# 55:041 Electronic Circuits The University of Iowa Fall Final Exam

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1 Final Exam Name: Score Max: 135 Question 1 (1 point unless otherwise noted) a. What is the maximum theoretical efficiency for a class-b amplifier? Answer: 78% b. The abbreviation/term ESR is often encountered in data sheets of capacitors. What does ESR stand for? Answer: Equivalent Series Resistance c. When researching part numbers for three-terminal regulators, an engineer encounters the term LDO. What does LDO stand for? Answer: Low Drop Out d. Name one important difference between an operational amplifier and a comparator. Answer: Comparators are very similar to op-amps, but comparators don t need and generally don t have internal frequency compensation. 1

2 e. What is v o in the following circuit if v REF = 1.2 V, R 1 = 680 Ω, and R 2 = 200 Ω? (3 points) Answer: The current through R 2 is 1.2/200 = 6 ma, which also flows through R 1. Thus, the output voltage is = 5.28 V f. What is the purpose of R 3 in the circuit below, and what should the value be to be effective? (2 points) Answer: This compensates for the op-amp s input currents. The value should be R 1 R 2. g. The output voltage of a three-terminal voltage regulator is 5 5 ma load, and A load. What is the regulator s output resistance? (2 points) Answer: R = ΔV ΔI = = 27 mω 2

3 h. Give one phrase that describes the purpose of the capacitor, and give another phrase that describes the purpose of two series diodes in the circuit below. (2 points) Answer: Frequency compensation, and reduction of cross-over distortion i. What type of negative feedback (series-shunt, series-series,... ) is used in the following amplifier? Answer: Series-shunt. j. What type of negative feedback (series-shunt, series-series, ) is used in the following amplifier? Answer: Shunt-shunt 3

4 k. What type of negative feedback (series-shunt, series-series, ) is used in the following amplifier? Answer: Shunt-shunt. l. The output voltage of a three-terminal voltage regulator is 5 5 ma load, and A load. What is the regulator s load regulation? (2 points) Answer: Load Regulation = V o(nl) V o(fl) 100 = = 0.8% V o(nl) 5 4

5 Question 2 Consider the comparator circuit below. What is the threshold voltage (i.e., the voltage where the output changes state? (5 points) R 1 = 10 kω R 2 = 20 kω V REF = 1.2 V Solution (This problem comes from a Midterm exam) The voltage at the inverting input is v = V REF v I R 1 + R 2 R 2 + v I The circuit changes state when this voltage is 0 V. That is, when Substituting values yield the threshold voltage: v I = V REF v I R 1 + R 2 R 2 v I = R 2 R 1 V REF v I = 20 (1.2 V) = 2.4 V 10 5

6 Question 3 For the amplifier below, determine the differential-mode voltage gain A d = v o3 v d. State and very briefly justify simplifying assumptions you make. (15 points) β = 200 V A = 80 V V BE(on) = 0.7 V Hints: This is a two-stage amplifier that consists of a differential-input stage and a buffer/output stage. Consider computing the gains of these stages separately. Solution To simplify calculations note that β is large, so ignore base currents. This means I C I E for all transistors. Further, using the BJT scaling property, the resistance looking into the base of Q 3 is larger than β(3.3k) = 660K. This is in parallel with the 8K collector resistance of the differential stage and it is large enough to ignore. Finally, with reasonable bias currents, BJT output resistances are ~100K, so in this circuit it is reasonable to ignore. Now I 1 = K = ma, and I E1 = I 1 2 = ma, and g m1 = 40I E1 = 38.3 ma/v. The gain of the differential input stage is A 1 = (g m1 R C ) 2 = The gain of the output stage is A 2 R C2 R E = 1.2 The total gain is then A d = A 1 A 2 = (153.2)( 1.2) = 186 6

7 Question 4 Consider the following small-signal model of an amplifier. Determine (i.e., write down an expression) for the output resistance R o indicated in the figure. (10 points) Solution (This problem comes from a Midterm exam, 2007) 7

8 Question 5 Using the circuit below, design a current source such that I O = 0.7 ma. (10 points) Transistors are matched with And V A =, β = 80 V BE(on) = 0.7 V V + = 10 V R 2 = 10K Solution Q 1, Q 2 are matched, and in the circuit have the same base-emitter voltage, and have the same collector currents I O = 0.7 ma. They also have the same base currents I O β. Then I E3 = I O β + I O β K = 2( ) = 87.5 μa 80 10K I B3 = (17.5 μa) (1 + β) = 1.08 μa I REF = I B3 + I C1 = 1.08 μa ma = ma R 1 = = 12.27K

9 Question 6 A power MOSFET has thermal characteristics given below and dissipates 25 W. Design, by specifying the thermal resistance, a heat sink that will ensure the MOSFET does not overheat. The ambient temperature is 25 o C. Assume one can keep the thermal resistance between the MOSFET case and heat sink (θ case sink or θ CS ) below 1 o C/W. (5 points) θ juntion case = θ JC = 1.75 /W θ ase ambient = θ CA = 50 /W T j,max = 150 Solution (This problem comes from a Midterm exam, 2007) A thermal model that captures the information is shown below: T j = T A + P D θ JC + θ CS + θ SA 150 = ( θ SA ) θ A = 2.25 /W The heat sink s thermal resistance must be less than this value. Note: a number of students used θ CA in various ways (incorrectly) in their calculations. θ CA is not part of the picture, since we will replace it with a much lower θ SA 9

10 Question 7 Consider the feedback amplifier in (i) and the small-signal model in (ii). The feedback network is modeled with a y-parameter two-port as shown. A dc analysis shows I C = 1 ma β = 150 V A = I C = 1 ma (i) (ii) (a) Determine y 12 (4 points) If you can t do this, use y 12 = A/V and continue. (b) Determine the gain of the feedback amplifier, namely A = v o /i i. (10 points) (c) Determine the input impedance R if (2 points) (d) Determine R i (2 points) (e) Determine R of (2 points) 10

11 Solution (This is a simplified version of some a homework question) Part (a) y 12 i 1 = 1 v 2 v 1 120K =0 Two-port model for R f Calculation for y 12 Part (b) g m = 40I C = 0.04 A/V, and r π = β g m = 3.75 K, and r o =. We ignore y 21, and y 12 is the feedback transfer ratio β. Next turn off the feedback (set y 12 = 0). At the input, (1/y 11 R S r π ) = 2.1K, and at the output, (1/y 22 R L R C ) = 1.85K Now v o = (g m v π )(1.85K) = g m i i (2.1K) (1.85K) = i i V A = v o i i = V/A Now (1 + βa) = ( K) = 2.3 A f = A 1 + βa = = V/A Part (c) R if = 2.1K (1 + βa) = 2.1K 2.3 = 0.913K Part (d) 1 = = R i R if R s 0.913K 1 5K = 1.12K Part (e) R of = 1.85K (1 + βa) = 1.85K 2.3 = 0.904K (Over for more space) 11

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13 Question 8 Below is a screen dump of a SPICE simulation of a Wien bridge oscillator that uses an incandescent lamp for gain control. The graph depicts the lamp resistance as a function of the voltage across the lamp. Determine the voltage at which the output stabilizes. (10 points) Lamp Resistance (Ohm) Lamp Voltage (V) (a) Solution (This problem comes from a Midterm exam) (b) At startup R lamp < 5 Ω, and the loop gain T = R 4 /(R lamp +39) +1 > 3.7, enough for a Wien bridge oscillator to start oscillating. As the output amplitude grows, V lamp grows, R lamp increases. However when R lamp increases, the loop gain T = R 4 /(R lamp +39) + 1 decreases. The loop stabilizes when R T lamp = R4 = 21 ( R Ω lamp + 39) = 3 From the plot R lamp = 21 Ω when the V lamp 1.25 V. The current flowing through the lamp is 1.25/21 = 60 ma. The voltage at the output is (0.06)( ) = 10.8 V. 13

14 Question 9 For the switched-capacitor circuit below, the parameters are C 1 = 30 pf, C 2 = 5 pf, C F = 12 pf. The clock frequency is 100 khz. Determine the low-frequency gain and the cutoff frequency. (10 points) Solution (This was a homework problem) The switched capacitors C 1 and C 2 function as resistors with values R 1 = 1 f C C 1 = kω, and R 2 = 1 f C C 2 = 2 MΩ respectively. At low frequencies C F is an open circuit and the low-frequency gain is A V = R 2 R 1 = C 1 C 2 = 6 The cutoff frequency is determined by C F and the switched capacitor R 2 f 3dB = 1 1 = 2πR 2 C F 2π( = 6.63 khz )

15 Question 10 The loop gain function of an amplifier is given by T(f) = j f j f f j 10 5 (a) What is the frequency (to a good approximation) at which the phase is 180 o? (4 points) (b) What is the magnitude of T(f) at this frequency? (4 points) (c) Is the amplifier stable? (2 points) Solution (This was a homework problem) Part (a) One must determine f 180 such that φ = tan 1 f 10 4 f f tan 1 tan = 180o Program this value into a programmable calculator and try different values for f to find f Hz Part (b) 500 T(f) = = (8.063)(1.887)(1.28) = Part (c) The amplifier is unstable, because the magnitude of the loop gain is greater than 1 when the phase shift is 180 o. 15

16 Question 11 Consider the differential amplifier below. Determine the low-frequency, singleended differential-mode gain A vd = v O1 v d and the upper 3-dB frequency. (15 points) V A = I Q = 1 ma R C = 10K R B = 0.5K C π = 8 pf C μ = 2 pf β = 100 Hint: draw a half-amplifier and its small-signal equivalent, and compute the Miller capacitance. Solution (This problem comes from the Review Questions) Below is a schematic of the left half-amplifier and its small-signal model. For the small-signal model r π = β = = g m 40I C (40)( ) = 5K The gain for this amplifier is The Miller capacitance is A vd = v O1 v d = v O1 = 1 2v 1 2 r π ( g r π + R m R c ) B = (40)( )( ) = 91 16

17 And f 3 db = C M = (1 + g m R C )C μ = 1 + (40)( )( ) (2 pf) = 402 pf 1 2π(C π + C M )[r π R B ] = 1 2π( )( = 853 khz ) 17

18 Question 12 K n = 0.5 ma/v 2, V TN = 2 V, λ = 0, C gd = 0.1 pf, C gs = 1 pf (a) Show that V GS = 3.55 V. (4 points) (b) Draw a detailed small-signal model for the amplifier showing the numerical values for the components. (6 points) Solution (This comes from the Review Questions) 18

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