Lecture 12
BJT’s Differential Pair

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topics
• Ideal characteristics of differential
amplifier
–
–
–
–

Input differential resistance
Input common-mode resistance
Differential voltage gain
CMRR

• Non-ideal characteristics of differential
amplifier
– Input offset voltage
– Input biasing and offset current

• Differential Amplifier with active load
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Differential pair

Figure 7.12 The basic BJT differential-pair configuration.

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Common mode operation

Q Q1 = Q2
∴ vB1 = vB 2 = vCM
⇒ iE1 = iE 2

I
=
2

⇒ vC1 = vC 2 = VCC


I
− α RC
2

vC1 − vC 2 = 0

Reject common mode input

Figure 7.13 Different modes of operation of the BJT differential pair: (a) The differential pair with a common-mode input signal vCM.

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The differential pair with a “large” differential input
signal

(1)VB1 >> VB 2
VB1 = +1V , VB 2 = 0
Q1 on → VE1 = 0.3V = VE 2 → Q2

off

VC1 = VCC − αIRC , VC 2 = VCC
VC1 − VC 2 = −αIRC


Figure 7.13 Different modes of operation of the BJT differential pair:. (b) The differential pair with a “large” differential input signal.

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(2)VB1 << VB 2
VB1 = −1V , VB 2 = 0
Q2
+
0.7

on → VE 2 = −0.7V = VE 2 → Q1 off

VC 2 = VCC − αIRC , VC1 = VCC
VC1 − VC 2 = αIRC

−

Figure 7.13 (Continued) (c) The differential pair with a large differential input signal of polarity opposite to that in .

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(3)VB1 > VB 2
VB1 = small , VB 2 = 0
I
I
I E1 = + ΔI , I E 2 = − ΔI
2
2
I
VC1 = VCC − α RC − αΔIRC
2
I
VC 2 = VCC − α RC + αΔIRC
2
VC1 − VC 2 = vo = −2αΔIRC
⇒ vo = f (ΔI )
Figure 7.13 (Continued) (d) The differential pair with a small differential input signal vi. Note that we have assumed the bias current source I
to be ideal (i.e., it has an infinite output resistance) and thus I remains constant with the change in vCM.

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Exercise 7.7

let α ≈ 1, vBE = 0.7V
find vE , vC1 and vC 2
5 − 0.7
I=
= 4.3mA
1k
+
0.7
−

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Large signal operation

iE 1 =

IS

e ( vB1 −vE ) / VT

iE 2 =


IS

e ( vB 2 −vE ) / VT

α

α

iE 1
= e ( vB1 −vB 2 ) / VT
iE 2
iE1
1
=
iE1 + iE 2 1 + e ( vB 2 −vB1 ) / VT
iE 2
1
=
iE1 + iE 2 1 + e ( vB1 −vB 2 ) / VT
iE 1 + iE 2 = I
iE 1 =
iE 2

I

1 + e −vid / VT
I
=
1 + e vid / VT


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iC1
1
≈
I 1 + e −vid / VT
1 + e −vid / VT
iC 2
I
1
=
⇒
≈
vid / VT
I 1 + e vid / VT
1+ e
I

iE 1 =
iE 2

⇒

How to enhance linear

region？
Figure 7.14 Transfer characteristics of the BJT differential pair of Fig. 7.12 assuming α . 1.

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Re ↑→vE ↑→vBE ↓→iC ↓

Figure 7.15 The transfer characteristics of the BJT differential pair (a) can be linearized (b) (i.e., the linear range of
operation can be extended) by including resistances in the emitters.

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large signal analysis (AC+DC)

αI

iC 1 =


L (1)
1 + e − v id / VT
αI
iC 2 =
1 + e v id / VT
α Ie v id / 2VT
e v id / 2VT
(1) × v id / 2VT ⇒ iC 1 = v id / 2VT
e
+ e − v id / 2VT
e
let
v id << 2VT
v id
)
2VT
≈
v
v
1 + id + 1 − id
2VT
2VT

α I (1 +

⇒ iC 1

⇒ iC 1 =
⇒ iC 2 =
ic =

Figure 7.16 The currents and voltages in the differential amplifier when a
small differential input signal vid is applied.

αI
2

αI
2

α I v id
2VT 2

+

α I v id
2VT 2

−
=

α I v id
2VT 2
gm
v id
2

v BE

Q1 = V BE +


v id
2

v BE

Q 2 = V BE −

v id
2

Taylor
series

= IC +

I C v id
VT 2

= IC −

I C v id
VT 2

AC

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Small signal analysis (AC)

g m = I C / VT =

vid
2re

ic = αie =
ie =

VT

VT
I /2

re = VT / I E =
ie =

αI / 2

αvid
2re

= gm

vid
2re


αie

αie

RC

+
Figure 7.17 A simple technique for determining the signal currents in a
differential amplifier excited by a differential voltage signal vid; dc quantities
are not shown.

vid
2

ie

re

id

vid

RC

−
ie

re
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vc1

vo = vc1 − vc 2

vc2

RC

RC
g m vπ

+

+
vπ
−

re

vid

id


g m vπ

−
+
vπ
−

re

vid
2reie
=
= 2(1 + β )re Input differential resistance
Rid =
i
id
e
1+ β

v id
I C αI E α
2
gm =
=
=
VT
VT
re
v id
v C 2 = + g m RC

2
v c1 − v c 2
= − g m RC Differential voltage gain
Ad =
v id
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v C 1 = − g m RC

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Common mode

vc1

vc2

vo = vc1 − vc 2

αie 2

αie1

RC
g m vπ

+
vπ
−

RC

g m vπ

ib1
ie1
re

DC

vicm ib 2

+

ie 2

vπ
−

re

vc1 = −αie1 RC

if

RC1 ≠ RC 2 ⇒ vo ≠ 0


vc 2 = −αie2 RC
vo = vc1 − vc 2 = −αRc (ie1 − ie2 )
∴ ie1 = ie2 = 0 ⇒ vo = −αRc (ie1 − ie2 ) = o
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External emitter resistance

vc1
RC1

vo = vc1 − vc 2

αie
g m vπ

ie

re

RE

vc2


vid

ib

RE

−

RC 2
g m vπ

RE ↑⇒ Ad ↓
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ie
v /( 2re + 2 RE )
= id
β +1
β +1
v
Rid = id = ( β + 1)(2re + 2 RE )
ib

Input differential resistance
re
RE

Differential voltage gain

RE ↑⇒ Rid ↑


vid
2re + 2 RE

ib =

αie

+

ie =

vid
2
v
v C 2 = + g m RC id
2
Ic α
gm =
=
VT re
v C1 = − g m RC

Ad =

vc1 − vc 2
= − g m RC
vid
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Bartlett Bisection theorem
+
v1

N

−

1. v1 = v2 → I = 0

+
v2

+
v1

−

−

I
1
N
2

1

N
2

M

2. v1 = −v2 → V = 0
Differential Mode

+
v1

+
v1

−

M

I=0 open circuit
Common-mode

−

1
N
2

M

V=0 short circuit

Differential-mode
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−

V

Common Mode

1
N
2

+
v2

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Differential Mode

Common Mode

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Equivalence of the differential amplifier to a CE amplifier

Figure 7.19 Equivalence of the BJT differential amplifier in (a) to the two common-emitter amplifiers in (b). This equivalence applies
only for differential input signals. Either of the two common-emitter amplifiers in (b) can be used to find the differential gain,
differential input resistance, frequency response, and so on, of the differential amplifier.

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vc1
g
= − m (ro // Rc )
vid
2
vc 2 g m
=
(ro // Rc )
vid
2
vc1 vc 2 vc1 − vc 2
−

=
= − g m ( RC // ro )
Ad =
vid vid
vid

Differential half-circuit
i =
V

id

2
R
Figure 7.21 (a) The differential half-circuit and (b) its equivalent circuit model.

id

Vπ
rπ
= V
=

V

2(1 + β )re

π
id


i

=

2V
Vπ

π

rπ

Input differential resistance
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= 2 rπ

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Common mode gain et CMRR (I=0Ỉ open circuit)
vc1 = −αie Rc
vicm = ie (re + 2 REE )
Acm

1
2


=−

αRC
(re + 2 REE )

≈−

αRC
(2 REE )

1
g m RC
2
A
CMRR 1 = d ≈ g m REE
2
Acm
Ad

1
2

=

Common-mode half-circuit
Figure 7.22 (a) The differential amplifier fed by a common-mode voltage signal vicm. (b) Equivalent “halfcircuits” for common-mode calculations.

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Common mode gain at CMRR ( Asymmetric case)
vc1 = −αie RC
vc 2 = −αie ( RC + ΔRC )

Last page

vo = vc1 − vc 2 = −αie ΔRC

αRC

vc1 = −vicm

vicm = ie (2 REE + re )

αRC ΔRC
− αΔRC
ΔRC
Acm =
≈
=−
2 REE + re 2 REE
2 REE RC
v1 + v2
vicm ≡
2

vid ≡ v1 − v2

vc 2 ≈ −vicm

<<
Acm

1
2

=−

2 REE

αRC
2 REE

αRC
2 REE

=

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αRC


1
g m RC
1
2
2
A
CMRR = d ≈ g m REE
Acm
Ad

v1 + v2
)
vo = Ad (v1 − v2 ) + Acm (
2

2 REE + re

≈ −vicm

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Common-mode input resistance

V = ie ( re + 2 REE )
ie
1+ β
≈ ( β + 1)( 2 REE // ro )

I = ib =
2 Ricm


Ricm ≈ ( β + 1)( REE //

ro
)
2

g m vπ

I

ro

+
vπ
−

RC

re

2 REE
Figure 7.23 (a) Definition of the input common-mode resistance
Ricm. (b) The equivalent common-mode half-circuit.

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αie

ie − i − αie

g m vπ

ib

re

−

i

RC
ro

+

ie vπ

Ricm ≈ ( β + 1)( REE

ro
// )
2


ie − i

2 REE

2 Ricm =

V
,
ib

V = reie + 2 REE i

2 REE i = ro (ie − i ) + Rc (ie − i − αie )

2 Ricm

⇒ (2 REE + ro + Rc )i = roie + Rc (ie − αie ) = roib (1 + β ) + Rc ib
⇒i=

ro ib (1 + β ) + Rc ib
2 REE + ro + Rc

V = reie + 2 REE i = reib (1 + β ) + 2 REE
2 Ricm =

ro ib (1 + β ) + Rc ib
2 REE + ro + Rc

2 REE ro
2 REE Rc

V
= re (1 + β ) + (1 + β )
+
ib
2 REE + ro + Rc 2 REE + ro + Rc

2 Ricm ≈ ( β + 1)( 2 REE // ro )
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Example 7.1
1. Input differential resistance

β = 100

re1 = re 2 =

VA = 100

VT 25mV
=
= 50
I E 0.5mA

Rid = 2( β + 1)(re + RE ) = 40k


2. Differential voltage gain
Ad =

vo vid
Rid
= g m RC
= 40
vid vs
Rs + Rid

3. Common-mode gain in worst case
Acm =

4. Input common-mode resistance

2 REE

RC
ΔRC
+ (re + RE ) RC

ΔRC = 0.02 RC
VA
−4
ro =
= 200k
A
=
5

×
10
cm
I /2
1
A
Ricm ≈ ( β + 1)(2 REE // ro ) = 6.7 M
CMRR = 20 log d = 98dB
2
Acm
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