Lect. 19 Cascode amplifier - High-Speed Circuits & Systems...

High-Speed Circuits and Systems LaboratoryElectronic Circuits 1

CG amplifier can be fast because it does not suffer from Miller Effect

But its overall voltage gain is not large because Rin is small and Rout is large

A circuit configuration for large gain and high speed amplification?

Cascode configuration

Lect. 19: Cascode amplifier (6.8.1-6.8.2)

Q1: CS Q2: CG


Lect. 19: Cascode amplifier

Rin=?

Rout=ro2 + Avo2Rout1= ro2 + Avo2ro1

Since Avo2=1 + (gm2+gmb2)ro2

Rout can be approximated by gm2ro2ro1

Input, Output resistance

Small-signal analysis



Open-circuit voltage gain, Avo

( )

( )

( )

2 1 2 2 2 1

1 1 1 2 1 1

22

2

2 2 2 1 1

2 2 2 1 1

2 2 2 2 2 1 1

[1 ]

( || )

( )

[1 ]

[1 ]

If assume , If two transistors a

o o o m mb o o

o m o in i m o i

o Lin

vo

o m mb o m o i

ovo m mb o m o

i

o m o vo m o m o

A g g r

g r R v g r vr R

RA

g g r g r v

A g g r g r

A g r A g r g r

υ

υ

υ υ υ

υ

υ

υυ

= = + +

= − = −

+= = ∞

∴ = − + +

= = − + +

−

∵

1 2 1 2

2 20 0

re identical=> and

( ) ( )m m m o o o

vo m o m o

g g g r r r

A g r A A g r

= = =>

∴ − => − =

Voltage gain with RL,

20

L Lv vo

L out L o o

R RA A A

R R R A r= −

+ +



Compared with CS, Rout: ro Aoro

Cascode Amplifier

vi

Norton Equivalent



Frequency Response Cdb: capacitance between D and B

It can be shown that (HW)

( )( ) ( ) ( )

( ) ( )( ) ( )

1 1 1 1 1

1 2 1 2

1 1 1 1 1 1 1 2

2

1 12 2 2

12

1

1

1where

HH

H gs sig gd m d sig d

db gs d L gd L out

sig gs gd m d d gd db gs

L out L gd

Ld o

m mb o

f

C R C g R R R

C C R C C R R

R C C g R R C C C

R R C C

RR r

g g Aυ

πτ

τ

≅

⎡ ⎤= + + +⎣ ⎦

+ + + +

⎡ ⎤= + + + + +⎣ ⎦

+ +

⎡ ⎤= +⎢ ⎥+⎣ ⎦

+ Cdb2



1 12 2 2

1 Ld o

m mb o

RR r

g g Aυ

⎡ ⎤= +⎢ ⎥+⎣ ⎦



Comparison of Cascode and CS for Frequency Response

( ) ( )( ) ( )

1 1 1 1 1 1 1 2

2

12

For Cascode

1

For CS (Lect. 16)[ (1 ') '] ( ) '

= [ (1 ')] '( )

where '

HH

H sig gs gd m d d gd db gs

L out L gd

H gs sig gd sig m L L L db L

sig gs gd m L L gd L db

L L

f

R C C g R R C C C

R R C C

C R C R g R R C C R

R C C g R R C C C

R R

πτ

τ

τ

≅

⎡ ⎤= + + + + +⎣ ⎦

+ +

= + + + + +

+ + + + +

= || or

( || ) 1.25 10 12.5

= [ (1 ')] '( )

= 10 [20+5 (1+1.25 10)]+10 (5+5+5) = 875 + 150 = 1025

155 Mz

v m L o

H sig gs gd m L L gd L db

H

A g R r

R C C g R R C C C

f

τ

= − × = − × = −

+ + + + +

× × × ×

=

( ) ( )( ) ( )

2 2

1 1 1 1 1 1 1 2

2 2

( ) (1.25 20) 62520625 20

20 640

1

= 10 [20+5 (1+1.25 1.22)] 1.22 (5 5 20) +(20||640) (5+5+

vo m o

Lv vo

L out

H sig gs gd m d d gd db gs

L out L db gd

A g rR

A AR R

R C C g R R C C C

R R C C C

τ

− × = − × = −

= = − × −+ +

⎡ ⎤= + + + + +⎣ ⎦

+ + +

× × × + × + +×

∼

5) =326+37+291 =654ps

243MzHf =



Numerical comparison (Example 6.12a)W/L=7.2µm/0.36µm ID=100µA, gm =1.25mA/V, χ=0.2, ro=20kΩ, Cgs=20fF, Cgd=5fF, Cdb=5fF, CL=5fF, Rsig=10kΩ, RL= 20kΩ

Cascode

fH difference is mainly due to difference in Rds1 and RL’

CS



1 12 2 2

1 Ld o

m mb o

RR r

g g Aυ

⎡ ⎤= +⎢ ⎥+⎣ ⎦

Cascode

CS

RL’=ro||RL

Cascode is faster becauseRd1 is reduced by low input resistance of CGand Miller effect is minimized.



+ Cdb2

Homework (Due on Nov. 14 before Tutorial)

1. Derive expression for τH.2. Do Example 6.12 (b).

(fH for CS and Cascode for Rsig=0)Explain why CS has larger fH.

Lect. 19 Cascode amplifier - High-Speed Circuits & Systems...

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