DIFFERENTIAL AMPLIFIER using MOSFET

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DIFFERENTIAL AMPLIFIER
using MOSFET
DONE BY,
NITHYAPRIYA
PRASHANNA
S.PRAVEENKUMAR
PREETHI
SATHISH KUMAR
SHAGARI

Differential amplifier Amplifies the difference between the input signals
INPUTS:Differential input:
Vid = Vi1Vi2Common input:
Vic=( Vi1+Vi2)/2
OUTPUTS:Differential output:Vod = Vo1Vo2
Common output:Voc=( Vo1+Vo2)/2

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)

MOS differential amplifier

Modes of operation

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 VOVDrift velocity= nE= n (VDS/L)
The drain current is the product of charge per unit length and drift velocityID=[( nCox)(W/L) VOV] VDS
ID=[( nCox)(W/L) VGSVt] VDS
ID=[kn(W/L) VGSVt] 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)V
2OV

The MOS differential pair with a
commonmode input voltage vCM

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
221
21
III
QQ
DD
GSCMs Vvv

Neglecting the channel length modulation and using the relation between VGS and ID is,(at saturation)
Where,W=width of the channelL=length of the channelVGS =gain to source voltageVt =threshold voltageKn
= n Cox
2' )(2
1tGSnD VV
L
WkI

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 VGSVT when Q1 and Q2 each carry a current of I/2.
Thus In terms of VOV ,
2' )(2
1
2tGSnD VV
L
Wk
II
2' )(2
1
2OVn V
L
Wk
I
L
Wk
IVVV
n
tGSOV'

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

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 RI
VVv2
(max)
)((min) tGStCSssCM VVVVVv

PROBLEM based on common mode:
For an NMOS differential pair with a commonmode voltage Vcmapplied as Shown in Fig.
Let Vdd=Vss=2.5V,Kn(W/L)=3(mA/V2),Vt =0.7V,I=0.2mA,RD =5K 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 ?

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

SOLUTION:
VOV= ==0.26V
1) VS1= VS2= Vcm  VGS=00.96=0.96V
2) ID1=ID2=I/2=0.1mA3) VD1=VD2 =VDD (I/2)*RD=+2.5(0.1*2.5)=2.25V
c) If Vcm =+11)VS1= VS2= Vcm  VGS =10.96
=0.04V2) ID1=ID2=I/2=0.1mA3) VD1=VD2 =VDD (I/2)*RD
=+2.5(0.1*2.5)=2.25V.

Contd
d) If Vcm =1V1)VS1= VS2= Vcm  VGS =10.96
=1.96V2) 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.

Differential Amplifier Common Mode
Because of the symmetry, the commonmode circuit breaks into two identical halfcircuits .

Differential Amplifier Differential Mode
Because of the symmetry, the differentialmode circuit also breaks into two
identical halfcircuits.

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  VGS2we know that,
Vd1 = Vdd  id1RD Vd2 = Vdd  id2RD
case(i)Vid is positiveVGS1 > VGS2 id1 > id2 Vd1 < Vd2 Hence, Vd2  Vd1 is positive.

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.

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

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.

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

Disadvantages
Lower gain
Complexity
Need for negative voltage source for proper bias

Applications
Analog systems
DC amplifiers
Audio amplifiers
 speakers and microphone circuits in cellphones
Servocontrol systems
Analog computers