ECE 442 – Jose Schutt-Aine 1 ECE 342 Solid-State Devices & Circuits 14. CB and CG Amplifiers Jose...

ECE 442 – Jose Schutt-Aine 1

ECE 342Solid-State Devices & Circuits

14. CB and CG Amplifiers

Jose E. Schutt-AineElectrical & Computer Engineering

University of [email protected]

1

Common Gate Amplifier

Substrate is not connected to the source must account for body effect

Drain signal current becomes

D m gs mb bsi g v g v

And since gs bsv v

Body effect is fully accounted for by using m m mbg g g


i m mb i roi g g v i

m mb ioi o i i L

ro io o L

o

1g g v

rv v v i Ri i

r r R1

r

with i sv v



i o L

ini m mb o

v r RR

i 1 g g r

As o inm mb

1r , R

g g

If L iR , R

o ro i m mb o i iv i v g g r v v

vo m mb oA 1 g g r



whereo L Lin o m mb o

vo m mb o

r R R1R , A g g r

A g g A

Taking ro into account adds a component (RL/Ao) to the input resistance.

vo m mb oA 1 g g r

The open-circuit voltage gain is:

The voltage gain of the loaded CG amplifier is:

Lv vo

L o vo s

RG A

R r A R



x x m mb ov i g g v r v with x sv i R

orout o m mb o s out o vo sR r 1 g g r R R r A R

CG Output Resistance

CG Amplifier as Current Buffer

sis vo

out

RG G 1

R

Gis is the short-circuit current gain

- Include CL to represent capacitance of load- Cgd is grounded- No Miller effect

High-Frequency Response of CG


P1

gs sm mb

1f

12 C R ||

g g

P2

gd L L

1f

2 C C R

• fP2 is usually lower than fp1• fP2 can be dominant• Both fP1 and fP2 are usually much higher than fP in CS case

2 poles:

High-Frequency Response of CG

o Lin

o L

e e

r RR

r R1

r 1 r

vo m oA 1 g r 'out o m o eR r 1 g r R

'e eR R || re

rr

1

CB Amplifier


High-Frequency Analysis of CB Amplifier

in 3dBx

S E

1

r rC R || R ||

1 1

out 3dBL

1

C R

The amplifier’s upper cutoff frequency will be the lower of these two poles.

From current gain analysis


11

1o

o

Define gr

Common source amplifier, followed by common gate stage – G2 is an incremental ground

MOS Cascode Amplifier

22

1o

o

gr

LR current source impedance

1L

L

GR


• CS cacaded with CG CascodeVery popular configurationOften considered as a single stage

amplifier• Combine high input impedance and

large transconductance in CS with current buffering and superior high frequency response of CG

• Can be used to achieve equal gain but wider bandwidth than CS

• Can be used to achieve higher gain with same GBW as CS

MOS Cascode Amplifier


MOS Cascode Incremental Model

Lo

L

iv

G

2 2 2 2 2 2L mb s m s s o oi g v g v v v g

2 2 2 2 2 2L

L s mb m s o oL

ii v g g v g g

G


22 2 2 21 o

L s mb m oL

gi v g g g

G

22

2 2 2

1 /L o Ls

mb m o

i g Gv

g g g

KCL at vs2

1 2 1 2 2 2 2 2 2( ) 0m gs s o m s mb s o s og v v g g v g v g v v

MOS Cascode Analysis


1 21 1

2 2 2

1 /1 0o o L

m gs Lo m mb

g g Gg v i

g g g

1 1

1 2

2 2 2

1 /1

m gsL

o o L

o m mb

g vi

g g G

g g g

1

1 2

2 2 2

o m

in o L oL

o m mb

v g

v g G gG

g g g

Two cases



1 2 2 2 1 2 om o m mb o o

in

vg g g g r r

v

1 2 1 2 1 2 1 2 MB m o m m m mb o oA g g g g g g r r

1 1 2 2MB m o m oA g r g r

CASE 1

The voltage gain becomes

1: 0LCase If G



1 2

2 2 2

2 : o L oL

o m mb

g G gCase If G

g g g

1 o m

MBin L

v gA

v G

CASE 2

The voltage gain becomes



MOS Cascode at High Frequency

23 2

1

2ods

fr C

The upper corner frequency of the cascode can be approximated as:

2 2 2 3 3where db gd db gdC C C C C


• Capacitance Cgs1 sees a resistance Rsig

• Capacitance Cgd1 sees a resistance Rgd1

• Capacitance (Cdb1+Cgs2) sees resistance Rd1

• Capacitance (CL+Cgd2) sees resistance (RL||Rout) gd 1 m1 d 1 sig d 1R 1 g R R R L

d 1 o1m2 mb2 vo2

R1R r ||

g g A

H gs1 sig gd 1 m1 d 1 sig d 1C R C 1 g R R R

db1 gs2 d 1 L gds2 L outC C R C C R || R

HH

1f

2



If Rsig is large, to extend the bandwidth we must lower RL. This lowers Rd1 and makes the Miller effect insignificant

If Rsig is small, there is no Miller effect. A large value of RL will give high gain


Effect of Cascoding(when Rsig=0)


Cascode Example


Cascode Example

The cascode circuit has a dc drain current of 50 mA for all transistors supplied by current mirror M3. Parameters are gm1=181 mA/V, gm2=195 mA/V, gds1= 5.87 mA/V, gmb2=57.1 mA/V, gds2= 0.939 mA, gds3= 3.76 mA/V, Cdb2 = 9.8 fF, Cgd2 = 1.5 fF, Cdb3=40.9 fF, Cgd3= 4.5 fF. Find midband gain and approximate upper corner frequency

1 2

2 2 2

5.87 3.76 0.9390.109 /

0.939 195 57.1o L o

Lo m mb

g G gA V G

g g g

3 3.76 / L dsG g A V

Therefore, we use Case 2 to compute the gain

The internal conductance of the current source is:


Cascode Example

1 1

3

18148.1 /

3.76 m m

MBL ds

g gA V V

G g

Gain can be approximated by

Upper corner frequency is approximated by

2 15 33 2

1 110.6

2 2 56.7 10 266 10ods

f MHzr C


Common emitter amplifier, followed by common base stage – Base of Q2 is an incremental ground

BJT Cascode Amplifier


BJT Cascode Incremental Model

2 2o m Lv g R v

2 22 2

2 2 2

1x xx

x

v r rv v v

r r r


BJT Cascode Analysis

1 1 2 22 2

xm m

x

vg v g v

r r

Ignoring rx2

21 1 2 2

2m m

vg v g v

r

1 1 2 22

1m mg v v g

r



11

1 1in

s x

rv v

R r r

1 12

1 12

2

1

1m in

s xm

g r vv

R r rg

r

2 1 1

1 12

2

1

1m L m in

os x

m

g R g r vv

R r rg

r



2 2 1 1

1 1 2 21o m m L

in s x m

v g r g r R

v R r r g r

1 2

1 1 2 1o L

in s x

v R

v R r r

1 2v m LA g R

If Rs << rx1+rp1, the voltage gain can be approximated by


BJT Cascode at High Frequency

Define rcs as the internal impedance of the current source. R includes generator resistance and base resistance rx1 base of Q1


There is no Miller multiplication from the input of Q2 (emitter) to the output (collector).



1 1 2 2 3

1 1 1

1 || 1 1MBe out

A Aj C r R j r C j R C

22 2

1

2 e

fr C

1 1

1

2in highfr R C

1 1

1 1MB

in high out high

A Af f

j jf f


3 2 ||out csDefine R r r



3

1

2out highout

fC R

2out out outcsC C C

current source output capacitanceoutcsC


Cascode Amplifier – High Frequency

High-frequency incremental model


' 's i s 1 1 1 a 1 bG V G g s C C V sC V

m1 1 a 2 m2 2 1 b 2 m2 2 d0 g sC V g g s C C V g g sC V

2 2 b x2 2 2 2 d 2 o0 g sC V g g s C C V sC V

m2 b m2 2 d L 2 o0 g V g sC V G sC V

Applying Kirchoff’s current law to each node:

Find solution using a computer



gm=0.4 mhos b=100rp=250 ohms rx=20 ohmsCp=100 pF Cm=5 pfGL=5 mmhos GS’=4.5 mmhossa=8.0 ns-1 sd=-0.0806 nsec-1sb=-2.02 + j5.99 se=-0.644sc=-2.02 – j5.99 sf=-4.05 sg=-16.45

POLES

ZEROS

As an example use:



If one pole is at a much lower frequency than the zeros and the other poles, (dominant pole) we can approximate w3dB

3dB 0.0806 Grad / sec

3dBf 12.9 MHz

For the same gain, a single stage amplifier would yield:

3dB 0.0169 Grad / sec

3dBf 2.7 MHz

Second stage in cascode increases bandwidth



CE Cascade Amplifier

Exact analysis too tedious use computer

CE cascade has low upper-cutoff frequency


Cascode Amplifier

First stage is CE and second stage is CB

1 2v m LA g R

If Rs << rp1, the voltage gain can be approximated by



High-frequency incremental model


' 's i s 1 1 1 a 1 bG V G g s C C V sC V

m1 1 a 2 m2 2 1 b 2 m2 2 d0 g sC V g g s C C V g g sC V

2 2 b x2 2 2 2 d 2 o0 g sC V g g s C C V sC V

m2 b m2 2 d L 2 o0 g V g sC V G sC V

Applying Kirchoff’s current law to each node:

Find solution using a computer



gm=0.4 mhos b=100rp=250 ohms rx=20 ohmsCp=100 pF Cm=5 pfGL=5 mmhos GS’=4.5 mmhossa=8.0 sd=-0.0806sb=-2.02 + j5.99 se=-0.644sc=-2.02 – j5.99 sf=-4.05 sg=-16.45

POLES (nsec-1)ZEROS (nsec-1)

As an example use:



If one pole is at a much lower frequency than the zeros and the other poles, (dominant pole) we can approximate w3dB 9

3dB 0.0806 10 rad / sec

3dBf 12.9 MHz

For the same gain, a single stage amplifier would yield:

93dB 0.0169 10 rad / sec

3dBf 2.7 MHz

Second stage in cascode increases bandwidth


ECE 442 – Jose Schutt-Aine 1 ECE 342 Solid-State Devices & Circuits 14. CB and CG Amplifiers Jose...

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Transcript of ECE 442 – Jose Schutt-Aine 1 ECE 342 Solid-State Devices & Circuits 14. CB and CG Amplifiers Jose...