Download Homework Assignment 04 2. The op-amp in the circuit is ideal

55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Homework Assignment 04
Question 1 (2 points each unless noted otherwise)
1.
A 9-V dc power supply generates 10 W in a resistor. What peak-to-peak amplitude should an ac
source have to generate the same power in the resistor?
(a) 12.73 V
2.
(b) 25.5 V
(c) 18 V
(d) 12.73 V
Answer: The ac source’s effective (or rms) value should also be 9 V. This measn the peak value
should be 9√2 V, so the peak-to-peak value should be 18√2 = 25.5 V, so the answer is (b).
The op-amp in the circuit is ideal, except for non-zero input bias currents. Further, 𝑅1 = 10K
and 𝑅2 = 30K. What should 𝑅3 ′𝑠 value be?
(a)
(b)
(c)
(d)
10K
𝑅1 + 𝑅2 = 30K
0Ω
𝑅1 ||𝑅2 = 7.5K
Answer: R 3 compensates for input bias currents and a rule of thumb is to choose R 3 = R1 ||R 2, so the
answer is (d).
3.
A resistor has a nominal value of 10K, but its actual value is 9.98K. What is the percentage
error?
(a) 0.02K
4.
(b) −0.02K
(c) −0.2% (d)
(e) 0.2%
Answer: Percentage error is ((𝟗. 𝟗𝟖 − 𝟏𝟎)⁄𝟏𝟎) × 𝟏𝟎𝟎 = −𝟎. 𝟐%, so the answer is (d)
The op-amp in the circuit is ideal, except that the open-loop gain is finite, namely 𝐴𝑑 =
100,000. Further, 𝑅2 = 100K and 𝑅1 = 10K. If, for some input signals, the output is
𝑣𝑂 = −5 V, then
(a)
(b)
(c)
(d)
𝑣− = 𝑣+ = 0V (virtual ground concept)
𝑣− = +50 𝜇V
𝑣− = −50 𝜇V
Need additional information: 𝑅3 , 𝑣𝑠
Answer: The input bias current is zero and input offset voltage is zero, so v+ = 0 V. However, the
gain is finite, so v− ≠ v+ . For a −5 V output, v− must be v− = vO ⁄Ad = 5⁄100,000 = 50 µV, so
the answer is (b).
1
55:041 Electronic Circuits. The University of Iowa. Fall 2014.
5.
In the circuit below, the op-amp is ideal, except for an input bias current |𝐼𝑏 | = 1 nA.
Further, 𝑅𝐹 = 10K, 𝑅1 = 100 Ω and 𝐶 = 1 𝜇F. The switch is opened at 𝑡 = 0. What is the
output voltage after 5 seconds? (3 points)
(a)
(b)
(c)
(d)
6.
≈ +500 mV
≈ −500 mV
≈ ±500 mV
Need additional information
Answer: For 𝑡 ≥ 0, the voltage across the capacitor is 𝑣𝐶 = (±𝐼𝑏 Δ𝑡)⁄𝐶 which is
�(±1 × 10−9 ) (5)�⁄(1 × 10−6 ) = ±5 mV for 𝑡 = 5 s. The gain of the amplifier is 1 +
𝑅𝐹 ⁄𝑅1 = 101, so that the output voltage is ± 505 mV. Thus (c) is the answer.
What frequency is 2.2 decades higher than 500 Hz?
(a) 1.01 kHz
7.
(c) 522 Hz
(d) 79.24 kHz
Answer: 2.2 = log(𝑓𝑥 ⁄500), so that 𝑓𝑥 = 79.24 Hz, so (d) is the answer.
What is frequency is 3 decades down from 220 Hz?
(a) 22 mHz
8.
(b) 644 Hz
(b) 220 mHz (c) 6.4 mHz
(d) 190 Hz
Answer: 3 = log(220⁄𝑓𝑥 ), so that 𝑓𝑥 = 220 mHz, so (b) is the answer.
A signal with amplitude 𝑣 = 4 V at 4 kHz decreases as frequency increases at −2 dB/octave.
What is the amplitude in V at 13 kHz? (3 points)
Answer: There are log 2 (13⁄4) = 1.7 octaves between 4 kHz and 13 kHz. Thus, the
amplitude decreases by 1.7 × 2 = 3.4 dB. The new amplitude is 20 log 4 − 3.4 = 8.6 dB.
This is equivalent to 2.7 V.
9.
A 10 mW signal is equivalent to a power of
(a) −20 dB
(b) −40 dB
(c) 20 dB
Answer: 𝑃𝑑𝐵 = 10 log(10 × 10−3 ) = −20 dB, so the answer is (a)
2
(d) 0dB
55:041 Electronic Circuits. The University of Iowa. Fall 2014.
10.
A 9-V dc power supply generates 10 W in a resistor. What peak-to-peak amplitude should
an ac source have to generate the same power in the resistor?
(a) 12.73 V
(b) 25.5 V
(c) 18 V
(d) 12.73 V
Answer: The ac source’s effective (or rms) value should also be 9 V. This means the peak
value should be 9√2 V, so the peak-to-peak value should be 18√2 = 25.5 V, so the answer is
(b).
11.
In the circuit 𝑉𝑅𝐸𝐹 = 2.5 V, and 𝑉𝐼𝑁 = 1.5 V. The output voltage is
(a)
(b)
(c)
(d)
(e)
1.5 V
2.5V
Close to the positive supply rail
Close to the negative supply rail
Need additional information
Answer: |𝑉+ | > |𝑉− |, so that the op-amp will pull 𝑉𝑂 > 0, and turn the diode on. The
cathode (𝑉𝑂𝑈𝑇 ) will follow until 𝑉− reaches 𝑉+ . Any further increase of will reduce 𝑉𝑂 and
𝑉𝑂𝑈𝑇 . Thus, the output stabilizes at 𝑉𝑂𝑈𝑇 = 2.5 V.
The T.A. was instructed not to grade this question.
12.
The voltage gain of the amplifier shown is
(a)
(b)
(c)
(d)
13.
≈ −5.7
≈ 5.7
≈ 6.77
≈ 13.4
Answer: 𝐴𝑣 = −𝑅𝑓 ⁄𝑅1 = 68K⁄12K = 5.7, so the answer is (a)
If the feedback/input resistor ratio of An op-amp feedback amplifier is 4.6 with 1.7 V applied
to the noninverting input, what is the output voltage value?
(a) 7.82 V
(b) Saturation
(c) Cutoff
(d) 9.52 V
Answer: 𝐴𝑣 = �1 + 𝑅𝑓 ⁄𝑅1 � = (1 + 4.6) = 5.6. The output voltage is then 𝑉𝑂 = 1.7 × 5.6 =
9.52 V. Thus, the answer is (d)
3
55:041 Electronic Circuits. The University of Iowa. Fall 2014.
14. Consider the amplifier below. 𝑉𝑖𝑛 = 1.5 V, what is 𝑉𝑜𝑢𝑡 ?
15.
Answer: The gain of the first amplifier is 𝐴𝑓 = − (30)⁄10 = −3 and the gain of the second
amplifier is −1, gving an overall gain of (−3)(−1) = 3. The output voltage is thus 𝑉𝑜𝑢𝑡 =
1.5 × 3 = 4.5 V.
In the circuit shown, the output voltage is
(a)
(b)
(c)
(d)
(e)
5(1 + 8⁄2) = 25 V
5(8⁄2) = 20 V
≈ 15 V
≈ −15 V
(8⁄2) = 120 V
Answer: This is a non-inverting amplifier with gain (1 + 8⁄2) = 5, so with a 5-V input the
output should be 25 V. However, the op-amp is powered by a +15-V power supply, so that
the output will be clamped to a value close to +15 V, so the answer is (c).
16.
Consider the circuit shown. Assume ideal op-amp behavior.
(a)
(b)
(c)
(d)
V− = 𝑉+ = 5 V (op-amp operation)
𝑉− = 10 × 2⁄(2 + 8) = 2 V (voltage division)
V− = 0 (op-amp input current = 0)
Need additional information
Answer: These is no feedback in the circuit to create a virtual short (𝑉− = 𝑉+ ). No current
flows into the input terminals so that 𝑉− follows from voltage division, so the answer is (b).
4
55:041 Electronic Circuits. The University of Iowa. Fall 2014.
17.
An engineer measures the (step response) rise time of an amplifier as 𝑡𝑟 = 0.7 𝜇s. Estimate
the 3 dB bandwidth of the amplifier.
Answer:
0.35
𝑡𝑟
0.35
=
0.7 × 10−6
= 500 kHz
𝐵𝑊 ≅
18.
Define the CMRR for a differential amplifier. What is the ideal value?
19.
Answer: CMMR = 20log10 |𝐴𝑑 ⁄𝐴𝑐𝑚 | ). Ideally, CMMR → ∞
An amplifier has a differential gain of -50,000 and a common-mode gain of 2. What is the
common-mode rejection ratio?
(a) –87.96 dB
(b) 44 dB
(c) -44 dB
(d) 87.96 dB
Answer: CMMR = 20 log10 |𝐴𝑑 ⁄𝐴𝑐 | = 20 log10 |50 × 103 ⁄2| = 87.96 dB, so the answer
is (d).
20. A differential amplifier has a common-mode gain of 0.2 and a common-mode rejection ratio of
3250. What would the output voltage be if the single-ended input voltage was 7 mV rms?
(a) 1.4 mV rms
21.
(b) 650 mV rms
(c) 4.55 mV rms
(d) 0.455 V rms
Answer: 𝐶𝑀𝑅𝑅 = |𝐴𝑑 ⁄𝐴𝑐𝑚 | so that 3250 = 𝐴𝑑 ⁄0.2 and 𝐴𝑑 = 650. The output voltage is
650 × 7 = 4.55 mV rms, so the answer is (c).
What is 𝑣𝑜 in the following circuit if 𝑣𝑅𝐸𝐹 = 1.2 V, 𝑅1 = 680 Ω, and 𝑅2 = 200 Ω?
Answer: The current through 𝑅2 is 1.2/200 = 6 mA, which also flows through 𝑅1 . Thus,
the output voltage is 1.2 + 0.006× 680 = 5.28 V
5
55:041 Electronic Circuits. The University of Iowa. Fall 2014.
22.
Consider the voltage regulator below, implemented with a reference voltage 𝑉𝑅𝐸𝐹 = 1.25 V,
an ideal op-amp and BJT with very large gain (𝛽 → ∞). Determine the output voltage to 4
significant figures. (5 points)
Answer: The op-amp and feedback loop maintains a voltage 𝑉𝑅𝐸𝐹 across 𝑅1 so the current
through both resistors is 𝐼 = 𝑉𝑅𝐸𝐹 ⁄𝑅1. The output voltage is then
𝑉𝑂 = 𝐼(𝑅1 + 𝑅2 ) =
23.
𝑉𝑅𝐸𝐹
(𝑅1 + 𝑅2 ) = 4.959 V
𝑅1
When researching part numbers for three-terminal regulators, an engineer encounters the
term “LDO”. What does “LDO” stand for?
Answer: “Low Drop Out”
24.
In the circuit below 𝑅1 = 10K, 𝑅2 = 15K, and 𝑅3 compensates for the op-amp’s input bias
current. What should it’s value be to be effective?
(a)
(b)
(c)
(d)
(e)
10K
15K
6K
25K
Need 𝐼𝑂𝑆
Answer: Choose 𝑅3 = 𝑅1 ‖𝑅2 = 6K, so (c) is the answer.
6
55:041 Electronic Circuits. The University of Iowa. Fall 2014.
25.
Which of the following depicts the correct current direction? Circle one. (1 point)
26.
This is a current-to-voltage converter with
𝑣𝑂 = −𝑖𝑆 𝑅𝐹 = (10 × 10−6 )(1 × 106 ) = −10 V
27.
𝑖𝑆 = 10 µA, 𝑅𝐹 = 1 MΩ V, 𝑣𝑜 =?
This is a follower where 𝑣𝑂 = 𝑣+ .
Thus
28.
𝑣𝑂 = 𝑣+ =
𝑣𝐼 = 6 V, 𝑣𝑜 =?
20
6 =2V
20 + 40
This is a noninverting amplifier where
𝑣+ =
Thus
𝑣𝐼1 𝑣𝐼2
+
=1+2=3
2
2
𝑣𝑂 = �1 +
𝑣𝐼1 = 2 V, 𝑣𝐼2 = 4 𝑉, 𝑣𝑜 =?
7
50
� 𝑣 = 2𝑣+ = 6 V
50 +
55:041 Electronic Circuits. The University of Iowa. Fall 2014.
29.
A current source supplies a nominal current 𝐼𝑅𝐸𝐹 = 1 mA. When connected to a 5K load,
only 0.95 mA flows through the load. What is the internal resistance of the current source?
Answer: The voltage across the load is (5 × 103 )(0.95 × 10−3 ) = 4.750 V. A current
0.05 mA flows through the current source’s internal resistance, which has value
4.75⁄(0.05 × 10−3 ) = 95K
30.
A bench power supply is set to an output voltage of 5 V. When it is connected to a circuit
that draws 𝐼𝑂 = 2.5 A, the output voltage drops to 4.95 V. What is the output resistance 𝑅𝑂
of the power supply?
(a) ≈ 20 mΩ
(b) ≈ 1.98 Ω
(c) Need additional information
31.
Answer: 𝑅𝑂 = Δ𝑉⁄Δ𝐼 = 0.05⁄2.5 = 20 mΩ, so (a)
An AAA cell has a no-load voltage of 1.605 V. When a 100 Ω resistor is connected across
its terminals, the voltage drops to 1.595 V. What is the cell’s internal resistance?
a) ≈ 620 mΩ
b) ≈ 10 mΩ
c) Need additional information
Answer: The current flowing through the load resistance is 𝐼𝐿 = 1.595⁄100 = 15.95 mA.
The internal resistance is 𝑅𝑂 = Δ𝑉⁄Δ𝐼 = (1.605 − 1.595)⁄(15.95 × 10−3 ) = 0.627 Ω.
Thus, (a) is the answer.
32.
In the circuit 𝑉𝐼𝑁 = 10 V, 𝑅1 = 10𝐾, and 𝑅𝐿 = 5𝐾. What current flows through 𝑅𝐿 ?
Answer: By op-amp action the voltage across 𝑅1 is 𝑉𝑖𝑛 and the current through 𝑅1 and 𝑅𝐿 is
10⁄10K = 1 mA.
8
55:041 Electronic Circuits. The University of Iowa. Fall 2014.
33.
A 100-mV source with internal resistance 𝑅𝑠 = 1K drives an amplifier with gain 𝐴𝑣 =
𝑣𝑜 ⁄𝑣𝑖 = 10 (see figure). The output voltage is 750 mV. What is the amplifier’s input
resistance 𝑅𝑖 ?
(a) ∞
(b) 1K
(c) 3K
(d) Need additional information
(e) 0 Ω
Answer: The source’s and amplifier’s internal resistances form a voltage divider and the
output voltage is 𝑣𝑂 = 𝐴𝑣 𝑣𝑠 (𝑅𝑖 ⁄𝑅𝑖 + 𝑅𝑠 ). Substituting for 𝑣𝑂 , 𝑣𝑠 , 𝐴𝑣 , and 𝑅𝑠 and solving for
𝑅𝑖 yields 𝑅𝑖 = 3K.
9
55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Question 2 The op-amp in the circuit is ideal except for nonzero input bias current 𝐼𝐵 = 10 nA. In the circuit, 𝑅𝐹 =
10K, 𝑅1 = 1K, and 𝑅3 = 100 Ω.
Determine the maximum and minimum output voltage 𝑉𝑂
resulting from 𝐼𝐵 . Remember that 𝐼𝐵 could be positive or
negative. (6 points)
Solution The op-amp is ideal with respect to gain so that 𝑉𝑃 = 𝑉𝑁 and we assume the input bias
current is the same for both the non-inverting and the inverting inputs. For the case where 𝐼𝐵
flows into the op-amp
𝑉𝑁 = 𝑉𝑃 = −𝑅3 (𝐼𝐵 ) = −(100)(10 × 10−9 ) = −1 𝜇V.
KCL at the inverting input, assuming current flows away from the node gives
Substitution of 𝑉𝑁 = −1 𝜇V yields
𝑉𝑁 𝑉𝑁 − 𝑉𝑂
+
+ 𝐼𝐵 = 0
1K
10K
−1 × 10−6 −1 × 10−6 − 𝑉𝑂
+
+ 10 × 10−9 = 0
1K
10K
⇒ 𝑉𝑂 = 89 𝜇V
For the case where 𝐼𝐵 flows out of the op-amp, 𝑉𝑂 = −89 𝜇V. Thus, the maximum output
voltage is −89 𝜇V and the minimum output voltage is +89 𝜇V.
I gave some students incomplete guidance on this question. The T.A. was instructed to
make this an extra-credit question by lowering the total score by 6 points. Thus, even if a
student did not do this question, he/she could still get 100% on the homework assignment.
Question 3 (Own, non-ideal op-amp) In the circuit the opamp is ideal, except for an input bias current |𝐼𝑏 | = 10 nA.
Further, 𝑅𝐹 = 10K, 𝑅1 = 100 Ω and 𝐶 = 6.8 𝜇F. The switch
is opened at 𝑡 = 0.
What is the output voltage after 10 seconds? (3 points)
Solution
For 𝑡 ≥ 0, the voltage across the capacitor is 𝑣𝐶 = (±𝐼𝑏 Δ𝑡)⁄𝐶 which is
�(±10 × 10−9 ) (10)�⁄(6.8 × 10−6 ) = ±14.71 mV for 𝑡 = 10 s.
The gain of the amplifier is1 + 𝑅𝐹 ⁄𝑅1 = 101, so that the output voltage is ± 1.485 V.
10
55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Question 4 (Varactor) For the circuit shown, 𝑉𝑃𝑆 = 5 V and = 10K . The varactor
characteristics are shown in the graph. What is the bandwidth of the circuit in Hz? (5 points)
Solution
From the graph, the varactor has a capacitance of 100 pF with a 5-V reverse voltage. The timeconstant of the circuit is 𝜏 = 𝑅𝐶 = (10 × 103 )(100 × 10−12 ) = 1 𝜇s. The bandwidth is then
𝐵 = 1⁄(2𝜋𝜏) = 159 kHz.
Question 5 With inputs 𝑣𝐼1 = −50 mV, and 𝑣𝐼2 = +50 mV, a difference amplifier has output
𝑣𝑂 = 1.0043 V. With inputs 𝑣𝐼1 = 𝑣𝐼2 = 5 V, the output is 𝑣𝑂 = 0.4153 V. Determine the
CMRR, expressed in dB. (4 points)
Solution
The differential input voltage is 𝑣𝐼2 − 𝑣𝐼1 = 100 mV, and the differential-mode gain
is 1.0043⁄0.1 = 10.043
With 𝑣𝐼1 = 𝑣𝐼2 = 5 V the common-mode voltage gain is 𝐴𝑐𝑚 = 0.4152⁄5 = 0.083
The common-mode rejection ratio is
Expressed in dB
CMMR = �
𝐴𝑑
10.043
�=�
� = 120.85
𝐴𝑐𝑚
0.083
CMMR dB = 20 log10 120.85 = 41.65 dB
11
55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Question 6 Consider the circuits below, and then complete the table (9 points).
Voltage follower
Simple difference amplifier
Current-to-voltage converter
Voltage to current converter
Logarithmic amplifier
Exponential amplifier
Differentiator
Instrumentation amplifier
Integrator
Summing amplifier
Circuit (a)
(c)
(b)
(d)
(e)
(h)
(f)
(j)
(g)
(i)
12
55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Question 7 We would like to measure the voltage 𝑉 = 𝑉1 − 𝑉2 in the circuit below with a
voltmeter. What is the value of 𝑉, and what is the common-mode voltage 𝑉𝑐𝑚 associated with
𝑉? What CMMR is required of the voltmeter if we are to measure 𝑉 to within 0.01%? Express
you answer in dB. (8 points)
Solution
The current through the resistance is 𝐼 = 15⁄(40K) = 0.375 mA. The voltage across the 10K
resistor is therefore 3.75 V.
Further, 𝑉1 = 15 − (0.375 × 15) = 9.37 V, and 𝑉2 = 0.375 × 15 = 5.625. The common-mode
voltage is then
𝑉𝑐𝑚 =
(𝑉1 + 𝑉2 )
= 7.5 V
2
The error must be less than 0.01% or 0.01% of 3.75 V, which is 0.375 mV. Thus, the
multimeter must suppress the 7.5 V common-mode voltage to less than 0.375 mV. In other
words, the CMMR must be at least
This is equivalent to 86 dB.
7.5
= 20 × 103
0.375 × 10−3
13
55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Question 8 (N TYU 9.13) An integrator is driven by a series of pulse shown below. At the end
of the 10th pulse, the output voltage is to be 𝑣𝑂 = −5 V. Assume 𝑉𝐶 = 0 at 𝑡 = 0. Determine the
time constant and values for 𝑅 and 𝐶 that will meet these specifications. (6 points)
Solution
1
𝑡
The output of the integrator is 𝑣𝑂 = − 𝑅𝐶 ∫0 𝑣𝐼 (𝑡)𝑑𝑡. During first pulse the output voltage
decreases linearly and at the end of the first pulse the output voltage is
10 𝜇𝑠
𝑣𝑂 = −
𝑅𝐶
The circuit holds this voltage until the next pulse, during which it again increases linearly. At
the end of n pulses, the voltage is
𝑣𝑂 = −𝑛
10 𝜇𝑠
𝑅𝐶
We have to design the circuit so that this voltage is -5 V when 𝑛 = 10. Thus
10 𝜇𝑠
𝑅𝐶
⇒ 𝑅𝐶 = 20 𝜇s
−5 = −10
Thus, the time constant is 20 𝜇s. If we pick 𝐶 = 0.01 𝜇F, then 𝑅 = 2 kΩ
14