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Unformatted text preview: 1 Problem 1 a. See Figure 1. R B = R B 1 k R B 2 = 21 k V th = 10 R B 2 R B 1 + R B 2 = 3 V (1) Using (1), we write the voltage equation around the BE loop V th I B R B . 7 I E R E = 0 V th I B R B . 7 101 I B R E = 0 I B = V th . 7 R B + 101 R E = 1 . 03 10 5 A (2) Other DC currents and voltages are I C = I B = 0 . 001 A I E = ( + 1) I B = 0 . 001 A V C = 10 I C R C = 8 V V E = I E R E = 2 . 08 V V B = 3 21 kI B = 2 . 78 V (3) We implicitly made the assumption that the BJT is in FA mode. Now is the time to check V BC = 2 . 78 8 < . 4 V (4) Assumption is verified. b. g m = I C V T = 0 . 040 A/V r o = V A I C = 20 k r = V T I B = 2 . 42 k (5) 1 c. Figure 2. d. R i = R B k r = 2 . 17 k R o = r o k R C = 1 . 82 k (6) v o = ( g m v be )( R o k R L ) v be = v s R i R i + R S A v = v o v s = g m R i R i + R S ( R o k R L ) = 37 . 90 (7) e. Let A v 1 = g m ( R o k R L ) = 46 . 6, then V B + v b ( V C  A v 1  v b ) . 4 v b . 4 V B + V C 1+  A v 1  = 0 . 12 V (8) V CC V C +  A v 1  v b v b V CC V C  A v 1  = 0 . 04 V (9) Taking the minimum of (8) and (9), and using v b = v s R i R i + R S , we see that v s,max . 05 V f. This is similar to the highfrequency analysis example shown in class. (See Figure 3) 2 First we calculate C . f T = g m 2 ( C + C ) C = 1 . 57 pF (10) We can replace C with C 1 and C 2 using Millers Theorem C 1 = C (1 + g m R o k R L )) = 75 pF (11) C 2 = C (1 + 1 g m R o k R L ) = 1 . 6 pF (12) (13) Therefore, the poles are f H 1 = 1 2 C 2 R o k R L = 88 MHz (14) f H 2 = 1 2 ( C 1 + C ) R S k R i = 4 . 91 MHz (15) (16) Clearly, f H 2 is the dominant pole. f BW = f H 2 . 2 Problem 2 This question is strongly analogous to Problem 1, so the solution will be more abbreviated a. R G = R 1 k R 2 = 21 k V th = 10 R 2 R 1 + R 2 = 3 V (17) Using (17), we write the current equation I D = 1 2 k n W L ( V GS V t ) 2 since V S = I D R S I D = 0 . 8 mA (18) The second solution to I D...
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 Spring '08
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 Volt

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