Unit 3
Differential & Multistage Amplifiers
Second Year Electrical Eng.
Semester: II
Kuenzang Thinley
Electronics & Communication Engineering
Department (ECED)
College of Science and Technology
kuenzangthinley.cst@rub.edu.bt
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Unit_3: Differential Amplifier
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Introduction
Unit Objectives:
1. Sigle ended and Differential signalling MOSFET
2. Difference between normal amplifier and differential
amplifier
3. Differential Amplifiers: Terminology and qualitative
description
4. DC transfer characteristic, differential gain, common-mode
gain, and CMRR.
5. DC and AC analysis of differential amplifiers
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Unit_3: Differential Amplifier
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Differential Amplifier
• Greatly affected by
noise and interference
Single-ended amplifier
The
Differential
Amplifier
amplifies
the
difference
between two input voltage
signal. Hence it is also called
as difference amplifier.
Differential Circuit
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• Greatly reduces the noise and
interference
• Removes the need of bypass
and coupling capacitors
Unit_3: Differential Amplifier
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Differential Amplifier Terminology
1. Differential Input Signal/Voltage
• There exits either amplitude or phasedifference between the inputs.
Vd = Vin1 − Vin 2
2. Differential Mode Gain (Ad)
• The gain with which diffamp
amplifies the differential input
signals (Vd)
VOut1 − VOut 2
Ad =
Vin1 − Vin 2
Ad → (Ideally), Ad → High (Practically)
3. Common Mode Input Voltage (Vcm)
• The inputs (Vin1 & Vin2) applied to
the diffamp are of same magnitude
and phase.
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Unit_3: Differential Amplifier
Vin1 + Vin 2
Vcm =
2
4
Differential Amplifier Terminology
4. Common Mode Gain (Acm)
• The gain with which differential
amplifier amplifies the commonmode signal
VOut
Vd =
VCm
Acm → 0(Ideally), Acm → very low (Practically)
5. Common Mode Rejection Ratio (CMMR)
• The ability of differential amplifier to
amplify the differential mode signal
and reject common-mode signal.
(Ratio of differential mode gain to
the common-mode gain)
Ad
CMRR =
Acm
Ad
CMRRdB = 20log
Acm
CMRR → (Ideally), CMRR → very High (Practically)
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Unit_3: Differential Amplifier
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Differential Pair
Differential amplifier
• The MOSFETs operate in
the saturation region at all
times.
Tail Current Source
• The BJTs operate in the
active region at all times.
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Unit_3: Differential Amplifier
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Differential Amplifier Types
MOSFET Based
1. Dual input balanced output differential amplifier (DIBO)
2. Dual input unbalanced O/P differential amplifier (DIUO)
3. Single input balanced o/p differential amplifier (SIBO)
4. Single input unbalanced o/p differential amplifier (SIUO)
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Unit_3: Differential Amplifier
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Differential Amplifier Types
1. Dual input balanced output
differential amplifier (DIBO)
VOut = Vout1 − Vout 2
Tail Current Source
2. Dual input unbalanced O/P differential
amplifier (DIUO)
VOut = Vout 2
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Unit_3: Differential Amplifier
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Differential Amplifier Types
3. Single input balanced o/p
differential amplifier (SIBO)
VOut = Vout1 − Vout 2
Tail Current Source
4. Single input unbalanced o/p differential
amplifier (SIUO)
VOut = Vout 2
Tail Current Source
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Unit_3: Differential Amplifier
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Differential Pair with Common-Mode Input Voltage
At saturation
1 'W
2
I D = kn (VGS − Vth )
2 L
I 1 'W
2

= kn (VGS − Vth )
2 2 L
The equilibrium overdrive voltage
VOV = VGS − Vth = I kn' (W L )
Applying KCL at “P”
Voltage at drain (Output Voltage)
I
VD1 = VD 2 = VDD − RD
2
I D1 + I D 2 = I
The circuit follows symmetry
I
I D1 = I D 2 =
2
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Voltage at source
VS = VCM − VGS
Unit_3: Differential Amplifier
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Differential Pair with Common-Mode Input Voltage
VinCM  VTh
1
VinCM = 0 → M 1 & M 2 is OFF I D1D 2 = 0
I D 3  0 (quit small )
M 3 → Operates in deep triode region
i.e VDS 3 becomes low, (VDS 3 = VP )
VOUT 1 = VOUT 2 = VDD
No amplification of signal
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Unit_3: Differential Amplifier
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Differential Pair with Common-Mode Input Voltage
2
VinCM  VTh
VinCM  0 → M 1 & M 2 is ON I D1D 2  0
VGS 1 & VGS 2 , VP 
VGS 1 = VGS 2 = VinCM
For the high value of VinCM
i.e VDS 3 >(VGS 3 − VTh )
M 3 Enters saturation region
I D1 + I D 2 = I = I D 3
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Unit_3: Differential Amplifier
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Differential Pair with Common-Mode Input Voltage
Applying KVL at input
VinCM − VGS 1 − VDS 3 = 0
VinCM = VGS 1 + VDS 3
For M3 to be in saturation
VinCM = VGS 1 + (VGS 3 − VTh 3 )
For proper operation
VinCM ( min )  VGS 1 + (VGS 3 − VTh 3 )
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Unit_3: Differential Amplifier
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Differential Pair with Common-Mode Input Voltage
3
VinCM  Further M 1 & M 2 → triode region
VDS  (VGS − VTh )
VinCM  VOUT 1 + VTh
I
VOUT 1 = VDD − I D1 RD = VDD − RD
2
I
VinCM (max)  VDD − RD + VTh
2
If there are any input perturbation the
differential pair does not respond (reject) to
the common-mode input signal.
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Unit_3: Differential Amplifier
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Differential Pair with Common-Mode Input Voltage
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Unit_3: Differential Amplifier
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Differential Pair with Differential Input Voltage
Input voltages are different
(Magnitude and Phase )
Vin1  Vin 2
Differential pair with Differential
Input
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Unit_3: Differential Amplifier
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Differential Pair with Differential Input Voltage
Input voltages are different
(Magnitude and Phase )
Vin1  V ,Vin 2 = 0
Vin1 − Vin 2
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+ve
1
Vin1  VTh → M 1 is ON I D1 = I
2
Vin 2  VTh → M 2 is OFF I D 2 = 0
 Vout1 = VDD − I D1 RD , Vout 2 = VDD
Unit_3: Differential Amplifier
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Differential Pair with Differential Input Voltage
Input voltages are different
(Magnitude and Phase )
Vin1 = 0,Vin 2  0
Vin1 − Vin 2
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−ve
3
Vin 2  VTh → M 2 is ON I D 2 = I
4
Vin1  VTh → M 1 is OFF I D1 = 0
 Vout1 = VDD − I D1 RD , Vout1 = VDD
Unit_3: Differential Amplifier
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Differential Pair with Differential Input Voltage
Input voltages are different
(Magnitude and Phase )
Vin1  VTh1 & Vin 2  VTh 2 & Vin1  Vin 2
I SS
5 M 1 &M 2 both ON I D1 = I D 2 =
2
I SS
6  Vout1 = Vout 2 = VDD − RD
2
If Vin1 = Vin2, the differential pair is in equilibrium
The equilibrium overdrive voltage :
I SS
VOV = VGS − Vth =
kn' (W L )
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Unit_3: Differential Amplifier
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Differential Pair with Differential Input Voltage
Differential input-output characteristic of differential pair
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Unit_3: Differential Amplifier
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Differential Pair with Differential Input Voltage
Note:
1. The maximum (VDD) and minimum (VDD - RDISS) output levels are
independent of the input CM level VinCM.
2. The difference between the input signal should be minimum
(Linear)
3. Circuit become non-linear as the I/P swing increases
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Unit_3: Differential Amplifier
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Large Signal Operation of Differential Pair
Large signal operation refers to the condition where the input signal is
sufficiently large enough to cause significant variations in the operating
point (bias conditions) of the amplifier's active devices.
• Objective: Derive equations for 𝐼𝐷1 , 𝐼𝐷2 , 𝑉𝑋 , 𝑉𝑌 , 𝑉𝑋 − 𝑉𝑌 as a function of
𝑉𝐼𝑁1 − 𝑉𝐼𝑁2
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Unit_3: Differential Amplifier
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Large Signal Operation of Differential Pair
1. Derive equation for 𝐼𝐷1 − 𝐼𝐷2 as a function of 𝑉𝐼𝑁1 − 𝑉𝐼𝑁2
Assuming both the transistor are ON and Vin1 is larger than Vin2
Vin1 − Vin 2 = VGS 1 − VGS 2
VX − VY = − RD ( I D1 − I D 2 )
I D1 + I D 2 = I SS
1
2
I D = n cox (W L) (VGS − VTh )
2
2I D
 VGS = VTh +
nCox (W L )
Substituting in equation 1
Vin1 − Vin 2 =
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2
nCox (W L )
( I − I )
D1
Unit_3: Differential Amplifier
D2
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Large Signal Operation of Differential Pair
Vin1 − Vin 2 =
( I − I )
2
nCox (W L )
Squaring both side
D1
D2
(
2
I D1 + I D 2 − 2 I D1 I D 2
(Vin1 − Vin 2 ) =
nCox (W L )
2
=
(
2
I SS − 2 I D1 I D 2
nCox (W L )
)
)
 4 I D1 I D 2 = 2 I SS − nCox (W L )(Vin1 − Vin 2 )
2
Squaring both side
 16 I D1 I D 2 =  2 I SS − nCox (W L )(Vin1 − Vin 2 ) 


2
2
 16 I D1 ( I SS − I D1 ) =  2 I SS − nCox (W L )(Vin1 − Vin 2 ) 


2
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Unit_3: Differential Amplifier
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Large Signal Operation of Differential Pair
2
2

 16 I − 16 I D1 I SS + 2 I SS − nCox (W L )(Vin1 − Vin 2 )  = 0


2
D1
Solving this quadratic equation
2
I SS 1
2
2


 I D1 =

4 I SS − nCox (W L )(Vin1 − Vin 2 ) − 2 I SS


2 4
=
 ID2 =
I SS Vin1 − Vin 2
+
2
4
nCox (W L )  4 I SS − nCox (W L )(Vin1 − Vin 2 ) 
2


I SS Vin 2 − Vin1
2
+
nCox (W L )  4 I SS − nCox (W L )(Vin 2 − Vin1 ) 


2
4

4 I SS
1
2
I D1 − I D 2 = nCox (W L )(Vin1 − Vin 2 ) 
− (Vin1 − Vin 2 ) 
2
 nCox (W L )

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Unit_3: Differential Amplifier
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Large Signal Operation of Differential Pair

4 I SS
1
2
I D1 − I D 2 = nCox (W L )(Vin1 − Vin 2 ) 
− (Vin1 − Vin 2 ) 
2
 nCox (W L )

Observation:
1
2
I D1 − I D 2 = 0 if (Vin1 = Vin 2 )
The above equation is valid
only if M1 & M2 are ON or
at the edge of the turning
off.
2 I SS
Vin1 − Vin 2 
nCox (W L)
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Unit_3: Differential Amplifier
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Large Signal Operation of Differential Pair
Edge of
Conduction
I D 2  0,VGS 2 = VTh
VGS 1 = VTh + 2 I SS kn' (W L )
2 I SS
= 2VOV
(Vin1 − Vin 2 )max = '
kn (W L)
➢ There exists a finite differential input voltage that completely
steers the tail current from one transistor to the other. This
value is known as the maximum differential input voltage.
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Unit_3: Differential Amplifier
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Large Signal Operation of Differential Pair
The minimum and maximum value of 𝑉𝑖𝑛1 − 𝑉𝑖𝑛2 at which one
transistor turns ON and OFF
1 ' W 
2
I SS =  kn  (VGS 1 − VTh )
2 L 
VGS 1 = VTh + 2 I SS kn' (W L )
Vin1 − Vin 2 = VGS 1 − VGS 2
I D 2  0,VGS 2 = VTh
VOV = VGS − Vth = I kn' (W L )
2 I SS
Vind =
kn' (W L)
Vind (min) = 2VOV
− 2V0V  Vind  2V0V
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Unit_3: Differential Amplifier
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Large Signal Operation of Differential Pair
2. Derive equation for 𝑉𝑋 − 𝑉𝑌 as a function of 𝑉𝐼𝑁1 − 𝑉𝐼𝑁2
VX − VY = − RD ( I D1 − I D 2 )
I D1 − I D 2 =

4 I SS
1
2
nCox (W L )(Vin1 − Vin 2 ) 
− (Vin1 − Vin 2 ) 
2
 nCox (W L )


4 I SS
RD
2
VX − VY = −
nCox (W L )(Vin1 − Vin 2 ) 
− (Vin1 − Vin 2 ) 
2
 nCox (W L )

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Unit_3: Differential Amplifier
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Large Signal Operation of Differential Pair
2. Derive equation for 𝑉𝑋 − 𝑉𝑌 as a function of 𝑉𝐼𝑁1 − 𝑉𝐼𝑁2

4 I SS
RD
2
VX − VY = −
nCox (W L )(Vin1 − Vin 2 ) 
− (Vin1 − Vin 2 ) 
2
 nCox (W L )

If (Vin1 − Vin 2 )
VX − VY  −
2
4 I SS
nCox (W L )


4 I SS
RD
W 
nCox   (Vin1 − Vin 2 ) 

2

C
W
L
(
)
L
 n ox

VX − VY  − RD
W 
nCox   I SS (Vin1 − Vin 2 )
L
VX − VY
W 
slope

−
R

C
(
)
D
n ox 
 I SS
(Vin1 − Vin 2 )
L
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Unit_3: Differential Amplifier
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Thank You
Questions ??
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Unit_3: Differential Amplifier
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