Lecture 13
MOSFET Differential
Amplifiers

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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

• Frequency response
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MOS differential pair

Figure 7.1 The basic MOS differential-pair configuration.

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Common mode operation

Q Q1 = Q2

BJT’s differential pair VCM no bound

∴ I D1 = I D 2 =


I
2

vs = vCM − VGS
I 1 'W
= k n (VGS − Vt ) 2
2 2 L
vD1 = vD 2 = VDD − I D RD
ID =

Q1 and Q2 in saturation mode
vDS > vGS − Vt
(VDD − I D RD ) − vS > vCM − vs − Vt

+

I
vCM (max) = Vt + VDD − RD
2
vCM (min) = −Vss + VCS + Vt + (VGS − Vt )

VCS
−
Figure 7.2 The MOS differential pair with a common-mode input voltage vCM.

vGS − Vt = vCM − vs − Vt
= vCM − (VCS − VSS ) − Vt
vCM = VCS − VSS + vGS

Make sure current source is working

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Exercise 7.1

ID =

VDD = VSS = 1.5V , k n'

1 'W
k n (VGS − Vt ) 2
2 L

I
= 0.2 = 2m(VGS − 0.5) 2
2
VGS = 0.816V

W
= 4mA / V 2 , Vt = 0.5V , I = 0.4mA, RD = 2.5k
L

VCM = 1V
VDS > VGS − Vt
VD − VS > VG − VS − Vt


VDS = 1.5 − 0.2m × 2.5k = 1V

VD > VG − Vt ⇒ VD > VCM − Vt

VDS > VGS − Vt = 0.816 − 0.5

⇒ VD = 1 > 1 − 0.5

Saturation mode

Saturation mode

VCM (max) = 1.5V
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+
0.48

VCS (min) = 0.48

−

VCM (min) = 0.82 + 0.48 − 1.5 = −0.2V


Figure 7.3 (Continued)

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vid = vGS 1 − vGS 2 = vG1 − vG 2

Large signal operation

iD1 − iD 2 =

1 'W
k n vid
2 L

i D1 +iD 2 = I
1 W 2
2 iD!iD 2 = I − k n'
vid
2 L
⎞
⎟
⎟
⎟

⎟
⎠

2

⎛
⎞
⎜
⎟
/
2
v
1 ' W ⎛ vid ⎞
I
⎟
iD1 = −
k n I ⎜ ⎟ 1 − ⎜ id
⎜ 1 'W ⎟
2
2 L ⎝ 2 ⎠
kn
⎜
⎟
2
L
⎝
⎠

2


⎛
⎜
v /2
1 ' W ⎛ vid ⎞
I
iD1 = +
k n I ⎜ ⎟ 1 − ⎜ id
⎜ 1 'W
2
2 L ⎝ 2 ⎠
kn
⎜
⎝ 2 L
Figure 7.5 The MOSFET differential pair for the purpose of deriving the
transfer characteristics, iD1 and iD2 versus vid = vG1 – vG2.

1 'W
iD1 = kn (vGS1 − Vt ) 2
2 L
1 'W
iD 2 = kn (vGS 2 − Vt ) 2
2 L

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Figure 7.6 Normalized plots of the currents in a MOSFET differential pair. Note that VOV is
the overdrive voltage at which Q1 and Q2 operate when conducting drain currents equal to
I/2.

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⎛
⎜
I
v /2
1 ' W ⎛ vid ⎞
iD1 = +
k n I ⎜ ⎟ 1 − ⎜ id
⎜ 1 'W
2
2 L ⎝ 2 ⎠
kn
⎜
⎝ 2 L

⎞
⎟
⎟

⎟
⎟
⎠

2

More k is bigger more linear
range of vid

Figure 7.7 The linear range of operation of the MOS differential pair can
be extended by operating the transistor at a higher value of VOV.

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7-2.1 Small signal operation (differential gain)
1
vG1 = vCM + vid
2
1
vG 2 = vCM − vid
2

v1 + v2
vCM =

2
vid = v1 − v2

Figure 7.8 Small-signal analysis of the MOS differential amplifier: (a) The circuit with
a common-mode voltage applied to set the dc bias voltage at the gates and with vid
applied in a complementary (or balanced) manner. (b) The circuit prepared for smallsignal analysis. (c) An alternative way of looking at the small-signal operation of the
circuit.

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vid
vo1 = − g m
( RD // ro1 )
2
vid
vo 2 = g m
( RD // ro 2 )
2
vo1
v
1
1
= − g m ( RD // ro1 ), o 2 = g m ( RD // ro 2 )
vid

2
vid
2
vo 2 − vo1
= g m ( RD // ro )
Ad ≡
vid

vid
2

vo1
+
v gs
−

g m v gs

Rid = ∞
Ro 1 = ro // RD
2

ro R
D

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ro effects

vo1 = − g m ( RD // ro )(vid / 2)
vo 2 = g m ( RD // ro )(vid / 2)
vo = vo 2 − vo1 = g m ( RD // ro )vid
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vo1 = − g m ( RD // ro )(vid / 2)
vo 2 = g m ( RD // ro )(vid / 2)
vo = vo 2 − vo1 = g m ( RD // ro )vid
Ad =

vo 2 − vo1
= g m ( RD // ro )
vid

Differential-mode equivalent circuit
V id
2

V o1


−

Vo 2

+

V gs

gm

Vid
2

ro1 R D

RD

ro 2

−

gm

−Vid
2

V id
2


+

V gs
−

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id

Common-mode gain et CMRR
vicm
1
gm

id

vo1
RD

2 Rss
2 Ricm

(1) Half circuit of differential pair
Acm 1 = RD / 2 Rss , Ad

2

CMRR ≡

1
g m RD
2

Ad
= g m Rss
Acm

(2) Full circuit

vo1 vo 2
RD
=
=−
vicm vicm
1 / g m + 2 RSS
Q RSS >> 1 / g m ⇒

1
2

=

vo1
v
R

= o1 = − D
vicm vicm
2 RSS

Acm = (vo2 −vo1) / vicm = 0, Ad = (vo2 −vo1) / vid = gm RD
CMRR≡

Ad
=∞
Acm

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+ VDD

+ VDD

1 'W
I D = k n (vGS − Vt ) 2
2 L
vGS1 = vGS 2 ⇒ I D1 = I D 2 = I ref

I Re f
I D2


VDD + VSS − VGS
I D1 =
R

I D1

Ro = ro = RSS
− VSS

g m v gs
ro

vo1

+
v gs
−

+
R

v gs
−

g m v gs

ro

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Non zero common gain due to RD mismatch
id

vicm

id

consider

vo1

RD
vicm
vo1 ≅ −
2 Rss

RD

1
gm

RD1 ≠ RD 2


RD + ΔRD
vicm
vo 2 ≅ −
2 Rss

2 Rss
2 Ricm

vo 2 − vo1 = −

vo1
v
RD
= o1 = −
vicm vicm
1 / g m + 2 RSS
Q RSS >> 1 / g m ⇒

Acm = −

vo1
v
R
= o1 = − D
vicm vicm
2 RSS

ΔRD
vicm
2 Rss


R ΔRD
ΔRD
=− D
2 Rss
2 Rss RD

Ad = − g m RD
CMRR ≡

Ad
ΔR
= 2 g m Rss / D
RD
Acm
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Non zero common gain due to gm mismatch
id =

consider g m1 ≠ g m 2
id 1 g m1
Q vgs1 = vgs 2 ⇒
=

id 2 g m 2

μCoxW

(v gs − Vt ) 2

2L
μCoxW
gm =
(v gs − Vt )
L

vs
vs = (id 1 + id 2 ) Rss ⇒ (id 1 + id 2 ) =
Rss
vs ≈ vicm ⇒ (id 1 + id 2 ) =
id 1 =

g m1vicm
( g m1 + g m 2 ) Rss

id 2 =

g m 2vicm
( g m1 + g m 2 ) Rss

Figure 7.11 Analysis of the MOS differential amplifier to determine the
common-mode gain resulting from a mismatch in the gm values of Q1 and
Q2.


vicm
Rss

let Δg m = g m1 − g m 2
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⇒ id 1 =
⇒ id 2

g m1vicm
2 g m Rss

g m 2 vicm
=
2 g m Rss

Δg m RD vicm
vo 2 − vo1 = −id 2 RD + id 1 RD =
2 g m Rss
Acm =

RD Δg m
Rss 2 g m


Ad = − g m RD
CMRR ≡

Ad
Δg
= 2 g m Rss / m
Acm
gm

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Input offset voltage

1. RD1 ≠ RD 2
W1
W2
2. Q1 ≠ Q2
≠
L1
L2
3. Q1 ≠ Q2 Vt1 ≠ Vt 2

Figure 7.25 (a) The MOS differential pair with both inputs grounded. Owing to device and
resistor mismatches, a finite dc output voltage VO results. (b) Application of a voltage equal to the

input offset voltage VOS to the terminals with opposite polarity reduces VO to zero.

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consider

RD1 ≠ RD 2

ΔR
RD1 = RD + D
2
ΔR
RD 2 = RD − D
2
I
ΔR
VD1 = VDD − ( RD + D )
2
2
I
ΔR
VD 2 = VDD − ( RD − D )
2
2

I
VO = VD 2 − VD1 = ΔRD
2
I
ΔRD
VO
VO
Vos =
=
=2
Ad g m RD g m RD
μCoxW

consider

VOV ΔRD
)(
)
RD
2
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W
W
W
1
+ Δ( )
)1 =
L
L

L
2
W
W
W
1
− Δ( )
( )2 =
L
L
L
2
I
I Δ (W / L )
I1 = +
2 2 (W / L )
I
I Δ (W / L )
I2 = −
2 2 (W / L )
I Δ (W / L )
ΔI =
2 2 (W / L )
V
Δ (W / L )
V os = ( OV )(
)
2
(W / L )
(


VoV

2
V
V
(
)
−
(VGS − Vt ) ΔRD
ΔRD
GS
t
2L
)(
)
= μCoxW
=(
R
R
2
(VGS − Vt ) D
D
L

Vos = (

Q1 ≠ Q 2

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consider Vt1 ≠ Vt 2
ΔVt
2
ΔVt
Vt 2 = Vt −
2
ΔV
ΔVt
1 W
1 W
(VGS − Vt − t ) 2 = k n'
(VGS − Vt ) 2 [1 −
]2
I1 = k n'
2 L
2
2 L
2(VGS − Vt )
Vt1 = Vt +

I1 ≈

ΔVt
1 'W
)

k n (VGS − Vt ) 2 (1 −
2 L
VGS − Vt

ΔVt
1 'W
2
)
I12 ≈ k n (VGS − Vt ) (1 +
2 L
VGS − Vt
I
1 'W
2
k n (VGS − Vt ) =
2
2 L
I 2ΔVt
I 2ΔVt
ΔI =
=
2 VGS − Vt 2 VOV
Vos = ΔVt
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Vos = (

VOV ΔRD 2 VOV Δ (W / L) 2
) +(
) + (ΔVt ) 2

2 RD
2 (W / L)

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Differential amplifier with active load
Active load

1. Differential gain
2. Common-mode gain et CMRR

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Differential-mode equivalent circuit with active load
Vo
Vid
2

Q4

Q1
V
g m id
2

ro1

ro3 //

1
gm

− Vid
2

Q2

gmVgs4 ro 4

ro 2

gm (


− Vid
)
2

Q3

Ro = ro 2 // ro 4
active load

Ro = ro = RSS
Passive load

active load

vo = − g m ( − v2id + v gs 4 )(ro 2 // ro 4 )
v gs 4 = − g

vid
m 2

1
(ro1 // ro 3 // ) ≈ − g m
gm

vid
2

1
= − v2id

gm

vo − g m ( −v2id + −v2id )(ro 2 // ro 4 )
Ad ≡
=
= g m (ro 2 // ro 4 )
vid
vid
when ro 2 = ro 4 = ro
Ad =

gm
ro
2

Passive load

Ad = g m ( RD // ro )
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Common-mode equivalent circuit with active load

id


ViCM

id

Vo
i1

1
gm

2 R ss

ro 1

ro 3 //

1
gm

gmVgs4

ro 4

id
ro 2

i1

vo = −ro 4 ( g m v gs 4 + i2 )
v gs 4


i2

V gs 4

1
= −i1 (ro 4 // )
gm

vicm ≈ i1 2 RSS
vicm ≈ i2 2 RSS

i2
v
vo = −ro 4 ( g m v gs 4 + icm )
2 RSS
v gs 4

ViCM
1
gm

2 R ss
⇓

(Rss ≈ ro )

vicm
1
(ro 3 // )

=−
gm
2 RSS

vo = −ro 4 [−
Acm =

g m vicm
v
1
(ro 3 // ) + icm ]
gm
2 RSS
2 RSS

vo
r
1
= − o4
vicm
2 RSS 1 + ro 3 g m
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