Final Exam. 55:041 Electronic Circuits. The University of Iowa. Fall 2013.

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1 Final Exam Name: Max: 130 Points Question 1 In the circuit shown, the op-amp is ideal, except for an input bias current I b = 1 na. Further, R F = 10K, R 1 = 100 Ω and C = 1 μf. The switch is opened at t = 0. What is the output voltage after 5 seconds? (5 points) Solution For t 0, the voltage across the capacitor is v C = (±I b Δt) C which is ((± ) (5)) ( ) = ±5 mv for t = 5 s. The gain of the amplifier is 1 + R F R 1 = 101, so that the output voltage is ± 501 mv. Question 2 Consider the voltage regulator. The input voltage is V C. What should V REF be for an output voltage of 5 V? (4 points) Solution The amplifier and feedback loop maintains a voltage V REF across R 1 so the current through both resistors is I = V REF R 1. The output voltage is then V O = I(R 1 + R 2 ) = V REF R 1 (R 1 + R 2 ) = 5.0 Substituting R 1 and R 2 and solving for V REF yields V REF = 2 V 1

2 Question 3 Determine the h 11 and h 21 parameters for the circuit. Be sure to supply the units and proper sign for each parameter. (8 points) Solution Setting v 2 = 0 h 11 = v 1 i 1 v 2 =0 h 21 = i 2 i 1 v 2 =0 With v 2 = 0, the input voltage is equal to the voltage across the 20K resistor, and Use KCL to find i 2 : v 1 = (20K)i 1 h 11 = (20K)i 1 i 1 = 20K i i 1 + i 2 = 0 i 2 = 51i 1 and h 21 = 51i 1 i 1 = 51 A/A Question 4 For the circuit, assume I DQ = 28 ma, and you may ignore r o. Further, assume the capacitors are shorts at the operating frequency. Determine the voltage gain A v = v i v o. (5 points) K n = 93 ma V 2 V TN = 2 V λ = 0 C S = 100 μf C C = 1 μf The small-signal model is shown where R G = 100K 56K and g m = 2 K n I D = = A/V. From the model the voltage gain is A V = g m ( K) =

3 Question 5 Consider the 555-based astable oscillator, which has component values so that the duty cycle and frequency are 60% and 5 khz respectively. Assume you have 2N7000 FETs, 2N2222 BJTs, and a collection of standard resistors and capacitors available. Provide a circuit along with the component values that will use the 555 timer s output and drive an IR LED so that the average current through the IR LED is 30 ma. (8 points) Hint: You encountered the circuit in both the IR Tx and Rx labs The peak- and average currents, and the duty cycle D in % are related as follows I ave = D 100 I peak I peak = 100 D I ave = 100 (30 ma) = 50 ma 60 Assume that for the BJT, V BE(ON) = 0.7 V, then R limit = 0.7 (50 ma) = 14 Ω. Use the closest standard value of 15 Ω. 3

4 Question 6 The diagram below shows part of the IR receiver of the final lab. Assume that all the capacitors appear/function as shorts at 5 khz. The exception is C c1, which has a capacitance of 1 nf. The BJT has a quiescent collector current of I C = 1 ma. (a) What is the voltage gain of the BJT amplifier? That is, the voltage gain from the transistor s base to its collector? Ignore loading effects at the collector. (2 points) A v = g m R C = 40I C R C = 40( )( ) = 224 (b) Calculate a value for C E so it appears as a short. (3 points) (c) What is the 3-dB bandwidth of the circuit shown? (4 points) C c1 sees a resistance R T of (R ph + R 1 R 2 ) or 10K + 135K = 145K. The time constant of the circuit is τ = R T C C1 = ( )( ) = 145 μs. The 3-dB bandwidth is B = 1 (2πτ) = 1.1 khz. 4

5 (d) Assume 5-kHz signal photocurrent is 10 μa rms. Estimate the 5-kHz signal rms voltage at B. (2 points) v B ( )(10K) = 100 mv rms (e) Assume an incandescent lamp generates a 100 μa rms, 120-Hz noise signal in the photodetector. Estimate the noise voltage at B. (Hint: use the bandwidth result from (c) above) (4 points) The 120-Hz noise signal is log 10 ( ) = 0.92 decades below the 1-kHz, 3-dB cutoff of the high-pass network formed by C c1, R 1, R 2, and R ph. Thus, the 120-kHz signal is attenuated by 3 + (0.92)(20) = db. This is equivalent to a factor 11.77, so that the 120-Hz signal at B will be v B ( )(10K) mv (f) In order to extend the range, a student places another photodiode in series with the one already in place. Explain very briefly why this does not work. (1 point) The photodiode functions as a current source, so placing two in series is nonsensical, and probably limits the current to the smaller of the two current sources. 5

6 Question 7 For the BJT in the amplifier shown, β = 100. Ignore the BJT s parasitic capacitances. Further, C C2, and R S = 100K. (a) Find I CQ. (3 points) If you can t do this use I CQ = 1 ma for the rest of the problem. For a dc analysis, we open the capacitors. KVL gives I B R B (1 + β)i B R E 10 = 0 Solving yields I B = 8.38 μa and consequently I B (1 + β)i B I C = βi B = ma. (b) Estimate the lower 3-dB bandwidth if C c1 = 10 nf. (5 points) g m = 40I C, r π = β g m = 2.5K. Using BJT scaling, C C1 sees a resistance R i to its right where R i = (r π + (1 + β)(10k 10K)) 100K = 84K The total resistance the capacitor sees is R S + R i = 184K, and the time constant is τ = (184K)( ) = 1.84 ms. The 3-dB bandwidth is thus B = 1 (2πτ) = 86.5 Hz. (c) Estimate the overall midband voltage gain A v = v o v i. (2 points) This is an emitter follower with voltage gain from base to emitter of slightly less than 1. Further, R i, the resistance to the right of C 1 is 84K and this forms a voltage divider with R s so that the overall gain is R A v (1) i 84K = R i + R s 100K + 84K =

7 Question 8 The open-loop gain and input resistance of the opamp below is 10 6 and 1 MΩ respectively. The op-amp s output resistance is 100 Ω. Further, R 1 = 99K, R 2 = 1K. What is the closed-loop gain and input resistance? (5 points) Solution This is series-shunt (voltage-voltage) feedback, with β = R 2 (R 1 + R 2 ) = Further, 1 + βa OL = 1 + (0.01)(10 6 ) Thus A f = A OL 1 + βa OL = = 100 R if = R i (1 + βa OL ) = (10 6 )(10 4 ) = 10 4 MΩ R of = βa OL = 10 mω Question 9 For the circuit below β = 300 and I C = 1 ma R 1 = 120 kω R 2 = 75 kω V BE(ON ) = 0.62 V Determine the output resistance R o using BJT scaling. You may ignore r o. (4 points) g m = 40 I C = 40 ms, and r π = β g m = 7.5K r π R o = R E ( ) = 4K (7.5K) = 24.7Ω 25 Ω 1 + β 301 7

8 Question 10 Show that the output resistance R o in the circuit is R o = 1 g m R S You may ignore r o (6 points) R o To find R O, follow the standard procedure: turn off all independent sources (v i in this case), add a test source V x and determine I x. Then, R O = V x I x. The small-signal model for the amplifier is shown in (a) below. The gate current is zero, so that v g is zero, and the small-signal model becomes as is shown in (b). Note that v gs = V x and that r o is in parallel with R O. (a) (b) KCL at the source gives Rearranging gives g m v gs I X + V x R s = 0 g m ( V x ) I x + V x R s = 0 R o = V X I X = 1 g m R S 8

9 Question 11 Consider the amplifier below. λ = 0, C gd = 0.3 pf, C gs = 2 pf V GS = 3.55 V, and g m = 1.55 ma/v (a) Draw a complete (showing voltage polarities etc.) small-signal model for the amplifier that includes the FET capacitances. (3 points) (b) Calculate the Miller capacitance. (4 points) C M = C gd (1 + g m (R D R L )) C M = 0.3( (4 20)) = 1.85 pf (c) Calculate the 3-dB frequency of the small-signal gain (3 points) R G = R 1 R 2 = 97.1 kω τ = (R G R i )(C gs + C M ) = ( )( ) = 34.9 ns f 3dB = 1 = 4.56 MHz 2πτ (d) Calculate the overall midband gain A v = v o v i (2 points) R 1 R 2 A v = g m (R D R L ) = R 1 R 2 + R i 9

10 Question 12 Consider a BJT with a rated power of 20 W, and a maximum allowable junction temperature T j,max = 175. The transistor is mounted on a heat sink with parameters θ case sink = 1 /W, and θ sink amb = 5 /W. Determine how much power the BJT can safely dissipate. Assume an ambient temperature of T A = 25. (10 points) Hint: first determine θ dev case. Solution The thermal resistance from the device/junction to the case is not given explicitly, so we need to determine it before proceeding. The BJT is rated at 20 W at T j,max = 175, and an ambient temperature of T A = 25 is assumed. A thermal model and the calculation of θ dev case is then T j = T A + P D (θ dev case ) 175 = (θ dev case ) θ dev case = 7.5 W Now we can determine the maximum allowable power dissipation when the BJT is mounted on a heat skink with the given parameters. A thermal model for the problem is shown below. T j = T A + P D (θ dev case + θ case sink + θ sink amb ) P D,max = = T j,max T A (θ dev case + θ case sink + θ sink amb ) = 11.1 W 10

11 Question 13 The maximum current, voltage, and power ratings for the MOSFET are 4 A, 60 V, and 40 W, respectively. (a) Sketch and label the SOA for the MOSFET using linear voltage and current scales. (3 points) (b) For the amplifier, determine R D and sketch the load line that produces maximum power in the transistor for V DD = 55 V. Be sure to clearly indicate all important points on the load line. (4 points) Solution The plot below shows the SOA and a load line that is anchored at V DD = 55 V and touches the maximum power hyperbola but still stays in the SOA. It touches the SOA at V DD 2 = 27.5 V. The power dissipation is 40 W so that I D at this point is = A. From this it follows that I D(max) = A as indicated. The slope of the load line is and this equals 1 R D so that R D = = 18.9 Ω. I D (A) 5 4 SOA Boundary A Load line V DS (V) 11

12 Question 14 The open loop gain A(f) of an amplifier is shown below. What is the phase margin and bandwidth if closed loop gain = 12 db? What will the rise time for a step input be? (10 points) Using the graphical subtraction method, draw 1 β, which is practically the same as the closedloop gain. Read the phase as 95, so the phase margin is about 85. The closed-loop bandwidth is about 100 MHz, so that the rise time is t r 0.35 ( ) = 3.5 ns. 12

13 Question 15 The figure is a plot of the open-loop gain function for the LT1007 voltage amplifier. An engineer will use the amplifier as a non-inverting amplifier with a mid-frequency voltage gain of 10. (a) What is the GBP of the LT1007? (2 points) (b) Use the plot and estimate the bandwidth of the feedback amplifier. (2 points) (c) Write an expression for the gain A(f) for the feedback amplifier. (2 points) (d) (d) By how much (μs) does the amplifier delay a 250-kHz sine wave? (2 points) Solution (a) The open loop gain is 120 db ( ) at f = 10 Hz, so the GBP is 10 MHz. Alternatively, the BW is about 10 MHz when the open loop gain is 0, so the GBP is 10 MHz. (b) A voltage gain of 10 is equivalent to a gain of 20 log 10 (10) = 20 db. A horizontal line at 20 db intercepts the LT1007 gain curve at 950 khz. Alternatively, from the GBP, with a gain of 10, the bandwidth is 1 MHz. (c) The closed-loop response is 10 A(f) = f 1 + j (d) The amplifier s phase is θ = tan 1 (f ) and at 250 khz this is 14. Further, the period of a 250-kHz sine wave is 4 μs and the delay is therefore: Δt = 14 4 μs = μs

14 Question 16 The open-loop gain of an amplifier is modeled by A(f) = 10 5 (1 + j f f 102) (1 + j 10 4) (1 + j f ) An engineer uses the amplifier and negative feedback so that the gain of the feedback amplifier is 220. (a) What is the feedback factor β? (4 significant figures). (1 points) β 1 = (b) Provide an expression for the loop gain T of the amplifier. (1 point) T = βa(f) = (1 + j f f 102) (1 + j 10 4) (1 + j f ) (c) Find the crossover frequency f x. That is, the frequency where the magnitude of the loop gain is 1. If you cannot do this, take f x = 10 khz for the rest of the problem. (8 points) T(f) = ( f 10 2) ( f ) f 1 + ( ) Using trail-and-error we find that f x = 20 khz results in T(f) (d) Determine, by calculating the phase, if the amplifier is stable. (5 points) The phase at f x = 20 khz is φ = tan 1 f x 10 2 tan 1 f x 10 4 tan 1 f x 10 5 Substituting f = 20 khz shows that φ = The phase margin is φ = = The amplifier is stable. 14

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