Homework Assignment 09

Homework Assignment 09 Question 1 (Short Takes) Two points each unless otherwise indicated. 1. What is the 3-dB bandwidth of the amplifier shown below if r π = 2.5K, r o = 100K, g m = 40 ms, and C L = 1 nf? (a) 65.25 khz (b) 10 khz (c) 1.59 khz (d) 10.4 khz Answer: The 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 timeconstant is τ = RC = 100 μs. The bandwidth is 1 (2πτ) = 1.59 khz, so the answer is (c). 2. Explain why one cannot use BJT scaling to determine R O in the circuit below. Answer: The transistor is a MOSFET and not a BJT so one cannot use BJT scaling. (Actually, with the appropriate interpretation of the BJT scaling equations, one could use them, but we did not cover this.) 3. Consider the following drive circuit for an IR remote control. The drive signal is a 0 5 V square wave and V CC = 9 V. The MOSFET is replaced with another MOSFET with a V TN that is 20% higher. The new peak current through the IR diode will be a) Increased by 20% b) Decreased by 20% c) Unchanged d) Decreased much more than 20%, since I D = K n (V T V GS ) 2 Answer: (c) 1

4. Consider the following drive circuit for an IR remote control. The drive signal is a 0 5 V square wave and V CC = 9 V. The battery voltage drops from 9 V to 5 V, i.e., by a factor 1.8. The average IR diode current will be a) Unchanged b) Increased by about 18% c) Decreased by 18% d) Decreased significantly more than 18%, since I D = K n (V T V GS ) 2 Answer: (a) 5. Consider the following drive circuit for an IR remote control. The drive signal is a 0 5 V square wave and V CC = 9 V. The IR diode is replaced with another IR diode that has a turn-on voltage that is 20% lower. The new peak current through the IR diode will be a) Unchanged b) Increased by 20% c) Decreased by 20% d) Decreased much than 20%, since I D = I S e V D/V T Answer: (a) 6. What is the output voltage V OUT at the end of the 2.82 ms pulse? (3 points) (a) 0 V (b) 0.75 V (c) 5.55 V (d) 14.25 V Answer: On the rising edge the capacitor is uncharged and 15V appears across R 1. The t τ voltage across the capacitor is V C = 15 1 e where τ = RC = 940 μs is the time t τ constant. The voltage across R 1 is 15e 2.82 ms 940 μs. At t = 2.82 ms, this is 5e = 15e 3 = 0.747 V, so the answer is (b). 2

Question 2 Consider the circuit shown. Determine the current through R L when R L is 1K, 3K, and 5K. Assume V BE(ON) = 0.64 V, β = 200, and V CC = 9 V. (5 points) Since β is large, we can neglect the base current. The voltage at the base is then V B = (9)(1 11) = 0.818 V. Then V E = 0.818 0.64 = 0.178 V. The emitter current is then 0.178 180 1 ma. Not 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 9 0.178 0.2 = 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. Question 3 The threshold voltage for each transistor below is 0.4 V. Determine the region of operation (Ohmic, saturation, etc.) of the transistor in each circuit. (6 points) These are n-mosfets, so that the transition between Ohmic and saturation region is at v DS (sat) = v GS V TN. Circuit (a): v DS (sat) = 2.0 0.4 = 1.8 V. Note that in the circuit, v DS = 2.2 V which is more than v DS (sat) MOSFET is in the saturation region. Circuit (b): v DS (sat) = 1 0.4 = 0.6 V. Note that in the circuit, v DS = 0.4 V wich is less than v DS (sat) MOSFET is in the Ohmic region. Circuit (c): v GS = 0 V MOSFET is off. 3

Question 4 In the amplifier shown, R 1 = R 2 = 470K, R Sig = 10K, R S = 10K, and I DQ = 0.4 ma. For the transistor K n = 50 ma V 2, V TN = 2 V, and λ = 260 10 6. Show that an expression for the output resistance is (8 points): 1 1 R o = g m + 1 = R s r o R s r o Determine the numerical value for the output resistance. (8 points) g m A small-signal model for the amplifier is shown. A test source V x was added to determine the output resistance R o. Assuming v i is turned off, then R o = V x I x. KCL at the source (using the convention that currents into the node is positive) gives g m v gs + I x V x V x = 0 R s r o g m v gs + I x V x = 0 R s r o From the circuit, with v i turned off R sig is in parallel with R 1 R 2. However, since the gate current is zero, v g = 0. That is, the gate is at signal ground. Thus, v gs = V x, so that V x g m V x + I x = 0 R s r o R o = V x 1 1 = I x g m + 1 = R s r o R s r o Next, determine numerical values for r o, g m, and R o : r o = 1 1 = λi IDQ (260 10 6 )(0.4 10 3 ) = 9.6M g m = 2 K n I DQ = 2 (50 10 3 )(0.4 10 3 ) = 8.9 10 3 A V R o = R s r o 1 = (10K) (9.6M) (112) 112 Ω g m g m 4

Checking results with SPICE (Not Required) The following circuit was used to check the calculation with SPICE. For the MOSFET the SPICE parameters were set as follows: KP = 0.1, W = L = 100μ and λ = 260 10 6. This translates to K n = 50 10 3 A V 2. The amplitude for v i was set to zero, and the amplitude for V X was 10 mv. Further, C C = C X = 1F which are essentially shorts at V X s frequency of 1 khz. A transient analysis was performed and the ratio RMS(V(V X )) RMS(I(V X )) gives the output resistance. The value measured was 111.5 Ω 5

Question 5 In the amplifier below, determine R O as indicated. Use the results from the previous question. (5 points) R i = 2.5K R S = 10K R 1 = R 2 = 470K R Sig = 10K I D = 0.4 ma K N = 50 ma V 2 r o The MOSFET is a follower and its output resistance is from the previous question R o = R S 1 g m. Now g m = 2 K n I DQ = 8.9 10 3 A V, so that 1 g m 112 Ω, and consequently R o 112 Ω. Question 6 Shown is the symbol a popular SPICE computer simulation program uses for an n-channel MOSFET, along with labels indicating the drain, source and gate terminals. There is also a 4 th terminal, indicated with an arrow. The physical device has three, not four terminals. A perplexed student asks her professor what is this terminal, and what should she do with in when she builds her circuit in SPICE. Provide a short (3 4 sentence) answer to the student. (3 points) 6

Question 7 The transistor in the circuit shows has parameters V TN = 0.8 V and K n = 0.5 ma V 2. Write an expression and sketch the load line for (a) V DD = 4 V, R D = 1K and (b) V DD = 5 V, R D = 3K. Additionally, calculate the Q-point for each case and indicate these on the plot. Finally, for each case determine if the transistor is operating in the saturation or non-saturation region. (12 points) Part (a) Load line is I D = V DS R D + V DD R D = V DS 1K + 4 1K. Assume saturation region operation, then I D = K n (V GS V TN ) 2. From the circuit V GS = V DD = 4 V, so that I D = (0.5)(4 0.8) 2 = 5.12 ma. This current will result in V DS = 4 (5.12)(1) = 1.12 V. This is not a valid so solution assumption of saturation region operation wrong and MOSFET is in Ohmic region. Thus V DD = I D R D + V DS = K N R D [2(V GS V T )V DS V 2 DS ] + V DS Substituting values and simplifying results in 0.5V 2 DS 4.2V DS + 4 = 0 Valid solution is V DS = 1.095 1.1 V, I D = (V DD V DS ) R D = (4 1.1) 1K = 2.92 ma. Part (b) Similar to Part (a), load line is I D = V DS 3K + 5 3K. Assume saturation region operation and find V DS = 11.36 V, which is not valid. Thus, MOSFET is in Ohmic region. Following the same procedure as for Part (a), we find 1.5V 2 DS 13.6V DS + 5 = 0 The valid solution is V DS = 0.38 V, from which it follows that I D = 1.54 ma. 5 5 3 I D (ma) Q-point Part (b) 4 3 2 1 Q-point, Part (a) 1 2 3 4 5 V DS (V) 7

Question 8 Find v E, v C1, and v C2 in the circuit shown. Also, find the current though the top 1K resistor. Assume that v BE(on) = 0.7 V, and that β is large. (6 points) The base of Q 1 is at 0.5 V, while the base of Q 2 is at 0 V, so that Q 2 s base-emitter voltage is much larger than that of Q 1, and Q 2 is turned on hard, while Q 1 is essentially off. From this it follows that v C1 = 5 V, and v E = 0.7 V. Further, the current through the top 1K resistor is = (5 0.7) 1K = 4.3 ma v C2 = 5 + (1K)(4.3 ma) = 0.7 V. Question 9 The transistor in the amplifier shown has β = 350 and V BE(ON) = 0.65 V. (a) Make reasonable assumptions and show that I CQ 1 ma (3 points) (b) Show that R i 13.7K (5 points) Part (a) Since β is large, ignore I BQ so that V B = (9)(27 (100 + 27) ) = 1.9V. Since V BE(ON) = 0.65 V, then V RE = 1.9 0.64 = 1.25 V. Consequently, I CQ I E = 1.25 1.3K = 0.962 ma 1 ma. Part (b) r π = β g m = 350 40I CQ = 8.75K. Using BJT scaling, R i = 65K 18K r π + (1 + β)(1.3k) = 13.68K 8

Question 10 Shown is the functional diagram of dual current source IC, REF200. In addition to two 100-μA sources, the IC incorporates a current mirror. Below are a collection of circuits that use the IC. For each of the circuits, determine I out. (12 points) I out I out I out (a) (b) (c) I out I out I out (d) (e) (f) (a) 50 μa, (b) 400 μa, (c) 50 μa, (d) (N + 1)100 μa (e) (N + 1) 100 μa (f) 300 μa 9

Question 11 Consider the amplifier below, which amplifies the signal from a sensor with an internal resistance of 1K. Ignore BJT s output resistance, and assume C 1 = C 2 = C 3. β = 100 I C = 0.245 ma (a) Determine g m, r π (4 points) (b) Using BJT scaling, determine R i see figure (4 points) (c) Using the ratio of the collector and emitter resistors, estimate the overall voltage gain A v = v o v s (4 points) (d) Calculate the voltage gain A v = v o v s, but do not use the approximation that involves the ratio of the collector and emitter resistors, but rather incorporate the β of the transistor (4 points) Part (a) Part (b) g m = 40 I C = 9.8 ms, r π = β g m = 10.2 K R i = R 1 R 2 [r π + (1 + β)r E ] = 300K 160K [10.2K + 101 3K] = 78.3K Part (c) The effective collector resistance is R C resistance and R i form a voltage divider. Thus = 22K 100K = 18K and the sensor s internal A v = v o R i R C = 78.3 v s R S + R i R E 1 + 78.3 18K 3K = 5.93 10

Question 12 For the amplifier below, R L = 500 Ω. Determine R o R ib, and estimate A v. (8 points) Hint, use BJT scaling. R S = 10K V + = 3 V V = 3 V I Q = 2 ma β = 300 V A = 100 V C C Since β is large, I C I E = I Q = 2 ma. Then g m = 40I C = 80 ma V and r π = β g m = 3.75K. Using BJT scaling: and R o = R S + r π 1 + β This is an Emitter Follower, so A v 1. = 10K + 3.75K 301 = 45.7 Ω R i = r π + (β + 1)R L = 3.75K + (301)(500) = 154.3K 11

Question 13 The figure is a plot of the open-loop gain function for the LF357 voltage amplifier. An engineer will use the amplifier as an inverting amplifier with a mid-frequency voltage gain at 100. Use the plot and estimate the bandwidth of the feedback amplifier. (3 points) Write expressions for the transfer function A(f) for the open loop amplifier as well as the closed loop, inverting amplifier. (6 points). A gain of 100 is equivalent to a gain of 20 log 10 (100) = 40 db. A horizontal line at 40 db intercepts the LF357 gain curve at 100 khz Thus, the bandwidth ~ 100 khz and the GBP is (100)(100 10 3 ) = 10 10 6. The pole for the open-loop amplifier is about 90 Hz, and the low-frequency gain is 105 db, so the open-loop gain is 20 10105 178 103 A(f) Open Loop = 1 + j f = 90 1 + j f 90 The closed-loop gain is 100 A(f) Closed Loop = f 1 + j 100 10 3 12

Question 14 Use BJT impedance scaling and determine the input and output impedances of the flowing circuits. Assume β = 300. (12 points) (a) (b) Circuit (a) g m = 40I C = 40 ms and 1 g m = 25 Ω, r π = β g m = 7.5 K. Using BJT scaling R i = (R 1 R 2 ) (r π + (1 + β)r L ) = (33K) (7.5K + (301)R L ) R o = r π (1 + β) = 7.5K 301 25 Ω (6 points) Circuit (b) We need to find I C first. Since β is large, we will ignore I b and V B = 2.75 V using voltage division. Then V E 2 V and I E = 1 ma. Thus, g m = 40I C = 40 ms and 1 g m = 25 Ω, r π = β g m = 7.5 K. Using BJT scaling: R i = (R 1 R 2 ) r π + (1 + β)(r L 2K) = (15.27K) 7.5K + (301)(R L 2K) r π R o = (1 + β) R E = 25 2K 25 Ω (6 points) 13

Question 15 For the circuits below, assume β = 100 and use BJT impedance scaling to find the missing circuit parameters. (2 points for each parameter) I C = 1.7 ma A v (R C R L ) R E = 1.133 r π = β g m = 100 (40I C ) = 1.47K R ib = r π + (1 + β) = 304.5K I C = 15 ma r π = β g m = 100 (40I C ) = 166 Ω R ib = r π + (1 + β)(r E R L ) = 15.32K R o = R S R 1 R 2 + r π R 1 + β E = 9.7 Ω I C 2 ma A v 1 r π = β g m = 100 (40I C ) = 1.25K R o = R S + r π (1 + β) = 111.4 Ω 14