Question 1 (2 points each unless noted otherwise) 1. Is the following circuit an STC circuit? Homework Assignment 02 (a) Yes (b) No (c) Need additional information Answer: There is one reactive element (the capacitor) that sees a single equivalent resistance, so the answer is (a). 2. What is the 3-dB bandwidth of the circuit below? (d) 8 khz (e) 31.83 khz (f) 15.92 khz (g) 100 khz Answer: The capacitor sees an equivalent resistance R = R 2 = 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 output rise time of an STC circuit in response to a step input is 1 μs. Estimate the 3-dB bandwidth of the circuit. Answer: BW 0.35 t r = 0.35 1 10 6 = 350 khz. 4. For SPICE, Explain very briefly the difference between the multiplier M and Meg, as in a resistor has value 2M versus a resistor with value 2Meg? Answer: SPICE interprets M as milli and Meg as mega so that 2M is 0.002 Ω" and 2Meg is 2,000,000 Ω. 5. An STC circuit has a time constant of τ = 10 ms. Estimate the rise time of the output in response to a step input. Answer: t r 2.2τ = 22 ms. 1
6. The op-amp in the circuit is ideal, and R 1 = 10K, R 2 = 100K, and R 3 = 10K. The input resistance that the source sees is (a) R 1 = 10K (b) R 1 + R 3 = 20K (virtual short between + and ) (c) (Ideal op-amp has R i = ) (d) R 1 R 2 R 3 = 4.72K (KCL at terminal) Answer: R 1 = 10K, so (a) is the answer. 7. A resistor has a nominal value of 10K, but its actual value is 9.98K. What is the percentage error? (a) 0.02K (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) 8. True or false: the mobility of holes is greater than the mobility of electrons in semiconductor materials. Answer: False 9. 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). 10. 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). 11. True or false: a silicon diode is biased so that V D = 0.7 at 25 o C. V D changes with 2 mv/ o C, so that at 125 o C, V D will be 0.7 + 100 0.002 = 0.9 V. Answer: False. V D decreases with increasing temperature 2
12. True or false: a diode, forward biased at I D = 1 ma, has a small-signal or incremental resistance r d of about 260 Ω. Answer: False, because r d = V T 26 mv = = 26 Ω 260 Ω I DQ 1 ma 13. Which of the following depicts the correct current direction? Circle one. 14. 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 15. A bench power supply is set to an output voltage of 5 V. When it is connected to a circuit that draws I O = 2.5 A, the output voltage drops to 4.95 V. What is the output resistance R O of the power supply? (a) 20 mω (b) 1.98 Ω (c) Need additional information Answer: R O = ΔV ΔI = 0.05 2.5 = 20 mω, so (a) 16. 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 I L = 1.595 100 = 15.95 ma. The internal resistance is R O = ΔV ΔI = (1.605 1.595) (15.95 10 3 ) = 0.627 Ω. Thus, (a) is the answer. 3
17. The voltage gain of the amplifier shown is (a) 5.7 (b) 5.7 (c) 6.77 (d) 13.4 Answer: A v = R f R 1 = 68K 12K = 5.7, so the answer is (a) 18. In the context of diodes, the term PIV means: Answer: Peak Inverse Voltage Question Shown is the symbol a popular SPICE computer simulation program uses for an n-channel MOSFET, along with labels indicating the drain, source and gate terminals. There is also a 4 th terminal, indicated with an arrow. The physical device has three, not four terminals. A perplexed student asks her professor what is this terminal, and what should she do with in when she builds her circuit in SPICE. Provide a short (3 4 sentences) answer to the student. (3 points) 4
Question 2 For the current source in the circuit shown, I S (t) = (0.5)u(t) ma, where u(t) is the unit step function. The capacitor is initially uncharged. What is V O at t = 22 μs? (8 points) The capacitor sees an equivalent resistance R = R 2 = 10K (the current source has infinite internal resistance) and the time-constant is τ = RC = 10 μs. For t, V O = (0.5 ma)(10k) = 5 V. Further, V O (t) t τ = 5(1 e ). Substituting t = 22 μs gives V O = 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 (Principles) Using nodal analysis, find I o in the circuit below as follows. Write a KCL equation at the node that connects the 4K, 2K, and other 4K resistor. Use the convention that currents flow away from the node. (6 points) Call the voltage at the node that connects the 4K, 2K, and other 4K resistor V x. A KCL equation is then Solving for V x yields V x = 9.82 V. Then V x 24 4K + V x 6 2K + V x 6K = 0 I o = V x 6 2K = 1.91 ma 5
Question 4 For the following circuit the diodes are Si. Make reasonable assumptions and determine I D and V o. (6 points) Assume the diodes internal resistance is negligible and that V γ = 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 V O 2K + V O (10 0.7) 2K + V O (10 0.7) 2K Solving yields V O = 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 V O 2K = 3.1 ma. By symmetry, half of this current flows through each diode, so that I D = 1.55 ma = 0 6
Question 5 For the following circuit one diode is made from Ge and the other from Si semiconductor. Make reasonable assumptions and determine V O and I. (6 points) Assume V γ 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 10 0.3 I = = 9.7 ma 1K V O = (9.7 ma) 1K = 9.7 V 7
Question 6 For the following circuit the diodes are made from Si semiconductor. Make reasonable assumptions and determine V O and I. (6 points) Assume V γ 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 V O = 16 0.7 0.7 = 14.6 V I = V o 12 14.6 12 = = 0.553 ma 4.7K 4.7K 8
Question 7 Consider the circuit below. The diodes D 1 and D 2 are Si diodes, and R = 10 K. The Zener diodes have V Z1 = 4.3 V, and V Z2 = 6.3 V. The input voltage is v I = 10 sin (ωt). Use a piecewise linear diode model with V γ = 0.7 V and diode series resistance r s = 0. Sketch the output voltage for one cycle, carefully labeling important features of the plot such as maxima and minima. (8 points) Positive part of the input. D 2 is reverse biased and D 2 and Z 2 are effectively removed from the circuit. The combination of D 1, Z 1 do not conduct until the input voltage reaches 4.3+0.7 = 5 V. For v I 5 V, the output is clipped at 5 V, otherwise the output is identical to the input. Negative part of the input. D 1 is reverse-biased and D 1 and Z 1 are effectively removed from the circuit. The combination D 2, Z 2 do not conduct until v I 6.3 0.7 = 7 V. For v I 7 V, the output is clipped at 7 V. Note: points are subtracted if students do not account for V γ and if v max, v min are not shown on the plot, and plot axis are not labeled. 9
Question 8 An amplifier has an input resistance R i = 1K, and has a voltage gain of A v = 100 when driven from a signal with internal resistance R s 0. The amplifier is used to amplify a v s = 1 mv signal from a sensor that has an internal resistance of R s 20K. What is the output amplitude? (6 points) The sensor s internal resistance and the amplifier s input resistance form a voltage divider so that the effective input voltage is v i = 1 1 mv = 47.6 μv 1 + 20 The output voltage is v o = A v v i = 100 47.6 10 6 = 4.76 mv 10
Question 9 Consider the circuit below. Use a piecewise linear diode model with V γ = 0.7 V and diode series resistance r s = 0, and assume the capacitor is initially uncharged. a) Sketch the output v O 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) v S = 5 sin(ωt) V B = 2 V For small v S the diode is reverse-biased and it is open, and v O = 5 sin(ωt). When v S 2.7 V, the diode conducts, C charges, and the output is at v O = 2.7 V. C continues charging until v S reaches its peak of 5 V. At this time v C = 5 2.7 = 2.3 V. When v S decreases, the diode becomes reverse-biased opens. v C subtracts from the input voltage, and v O = 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 V γ and if v max, v min were 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. 11
Question 10 Consider the circuit below. Assume V PS = 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 I D and V D (6 points). 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 I D 8 ma and V D 2.25 V. 12
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 junctionand diffusion capacitances. (5 points) 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. 13