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Unformatted text preview: April 9, 2008 ECE315 Homework # 6 Solutions (Rev. 0) Spring 2008 1. Textbook Chapter 10, Problem 32. BJT Differential Amplifier CIrcuits. (a) First consider in each collector circuit for the diode connected resistor BJT combina tion. Its small signal equivalent is shown below. • Using this circuit it is easy to show that: R eff D = v x i x = r 03 k ( R 1 + r π 3 ) k ( R 1 + r π 3 g m 3 r π 3 ) . • The full differential voltage gain (non inverting) is then: A d v = v out 2 v out 1 v in 1 v in 2 = g m 1 h r o 1 k R eff D i = g m 1 r o 1 k r 03 k ( R 1 + r π 3 ) k ( R 1 + r π 3 g m 3 r π 3 ) . (b) The additional resistors ( R 1 and R 2 ) appear in parallel with the collectoremitter cir cuits of Q 1 and Q 2 . So we can lump them in parallel with the output resistance of each transistor. As a result for the inverting differential gain (determined by v out 1 v out 2 ): A d V = g m 1 2 ( R 1 k R C k r 01 ) g m 2 2 ( R 2 k R C k r 02 ) . With matched transistors and R 1 = R 2 , then: A d V = g m 1 ( R 1 k R C k r 01 ) . (c) Again we must consider the equivalent circuit of what is placed in the drain circuit. Its small signal circuit is shown below. ECE315 April 9, 2008 Page 2 • Using this circuit it can be shown that: R o = v x i x = r o 3 ( R 1 + r π 3 ) (1 + R 1 r π 3 ) 1 g m 3 , and v out = R 1 R 1 + r π 3 v D 1 . • So the voltage gain becomes: A v = v D 1 v in v out v D 1 = g m 1 ( r o 1 k R o ) R 1 R 1 + r π 3 . (d) Same circuit as part c) as far as output resistance ( R o ) is concerned. The difference is that the output is taken directly off the collector of Q 1 . As a result: A v = v D 1 v in = v out v in = g m 1 ( r o 1 k R o ) . 2. Textbook Chapter 10, Problem 51. MOSFET Differential Amplifier. Assume λ = 0 for each MOSFET and the source I SS is ideal. (a) Treat M 1 as a CS stage. • From the MOSFET small signal analysis summary, the CS voltage gain is given when r o → ∞ as: A v = g m ( R D R L ) 1 + g m R s = R D 1 g m + R s . • The effective source resistance is the input resistance of the common gate M 2 with a drain resistor R D . From the MOSFET small signal analysis summary, the CG input resistance when R S → ∞ is given by: R in = 1 g m + 1 r o 1 r o ( 1 r o + g m )( R D k r o ) ≈ 1 g m . ECE315 April 9, 2008 Page 3 • Combining these, we obtain: A v = R D 1 g m 1 + 1 g m 2 . With M 1 and M 2 matched then: A v = g m R D 2 . (b) Treat M 1 as a CD (source follower) stage. From the MOSFET small signal analysis summary, the CD voltage gain is given when r o → ∞ as: A v = v out v in = g m R S k R L g m R S k R L + 1 = R S R S + 1 g m ....
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 Spring '07
 SPENCER
 Microelectronics, Trigraph, Electronic amplifier, voltage gain

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