elec 2301 bjt diff

8/2/2019 Elec 2301 Bjt Diff

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Differential Amplifiers:

-

Differential input signal is the difference of 2 input signals

Common-mode in ut si nal is the avera e of 2 in ut

12vvv

Id=

signal)(

2

112

vvvIcm

+=

Expressing input signals v1, v2 in terms of differential and

common-mode si nals:

2/1 IdIcm

vvv =

2/2 IdIcm

vvv +=

1

Pictorial representation

2


2/14

Most widely-used building block in analog electronics

Input stage of every op-amp is a differential amp

Wh differential?

(1) Much less sensitive to noise and interference than

sin le-ended circuits

(2) No need for bypass or coupling capacitors

3

BJT Differential Pair

2 matched transistors , emitters connected together and

biased by a constant current sourceI

4


3/14

-

Vary the value of the common-

mode input voltage vcm

As long as Q1 and Q2 remainin the active region current will

still divide equally between Q1an 2 an t e vo tages at

collectors will not change

Thus differential pair does not

respond (rejects) common-

mo e npu s gna s

5

Let vB2 be grounded and let vB1= +1 V

Q1 will be ON and conducting

all of the current I and Q2 willbe OFF

For Q1 to be ON (with VBE1 =

0.7 V) emitter has to be at ~

+0.3 V, keeping EBJ of Q2reverse biased

6


4/14

Let vB2 be grounded and let vB1= -1 V

Q1 will be OFF and Q2 will beconducting all of the current I

Common emitter will be at -

0.7V, EBJ of Q1 will be

reverse biased by 0.3V

7

From the previous analysis the differential pair responds

to large differential signals

With relatively small difference voltages we are able to

steer the entire bias current from one side to another To use the BJT differential pair as a linear amplifier we

apply a small differential signal (a few mV), which will

result in one of the transistor conducting a current ofI/2 +

I; the current in the other transistor will beI/2 - I, with

Ibeing proportional to the differential input voltage

Output voltage taken between two collectors will be

2IRC, which is proportional to differential input signal

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5/14

9

TBE VV /=TEB VvvsI /1= TEB Vvvs

I /2=

Combinin ,

sCE1

E2

( ) TBB Vvv

E

E

ei

/

2

1 21

=

( ) TBB Vvv

Ei

/

1

12

1

=EE 21

Ei

21

=( ) TBB Vvv

EEeii

/

21

211

++

10


6/14

The circuit imposes a constraint

Therefore

IiiEE=+

21

Tid VvE e

Ii

/1

1

+=

Tid VvE e

Ii

/2

1+=

11

A relatively small difference voltage causesIto flow

~ T

12


7/14

Include two equal

resistances Re in

series with emitterso 1 an 2

Reduced gm and

overa ga n

13

-

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8/14


9/14

Assume current source is ideal; its incremental resistance will

then be infinite

Thus the voltage vidappears across a total resistance of2re,w ere

2/I

V

I

Vr T

E

T

e==

orrespon ng y t ere w e a s gna current e g ven y

e

id

e

r

vi

2=

Thus the collector of Q1 will exhibit increment ic and the

collector of Q2 will exhibit a current decrement ic:

22

id

m

e

id

ecg

rii ===

17

18


10/14

This method of analysis is particularly useful when resistances

are included in the emitters

ee

id

e

Rr

vi

22 +=

19

( )( )eeid

RrR 221 ++=

The resistance seen between the 2 bases is equal to the

total resistance in the emitter circuit multiplied by (+1):

resistance reflection rule

20


11/14

For small difference inputs

21

id

mCC

vgIi +=

idv

where2

2 mCC

II

c

=

Thus the total voltages at the collectors will be

21

id

CmCCCCCRgRIVv =

( )2id

CmCCCCC

v

RgRIVv +=

21

Differential gain (taken between 2 collectors):

Cm

d

CC

dRg

v

vvA =

= 21

Single-ended gain (between 1 collector and GND):

v 1Cm

d

C

dg

v 2==

( )CC

RR=

2

eeee

d

RrRr ++ 22

22


12/14

-

EE

C

icm

eEE

C

icmC

R

Rv

rR

Rvv

221

+=

EE

C

icmC

R

Rvv

22

=

23

Common-mode voltage vo = (vC1- vC2) = 0

=> common-mode gain = 0

For sin le-ended out ut, common-mode ain

EE

C

cm

RRA

2=

CmdRgA

2

1=

-

EEm

cm

d RgA

ACMRR =

Expressed in decibelsd

A

ACMRR log20=

24

cm


13/14

The differential amp uses transistors with =100. Evaluate (a)

id ,

gain (neglect the effect or ro) and (c) the CMRR, in dB.

25

(a) Each transistor is biased at an emitter current of 0.5 mA. Thus25mVV

The input differential resistance

==== 505.0

21

mAIrr

E

T

ee

(b) Voltage gain from signal source to bases of Q1 and Q2:

eeid

Rv 40

Voltage gain from bases to output

RRvsigidsig

.4055

=++

=+

=

( ) ( )VV

Rr

R

v

v

Ee

C

id

o /5010150502

102

2

23=

+

=

+=

Overall differential voltage gain:

VVv

v

v

v

v

vA oido

d/40508.0 ====

26

idsigsig


14/14

(c) ACMRR dlo20=

dB

Acm

98105

40log20 4 ==

27

elec 2301 bjt diff

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