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55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Homework Assignment 02
Question 1 (2 points each unless noted otherwise)
1. What is the 3-dB bandwidth of the amplifier shown below if ππ = 2.5K, ππ = 100K, ππ = 40 mS, and πΆπΏ = 1 nF?
(a) 65.25 kHz (b) 10 kHz (c) 1.59 kHz (d) 10.4 kHz
Answer: The capacitor sees an equivalent resistance ππ = 100K. (If one turns off ππΌ, πππ£π = 0, and the current source is effectively removed from the circuit.) The time-constant is π = π πΆ =100 πs. The bandwidth is 1 (2ππ) = 1.59 kHzβ , so the answer is (c).
2. What is the 3-dB bandwidth of the circuit below?
(a) β 8 kHz (b) 31.83 kHz (c) 15.92 kHz (d) 100 kHz
Answer: The capacitor sees an equivalent resistance π = π 2 = 10K (the current source has infinite internal resistance) and the time-constant is π = π πΆ = 10 πs, so that the bandwidth is 1 (2ππ) = 15.92 kHzβ , and (c) is the answer.
3. The op-amp in the circuit is ideal, and π 1 = 10K, π 2 = 100K, and π 3 = 10K. The input resistance that the source sees is
(a) π 1 = 10K (b) π 1 + π 3 = 20K (virtual short between + and β) (c) β (Ideal op-amp has π π = β) (d) π 1||π 2||π 3 = 4.72K (KCL at β terminal)
Answer: π 1 = 10K, so (a) is the answer.
4. True or false: the mobility of holes is greater than the mobility of electrons in semiconductor materials.
Answer: False
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55:041 Electronic Circuits. The University of Iowa. Fall 2014.
5. 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 sourceβs effective (or rms) value should also be 9 V. This means the peak value should be 9β2 V, so the peak-to-peak value should be 18β2 = 25.5 V, so the answer is (b).
6. In the circuit shown, the output voltage is
(a) 5(1 + 8 2β ) = 25 V (b) 5(8 2β ) = 20 V (c) β 15 V (d) β β15 V (e) (8 2β ) = 120 V
Answer: This is a non-inverting amplifier with gain (1 + 8 2β ) = 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).
7. True or false: a silicon diode is biased so that ππ· = 0.7 at 25 oC. VD changes with 2 mV/ oC, so that at 125 oC, ππ· will be 0.7 + 100Γ0.002 = 0.9 V.
Answer: False. ππ· decreases with increasing temperature
8. True or false: a diode, forward biased at ID = 1 mA, has a small-signal or incremental resistance ππ of about 260 Ξ©.
Answer: False, because
ππ =πππΌπ·π
=26 mV1 mA
= 26 Ξ© β 260 Ξ©
9. Which of the following depicts the correct current direction? Circle one.
10. In the context of diodes, the term βPIVβ means:
Answer: Peak Inverse Voltage
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55:041 Electronic Circuits. The University of Iowa. Fall 2014.
11. True or false: in the circuit below, even though the diode equation is nonlinear, the photocurrent is essentially linear with photon flux density.
Answer: True
12. A bench power supply is set to an output voltage of 5 V. When it is connected to a circuit that draws πΌπ = 2.5 A, the output voltage drops to 4.95 V. What is the output resistance π π of the power supply?
(a) β 20 mΞ© (b) β 1.98 Ξ© (c) Need additional information
Answer: π π = Ξπ ΞπΌβ = 0.05 2.5 = 20 mΞ©β , so (a)
13. An AAA cell has a no-load voltage of 1.605 V. When a 100 Ξ© resistor is connected across its terminals, the voltage drops to 1.595 V. What is the cellβs internal resistance?
a) β 620 mΞ© b) β 10 mΞ© c) Need additional information
Answer: The current flowing through the load resistance is πΌπΏ = 1.595 100 = 15.95 mAβ . The internal resistance is π π = Ξπ ΞπΌ = (1.605β 1.595) (15.95 Γ 10β3)ββ = 0.627 Ξ©. Thus, (a) is the answer.
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55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Question 2 For the current source in the circuit shown, πΌπ(π‘) = (0.5)u(π‘) mA, where u(π‘) is the unit step function. The capacitor is initially uncharged. What is ππ at π‘ = 22 πs? (8 points)
Solution The capacitor sees an equivalent resistance π = π 2 = 10K (the current source has infinite internal resistance) and the time-constant is π = π πΆ = 10 πs. For π‘ β β, ππ = (0.5 mA)(10K) = 5 V. Further, ππ(π‘) = 5οΏ½1 β ππ‘ πβ οΏ½. Substituting π‘ = 22 πs gives ππ = 4.45 V.
Alternatively, recognize that 22 ππ is 2.2π which is the well-known approximation for the 90% rise time. Thus, the output will be 0.9 Γ 5 = 4.5 V which is close to 4.45 V.
Question 3 For the circuit shown, determine ππ and πΌ, assuming that the diode is a Si diode. (4 points)
ππ =? , πΌ = ?
Solution
πΌ =2 β 0.7 β (β8)
(5K + 20K)= 0.37 mA
ππ = β8 + (20K)(πΌ) + 0.7 = 0.14 V
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55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Question 4 For the following circuit the diodes are Si. Make reasonable assumptions and determine πΌπ· and ππ. (6 points)
Solution
Assume the diodesβ internal resistance is negligible and that ππΎ = 0.7 V. Assume that both diodes are forward-biased. Replace the diodes with linear models as shown below. This is now a linear circuit that one can solve using nodal analysis, KCL, KVL, superposition, Thevenin or Norton equivalent circuits, etc.
A KCL equation for the output node is
ππ2K
+ππ β (10 β 0.7)
2K+ππ β (10 β 0.7)
2K= 0
Solving yields ππ = 6.2 V. The sign of the voltage is consistent with our assumption: the diodes are forward biased.
The current through the output resistor is ππ 2Kβ = 3.1 mA. By symmetry, half of this current flows through each diode, so that
πΌπ· = 1.55 mA
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55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Question 5 For the following circuit one diode is made from Ge and the other from Si semiconductor. Make reasonable assumptions and determine ππ and πΌ. (6 points)
Solution
Assume ππΎ for Si is 0.7 V and for Ge is 0.3 V, and the diodesβ internal resistance is negligible. Given the battery and diodesβ polarities, assume the Ge diode is forward-biased. The voltage across the Si diode is then less that its turn on-voltage so it is off. Replace the two diodes with their corresponding piecewise linear models as shown.
Then
πΌ =10 β 0.3
1K= 9.7 mA
ππ = (9.7 mA) Γ 1K = 9.7 V
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55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Question 6 For the following circuit the diodes are made from Si semiconductor. Make reasonable assumptions and determine ππ and πΌ.
Solution
Assume ππΎ for Si is 0.7 V and the diodesβ internal resistance are negligible. Given the battery and diodesβ polarities, assume diodes are forward-biased. Replace the diodes with piecewise linear models as shown.
Then
ππ = 16 β 0.7 β 0.7 = 14.6 V
πΌ =ππ β 12
4.7K=
14.6 β 124.7K
= 0.553 mA
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55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Question 7 Consider the circuit below. The diodes π·1 and π·2 are Si diodes, and π = 10 K. The Zener diodes have ππ1 = 4.3 V , and ππ2 = 6.3 V. The input voltage is π£πΌ = 10 sin (ππ‘). Use a piecewise linear diode model with ππΎ = 0.7 V and diode series resistance ππ = 0.
Sketch the output voltage for one cycle, carefully labeling important features of the plot such as maxima and minima. (8 points)
Solution
Positive part of the input. π·2 is reverse biased and π·2 and π2 are effectively removed from the circuit. The combination of π·1, π1 do not conduct until the input voltage reaches 4.3+0.7 = 5 V. For π£πΌ β₯ 5 V, the output is clipped at 5 V, otherwise the output is identical to the input.
Negative part of the input. π·1 is reverse-biased and π·1 and π1 are effectively removed from the circuit. The combination π·2, π2 do not conduct until π£πΌ β€ β6.3β 0.7 = β7 V. For π£πΌ β€ β7 V, the output is clipped at β7 V.
Note: points are subtracted if students do not account for ππΎ and if π£πππ₯, π£πππ are not shown on the plot, and plot axis are not labeled.
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55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Question 8 For the following circuit, sketch the output for one cycle of the input voltage. Use a piecewise linear diode model with ππΎ = 0.7 V and diode series resistance ππ = 0. Indicate important features of the plot such as maxima and minima. (12 points)
Solution
Negative input. π·1 is reverse-biased and effectively removed from the circuit, and the output is identical to the input: π£π = 6 sin (ππ‘). The minimum value is β6 V.
Positive input. For small input voltages, π·1 is reverse-biased and effectively removed from the circuit, and the output is identical to the input. When the input voltage β₯ 2 + 0.7 = 2.7 V, π·1 is forward-biased, π·1 conducts, and the bottom of π 2 is 2.7 V. For this part of the cycle,
π£π = (6 β 2.7)10
10 + 10sin(ππ‘) + 2.7 V
= 1.65 sin(ππ‘) + 2.7 V
The maximum value is 1.65 + 2.7 = 4.35 V
Note: points were subtracted if students did not account for ππΎ and if π£πππ₯, π£πππwere not shown on the plot, and plot axis were not labeled. A key aspect of the plot is the fact that it is not a clipper/clamp, but a shaper: for π£πΌ β₯ 2.7 V, the output is attenuated.
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55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Question 9 Consider the circuit below. Use a piecewise linear diode model with ππΎ = 0.7 V and diode series resistance ππ = 0, and assume the capacitor is initially uncharged.
a) Sketch the output π£π for two cycles of the input voltage starting at π‘ = 0. Indicate important features of the plot such as maxima and minima. (4 points)
b) What is the steady-state (long term) maximum output voltage? (1 point) c) What is the steady-state (long term) minimum output voltage? (1 point)
π£π = 5 sin(ππ‘) ππ΅ = 2 V
Solution
For small π£π the diode is reverse-biased and it is open, and π£π = 5 sin(ππ‘). When π£π β₯ 2.7 V, the diode conducts, C charges, and the output is at π£π = 2.7 V. C continues charging until π£π reaches its peak of 5 V. At this time π£πΆ = 5 β 2.7 = 2.3 V. When π£π decreases, the diode becomes reverse-biased opens. π£πΆ subtracts from the input voltage, and π£π = 5 sin(ππ‘) β 2.3 V.
The steady-state maximum voltage is 2.7 V The steady-state minimum voltage is β5β 2.3 = β7.3 V
Note: points were subtracted if students did not account for ππΎ and if π£πππ₯, π£πππwere not shown on the plot, and plot axis were not labeled. Two key areas of the plot are the initial charging of the C, and the clipping at 2.7 V.
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55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Question 10 Consider the circuit below. Assume πππ= 3.3 V, and π = 150 Ξ©. Also shown, are the LEDβs voltage-current characteristics. Draw the circuitβs dc load line on the characteristics and find πΌπ· and ππ· (6 points)
Solution. On the voltage axis, mark the supply voltage: 3.3 V. On the current axis, mark the maximum current that can flow through the resistor: πΌ = 3.3 150β = 22 mA. Connect the two points to get the dc load line. The dc load line intersects the diode V-I curve at around πΌπ· β 8 mA and ππ· β 2.25 V.
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55:041 Electronic Circuits. The University of Iowa. Fall 2014.
Question 11 In the circuit shown, the constant current source forces a dc current of 1 mA through π and π·. The coupling capacitor is large enough so that it is effectively a short at the ac sourceβs frequency. The amplitude of the ac source is 10 mV. Determine the amplitude of the (ac) output voltage. The frequency is low enough so that one can ignore the diode junction- and diffusion capacitances. (5 points)
Solution
The diodeβs small-signal resistance is 1 (40πΌ) = 25 Ξ©β . This forms a voltage divider with the π so that the ac output voltage is 5 mV.
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