HO18_214W09_WidebandAmp-Part2_2pp

HO18_214W09_WidebandAmp-Part2_2pp - Handout #18 EE 214...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
B. Murmann, B. Wooley EE214 Winter 2008-09 1 Wideband Amplifiers (Part II) B. Murmann and B. A. Wooley Stanford University Handout #18 EE 214 Winter 2009 B. Murmann, B. Wooley EE214 Winter 2008-09 2 Emitter Peaking The bandwidth of the series f/b stage can be increased by introducing an “emitter peaking” capacitor in shunt with R E : To analyze this circuit, substitute Z E for R E in the preceding analysis, where Z E = 1 Y E = R E 1 + sR E C E ~ Q 1 R S R E C E i o Not a large bypass capacitor + v i
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
B. Murmann, B. Wooley EE214 Winter 2008-09 3 Then,defining ! E ! R E C E and assuming ! T " C # g m since g m >> 1 r ! i o v i = ! g m 1 + g m Z E + (R S + Z E )( 1 r " + sC " ) = ! g m 1 + R S r " + sC " R S + R E 1 + s # E $ % & ( ) g m + 1 r " + sC " $ % & ( ) ! ! g m 1 + R S r " + sC " R S + g m R E 1 + s # E $ % & ( ) 1 + s C " g m $ % & ( ) B. Murmann, B. Wooley EE214 Winter 2008-09 4 If C E is chosen so that ! E = ! T ! i o v i " # g m 1 + R S r $ + sC $ R S + g m R E 1 + s % T 1 + s % E & ( ) * + and it is true that g m R E >> R S r ! = g m R S " 0 Then i o v i ! " g m 1 + g m R E # $ % & ( 1 1 + sC ) R S 1 + g m R E # $ % & ( * + , , , , , - . / / / / /
Background image of page 2
B. Murmann, B. Wooley EE214 Winter 2008-09 5 The result is a single-pole response with p 1 = ! 1 + g m R E R S C " # !$ T R E R S % & ( ) * In this case, if R E > R S , then |p 1 | > ! T Input impedance with emitter peaking : Z in = 1 + g m Z E Y ! + r b + Z E = 1 + g m Y ! " # $ % & Z E + 1 Y ! + r b = g m + g ! + sC ! g ! + sC ! " # $ % & R E 1 + s ( E " # $ % & + 1 Y ! + r b B. Murmann, B. Wooley EE214 Winter 2008-09 6 Since g m >> g " , if # E = # T , then Z in ! r b + 1 (g " + sC " ) + g m + sC " g " + sC " # $ % & ( R E 1 + s ) E # $ % & ( = r b + 1 g " + sC " # $ % & ( 1 + g m R E 1 + s ) T 1 + s ) E # $ % & ( * + , , - . / / = r b + 1 + g m R E g " + sC " = r b + r " (1 + g m R E ) 1 + sr " C " = r b + r " + g m R E ) 1 + s C " 1 + g m R E # $ % & ( r " + g m R E ) * + - .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
B. Murmann, B. Wooley EE214 Winter 2008-09 7 In this case, the equivalent input circuit is: where r ! eq = r ! (1 + g m R E ) C ! eq = C ! 1 + g m R E " C ! g m R E = # T R E = C E since # E = # T r " eq C " eq r b B. Murmann, B. Wooley EE214 Winter 2008-09 8 # E > # T results in a pole-zero separation that can cause peaking in the frequency response p 2 p 1 j ! $ z 1 |p 1 | |z 1 | |p 2 | ! |i o /v i |
Background image of page 4
B. Murmann, B. Wooley EE214 Winter 2008-09 9 Shunt Feedback Stage Small-signal equivalent circuit: Include transistor C μ in C F Neglect R S or include it in r " Neglect r b , r c , r e , and r μ Include r o and C cs in R L and C L r " C " C F g m v 1 R F R L i i + v 1 + v o C L C L is important because we want a large Z L (small C L ) B. Murmann, B. Wooley EE214 Winter 2008-09 10 Define ! F ! R F C F ! L ! R L C L and Y F = 1 R F + sC F = 1 R F (1 + sC F R F ) = 1 R F + s ! F ) Y L = 1 R L + sC L = 1 R L + s ! L ) Y " = 1 r " + sC "
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
B. Murmann, B. Wooley EE214 Winter 2008-09 11 From the equivalent circuit: i i + (v o ! v 1 )Y F = v 1 Y " (1) v o Y L + g m v 1 + (v o ! v 1 )Y F = 0 (2) Then from (2) v 1 = ! v o Y L + Y F g m ! Y F " # $ % & Substituting in (1) i i + v o Y F + Y L + Y F g m ! Y F " # $ % & (Y ( + Y F ) ) * + + , - . . = 0 ! v o i i = " g m " Y F (g m " Y F )Y F + (Y L + Y F )(Y # + Y F ) B. Murmann, B. Wooley EE214 Winter 2008-09 12 Assuming that g m >> 1/R F = G F g m ! Y F = g m ! (G F + sC F ) " g m ! sC F Then v o i i ! " g m " sC F (g m " sC F )(G F + sC F ) + (G L + G F ) + s(C L + C F ) # $ % & (g + G F ) + s(C + C F ) # $ % & Thus v o i i ! " g m " sC F a 0 + a 1 s + a 2 s 2 where a 0 = g m G F + (G L + G F )(g ! + G F ) a 1 = (g m " G F )C F + (G L + G F )(C !
Background image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 08/13/2009 for the course EE EE214 taught by Professor Borismurmann during the Winter '08 term at Stratford.

Page1 / 24

HO18_214W09_WidebandAmp-Part2_2pp - Handout #18 EE 214...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online