Homework Assignment 06

Transcription

1 Question 1 (2 points each unless noted otherwise) Homework Assignment True or false: when transforming a circuit s diagram to a diagram of its small-signal model, we replace dc constant current sources with open circuits. 2. Tue or false: when transforming a circuit s diagram to a diagram of its small-signal model, we replace dc voltage sources with shorts. 3. The single-supply op-amp amplifier shown has a serious flaw. What is it? Answer: There is no dc path to bias the non-inverting input.. 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 Answer: False. V D decreases with increasing temperature 5. An engineer measures the gain of the amplifier below, and finds that with an input voltage v i = 3 V, the output voltage is v o = 18 V, so that the gain of the amplifier is 6. However, op-amp theory suggests the gain is = 7.2. Give one phrase that could explain the difference. Answer: Slew rate. 6. True or false: the diffusion C d capacitance of a forward-biased pn junction is generally much larger than the junction capacitance C j. 7. True or false: the diffusion capacitance C d of a pn junction is negligible when the junction is reversebiased. 1

2 8. 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. 9. True or false: the turn-on voltage of red LEDs is larger than the turn-on voltage of blue LEDs. Answer: False 10. A single-pole op-amp has an open-loop low-frequency gain of A = 10 5 and an open loop, 3-dB frequency of Hz. If an inverting amplifier with closed-loop low-frequency gain of A f = 50 uses this op-amp, determine the closed-loop bandwidth. Answer. The gain-bandwidth product is 10 5 Hz. The bandwidth of the closed-loop amplifier is then is 10 5 /50 = 8 khz. 11. 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? 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. 12. 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? 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. 13. A 100-mV source with internal resistance R s = 1K drives an amplifier with gain A v = v o v i = 10 (see figure). The output voltage is 750 mv. What is the amplifier s input resistance R i? (a) (b) 1K (c) 3K (d) Need additional information (e) 0 Ω Answer: The source s and amplifier s internal resistances form a voltage divider and the output voltage is v O = A v v s (R i R i + R s ). Substituting for v O, v s, A v, and R s and solving for R i yields R i = 3K., so the answer is (c). 2

3 1. What is the 3-dB bandwidth of the circuit below? (a) 8 khz (b) 318 Hz (c) khz (d) 2 khz Answer: The current source has an infinite internal resistance, so that the capacitor sees an equivalent resistance R = (R 1 + R 2 ) R 3 = 5K so that the time-constant is τ = RC = 500 μs, so that the bandwidth is 1 (2πτ) = 318 Hz, and (b) is the answer. 15. Consider a linear power supply consisting of a transformer, a half-wave, 1-diode rectifier, and a smoothing capacitor. The rectifier diode is now replaced with a bridge (-diode) rectifier. Neglecting the diodes turn-on voltages, the ripple voltage will a) Decrease by a factor 2 b) Decrease by a factor c) Stay the same d) Increase by a factor Answer: (a) Question 2 What is the 3-dB bandwidth of the amplifier shown below if r π = 2.5K, r o = 100K, g m = 0 ms, and C L = C F = 1 nf? (5 points) Solution If one turns off V I, it is shorted to ground and C L is in parallel with C F so the effective capacitance is 2 nf. This capacitor sees an equivalent resistance r o = 100K. (If one turns off V I, g m v π = 0, and the current source is effectively removed from the circuit.) The time-constant is τ = RC = (2 nf)(100k) = 200 μs. The bandwidth is 1 (2πτ) = Hz. Question 3 Shown is the gain frequency response plot for an amplifier. Can we use the constant GBP model for this amplifier? Briefly motivate your answer. (2 points) Solution: No, this is not a first-order system, because the slope where the gain decreases with frequency is not 20 db/decade. 3

4 Question Consider a first-order RC low-pass filter with corner frequency f = 1 khz. What is the phase shift in degrees at 15 khz? (3 points) Solution The phase shift at 1 khz is 5 and increases at 5 / decade. 15 khz is log(15 1) = 1.18 decades higher than 1 khz. Thus, the phase shift is = However, the phase shift for a first-order RC network must always be less than 90, so the answer is 90. A more accurate calculation gives the phase shift as tan 1 (15 1) = Question 5 The circuit shown has a time-constant of τ = 1 ms. What is the attenuation (in db) at a frequency of 1.6 khz? ( points) Solution 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. Question 6 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. ( points) Solution 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 ) = db.

5 v O (V) ECE:310 Electronic Circuits. The University of Iowa. Fall Question 7 Consider a first-order RC low-pass filter with 3-dB frequency f = 60 Hz. By how much does it delay a 50 Hz sine wave? Express you answer in ms. (3 points) Solution The phase shift at 60 Hz is 5 and increases at 5 / decade. 50 Hz is log(50 60) = 0.08 decades higher than 60 Hz. (The negative sign implies 50 Hz is 0.08 decades before 60 Hz.) Thus, the phase shift is = 1.. The period of a 50 Hz sine wave is 20 ms, so the delay is (20)(1. 360) = 2.3 ms An alternate and more accurate calculation for the phase is tan 1 (50 60) = 39.8 and delay of 2.2 ms. Question 8 For the op-amp in the circuit show below, the supply voltage is ± 15 V and the slew rate is 1.5 V/μs. Sketch the output voltage v O. Label the each axis and clearly indicate important points on your plot. (5 points) Solution Time (μs) The gain of the amplifier is A v = K 10K = 9. The input signal ramps up/down at the start/end with a rate ±0.5 V μs. The desired leading ramp at the output is =.5 V μs. However, the amplifier has a slew rate of 1.5 V/μs, so that it will reach.5 V after 3 s. Similarly, starting at 19 s, the output ramps down at 1.5 V/μs and reaches 0 at 22 s. 5

6 Question 9 For the circuit below, derive an expression for the output voltage as a function of V OS, I N, C, R, and time. Apart from the input-offset voltage and input-bias current, the op-amp is ideal. (8 points) Solution Below is a model of an inverting integrator that incorporates I N and V OS. The voltage at the inverting input is V OS so that a KCL equation at this node is V OS R + I N + C dv C dt = 0 where v C = V OS v O (t) is the voltage across the capacitor. Since V OS is a constant V OS R + I N C dv O dt = 0 t v O = + 1 C (V OS R + I N) dt = + 1 C (V OS R + I N) t 0 6

7 Question 10 An amplifier is designed to provide a 12 V peakto-peak swing across a Ω load. Assume sinusoidal signals. (a) Assuming the amplifier has output resistance R o 0 Ω, how much power will the load dissipate? (2 points) (b) Assuming the amplifier has output resistance R o = 0. Ω, how much power will the load dissipate? (3 points) Solution Part (a) P = V 2 rms R L A 12 V peak-to-peak sinusoidal signal has an amplitude of 6 V, and an rms value of 6 2 =.2 V. Thus Part (b) P = (.25)2 With R o = 0. Ω, the signal amplitude across the load is V L = The rms value is = 3.86 V, and the power is =.5 W = 5.6 V P = (3.86)2 = 3.72 W 7

8 Question 11 An amplifier has an open-loop gain of 100,000 and an open-loop bandwidth of 25 Hz. Assume that the dominant-pole model is valid for the amplifier. (a) Write an equation for the open-loop transfer function. (3 points) (b) Using negative feedback, the amplifier is configured for a gain of 100. Write an equation for the transfer function of the feedback amplifier. ( points) (c) Calculate the delay (in μs) that a 20 khz sine wave will experience when amplified by the feedback amplifier. (3 points) Solution The phrase dominant-pole implies we can approximate the amplifier with a single pole and use the constant GBP concept. Part (a) H(f) = 1 + j f 25 Part (b) The gain is 1,000 times smaller so the bandwidth is 1,000 times larger, so that Part (c) This corresponds to a delay of 100 H(f) = f 1 + j θ = tan 1 ( ) = Δt = = 5.37 μs