lecture_07_multi-stage_amplifiers

# Lecture_07_multi-sta - Handout#7 EE 214 Winter 2009 Multi-Stage Amplifiers B Murmann and B A Wooley Stanford University B Murmann B Wooley EE214

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B. Murmann, B. Wooley EE214 Winter 2008-09 1 Multi-Stage Amplifiers B. Murmann and B. A. Wooley Stanford University Handout #7 EE 214 Winter 2009 B. Murmann, B. Wooley EE214 Winter 2008-09 2 Dominant-Pole Approximation The response of a multi-stage amplifier, or a more exact representation of a single stage, has the form A(s) = A(0) N(s) D(s) where N(s) = 1 + a 1 s + a 2 s 2 + ! + a m s m D(s) = 1 + b 1 s + b 2 s 2 + ! + b n s n The roots of D(s) are the POLES of A(s), while the roots of N(s) are the ZEROS. In an “open loop” amplifier with a low-pass response a DOMINANT POLE condition may exist, thereby permitting a simple approximation of the amplifier’s response.

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B. Murmann, B. Wooley EE214 Winter 2008-09 3 If the zeros of A(s) are large enough to be neglected, then A(s) ! A(0) 1 + b 1 s + b 2 s 2 + ! + b n s n = A(0) 1 " 1 ( ) 1 " 2 ( ) ! 1 " n ( ) From the above expressions it is apparent that b 1 = ! 1 p i " # \$ % & i = 1 n ( If |p 1 | << |p 2 |,|p 3 |, … |p n |, then p 1 is a dominant pole and b 1 ! " 1 p 1 = 1 p 1 B. Murmann, B. Wooley EE214 Winter 2008-09 4 Under this condition A(j ! ) = A(0) 1 + ! p 1 " # \$ % & 2 and ! " 3dB # p 1 # 1 b 1
B. Murmann, B. Wooley EE214 Winter 2008-09 5 For the case of a transfer function with widely separated poles on the negative real axis, each of the poles an be approximated in succession. For example, if D(s) = 1 + b 1 s + b 2 s 2 + b 3 s 3 = 1 ! s p 1 ( ) 1 ! s p 2 ( ) 1 ! 3 ( ) then b 1 = ! 1 p 1 + 1 p 2 + 1 p 3 " # \$ % & b 2 = 1 p 1 p 2 + 1 p 1 p 3 + 1 p 2 p 3 b 3 = ! 1 p 1 p 2 p 3 B. Murmann, B. Wooley EE214 Winter 2008-09 6 If |p 1 | << |p 2 | << |p 3 |, then b 1 ! " 1 p 1 b 2 ! 1 p 1 p 2 ! " b 1 p 2 b 3 ! " 1 p 1 p 2 p 3 ! " b 2 p 3

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B. Murmann, B. Wooley EE214 Winter 2008-09 7 Example R S * r ! C ! C μ g m v 1 R L v i ~ + v o Consider the equivalent circuit for a common-emitter stage The nodal admittance equations for this circuit are G S * + g ! + s(C ! + C μ ) " sC μ g m " sC μ G L + sC μ # \$ % % & ( ( v 1 v o # \$ % % & ( ( = G S * v i 0 # \$ % % & ( ( ! A V (s) = v o (s) v i (s) = " G S * (g m " sC μ ) G S * + g # + s(C # + C μ ) [ ] (G L + sC μ ) " (g m " sC μ )( " sC μ ) + v 1 B. Murmann, B. Wooley EE214 Winter 2008-09 8 A V (s) = A V (0) 1 ! s z 1 ( ) 1 + b 1 s + b 2 s 2 where A V (0) = ! g m G S * G L (G S * + g " ) = ! g m R L r " R S * + r " # \$ % & ( z 1 = + g m C μ b 1 = R L C μ + RC μ + RC ! + g m R L RC μ b 2 = R L RC !
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## This note was uploaded on 08/23/2009 for the course EE 214 taught by Professor Murmann,b during the Spring '04 term at Stanford.

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Lecture_07_multi-sta - Handout#7 EE 214 Winter 2009 Multi-Stage Amplifiers B Murmann and B A Wooley Stanford University B Murmann B Wooley EE214

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