DIFFERENTIAL AMPLIFIER using MOSFET, Modes of operation, The MOS differential pair with a common-mode input voltage ,Common mode rejection,gain, advantages and disadvantages.

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