DIFFERENTIAL AMPLIFIER using MOSFET

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 = Vi1-Vi2

Common input:Vic=( Vi1+Vi2)/2

OUTPUTS:Differential output:Vod = Vo1-Vo2

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 VOV

Drift velocity= μn|E|= μn (VDS/L)

The drain current is the product of charge per unit length and drift velocityID=[( 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)V2OV

The MOS differential pair with a

common-mode 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

’ = µn 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 VGS-VT 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 common-mode voltage Vcmapplied 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 ?

GIVEN:

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

SOLUTION:

VOV= ==0.26V

1) VS1= VS2= Vcm - VGS

=0-0.96=-0.96V2) ID1=ID2=I/2=0.1mA3) VD1=VD2 =VDD -(I/2)*RD

=+2.5-(0.1*2.5)=2.25Vc) If Vcm =+1

1)VS1= VS2= Vcm - VGS =1-0.96=0.04V

2) 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 =-1-0.96

=-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 = -1.24VVSMIN = VCMIN -VGS

= -1.24 - 0.96 = -2.2V.

Differential Amplifier – Common Mode

Because of the symmetry, the common-mode circuit breaks into two identical “half-circuits” .

Differential Amplifier – Differential Mode

Because of the symmetry, the differential-mode circuit also breaks into two

identical half-circuits.

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 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’ (

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( µ𝑛) and oxide capacitance (𝐶𝑜𝑥).

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