The document discusses the design and operation of a MOSFET differential amplifier, detailing its input and output signals, regions of operation, and common-mode rejection capabilities. It explains the equations governing drain current, overdrive voltage, and the implications of common-mode and differential modes of operation. Additionally, it highlights the advantages and disadvantages of using differential amplifiers in various applications, including analog systems and audio amplifiers.

1. DIFFERENTIALAMPLIFIER
using MOSFET
DONEBY,
NITHYAPRIYA
PRASHANNA
S.PRAVEENKUMAR
PREETHI
SATHISHKUMAR
SHAGARI

2. Differential amplifier
• Amplifies the difference between the input signals
INPUTS:
Differential input:
Vid = Vi1-Vi2
Common input:
Vic=( Vi1+Vi2)/2
OUTPUTS:
Differential output:
Vod = Vo1-Vo2
Common output:
Voc=( Vo1+Vo2)/2

3. Why differential amplifiers are
popular?
• Less sensitive to noise(CMRR>>1)
• Biasing:
1) Relatively easy direct coupling of stages
2) Biasing resistor doesnot affect the
differential gain(no need for bypass capacitor)

4. MOS differential amplifier

5. Modes of operation

6. Regions of operation
• Cut off region- VGS ≤ Vt
• Active region- VDS ≤ VOV
• Saturation region- VDS ≥ VOV
TO DERIVE DRAIN CURRENT EQUATION
|Q|/unit channel length = Cox W VOV
Drift velocity= μn|E|= μn (VDS/L)
The drain current is the product of charge per unit length and drift velocity
ID=[( μnCox)(W/L) VOV] VDS
ID=[( μnCox)(W/L) VGS-Vt] VDS
ID=[kn’(W/L) VGS-Vt] VDS
Replacing VOV by (VOV-(1/2) VDS)
ID=kn’(W/L) (VOV-(1/2) VDS) VDS
At saturation mode, VDS ≥ VOV,
ID=(1/2) kn’(W/L)V2
OV

7. The MOS differential pair with a
common-mode input voltage vCM

8. Operation with common mode input
• The two gate terminals are connected to a
voltage VCM called common mode voltage.
• So VG1 = VG2 =VCM
• The drain currents are,
• Voltage at sources, will be
2
21
21
I
II
QQ
DD 

GSCMs Vvv 

9. • Neglecting the channel length modulation and
using the relation between VGS and ID is,(at
saturation)
• Where,
W=width of the channel
L=length of the channel
VGS =gain to source voltage
Vt =threshold voltage
Kn
’ = µn Cox
2'
)(
2
1
tGSnD VV
L
W
kI 

10. Overdrive Voltage
• Substituting for ID we get,
• The equation can be expressed in terms of
overdrive voltage as, VOV = VGS -VT .
• overdrive voltage is defined as VGS-VT when Q1
and Q2 each carry a current of I/2.
• Thus In terms of VOV ,
2'
)(
2
1
2
tGSnD VV
L
W
k
I
I 
2'
)(
2
1
2
OVn V
L
W
k
I

 
L
W
k
I
VVV
n
tGSOV
'


11. Common mode rejection
• Voltage at each drain will be,
• Since the operation is common mode the voltage
difference betwee.n two drains is zero.
• As long as, Q1 and Q2 remains in saturation region
the current I will divide equally between them.
And the voltage at drain does not changes.
• Thus the differential pair does not responds to
common mode input signals.
DDDDDD RIVvv  21

12. Input common mode range
• The highest value of VCM ensures that Q1 and
Q2 remains in saturation region.
• The lowest value of VCM is determined by
presence of sufficient voltage VCS across
current source I for its proper operation.
• This is the range of VCM over which the
differential pair works properly.
DDDtCM R
I
VVv
2
(max) 
)((min) tGStCSssCM VVVVVv 

13. PROBLEM based on common mode:
For an NMOS differential pair with a common-mode voltage Vcm
applied as Shown in Fig.
Let Vdd=Vss=2.5V,Kn’(W/L)=3(mA/V2),Vt =0.7V,I=0.2mA,RD =5KΩ and
neglect channel length modulation.
a)Find Vov and VGS for each transistor.
b)For Vcm =0 find Vs,iD1,iD2,VD1 and VD2.
c)For Vcm =+1V.
d)For Vcm =-1V.
e)What is the highest value of Vcm for which Q1 and Q2 remain in
saturation?
f)If current source I requires a minimum voltage of 0.3v to operate
properly, what is the lowest value allowed for Vs and hence for Vcm ?

14. GIVEN:
VDD=VSS=2.5V, Kn’(W/L)=3(mA/V2) , Vt=0.7V , I=0.2mA,
RD=5KΩ

15. SOLUTION:
VOV=
=
=0.26V
1) VS1= VS2= Vcm - VGS
=0-0.96=-0.96V
2) ID1=ID2=I/2=0.1mA
3) VD1=VD2 =VDD -(I/2)*RD
=+2.5-(0.1*2.5)=2.25V
c) If Vcm =+1
1)VS1= VS2= Vcm - VGS =1-0.96
=0.04V
2) ID1=ID2=I/2=0.1mA
3) VD1=VD2 =VDD -(I/2)*RD
=+2.5-(0.1*2.5)=2.25V.

16. Contd…
d) If Vcm =-1V
1)VS1= VS2= Vcm - VGS =-1-0.96
=-1.96V
2) ID1=ID2=I/2=0.1mA
3) VD1=VD2 =VDD -(I/2)*RD
=+2.5-(0.1*2.5)=2.25V.
e)VCMAX = Vt +VDD -(I/2)*RD
= 0.7+2.5-(0.1*2.5)=+2.95V.
f)VCMIN = -VSS + VCS +Vt +VOV
=-2.5+0.3+0.7+0.26 = -1.24V
VSMIN = VCMIN -VGS
= -1.24 - 0.96 = -2.2V.

17. Differential Amplifier – Common
Mode
Because of the symmetry, the common-mode circuit breaks into two
identical “half-circuits” .

18. Differential Amplifier – Differential
Mode
Because of the symmetry, the differential-mode circuit also breaks into two
identical half-circuits.

19. OPERATION OF MOS DIFFERENTIAL
AMPLIFIER IN DIFFERENCE MODE
Vid is applied to gate of Q1 and gate of Q2 is grounded.
Applying KVL,
Vid = VGS1 - VGS2
we know that,
Vd1 = Vdd - id1RD
Vd2 = Vdd - id2RD
case(i)
Vid is positive
VGS1 > VGS2
id1 > id2
Vd1 < Vd2
Hence, Vd2 - Vd1 is positive.

20. case(ii)
Vid is negative
VGS1 <VGS2
id1 < id2
Vd1 > Vd2
Hence, Vd2 - Vd1 is negative
Differential pair responds to difference mode or
differential input signals by providing a
corresponding differential output signal between
the two drains.

21. If the full bias current flows through the Q1 , VG2 is reduced to Vt
, at which point VS = - Vt , id1 = I.
I =
1
2
kn’ (
𝑊
𝐿
)(𝑉𝐺𝑆1 − 𝑉𝑡)2
by simplyfication, VGS1 = Vt+ 2𝐼/kn’(
𝑊
𝐿
)
But, VOV = 𝐼/kn’(
𝑊
𝐿
)
hence, VGS1 = Vt+ 𝟐 VOV
Where, kn’-process transconductance parameter which is the product of electron
mobility( µ 𝑛) and oxide capacitance (𝐶 𝑜𝑥).
where VOV is the overdrive voltage corresponds to the drain current of I/2.

22. Thus the value of Vid at which the entire bias current I is
steered into Q1 is,
Vidmax = VGS1 +VS
= Vt+ 2 VOV - Vt
Vidmax = 2 VOV
(i) Vid > 2 VOV
id1 remains equal to I
VGS1 remains Vt+ 2 VOV
VS rises correspondingly(thus keeping Q2 off)
(ii)Vid ≥ - 2 VOV
Q1 turns off, Q2 conducts the entire bias current I. Thus
the current I can be steered from one transistor to other by
varying Vid in the range,
- 2 VOV ≤ Vid ≤ 2 VOV
Which is the range of different mode operation.

23. Advantages
• Manipulating differential signals
• High input impedance
• Not sensitive to temperature
• Fabrication is easier
• Provides immunity to external noise
• A 6 db increase in dynamic range which is a
clear advantage for low voltage systems
• Reduces second order harmonics

24. Disadvantages
• Lower gain
• Complexity
• Need for negative voltage source for proper
bias

25. Applications
• Analog systems
• DC amplifiers
• Audio amplifiers
- speakers and microphone circuits in
cellphones
• Servocontrol systems
• Analog computers