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
20
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
22
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.
23
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.
24
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.
29
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.
30
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.
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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.
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