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55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Homework Assignment 01
In this homework set students review some basic circuit analysis techniques, as well as review
how to analyze ideal op-amp circuits. Numerical answers must be supplied using engineering
notation. For example, 𝐼𝑜 = 19 mA, and not 𝐼𝑜 = 0.019 A,or 𝐼𝑜 = 1.9 × 10−2 A.
Question 1 (2 points each unless noted otherwise)
1.
What is the 3-dB bandwidth of the amplifier shown below if 𝑟𝜋 = 2.5K, 𝑟𝑜 = 100K, 𝑔𝑚 = 40 mS,
and 𝐶𝐿 = 1 nF?
(a)
(b)
(c)
(d)
65.25 kHz
10 kHz
1.59 kHz
10.4 kHz
Answer: The capacitor sees an equivalent resistance 𝑟𝑜 = 100K. (If one turns off 𝑉𝐼 , 𝑔𝑚 𝑣𝜋 = 0, and the
current source is effectively removed from the circuit.) The time-constant is 𝜏 = 𝑅𝐶 = 100 𝜇s. The
bandwidth is 1⁄(2𝜋𝜏) = 1.59 kHz, so the answer is (c).
2.
What is the 3-dB bandwidth of the circuit below?
(a)
(b)
(c)
(d)
63.6 Hz
31.8 Hz
5.07 Hz
10.1 Hz
Answer: The capacitor sees an equivalent resistance 𝑅𝑆 ||𝑅𝐿 = 5K. The time-constant is 𝜏 = 𝑅𝐶 =
5 ms, so that the bandwidth is 1⁄(2𝜋𝜏) = 31.82 Hz, and (b) is the answer.
3. Consider the amplifier below. 𝑉𝑖𝑛 = 1.5 V, what is 𝑉𝑜𝑢𝑡 ?
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.
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55:041 Electronic Circuits. The University of Iowa. Fall 2014.
4.
Decreasing the magnitude of the gain in the given circuit could be achieved by
(a) Reducing amplitude of the input voltage
(b) Increasing 𝑅𝑓
(c) Removing 𝑅𝑓
(d) Increasing 𝑅𝑖
5.
Answer: 𝐴 = − 𝑅𝑓 ⁄𝑅𝑖 so one should increase 𝑅𝑖 to reduce the voltage gain.
Assuming ideal op-amp behavior, the input resistance 𝑅𝑖 of the amplifier is
(a)
(b)
(c)
(d)
(e)
6.
𝑅𝑥
∞Ω
0Ω
𝑅1
𝑅1 + 𝑅𝑥
Answer: For an ideal op-amp, no current flows into the + input terminal, so the the input resistance
is ∞. This, the answer is (b).
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 ≅
7.
8.
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
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
Answer: 𝑅𝑂 = Δ𝑉⁄Δ𝐼 = 0.05⁄2.5 = 20 mΩ, so (a)
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55:041 Electronic Circuits. The University of Iowa. Fall 2014.
9.
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.
10.
What is the impedance of a 0.1 𝜇F capacitor at 𝑓 = 1 kHz?
(a) ≈ −𝑗1.6 × 103 Ω
(b) 𝑗10 × 103 Ω
(c) ≈ +𝑗1.6 × 103 Ω
(d) −1.6 × 103 Ω
(e) 10K
Answer: 𝑍𝐶 = −𝑗⁄(2𝜋𝑓𝐶) = −𝑗⁄(2𝜋 × 1 × 103 × 0.1 × 10−6 ) = − 𝑗1.592K. Thus, (a) is the
answer.
11.
12.
A 𝐼𝑅𝐸𝐹 = 1 mA current source has an output resistance 𝑅𝑜 = 100 kΩ and drives a 1 kΩ load. What
current flows through the load?
Answer: 𝐼𝑙𝑜𝑎𝑑 = 𝐼𝑅𝐸𝐹 [100⁄(100 + 1)] = 0.99 mA.
What is the impedance of a 10 mH inductor 𝑓 = 100 kHz?
(a)
(b)
(c)
(d)
(e)
13.
1 × 103 Ω
𝑗6.28K
−𝑗(6.2.8 × 103 Ω)
−6.28 × 103 Ω
1K
Answer: 𝑍𝐶 = 𝑗2𝜋𝑓𝐿 = 𝑗(2𝜋)(10 × 10−3 )(100 × 103 ) = 6.282K Thus, (b) is the answer.
Four resistor in ascending order are
(a) 22R, 270K, 2K2 1M
(b) 4K7, 10K, 47R, 330K
(c) 3R3, 4R7, 22R, 5K6
(d) 100R, 10K, 1M, 3K3
Answer: Option (c).
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55:041 Electronic Circuits. The University of Iowa. Fall 2014.
14.
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
15.
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.
A schematic shows a capacitor’s value as 100n. This is equivalent to a capacitor with value
(a) 0.1 𝜇F
Answer: Option (a).
16.
(b) 1,000 pF
(c) 0.0001 𝜇F
(d) 0.01 𝜇F
A silicon diode measures a low value of resistance with the meter leads in both positions. The
trouble, if any, is that
(a)
(b)
(c)
(d)
The diode is broken and internally open
The diode is broken and internally shorted
The diode is working but shorted to ground
The diode is working correctly
Answer: Option (b).
17.
An engineer tests a silicon diode with a multimeter using the Ohm-meter function. The meter
measures a low value of resistance with the meter leads in both positions. The trouble, if any, is that
(a)
(b)
(c)
(d)
The diode is broken and internally open
The diode is broken and internally shorted
The diode is working but shorted to ground
The diode is working correctly
Answer: Option (b).
18.
A diode conducts when it is forward-biased, and the anode is connected to the ________ through a
limiting resistor.
(a) Anode
(b) Positive supply
Answer: Option (b).
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(c) Negative supply
(d) Cathode
55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Question 2 For the circuit shown, 𝑅1 = 20 Ω, 𝑅2 = 10 Ω, and
𝐶 = 10 𝜇F . Determine the equivalent resistance the capacitor
sees. In other words, determine the Thevenin resistance of the
network to the left of the capacitor. (8 points)
Solution To determine the Thevenin equivalent resistance, inject a current 𝐼𝑥 and determine the
voltage 𝑉𝑥 , see below. Then, 𝑅𝑇𝐻 = 𝑉𝑥 ⁄𝐼𝑥
KCL at 𝐴, using the convention that currents flowing away from the node is positive, gives
𝐼1 − 1.5𝐼1 − 𝐼𝑥 = 0
⇒ 𝐼𝑥 = −0.5𝐼1
Further, Ohm’s law gives 𝐼1 = 𝑉𝑥 ⁄30, so that
𝐼𝑥 = −
0.5𝑉𝑥
𝑉𝑥
⇒ 𝑅𝑇𝐻 = = −60 Ω
30
𝐼𝑥
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55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Question 3 (Op-Amps) The input voltage is 𝑣𝐼 for each ideal op-amp below. Determine each
output voltage. Assume 𝑣𝐼 = 6 V. (2 points each)
Solution
(a) This is a follower where 𝑣𝑂 = 𝑣+ . Thus
𝑣𝑂 = 𝑣+ =
20
6= 2V
20 + 40
(b) Same answer as (a)
(c) This is a noninverting amplifier where 𝑣𝑂 = (1 + 10⁄10)𝑣+ = 2𝑣+ . Thus
𝑣𝑂 = 2𝑣+ = 2 �
6
6� = 1.333 V
6 + 48
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