Homework Assignment 04

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1 Question 1 (Short Takes) Homework Assignment Consider the single-supply op-amp amplifier shown. What is the purpose of R 3? (1 point) Answer: This compensates for the op-amp s input bias current. The value should be R 1 R The single-supply op-amp amplifier shown has a serious flaw. What is it? (1 point) Answer: There is no dc path to bias the non-inverting input. 3. An engineer measures the (step response) rise time of an amplifier as t r = 0.7 μs. Estimate the 3 db bandwidth of the amplifier. (2 points) Answer: BW 0.35 t r 0.35 = = 500 khz 1

2 4. A I REF = 1 ma current source has an output resistance R o = 100 kω and drives a 1 kω load. What current flows through the load? (2 point) Answer: I load = I REF [100 ( ) ] = 0.99 ma 5. True or false: the mobility of holes is greater than the mobility of electrons in semiconductor materials. (1 point) Answer: False, electrons are more mobile. 6. True or false: the diffusion C d capacitance of a pn junction is generally much larger than the junction capacitance C j. (1 point) Answer: False. 7. True or false: the diffusion capacitance C d of a pn junction is negligible when the junction is reverse-biased. (1 point) Answer: False. 8. 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 = 0.9 V (2 points) Answer: False. V D decreases with increasing temperature 9. 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. (2 points) r d = V T 26 mv = = 26 Ω 260 Ω I DQ 1 ma 10. True or false: the turn-on voltages of Schottky diodes are less than that of Si diodes. However, their reverse leakage/saturation currents are also higher. (1 point) Answer: True 11. True or false: The turn-on voltage of red LEDs is larger than the turn-on voltage of blue LEDs. (1 point) Answer: False. 2

3 12. Which of the following depicts the correct current direction? Circle one. (1 point) Answer: (a) 13. Consider the small-signal equivalent model for a diode below. A typical value for the series resistance r S for small silicon diode is (circle one). (1 point) (a) 20 μω 20 mω (b) 20 mω 20 Ω (c) 20 Ω 200 Ω 14. The reverse saturation current for a Si diode is I S = A at room temperature (25 ). Estimate the value of I S at 5. (2 points) Answer: I S halves for every 10 drop in temperature. Between 25 and 15, there are three 10 drops, so I S is 1 8 of the value at 25, or I S = A 15. In the context of diodes, the term PIV means: (1 point) Answer: Peak Inverse Voltage 16. True or false: in the circuit below, even though the diode equation is nonlinear, the photocurrent is essentially linear with photon flux density. (1 point) Answer: True 3

4 17. A single-pole op-amp has an open-loop low-frequency gain of A = 10 5 and an open loop, 3-dB frequency of 4 Hz. If an inverting amplifier with closed-loop low-frequency gain of A f = 50 uses this op-amp, determine the closed-loop bandwidth. (2 points) Answer: The gain-bandwidth product is Hz. The bandwidth of the closed-loop amplifier is then is /50 = 8 khz. 18. A single-pole op-amp has an open-loop gain of 100 db and a unity-gain bandwidth frequency of 2 MHz. What is the open-loop bandwidth of the op-amp? (2 points) Answer: A gain of 100 db corresponds to 10 5 and the gain-bandwidth product is 2 MHz. Thus, the open-loop bandwidth is (2 MHz) 10 5 = 20 Hz 19. A single-pole op-amp has an open-loop gain of 100 db and a unity-gain bandwidth frequency 5 MHz. What is the open-loop bandwidth of the amplifier? The amplifier is used as a voltage follower. What is the bandwidth of the follower? (2 points) Answer: A gain of 100 db corresponds to 10 5 and the gain-bandwidth product is 5 MHz. Thus, the open-loop bandwidth is (5 MHz) 10 5 = 50 Hz. A unity follower will have a bandwidth of 5 MHz. 20. Consider a frequency f 1 = 2.4 Hz. How many octaves higher is the frequency f 2 = 10 Hz? (2 points) Answer: Each octave means a doubling in frequency. Thus, we have to find n in f 2 = 2 n f 1. Substituting values gives n = log(10 2.4) log(2) = 2.06 octaves. 21. Consider a frequency f 1 = 2.4 Hz. How many decades higher is the frequency f 2 = 10 Hz? (2 points) Answer: Each decade means a frequency 10 higher. Thus, we have to find n in f 2 = 10 n f 1. Substituting values gives n = log(10 2.4) = 0.62 decades. 4

5 22. Consider a first-order RC low-pass filter with 3-dB frequency f = 25 Hz. What is the phase shift in degrees at 75 Hz? (3 points) Answer: The phase shift at 25 Hz is 45 and increases at 45 / decade. 75 Hz is log(75 25) = 0.48 decades higher than 25 Hz. Thus, the phase shift is = 67. A more accurate calculation gives the phase shift as tan 1 (75 25) = The following circuit has a time-constant of τ = 1 ms. What is the attenuation (in db) at a frequency of 1.6 khz? (4 points) Answer. This is a 1 st order low-pass network with a corner frequency of f 3dB = 1 (2πτ) = Hz. The attenuation is 20 db/decade above f 3dB and 1.6 khz is 1 decade higher than khz. Thus, the network will attenuate at 20 db at 1.6 khz. An alternate calculation is 20 log 1 + ( ) 2 = 20.1 db. 24. Consider the Bode plot of a 1 st order RC network. What is the attenuation of the network at f = 60 Hz? Provide your answer in db. (4 points) Answer: 60 Hz is log(60 2.5) = 1.38 decades higher than the 2.5 Hz corner frequency. The attenuation increases by 20 db per decade, so that at 60 Hz v o v i (in db) is = 31.1 db. The attenuation is 31.1 db. An alternate calculation is log 1 + (60 2.5) 2 = 3.11 db. 25. Define the CMRR for a differential amplifier. What is the ideal value? (2 points) Answer: CMMR = 20log 10 A d A cm ). Ideally, CMMR 5

6 Question 2 An engineer designs a power supply that consists of a transformer, a full-wave, 4- diode bridge rectifier and a smoothing capacitor. The nominal load current is 1.2 A. By what percentage will the ripple voltage increase/decrease if the supply is used with an actual load of 1.5 A? Assume the transformer and rectifier diodes are capable of handling the increase in load current. (5 points) Solution The ripple voltage for a full-wave rectifier is (see chapter 2 of 4 th edition Neaman s text book): V r = V M 2fRC = 1 2fC V M R The V M R term has units of current and represents the peak current though the load resistance R, f is the input sine wave frequency, and C is the capacitance of the smoothing capacitor. Clearly, increasing the load from 1.2 A to 1.5 A (a 25 % increase) will increase the ripple voltage by the same percentage, or 25 %. 6

7 Question 3 For the circuit shown, determine I D and V D if the diode has I S = A. Assume V T = 26 mv. Your answer should be correct to three decimal places. Hints: consider replacing the linear part of the circuit with a Thevenin equivalent; use trial and error for the numerical solution. (10 points) Solution The Thevenin equivalent voltage and resistance seen by the diode are V TH = = 4.5 V, R TH = R 1 R 2 = 18.75K Original circuit Linear part of circuit replaced with Thevenin equivalent KVL for the equivalent circuit (see figure above) is V TH + I D R TH + V T ln I D I S 4.5 = I D (18.75K) ln By trying different values for I D we find that with I D = μa the right hand side of the equation above is V, so that I D = μa is the solution for the current. Then V D = V T ln I D = ln I S = V One can also find the diode voltage from the Thevenin equivalent circuit: V D = V TH = I D R TH = 4.5 ( )( ) = V I D 7

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