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Unformatted text preview: The equivalent resistance looking into the collector/base of Q 2 is 1 g m 2 b r 2 1 g m 2 . Thus, this is just like a common emitter amplier with an emitter resistor of size 1 g m 2 . A v = g m 1 /g m 2 1 + g m 1 /g m 2 R in = r 1 + (1 + ) 1 g m 2 R out = 1 g m 1 b 1 g m 2 Problem 3. Emitter Follower Analysis The equivalent resistance at the emitter (assuming C 2 is a short in small signal analysis) is 1 k b 100 . The 10 k resistor at the base doesnt aect the voltage gain, so we have: A v = g m (1000 b 100) 1 + g m (1000 b 100) 1 We need to fnd I C so that we can calculate g m . This is a biasing problem similar to those From homework 4. V BE = V CC I B (10 k) I E (1 k) = V CC I C (10 k) 1 + I C (1 k) = V T ln p I C I S + 1 P I C = 1 . 586 mA g m = 61 mS A v = . 847 2...
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- Fall '07
- Electrical Engineering