Sheet 1 of 3

Common-Base/Gate Circuits Common-Base BJT circuit The figure below shows the simplied ‘Pi’ model of a common-base BJT.

VoutVin

B

E C

rbe

rce

β.ib or gm.Vbe

ib(β+1)

ZoutRL

c

Vout

b

Vin

Zin

e

ib

VI e wher

gm1 r

1)(βi1).r(βi

IVR

T

CQbe

b

beb

IN

ININ ===

++

== gm

)rgm1 (as

VV

IV.

VI

rr

1)r(βr.β

1)r(βir.i.β

VVA ce

T

A

CQ

A

T

CQ

be

ce

be

ce

beb

ceb

IN

OUTV ===≈

+=

+==

( ) 1 1β

β )1βi

β.i I

I Ab

b

IN

OUTi ≈

+=

+==

Sheet 2 of 3

Vout

B

E C

rbe

rce

β.ib

ib

ZoutRL

c

Vout

b Vin

e

iT-ib

Vs

Rs

iT

iT+β.i

iT

To determine the Output impedance of the circuit we can connect a voltage source (Vs + Rs) to the base and ground the input ie the emitter. We then have to resistances in parallel connected to the current source β.ib.

( )

equation above into sub Rr

Ri i Also

r.iri.i V

iV R

sbe

sTb

bebcebTOUT

T

OUTOUT

+=

++=

=

β

sbe

bes

sbe

scece

T

OUTOUT

besce

sTce

sbe

sTTOUT

Rrr.R

RrRr.r

iVR

r.Rr

RirRr

Ri.i V

++

++==

++⎟⎟

⎠

⎞⎜⎜⎝

⎛+

+=

β

β

( ) cececeOUTs

ceOUTs

r.1 1r.r R then large R If

r R then 0 R If

+=++==

==

ββ

Sheet 3 of 3

Common-Gate MOSFET Circuit

Vin

s

gmVsg

g

d

rds

rdg

Io

Vout

RL

gate

source drain

Vout Vin

IIN = gmVsg

Vsg rIN

RL

Voltage Gain Av

( )

Rr.Rr

gm A //RrgmVsg V

Vsg V VV A

Ldg

LdgVLdgO

ININ

OV

⎟⎟⎠

⎞⎜⎜⎝

⎛

+==

==

Input Resistance

λI gm1

gmVgsVgs

IV R D

IN

ININ ====

Output Resistance As the source is low impedance ie close to ground for ROUT – rds appears to be connected across rds to ground.

dsdgO

OOUT //rr

IV R ==

Current Gain Ai Ai = 1