![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](https://image.slidesharecdn.com/mos-170622151906/85/DIFFERENTIAL-AMPLIFIER-using-MOSFET-6-320.jpg)
DIFFERENTIAL AMPLIFIER using MOSFET
- 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)
- 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