Homework Assignment Consider the circuit shown. Assume ideal op-amp behavior. Which statement below is true?

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1 Question 1 (2 points each unless noted otherwise) Homework Assignment Consider the circuit shown. Assume ideal op-amp behavior. Which statement below is true? (a) V = VV + = 5 V (op-amp operation) (b) VV = 10 2 (2 + 8) = 2 V (voltage division) (c) V = 0 (op-amp input current = 0) (d) No of the above are true Answer: These is no feedback in the circuit to create a virtual short (VV = VV + ). No current flows into the input terminals so that VV follows from voltage division, so the answer is (b). 2. 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 RR = (RR 1 + RR 2 ) RR 3 = 5K so that the time-constant is ττ = RRRR = 500 μμs, so that the bandwidth is 1 (2ππππ) = 318 Hz, and (b) is the answer. 3. An STC circuit has a time constant of ττ = 10 ms. Estimate the rise time of the output in response to a step input. Answer: tt rr 2.2ττ = 22 ms. 4. Pick the word/phrase that best completes the following sentence. To obtain a frequency response plot of a circuit in SPICE, one must perform analysis Answer: (a) (a) an AC (b) Transfer function (b) Harmonic Distortion (c) Transient 5. A MOSFET-based class-ab amplifier to produces a 7.07 V peak voltage across a 4 Ω resistive load. What is power dissipated by the load? Answer: VV rrrrrr = VV pppppppp 2 = VV pppppppp 2 2 = 5 V. Thus, PP = VV rrrrrr RR = 25 5 = 6.25 W. 1

2 6. An engineer designs a power amplifier to deliver 2 W (sinusoidal) signal power to an 8 Ω resistive load. What is the required peak-to-peak voltage swing across the load? 2 Answer: PP = VV rrrrrr RR, so that VV rrrrrr = 4 V, so that VV pppp = 11.3 V 7. Consider a linear power supply consisting of a transformer, a full-wave, 4-diode bridge rectifier, smoothing capacitor, and a load current 1.2 A. By what percentage will the ripple voltage increase if the load current increases to 1.5 A? a) 100 % (b) 25% (c) Stay the same (d) 50% Answer: 25% 8. Consider a linear power supply consisting of a transformer, a full-wave, a bridge rectifier and a smoothing capacitor. Increasing the smoothing capacitor by 50% will (a) Reduce ripple voltage by 50% and increase maximum inrush current by 50% (b) Reduce both ripple voltage and maximum inrush current by 50% (c) Reduce ripple voltage by 50% and leave maximum inrush current unaffected (d) Reduce ripple voltage by 50% and increase maximum inrush current by 100% Answer: (a) 9. Consider a linear power supply consisting of a transformer, a half-wave, 1-diode rectifier, and a smoothing capacitor CC. The load is a resistor RR. The rectifier diode is now replaced with a bridge (4- diode) rectifier. Neglecting the diodes turn-on voltages (VV γγ ), the ripple voltage will: a) Decrease by a factor 4 b) Decrease by a factor 2 c) Stay the same d) Increase by a factor 4 Answer: The ripple voltage is VV rr 1 ff, where ff is the frequency. A full-wave rectifies effectively doubles the frequency of the output voltage that the smoothing must smooth. Thus, doubling the frequency, decreases the ripple voltage by a factor 2. Thus (b) is the current answer. 10. In an ac series RC circuit, if 20 VAC is measured across the resistor and 40 VAC is measured across the capacitor, the magnitude of the applied voltage is: (a) 60 VAC (b) 55 VAC (c) 50 VAC (d) 45 VAC Answer: The applied voltage is VV IIII = VV RR + jjvv CC, so that VV IIII = VV RR 2 + VV CC 2 = 2, VAC. Thus (d) is the answer. 2

3 11. What is the magnitude of the current phase angle for a 5.6 μμf capacitor and a 50-Ω resistor in series with a 1.1 khz, 5 VAC source? (a) 72.9 (b) 62.7 (c) 27.3 (d) 17.1 Answer: The impedance of the RC circuit is = RR 1 jj2ππππππ = 50 jj25.84 Ω. The magnitude of the phase angle is tan 1 ( ) = Thus, (c) is the answer. 12. Estimate the current through the red LED in the circuit shown. (a) 0 ma (b) 16 ma (c) 18 ma (d) 13 ma Answer: VV γγ for a Si diode is nominally 0.7 V, but LEDs are made from other semiconductor material and LEDs turn-on voltages are different. In particular, red LEDs have VV γγ 1.6 V. Thus, the current in the circuit is II = (6 1.6) ma, and the answer is (d). 13. A filtered full-wave rectifier voltage has a smaller ripple than does a half-wave rectifier voltage for the same load resistance and capacitor values because: (a) There is a shorter time between peaks (b) There is a longer time between peaks (a) The larger the ripple, the better the filtering action (b) None of the above Answer: Option (a). 14. The PIV across a nonconducting diode in a bridge rectifier equals approximately: (a) half the peak (b) 2 the peak (c) the peak (d) 4 the peak value of the transformer secondary voltage. Answer: Option (c) 3

4 15. What is the current through the ideal diode? (a) (b) (c) (d) 1 ma ma ma ma Answer: For an ideal diode there is no forward voltage drop, so II = 12 12K = 1 ma, so option (a) is the answer. 16. With a 12-V supply, a silicon diode, and a 370 Ω resistor in series, what voltage will be dropped across the diode? Answer: Option (b) (a) 0.3 V (b) 0.7 V (c) 0.9 V (d) (a) 1.4 V 4

5 Question 2 For the circuit shown, VV PPPP = 5 V, RR = 5K, vv γγ = 0.6 V. The input is vv ii = 0.1 sin(ωωωω) V. Draw a small-signal ac model and provide numerical values for the model parameters. Then determine and expression for the time-varying component of the output voltage vv OO (tt). Provide your answer to 3 significant digits. (6 points) The dc current through the diode is II DDDD = VV PPPP vv γγ RR = (5 0.6) (5K) = 0.88 ma. The smallsignal (incremental) resistance of the diode is rr dd = VV TT II DDDD = (26 mv) (0.88 ma) 30 Ω. A smallsignal ac model is shown below. The small-signal output voltage is vv OO (tt) = RR 5K vv RR + rr ii (tt) = (0.1) sin(ωt) = sin(ωt) V dd 5K + 30 Ω 5

6 Question 3 For the diode shown VV γγ = 0.7 V. Determine II and VV OO. (6 points) A KCL equation at VV OO using the convention that currents flow away from the node is Next, calculate II VV OO ( ) 20K + VV OO 2 5K II = K = 0 VV OO = 0.14 V = ma Another approach is to start with a KVL equation. Starting at the 2-V source and using the convention that if we step out of a voltage source, change the sign of the source, and if we step over a passive component or step into a source, retain the sign, one gets: 2 + (5K)(II) (20K)(II) 8 = 0 II = ma Next, calculate VV OO VV OO = 2 (5K)(I) = 0.14 V 6

7 Question 4 An engineer measures the bandwidth of the circuit below by driving it with a sinusoidal signal and measuring the attenuation at various frequencies. She uses a scope with an input impedance of 1 MΩ with a 1 probe, and then a 10 probe. Complete the following table (6 points) True BW in Hz Measured BW in Hz ( 1) probe Measured BW in Hz ( 10) probe This is a simple 1 st order system with bandwidth BB = 1 (2ππππππ). Here RR is the equivalent resistance that the capacitor CC sees. True bandwidth calculation: the Thevenin equivalent resistance that CC sees is simply RR 1, so BB TTTTTTTT = 1 2ππππππ = 1 2ππ( )( = 175 Hz ) 11 Probe bandwidth calculation: the probe+scope has a 1M resistance that is effectively in parallel with RR 1, so that the Thevenin equivalent resistance that CC sees is 1M RR 1, so that BB 1 pppppppppp = 1 2ππ(RR RR ssssssssss )CC = 1 2ππ( ) ( )( = 334 Hz ) 1111 Probe bandwidth calculation: 10 probes increase the 1 MΩ probe+scope resistance to 10 MΩ, so that the Thevenin equivalent resistance that CC sees is 10M RR 1, so that BB 1 pppppppppp = 1 2ππ(RR RR ssssssssss )CC = 1 2ππ( ) ( )( = 191 Hz ) True BW in Hz Measured BW in Hz ( 1) probe Measured B in Hz ( 10) probe 175 Hz 334 Hz 191 Hz 7

8 Question 5 What is the voltage across a capacitor after being charged from a 100 V source for a period of one time constant? The initial voltage across the capacitor is 0 V. (4 points) The voltage across the capacitor is vv cc (tt) tt ττ = ee. Thus, vv cc (ττ) = 100(1 1 ee 1 ) = 63.3 V. Thus, (c) is the answer. Note, it is a standard result that one should know by heart a capacitor charges to 63% of its final value after one time constant. Question 6 Determine absolute value of the peak current through the load resistor? Assume VV γγ = 0.7 V for the diodes. (4 points) When vv ii = 10 V, DD 1 is reverse-biased and an open ciruit. DD 2 is forward biased and has a 0.7 V voltage drop across it. It is in series with RR LL and the left 2K resistor, so the current that flows is II = (10 0.7) (2K + RR LL ) = 9.3 (4K) = ma. When VV ii = 10 V, then DD 2 is reverse-biased but DD 1 is forward biased. The current that flows is again ma, but now it flows in the opposite direction. Regardless the answer is ma. Question 7 Consider the circuit below. What is the output voltage VV OOOOOO at the end of the 2.82 ms pulse? (5 points) On the rising edge the capacitor is uncharged and 15V appears across RR 1. The voltage across the capacitor tt ττ is VV CC = 15 1 ee tt ττ where ττ = RRRR = 940 μμs is the time constant. The voltage across RR 1 is 15ee ms 940 μμs At tt = 2.82 ms, this is 5ee = 15ee 3 = V. 8

9 Question 8 (diodes, load line) Consider the circuit below. Assume VV PPPP = 3.5 V, and RR = 180 Ω. Also shown, are the LED s voltage-current characteristics. Draw the circuit s dc load line on the characteristics and find II DD and VV DD (6 points). On the voltage axis, mark the supply voltage:3.5 V. On the current axis, mark the maximum current that can flow through the resistor: II = = 19.4 ma. Connect the two points to get the dc load line. The dc load line intersects the diode V-I curve at around II DD 6 ma and VV DD 2.8 V. 9

10 Problem 9 A full-wave, 4-diode bridge rectifier circuit with a 1 kω load operates from a 120-V (rms) 60- Hz household supply through a 10-to-1 step-down transformer. It uses silicon diodes that one can model to have a 0.7-V drop for any current. (a) What is the peak voltage of the rectified output? (3 points) (b) For what fraction of the time does the diode conduct? (5 points) Part (a) The peak voltage after the 10-to-1 step down is VV PP = (12) 2 = V so the peak voltage of the rectified output is this, minus two diode drops, or VV PP(llllllll) = = V Part (b) Since the rectified wave is periodic, we need to consider only one cycle of the rectified wave. Conduction starts when the output voltage from the transformer (i.e., sin(ωωωω) equals two diode drops. In other words, when sin(ωωωω) = 1.4. Solving yields θθ = ωωωω = radian. The rectified half- cycle represents ππ radians, so conduction starts (0.083 ππ) 100 = 2.63% into the halfcycle. By symmetry, conduction stops at = 97.37% of the half-cycle, and the diodes conduct = 94.7% of the time. 10

11 Problem 10 An engineer designs a power supply that consists of a transformer, a full-wave, 4-diode bridge rectifier and a smoothing capacitor. She designed the supply to operate in the U.S. where the power line (mains) frequency and voltage is 60 Hz and 120 V respectively. The ripple voltage at full load is 20 mv. Estimate the ripple voltage when the unmodified supply is used in regions of Japan where the corresponding values are 50 Hz and 100 V respectively. Assume that the equivalent load resistance stays the same. (5 points) The ripple voltage for a full-wave, 4-diode bridge rectifier is (see chapter 2 of 4 th edition of Neaman s text book): VV rr = VV MM 2ffffff Here VV MM is the maximum (peak) voltage of the input sine wave, ff is the frequency, CC is the capacitance of the smoothing capacitor, and RR is the load resistance. With new mains voltage VV MM = ( )VV MM and new mains frequency ff = (50 60)ff, the new ripple voltage is VV rr = VV MM 2ff RRRR = ( )VV MM 2(50 60 That is, the ripple voltage will not change. )ffffff = VV MM 2ffffff = VV MM = 20 mv 2ffffff 11

12 Question 11 Consider the linear power supply shown. The step-down transformer has a secondary voltage of 9 VAC under load. The load is a 50-Ω resistor. The forward voltage (VV γγ ) for the 1N4001 rectifier diodes is 1 V, and smoothing capacitor CC has a value of 680 μμf. The capacitor ESR is 0.75 Ω. (a) Estimate dc value of the output voltage. That is, the voltage at AA (5 points). (b) Estimate the output ripple voltage. (3 points) (c) Estimate the maximum reverse voltage across any diode. (2 points) (d) Estimate the worst-case inrush current through the diodes. Ignore the transformer winding resistance. (4 points) Part (a) The 9 VAC transformer specification refers to the secondary rms voltage. Thus, the peak voltage at the transformer secondary is VV pp = 9 2 = V. The output voltage is therefore VV OO = VV AA = VV γγ = V Strictly speaking, to find the dc (i.e., average) value, one must subtract half of the ripple voltage from this value. The ripple voltage is 2.6 V (see b)) so the dc value is 9.4 V. However, the question says estimate, and V is an estimate. Part (b) The ripple voltage is (see text an lecture notes) VV rr = VV MM 2ffffff = (2)(60)(50)( ) = 2.6 V Part (c) The maximum reverse voltage a diode experiences is VV OO VV γγ = = V Part (d) The worst-case inrush current occurs when the smoothing capacitor is uncharged and power is applied right when the input voltage to the bridge rectifier crests. This voltage is 2 9 = V and the current through a pair of diodes and the capacitor is II inrush = VV DD 0.75 = = 14.3 A SPICE Simulation (not required but shown as an example how to use SPICE) Below is a screen capture of a Micro-Cap SPICE simulation. 12

13 Below are the output plots from a transient analysis. The annotations show that the simulation values match the hand-calculations well. VV dddd = 9.6 V 1 V VV rr = 2.5 V SPICE Transient Analysis output. The top panel shown the voltage across the load (solid line), and the average/dc value (broken line). The bottom panel shows the voltage across diode DD 4. 13

14 Question 12 Determine V D for the circuit shown to within 0.01 V. This is not necessarily a Si diode so you can t assume VV DD = 0.7 V. Rather, use the diode equation II DD = II SS ee VV DD VV TT 1 and assume that I S = A and TT = 300K. Use the bisection numerical method with the initial bracket values VV LL = 0.6 V and VV HH = 0.65 V. Organize you values neatly in a table. (8 points) KVL gives VV PPPP + II DD RR + VV DD = 0 For the diode II DD = II SS ee VV DD VV TT Substituting this into the KVL equation and reorganizing gives VV PPPP = RRII ss ee VV DD 1 Eq. 1 VV TT 1 + VV DD VV DD = 5 ( )( ) ee VV DD Now try different values for VV DD substituting in the RHS of the equation above. Iteration VV LL VV HH VV DD = (VV HH VV LL ) 2 1 Eq. 2 VV DD which is from RHS of Equation 2 VV DD VV DD % Matlab Script R = 2e3; Is = 1e-13; VT = 0.026; VH = 0.65; VL = 0.6; for i = 1:100 VD = (VH+VL)/2; VDP = 5 - Is*R*(exp(VD/VT)-1); err = VDP-VD; s = sprintf('%d, %7.5f, %7.5f, %7.5f, %7.5f, %6.4f',i,VL,VH,VD,VDP,err); disp(s) if (err < 0) VH = VD; else VL = VD; end if (abs(err) < 0.01) break; end end 14