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L14 Part 2 Exam1 Sp 2000 Grounded Emitter_1

# L14 Part 2 Exam1 Sp 2000 Grounded Emitter_1 - Ω C2 = 1(R...

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Exam 1 2000 Q9 & Q10 Grounded Emitter Circuit Zubair Rehman

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3 Question 9 Y 1 = Y(1-K) Y 2 = Y(1- 1 / K ) R = 1/Y R1 = R/(1-K) R2 = R/ (1 – 1/K) R1 M, IN = R1 / (1 - A V ) & R1 M,OUT = R1 / (1 - 1 /A V ) A V = v c / v b = -g m * [R C || 5 Meg || R1 M,OUT ] g m = I C / V T = 0.7 mA / 25 mV = 28 mS Key Step: Assume |A V | >> 1 Then (1 - 1 /A V ) ~ = 1 R1 M,OUT = R1 A V = -g m * (R C || 5Meg || R1) = -28 mS * 8.78 k || 44.8 k = -206 |A V | >> 1 So setting R1 M,OUT = R1 is justified R 2S = R L + R C || R1 M,OUT = 5 Meg + 8.78 k || 44.8 k = 5 Meg

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Unformatted text preview: Ω C2 = 1 / (R 2S ω L ) = 1 / [5 Meg * 2 π * 1 Hz] = 0.032 μ F (D) 4 Question 10 R 1S = R2 || R1 M,IN || R IN,Q R IN,Q = r π = β / g m = 50 / 28 mS = 1.785 k R1 M,IN = R1 / (1 - A V ) = 44.8 k / (1-(-206)) = 44.8 k / 207 = 0.216 k R 1S = 111 k || 0.216 k || 1.785 k = 0.193 k = 193 Ω C1 = 1 / (R 1S ω L ) = 1 / (193 * 2 π * 10) = 8.2 E-05 = 82 μ F...
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L14 Part 2 Exam1 Sp 2000 Grounded Emitter_1 - Ω C2 = 1(R...

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