Homework Assignment 06

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1 Question 1 (2 points each unless noted otherwise) Homework Assignment Typically, the C-E saturation voltage for a BJT, namely V CE(sat), is in the range of (circle one) Answer: (a) (a) V (b) 0.7 V (c) 2 3 V 2. 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. 3. Tue or false: when transforming a circuit s diagram to a diagram of its small-signal model, we replace dc voltage sources with shorts. 4. True or false: the β of a transistor decreases with decreasing temperature. Answer: False, it increases with increasing temperature. 5. True or false: the β of a transistor is a function of temperature, but essentially independent of collector current. 6. True or false: consider a BJT in the CE configuration, biased at I C = 1 ma. The small-signal input resistance r π is in the order of 500K Answer: False. r π = β g m = β (40I C ). With β = 100, r π = 2.5K 7. 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. 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 Answer: False. V D decreases with increasing temperature 1

2 9. An engineer measures the gain of the amplifier below, and finds that with an input voltage vv i = 3 V, the output voltage is vv 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. 10. True or false: the diffusion C d capacitance of a forward-biased pn junction is generally much larger than the junction capacitance C j. 11. True or false: the diffusion capacitance C d of a pn junction is negligible when the junction is reversebiased. 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. 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. 14. True or false: the turn-on voltage of red LEDs is larger than the turn-on voltage of blue LEDs. Answer: False 15. 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. Answer. The gain-bandwidth product is Hz. The bandwidth of the closed-loop amplifier is then is /50 = 8 khz. 16. 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. 2

3 17. 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. 18. 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 (4-diode) rectifier. Neglecting the diodes turn-on voltages, the ripple voltage will (1 points) a) Decrease by a factor 4 b) Decrease by a factor 2 c) Stay the same d) Increase by a factor 4 Answer: (b) 19. A 100-mV source with internal resistance R s = 1K drives an amplifier with gain A v = vv o vv 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 vv OO = A v vv s (R i R i + R s ). Substituting for vv OO, vv s, A v, and R s and solving for R i yields R i = 3K., so the answer is (c). 3

4 Question 2 Write down the dc load line equation for the BJT in the circuit shown. (3 points) Answer: I C = V CC R C V CE R C Question 3 Write down dc load line equation for the BJT in the circuit below, incorporating β. That is, do not assume that I E I C. (4 points) V CC = I C R C + V CE + (I C α)r E + V DD α = β β + 1 V CE I C = R C + R E α + V CC V DD R C + R E α Question 4 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) The phase shift at 1 khz is 45 and increases at 45 / 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) =

5 Question 5 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) 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. (4 points) 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 vv o vv 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

6 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) The phase shift at 60 Hz is 45 and increases at 45 / 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 = The period of a 50 Hz sine wave is 20 ms, so the delay is (20)( ) = 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 vv OO. Label the each axis and clearly indicate important points on your plot. (5 points) vv OO (V) 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 = 4.5 V μs. However, the amplifier has a slew rate of 1.5 V/µs, so that it will reach 4.5 V after 3 s. Similarly, starting at 19 s, the output ramps down at 1.5 V/µs and reaches 0 at 22 s. 6

7 Question 9 For the circuit below, derive an expression for the output voltage as a function of V OOS, I N, C, R, and time. Apart from the input-offset voltage and input-bias current, the op-amp is ideal. (8 points) Below is a model of an inverting integrator that incorporates I N and V OOS. The voltage at the inverting input is V OOS so that a KCL equation at this node is V OOS R + I N + C dvv C dt = 0 where vv C = V OOS vv OO (t) is the voltage across the capacitor. Since V OOS is a constant V OOS R + I N C dvv OO dt = 0 t vv OO = + 1 C V OOS R + I N dt = + 1 C V OOS R + I N t 0 7

8 Question 10 Consider a semiconductor diode, biased at I DQ. Show that the small-signal or incremental resistance is r d = V T I DQ. (5 points) The equation that relates the diode current to the diode voltage is The incremental conductance is i D = I S e v D V T 1 I S e v D V T g d = di D = di Se vd dvv D dvv D VT Apply the chain rule: V T g d = d I Se v D d(vv D V T ) d(vv D V T ) dvv D = I S e v D V T 1 V T = i D 1 V T At a bias current I DQ, the value is g d = I DQ V T The incremental resistance is the inverse of the conductance: r d = V T I DQ 8

9 Question 11 R 1 = 200 kω R 2 = 220 kω R C = 2.2 kω R E = 1 kω r s = 100 kω V CC = 5 V R L = 4.7 kω β o = 100 C μ = 2 pf C π = 10 pf V A = V BE(OON) = 0.7 V Show that I B = 9.3 μa. Note that you cannot assume I B = 0 (4 points) The Thevenin voltage and resistance for the V CC and the bias resistances are V TH = 5[R 1 (R 2 + R 2 )] = 2.62 V, and R 2 R 1 = 105K respectively. Then, using BJT scaling: V TH + I B (β + 1)I B R E = 0 I B = 9.3 μa Question 12 Consider the circuit shown. Determine the current through R L when R L is 1K, 3K, and 5K. Assume V BE(OON) = 0.64 V, β = 200, and V CC = 9 V. (6 points) Since β is large, we can neglect the base current. The voltage at the base is then V B = (9)(1 11) = V. Then V E = = V. The current is then ma. Note that this value is independent of R L as long as the BJT does not saturate. Taking V CE(SAT) = 0.2 V, saturation occurs when the drop across R L is around = 8.62 V. This will occur if when R L = 8.62K. All the R L values in the problem statement are less than this. Thus, the current for all the cases is 1 ma. 9

10 Question 13 An amplifier is designed to provide a 12 V peak-to-peak swing across a 4 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 = 0.4 Ω, how much power will the load dissipate? (3 points) 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 = 4.24 V. Thus Part (b) P = (4.25)2 4 With R o = 0.4 Ω, the signal amplitude across the load is V L = The rms value is = 3.86 V, and the power is = 4.5 W = 5.46 V P = (3.86)2 4 = 3.72 W 10

11 Question 14 For the BJT amplifier below, determine I B, I C, V RE. For the Si transistor, β = 50, and assume the transistor is operating in the forward active mode. (10 points) Replace the base bias network with its Thevenin equivalent network as shown below. For a Si transistor, V BE = 0.7 V. R TH = R 1 R 2 = 23.65K 36 V TH = = 3.46 V V BE = 0.7 V Now However, I E = (1 + β)i B so that V TH = I B R TH I E R E Solving yields I B = 1.78 μa. From this follows V TH = I B R TH (1 + β)i B R E 3.46 = (I B )(23.56K) (51)(I B )(30K) I C = βi B = 88.9 μa V RE = (I C + I B )R E = 2.72 V 11

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