Current Sources Slide

8/3/2019 Current Sources Slide

1/15

A simple two-transistor current source.


2/15

Ic1 = Ic2

Summing currents at the collector of Q1

yields

021

1!

F

Ccref

III

Fand thus

212

1

C

F

ref

C I

I

I !

!

F

ifFF is large, the collector current of Q2 is nearly equal to the

reference current:

R

VVII

onBECC

refC

)(

2

!$


3/15

Simple current source with current gain.


4/15

The emitter current of transistor Q3 is equal to

2

21

3

2

C

FF

C

F

CE I

III

FFF!!

The base current of transistor Q3 is equal to

2

3

3

1

2

1C

FFF

EB I

II

!

!

FFFFinally, summing currents at the collector of Q1, we obtain

0

1

221

!

CFF

Cref IIIFF

Since IC1 and IC2 are equal,

FF

ref

Co

III

FF

!!

2

22

1


5/15

Widlar current source


6/15

Summing voltages around the emitter-base loop, assuming that

VA is infinite, and neglecting base currents results in

022

2

2

1

1 ! RI

I

IInV

I

IInV C

S

CT

S

CT

For identical transistors, IS1 and IS2 are equal, the equation

becomes

22

2

1RI

IIInV CS

CT !

02221 ! RIVV CBEBE

and thus


7/15

Cascode current source with bipolar transistors.


8/15

Thus, cascode current sources can boost the output resistance

and equivalent open-circuit voltage approximatelyF0 / 2.

ForF0 = 100 and VA =130V, we have

VVIr

V AooThev 650022

00 !!!FF

At an output current Io = 1mA, we find that

;!! M

mA

VRo 5.6

1

6500


9/15

N-Channel mirror


10/15

In the most general case, the ration of iO to iI is

-

!

1

2

1

2

2

1

2

21

21

1

1

K

K

vv

VV

VV

LW

WL

i

i

DS

DS

TGS

TGS

I

O

PP

-

!1

2

21

21

1

1

DS

DS

I

O

v

v

LW

WL

i

i

P

P

If vDS2 = vGS1, then the ratio of iO/iIbecomes

!

21

21

LWWL

ii

I

O

-

!

1

2

11

DS

DS

I

O

vv

ii

PP


11/15

The Differential Amplifier

The emitter-coupled or differential pair stage


12/15

CmCo

DM

oDM Rg

r

R

V

VA !!!

T

F1

Small-signal models for(a) the differential mode and (b) the common mode.


13/15

Common-Mode Gain ACM

TFF

rR

R

V

V

A Eo

Co

CM

o

CM

!! 12

1

with Fo >> 1 and division by rT, reduces to

E

C

Em

CmCM

R

R

Rg

RgA

221

}

!

The 2gmRE >> 1. Because the same signal is applied to Q1 and Q2,

both Vo1 and Vo2 are 180o out of phase with VCM.

The Common-Mode Rejection Ratio:

CM

DM

A

ACMRR !

CMRR = 1 + 2gmRE } 2gmRE


14/15

Output for Arbitrary Input signals

22

21 d

DM

VVV

V !

! 2

21VV

VCM

!

The output voltage Vo1 is

CMCMDMDMO VAVAV !1

!CMRR

VVA CMDMDM

! CMRR

VVV

AV d

DMo

21

1

2

!

CMRR

VVV

AV

d

DMo

21

22


15/15

A source-coupled pair

For rd >> RD and Q >>1,

SgmSgm RRCMRR 221 }!

FETDifferential Amplifier

Dd

S

Rr

RCMRR

!

Q121

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Current Sources Slide