hw8_soln - [1 jωR S C π C μ(1 g m r o[1 jωr o C CS C...

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UNIVERSITY OF CALIFORNIA AT BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences Homework #8 Solutions Due Tuesday, October 23, 2007 EE 105 Fall 2007 Prof. Liu Problem 1. Impact of Early Effect on A v for a Common-Base Stage (Miller Approximation) At node X , we add a resistor to ground with value R X = r o 1 - A v = r o 1 - g m R C . At the output, we add a resistor to ground with value R o = r o 1 - 1 /A v = r o 1 - 1 gmR C = g m r o R C g m R C - 1 . Once we split r o in this manner, the circuit takes the form of the example in Lecture 11, Slide 7, so we can apply the gain equation listed there (where R E R X , R C R C bardbl R o , and R B 0): A v = R C bardbl R o 1 /g m + R S bardbl R E · R E R S + R E = R C bardbl g m r o R C g m R C - 1 1 /g m + R S bardbl r o 1 - g m R C · r o 1 - g m R C R S + r o 1 - g m R C Problem 2. Frequency Response of a Common-Emitter Stage a) ω p,in = 1 R S C in C in = C π + C μ (1 A v ) A v = g m r o C in = C π + C μ (1 + g m r o ) ω p,in = 1 R S ( C π + C μ (1 + g m r o )) ω p,out = 1 r o C out C out = C CS + C μ (1 1 /A v ) = C CS + C μ (1 + 1 /g m r o ) ω p,out = 1 r o [ C CS + C μ (1 + 1 /g m r o )] b) A 0 = g m r o r π R S + H ( ) = g m r o r π
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Unformatted text preview: [1 + jωR S ( C π + C μ (1 + g m r o ))] [1 + jωr o [ C CS + C μ (1 + 1 /g m r o )]] Problem 3. Common-Base Stage Design 1 a) Assuming the output node dominates, we have: 2 π × 10 10 ≤ ω p,out = 1 R C ( C μ + C CS ) R C ≤ 637 Ω b) We know that the DC gain is A v = g m R C 1+ g m R S . We can see that picking a larger R C will result in a larger gain but a smaller bandwidth and vice-versa. Thus, our trade-of is between gain and bandwidth. Problem 4. Frequency Response o± Emitter Follower C in = C μ + C π (1 − A v ) A v = R E 1 /g m + R E C in = C μ + C π p 1 − R E 1 /g m + R E P ≤ 50 ±F g m = I C /V T = 1 / 26 S R E ≥ 39 Ω 2...
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