HW6_Soln - April 9, 2008 ECE315 Homework # 6 Solutions...

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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 3 ) k ( R 1 + r 3 g m 3 r 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 3 ) k ( R 1 + r 3 g m 3 r 3 ) . (b) The additional resistors ( R 1 and R 2 ) appear in parallel with the collector-emitter 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 3 ) (1 + R 1 r 3 ) 1 g m 3 , and v out = R 1 R 1 + r 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 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 = 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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This note was uploaded on 11/02/2008 for the course ECE 3150 taught by Professor Spencer during the Spring '07 term at Cornell University (Engineering School).

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HW6_Soln - April 9, 2008 ECE315 Homework # 6 Solutions...

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