 10.5:

Vout

R1 −

+ Vin

8.1 kΩ

Additional Discussion • • • •

Calculate the bandwidth for this amplifier if the opamp chosen is a LM741 Calculate the bandwidth for this amplifier if the opamp chosen is a LM318 Why is the bandwidth different for the LM741 and LM318? If R1 is 5% larger than ideal, what happens to the gain of this amplifier? What happens to the bandwidth of this amplifier? Kuphaldt file 02724c

Question 26 Calculate all voltage drops and currents in this circuit, complete with arrows for current direction and polarity markings for voltage polarity. Then, calculate the overall voltage gain of this amplifier circuit (AV ), both as a ratio and as a figure in units of decibels (dB):

22 kΩ R2

47 kΩ R1

− Vout = ???

+ Vin = +3.2 volts

Kuphaldt file 02459

13

Question 27 Calculate the necessary values of R1 and R2 to limit the minimum and maximum voltage gain of this opamp circuit to 10 and 85, respectively:

R1

15 kΩ

R2 −

Vin

+

Vout

Additional Discussion • Calculate the maximum and minimum bandwidth for this amplifier if the opamp chosen is a LM741 Kuphaldt file 02771c

14

Question 28 Complete the table of voltages for this opamp ”voltage follower” circuit:

+15 V

− Vin

+

Vout

-15 V Vin 0 volts +5 volts +10 volts +15 volts +20 volts -5 volts -10 volts -15 volts -20 volts

Vout 0 volts

Additional Discussion • The output voltage values given in this table are ideal. A real opamp would probably not be able to achieve even what is shown here, due to idiosyncrasies of these amplifier circuits. Explain what would probably be different in a real opamp circuit from what is shown here. • Even though this circuit has negative feedback, why is the difference between the inverting and noninverting input not always equal to 0? Kuphaldt file 02289c

15

Question 29 Calculate the voltage gain of the following opamp circuit with the potentiometer turned fully up, precisely mid-position, and fully down:

10 kΩ

10 kΩ

Vin − +

10 kΩ

• AV (pot fully up) = • AV (pot mid-position) = • AV (pot fully down) = Kuphaldt file 03002

16

Vout

Question 30 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:

− Long transmission wire

+

. . . to rest of circuit

Signal source

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 a difference amplifier at the receiving end. Note the four waveforms shown, representing voltages at those points measured with reference to earth ground:

(clean signal)

(signal + noise)

(clean signal)

Signal source

(ground potential)

− Cable +

. . . to rest of circuit

(noise)

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. Kuphaldt file 02519

17

Question 31 Determine all current magnitudes and directions, as well as voltage drops, in this circuit:

8.33 kΩ

25 kΩ − Vout

+

5 kΩ 10 V

5 kΩ

5 kΩ

5 kΩ

3V

5V

11 V

Kuphaldt file 02515 Question 32 Determine the amount of current from point A to point B in this circuit:

1 kΩ 1 kΩ

2V 3.5 V

1 kΩ 1V

Kuphaldt file 02516

18

A

B I = ???

Question 33 Determine the amount of current from point A to point B in this circuit, and also the output voltage of the operational amplifier:

1 kΩ 1 kΩ

2V 3.5 V

A I = ???

1 kΩ

1 kΩ

B − +

1V

Vout Kuphaldt file 02517 Question 34 A student with nothing better to do decides to build this monster of a circuit:

Vin

+

−

− 10 kΩ

+ Vout

20 kΩ

Calculate the overall voltage gain of this circuit, both as a ratio and as a figure in units of decibels. AV (ratio) = Kuphaldt file 02723

AV (dB) =

19

Question 35 There is something wrong with this amplifier circuit. Note the relative amplitudes of the input and output signals as measured by an oscilloscope:

12 kΩ

+12 V

7.9 kΩ

− 0V

Vout

+ Vin

-12 V

0.4 V RMS

This circuit used to function perfectly, but then began to malfunction in this manner: producing a ”clipped” output waveform of excessive amplitude. Determine the approximate amplitude that the output voltage waveform should be for the component values given in this circuit, and then identify possible causes of the problem and also elements of the circuit that you know cannot be at fault. Kuphaldt file 02465

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Question 36 A student intends to connect a TL082 opamp as a voltage follower, to ”follow” the voltage generated by a potentiometer, but makes a mistake in the breadboard wiring:

+

-

12 V TL082

+

-

12 V

V

A

V

A OFF

A

COM

Draw a schematic diagram of this faulty circuit, and determine what the voltmeter’s indication will be, explaining why it is such. Kuphaldt file 01148

21

Question 37 Predict how the operation of this operational 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):

+ Vin

Vout

− R1

R2

• Resistor R1 fails open: • Solder bridge (short) across resistor R1 : • Resistor R2 fails open: • Solder bridge (short) across resistor R2 : • Broken wire between R1 /R2 junction and inverting opamp input: For each of these conditions, explain why the resulting effects will occur. Kuphaldt file 03774c

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Answers Answer 1 Vout = 100, 000(Vin(+) − Vin(−) ) Answer 2 Vout = 440 volts Follow-up question: is this voltage figure realistic? Is it possible for an op-amp such as the model 741 to output 440 volts? Why or why not? The differential input voltage necessary to drive the output of this op-amp to -4.5 volts is -20.455 µV. Follow-up question: what does it mean for the input voltage differential to be negative 20.455 microvolts? Provide an example of two input voltages (Vin(+) and Vin(−) ) that would generate this much differential voltage. Answer 3 ”Open-loop voltage gain” simply refers to the differential voltage gain of the amplifier, without any connections ”feeding back” the amplifier’s output signal to one or more of its inputs. A high gain figure means that a very small differential voltage is able to drive the amplifier into saturation. Answer 4 The output voltage of an opamp cannot exceed either power supply ”rail” voltage, and it is these ”rail” connections that either source or sink load current. Follow-up question: does this model realistically depict the input characteristics (especially input impedance) of an opamp? Why or why not? Answer 5 • • • • • • • •

Magnitude of input terminal currents: 0 Input impedance: infinite Output impedance: 0 Input voltage range: never exceeding +Vcc/-Vee Output voltage range: never exceeding +Vcc/-Vee Differential voltage (between input terminals) with negative feedback: 0 Output current: infinite This one is probably the biggest assumption on this list. Voltage Gain of opamp: infinite Ideal values are realistic.

Answer 6 Being able to ”swing” the output voltage ”rail to rail” means that the full range of an op-amp’s output voltage extends to within millivolts of either power supply ”rail” (+V and -V). Challenge question: identify at least one op-amp model that has this ability, and at least one that does not. Bring the datasheets for these op-amp models with you for reference during discussion time.

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Answer 7 Ideally, Vout = 0 volts. However, the output voltage of a real op-amp under these conditions will invariably be ”saturated” at full positive or full negative voltage due to differences in the two branches of its (internal) differential pair input circuitry. To counter this, the op-amp needs to be ”trimmed” by external circuitry. Follow-up question: the amount of differential voltage required to make the output of a real opamp settle at 0 volts is typically referred to as the input offset voltage. Research some typical input offset voltages for real operational amplifiers. Challenge question: identify a model of op-amp that provides extra terminals for this ”trimming” feature, and explain how it works. Answer 8

a inverting input – signal input terminal. Increasing the signal on this input causes the output signal to decrease or become more negative. b non-inverting input – signal input terminal. Increasing the signal on this input causes the output signal to increase or become more postive c positive power supply (VCC, VDD, V+). Positive DC power supply. This is required for operation d negative power supply (VEE, VSS, GND, V-). This is called the negative DC supply. In reality it just has to be a DC supply with a value ¡ VCC. It can be 0 or a positive power, but is usually 0 or -VCC. It is required for operation. e output terminal. This is the output. It is a function of the two input terminals but is bounded by the power supplies. This means that it can’t be larger or smaller than VCC and VEE, but where in between is controlled by the non-inverting and inverting inputs.

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Answer 9 The arrows shown in this diagram trace ”conventional” current flow, not electron flow:

+V +V + +V

−

+V Power supply

25

Answer 10 In these illustrations, I have likened the op-amp’s action to that of a single-pole, double-throw switch, showing the ”connection” made between power supply terminals and the output terminal.

+V

+V

+ −

+ (-)

−

-V

-V

+V

+V

+

+

−

−

-V

-V

+V

+V

+ −

(+)

+ (-)

−

-V

-V

26

(+)

Answer 11 I will provide answers for only one of the examples, the voltage regulator: • What variable is being stabilized by negative feedback? Output voltage. • How is the feedback taking place (step by step)? When output voltage rises, the system takes action to drop more voltage internally, leaving less for the output. • What would the system’s response be like if negative feedback were not present? Without negative feedback, the output voltage would rise and fall directly with the input voltage, and inversely with the load current. Answer 12 For the left-hand circuit:

Vout = Vin − 4

For the right-hand circuit:

Vout =

p

Vin

The result of placing a mathematical function in the feedback loop of a noninverting opamp circuit is that the output becomes the inverse function of the input: it literally becomes the value of x needed to solve for the input value of y:

y

+ x = f -1(y)

− x

Feedback network

y y = f(x)

Answer 13 GBW product is a constant value for most operational amplifiers, equal to the open-loop gain of the opamp multiplied by the signal frequency at that gain. Answer 14 ”Slew rate” is the maximum rate of voltage change over time ( dv dt ) that an op-amp can output. Answer 15 • The LM318 has a larger slew rate. • The steepest slope of the sine wave is a function of the frequency. Reducing this frequency reduces the input slope. If the input slope reduces then the output slope also reduces. • It means the input frequency is above the cutoff frequency of my circuit and I am seeing attentuation. fcutof f =

27

GBW A

Answer 16 The noninverting amplifier circuit has extremely high input impedance (most likely many millions of ohms), while the inverting amplifier circuit only has 5 kΩ of input impedance. Answer 17 ”Common-mode voltage” is the amount of voltage common to both inputs of a differential amplifier. Ideally, a differential amplifier should reject this common-mode voltage, only amplifying the difference between the two input voltages. Follow-up question: what does the phrase common-mode rejection ratio (CMRR) mean for a differential amplifier? Answer 18 The assumption is a problem if the opamp cannot produce enough output current to drive the load. As the load gets bigger (load resistor gets smaller) the opamp must work harder to provide enough current so that the output voltage is appropriate. There are limits to this. Usually the short circuit current gives some indication of the limit. The short circuit current indicates the maximum current output that the opamp can supply. Challenge question: How would you solve this problem so the output is correct? Answer 19

+12 V

+ − Load -12 V Ground Answer 20 Vout = -8.1 volts AV = 5.4 BW = 156kHz Answer 21 R1 = 1.467 kΩ

28

Answer 22 Vin = -10 V

Vout = 24 V

Follow-up question: how do we know that the input voltage in this circuit is negative and the output voltage is positive? Answer 23 Stage 1: • AV = 1.5 = 3.522 dB Stage 2: • AV = 1.545 = 3.781 dB Overall: • AV = 2.318 = 7.303 dB Answer 24 Iload = 1.818 mA If the load changes resistance, there will be no effect on the amount of current through it, although there will be an effect on the output voltage of the opamp! Answer 25 R1 = 76.95 kΩ BW using LM741 = 143kHz (GBW=1.5MHz) BW using LM318 = 1.43MHz (GBW=15MHz) Answer 26

IR2 = IR1 = 68.09 µA VR2 = 1.498 V VR1 = 3.2 V 22 kΩ 47 kΩ

− Vout = +4.698 V

+

Current arrows drawn in the direction of conventional flow notation.

Vin = +3.2 volts

AV = 1.468 = 3.335 dB Answer 27 R1 = 2 kΩ R2 = 153 kΩ Max BW = 150kHz, when gain is min. Min BW = 17.6kHz, when gain is max.

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Answer 28 Vin 0 volts +5 volts +10 volts +15 volts +20 volts -5 volts -10 volts -15 volts -20 volts

Vout 0 volts +5 volts +10 volts +15 volts +15 volts -5 volts -10 volts -15 volts -15 volts

Answer 29 • AV (pot fully up) = +1 • AV (pot mid-position) = 0 • AV (pot fully down) = -1 Follow-up question: can you think of any interesting applications for a circuit such as this? Challenge question: modify the circuit so that the range of voltage gain adjustment is -6 to +6 instead of -1 to +1. Answer 30 ”Common-mode” voltage refers to that voltage which is common to two or more wires as measured with reference to a third point (in this case, ground). The amplifier in the second circuit only outputs the difference between the two signals, and as such does not reproduce the (common-mode) noise voltage at its output. Challenge question: re-draw the original (one wire plus ground) schematic to model the sources of interference and the wire’s impedance, to show exactly how the signal could become mixed with noise from source to amplifier.

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Answer 31

7.25 V 8.33 kΩ 870 µA

(no current)

21.76 V 25 kΩ − +

870 µA

Vout +29.01 V

+7.25 V 5 kΩ

5 kΩ 550 µA

10 V

3V

5 kΩ 850 µA

5V

5 kΩ 450 µA

750 µA

11 V

Conventional flow notation used for all current arrows Follow-up question: what would be required to get this circuit to output the exact sum of the four input voltages? Answer 32 I = 6.5 mA Answer 33 I = 6.5 mA

Vout = -6.5 V

Answer 34 AV (ratio) = 0.333

AV (dB) = -9.542 dB

Answer 35 Vout (ideal) = 1.01 volts RMS I’ll let you determine possible faults in the circuit! From what we see here, we know the power supply is functioning (both +V and -V rails) and that there is good signal getting to the noninverting input of the opamp.

31

Answer 36 Circuit schematic, as wired:

+V +V − + V Voltmeter -

+ -V

-V

The output voltage will saturate at approximately +11 volts, or -11 volts, with the potentiometer having little or no effect. Answer 37 • Resistor R1 fails open: Output saturates positive. • Solder bridge (short) across resistor R1 : Vout = Vin . • Resistor R2 fails open: Vout = Vin . • Solder bridge (short) across resistor R2 : Output saturates positive. • Broken wire between R1 /R2 junction and inverting opamp input: Output voltage unpredictable.

Kuphaldt, Tony. ”Socratic Electronics.” Socratic Electronics. Ibiblio.org, n.d. Web. 28 Dec. 2014. This worksheet and all related files are licensed under the Creative Commons Attribution License, version 1.0. To view a copy of this license, visit http://creativecommons.org/licenses/by/1.0/, or send a letter to Creative Commons, 559 Nathan Abbott Way, Stanford, California 94305, USA. The terms and conditions of this license allow for free copying, distribution, and/or modification of all licensed works by the general public.

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