HW3_Soln - March 11, 2008 ECE315 Homework # 3 Solutions...

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Unformatted text preview: March 11, 2008 ECE315 Homework # 3 Solutions (Rev. 2) Spring 2008 1. Textbook Chapter 5, Problem 3. Input Resistance where r . (a) Looking into the base of the BJT with no r and R E = 0: r in 4 = v b i b = r - R in = R 1 + R 2 || r . (b) Looking into the emitter of the BJT with no r and R C = 0: r in 4 = v e i e = r + 1 r || 1 g m- R in = R 1 || r + 1 R 1 || r || 1 g m . (c) Looking into the collector of the BJT ( Q 2 ) with the base shorted to the collector (emitter grounded): r in 4 = v c i c = r 2 2 + 1 . Now looking into the base of Q 1 with an emitter resistance coming from the diode connected Q 2 , then r in 4 = v b i b = r 1 + ( 1 + 1) R E- R in = r 1 + ( 1 + 1) r 2 2 + 1 r 1 + ( 1 + 1) r 2 || 1 g m 2 . (d) Looking into the base of the Q 1 with no r : r in 4 = v b i b = r 1 + ( 1 + 1) R E- R in = r 1 + ( 1 + 1) r 2 . 2. Textbook Chapter 5, Problem 4. Output Resistance with r . ECE315 March 11, 2008 Page 2 (a) Looking into the collector of the BJT: r out 4 = v c i c = r- R out = R 1 || r . (b) Looking into the collector of the BJT, R B has no effect: r out 4 = v c i c = r- R out = r . (c) Looking into the collector of Q 1 with R B = 0, we modify our common base output resistance expression (assuming r >> r ): r out 4 = v c i c = r 01 1 + 1 R E || R s R E || R s + r 1 . Replacing R E || R s with the input resistance of diode connected Q 2 ( r 2 || 1 g m 2 || r o 2 r 2 || 1 g m 2 ): R out = r 01 " 1 + 1 ( r 2 || 1 g m 2 ) r 2 || 1 g m 2 + r 1 # . (d) Looking into the collector of Q 1 with R B = 0, we modify our common base output resistance expression (assuming r >> r ): r out 4 = v c i c = r 01 1 + 1 R E || R s R E || R s + r 1 . Replacing R E || R s with the input resistance of Q 2 ( r 2 ): R out = r 01 [1 + 1 r 2 r 2 + r 1 ] . 3. Textbook Chapter 5, Problem 11a. For Q 1 ignore the Early effect, = 100, V sat CE = . 25 volts and I S = 6(10- 16 ) amps....
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HW3_Soln - March 11, 2008 ECE315 Homework # 3 Solutions...

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