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
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
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