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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 = 2.5K, = 100K, = 40 mS, and = 1 nF? (a) 65.25 kHz (b) 10 kHz (c) 1.59 kHz (d) 10.4 kHz Answer: The capacitor sees an equivalent resistance = 100K. (If one turns off , = 0, and the current source is effectively removed from the circuit.) The time-constant is = = 100 s. The bandwidth is 1 (2) = 1.59 kHz , so the answer is (c). 2. What is the 3-dB bandwidth of the circuit below? (a) 8 kHz (b) 31.83 kHz (c) 15.92 kHz (d) 100 kHz Answer: The capacitor sees an equivalent resistance = 2 = 10K (the current source has infinite internal resistance) and the time-constant is = = 10 s, so that the bandwidth is 1 (2) = 15.92 kHz , and (c) is the answer. 3. The op-amp in the circuit is ideal, and 1 = 10K, 2 = 100K, and 3 = 10K. The input resistance that the source sees is (a) 1 = 10K (b) 1 + 3 = 20K (virtual short between + and ) (c) (Ideal op-amp has = ) (d) 1 || 2 || 3 = 4.72K (KCL at terminal) Answer: 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 1

Homework Assignment 02 Question 1 (2 points each unless ...s-iihr64.iihr.uiowa.edu/MyWeb/Teaching/ece_55041...Homework Assignment 02 . Question 1 (2 points each unless noted otherwise)

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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 𝑟𝜋 = 2.5K, 𝑟𝑜 = 100K, 𝑔𝑚 = 40 mS, and 𝐶𝐿 = 1 nF?

(a) 65.25 kHz (b) 10 kHz (c) 1.59 kHz (d) 10.4 kHz

Answer: The capacitor sees an equivalent resistance 𝑟𝑜 = 100K. (If one turns off 𝑉𝐼, 𝑔𝑚𝑣𝜋 = 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) ≈ 8 kHz (b) 31.83 kHz (c) 15.92 kHz (d) 100 kHz

Answer: The capacitor sees an equivalent resistance 𝑅 = 𝑅2 = 10K (the current source has infinite internal resistance) and the time-constant is 𝜏 = 𝑅𝐶 = 10 𝜇s, so that the bandwidth is 1 (2𝜋𝜏) = 15.92 kHz⁄ , and (c) is the answer.

3. The op-amp in the circuit is ideal, and 𝑅1 = 10K, 𝑅2 = 100K, and 𝑅3 = 10K. The input resistance that the source sees is

(a) 𝑅1 = 10K (b) 𝑅1 + 𝑅3 = 20K (virtual short between + and −) (c) ∞ (Ideal op-amp has 𝑅𝑖 = ∞) (d) 𝑅1||𝑅2||𝑅3 = 4.72K (KCL at – terminal)

Answer: 𝑅1 = 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 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).

6. In the circuit shown, the output voltage is

(a) 5(1 + 8 2⁄ ) = 25 V (b) 5(8 2⁄ ) = 20 V (c) ≈ 15 V (d) ≈ −15 V (e) (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).

7. True or false: a silicon diode is biased so that 𝑉𝐷 = 0.7 at 25 oC. VD changes with 2 mV/ oC, so that at 125 oC, 𝑉𝐷 will be 0.7 + 100×0.002 = 0.9 V.

Answer: False. 𝑉𝐷 decreases with increasing temperature

8. True or false: a diode, forward biased at ID = 1 mA, has a small-signal or incremental resistance 𝑟𝑑 of about 260 Ω.

Answer: False, because

𝑟𝑑 =𝑉𝑇𝐼𝐷𝑄

=26 mV1 mA

= 26 Ω ≠ 260 Ω

9. Which of the following depicts the correct current direction? Circle one.

10. In the context of diodes, the term “PIV” 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 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)

13. 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.

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55:041 Electronic Circuits. The University of Iowa. Fall 2014.

Question 2 For the current source in the circuit shown, 𝐼𝑆(𝑡) = (0.5)u(𝑡) mA, where u(𝑡) is the unit step function. The capacitor is initially uncharged. What is 𝑉𝑂 at 𝑡 = 22 𝜇s? (8 points)

Solution The capacitor sees an equivalent resistance 𝑅 = 𝑅2 = 10K (the current source has infinite internal resistance) and the time-constant is 𝜏 = 𝑅𝐶 = 10 𝜇s. For 𝑡 → ∞, 𝑉𝑂 = (0.5 mA)(10K) = 5 V. Further, 𝑉𝑂(𝑡) = 5�1 − 𝑒𝑡 𝜏⁄ �. Substituting 𝑡 = 22 𝜇s gives 𝑉𝑂 = 4.45 V.

Alternatively, recognize that 22 𝜇𝑠 is 2.2𝜏 which is the well-known approximation for the 90% rise time. Thus, the output will be 0.9 × 5 = 4.5 V which is close to 4.45 V.

Question 3 For the circuit shown, determine 𝑉𝑂 and 𝐼, assuming that the diode is a Si diode. (4 points)

𝑉𝑂 =? , 𝐼 = ?

Solution

𝐼 =2 − 0.7 − (−8)

(5K + 20K)= 0.37 mA

𝑉𝑂 = −8 + (20K)(𝐼) + 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 𝐼𝐷 and 𝑉𝑜. (6 points)

Solution

Assume the diodes’ internal resistance is negligible and that 𝑉𝛾 = 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

𝑉𝑂2K

+𝑉𝑂 − (10 − 0.7)

2K+𝑉𝑂 − (10 − 0.7)

2K= 0

Solving yields 𝑉𝑂 = 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 𝑉𝑂 2K⁄ = 3.1 mA. By symmetry, half of this current flows through each diode, so that

𝐼𝐷 = 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 𝑉𝑂 and 𝐼. (6 points)

Solution

Assume 𝑉𝛾 for Si is 0.7 V and for Ge is 0.3 V, and the diodes’ internal resistance is negligible. Given the battery and diodes’ 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

𝐼 =10 − 0.3

1K= 9.7 mA

𝑉𝑂 = (9.7 mA) × 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 𝑉𝑂 and 𝐼.

Solution

Assume 𝑉𝛾 for Si is 0.7 V and the diodes’ internal resistance are negligible. Given the battery and diodes’ polarities, assume diodes are forward-biased. Replace the diodes with piecewise linear models as shown.

Then

𝑉𝑂 = 16 − 0.7 − 0.7 = 14.6 V

𝐼 =𝑉𝑜 − 12

4.7K=

14.6 − 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 𝐷1 and 𝐷2 are Si diodes, and 𝑅 = 10 K. The Zener diodes have 𝑉𝑍1 = 4.3 V , and 𝑉𝑍2 = 6.3 V. The input voltage is 𝑣𝐼 = 10 sin (𝜔𝑡). Use a piecewise linear diode model with 𝑉𝛾 = 0.7 V and diode series resistance 𝑟𝑠 = 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. 𝐷2 is reverse biased and 𝐷2 and 𝑍2 are effectively removed from the circuit. The combination of 𝐷1, 𝑍1 do not conduct until the input voltage reaches 4.3+0.7 = 5 V. For 𝑣𝐼 ≥ 5 V, the output is clipped at 5 V, otherwise the output is identical to the input.

Negative part of the input. 𝐷1 is reverse-biased and 𝐷1 and 𝑍1 are effectively removed from the circuit. The combination 𝐷2, 𝑍2 do not conduct until 𝑣𝐼 ≤ −6.3− 0.7 = −7 V. For 𝑣𝐼 ≤ −7 V, the output is clipped at −7 V.

Note: points are subtracted if students do not account for 𝑉𝛾 and if 𝑣𝑚𝑎𝑥, 𝑣𝑚𝑖𝑛 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 𝑉𝛾 = 0.7 V and diode series resistance 𝑟𝑠 = 0. Indicate important features of the plot such as maxima and minima. (12 points)

Solution

Negative input. 𝐷1 is reverse-biased and effectively removed from the circuit, and the output is identical to the input: 𝑣𝑂 = 6 sin (𝜔𝑡). The minimum value is –6 V.

Positive input. For small input voltages, 𝐷1 is reverse-biased and effectively removed from the circuit, and the output is identical to the input. When the input voltage ≥ 2 + 0.7 = 2.7 V, 𝐷1 is forward-biased, 𝐷1 conducts, and the bottom of 𝑅2 is 2.7 V. For this part of the cycle,

𝑣𝑜 = (6 − 2.7)10

10 + 10sin(𝜔𝑡) + 2.7 V

= 1.65 sin(𝜔𝑡) + 2.7 V

The maximum value is 1.65 + 2.7 = 4.35 V

Note: points were subtracted if students did not account for 𝑉𝛾 and if 𝑣𝑚𝑎𝑥, 𝑣𝑚𝑖𝑛were 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 𝑣𝐼 ≥ 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 𝑉𝛾 = 0.7 V and diode series resistance 𝑟𝑠 = 0, and assume the capacitor is initially uncharged.

a) Sketch the output 𝑣𝑂 for two cycles of the input voltage starting at 𝑡 = 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)

𝑣𝑆 = 5 sin(𝜔𝑡) 𝑉𝐵 = 2 V

Solution

For small 𝑣𝑆 the diode is reverse-biased and it is open, and 𝑣𝑂 = 5 sin(𝜔𝑡). When 𝑣𝑆 ≥ 2.7 V, the diode conducts, C charges, and the output is at 𝑣𝑂 = 2.7 V. C continues charging until 𝑣𝑆 reaches its peak of 5 V. At this time 𝑣𝐶 = 5 − 2.7 = 2.3 V. When 𝑣𝑆 decreases, the diode becomes reverse-biased opens. 𝑣𝐶 subtracts from the input voltage, and 𝑣𝑂 = 5 sin(𝜔𝑡) − 2.3 V.

The steady-state maximum voltage is 2.7 V The steady-state minimum voltage is −5− 2.3 = −7.3 V

Note: points were subtracted if students did not account for 𝑉𝛾 and if 𝑣𝑚𝑎𝑥, 𝑣𝑚𝑖𝑛were 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 𝑉𝑃𝑆= 3.3 V, and 𝑅 = 150 Ω. Also shown, are the LED’s voltage-current characteristics. Draw the circuit’s dc load line on the characteristics and find 𝐼𝐷 and 𝑉𝐷 (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: 𝐼 = 3.3 150⁄ = 22 mA. Connect the two points to get the dc load line. The dc load line intersects the diode V-I curve at around 𝐼𝐷 ≅8 mA and 𝑉𝐷 ≅ 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 𝑅 and 𝐷. The coupling capacitor is large enough so that it is effectively a short at the ac source’s 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 diode’s small-signal resistance is 1 (40𝐼) = 25 Ω⁄ . This forms a voltage divider with the 𝑅 so that the ac output voltage is 5 mV.

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