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440_hmk4_soln

440_hmk4_soln - ECE 440 Electronics II Solutions to...

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ECE 440 – Electronics II Solutions to Homework #4 1. a. A f = A 1 + β A = 50 1 + .5(50) = 1.92 b. dA f A f = 1 1 + β A dA A = 1 1 + 25 (15%) = 0.58% c. dA f A f = β A 1 + β A d β β = (.5)(50) 1 + (.5)(50) (15%) = 14.4% 2. A f = A 1 + β A = 2 × 10 5 1 + jf 10 1 + β 2 × 10 5 1 + jf 10 = 2 × 10 5 1 + β 2 × 10 5 + jf 10 = 2 × 10 5 1 + β 2 × 10 5 1 + jf 10 1 + β 2 × 10 5 ( ) a. A of = 2 × 10 5 1 + β 2 × 10 5 = 2 × 10 5 1 + (0.5)2 × 10 5 = 2 b. BW = 10(2 β x 10 5 ) = 1MHz c. G-BW = 2 MHz 3. DC Analysis, open all of the capacitors: For Q1, find the Thevenin equivalent for the input circuit and write KVL around the input loop, resulting in: 47 K 47 K + 150 K (25) = (150 K || 47 K ) I C 1 β + 0.7 + 4.8 KI C 1 I C 1 = 0.964 mA g m 1 = 0.0367 r π 1 = 1362 Ω Similarly for Q2: 33 K 47 K + 33 K (25) = (33 K || 47 K ) I C 1 β + 0.7 + 4.7 KI C 1 I C 1 = 1.89 mA g m 1 = 0.073 r π 1 = 689 Ω The feedback network is: β = v f v o = 100 4800 = 0.0208 100 4700 + v o - + v f -

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redrawing the “A” amplifier with the feedback loading the circuit: In order to correctly use the feedback formulas we need to replace the circuit to the left of Q1 with its Thevenin equivalent: where V TH = 150 K || 47 K 150 K || 47 K + 5 K V
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440_hmk4_soln - ECE 440 Electronics II Solutions to...

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