DIFFERENTIAL AMPLIFIER using MOSFET

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

2. Differential amplifier Amplifies the difference between the input signals INPUTS: Differential input: V id = V i1 -V i2 Common input: V ic =( V i1 +V i2 )/2 OUTPUTS: Differential output: V od = V o1 -V o2 Common output: V oc =( V o1 +V o2 )/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- V GS ≤ V t Active region- V DS ≤ V OV Saturation region- V DS ≥ V OV TO DERIVE DRAIN CURRENT EQUATION |Q|/unit channel length = C ox W V OV Drift velocity= μ n |E |= μ n (V DS /L) The drain current is the product of charge per unit length and drift velocity I D =[( μ n C ox )(W/L) V OV ] V DS I D =[( μ n C ox )(W/L) V GS - V t ] V DS I D =[ k n ’ (W/L) V GS - V t ] V DS Replacing V OV by (V OV -(1/2) V DS ) I D = k n ’ (W/L) (V OV -(1/2) V DS ) V DS At saturation mode, V DS ≥ V OV, I D =(1/2) k n ’ (W/L)V 2 OV    

7. The MOS differential pair with a common-mode input voltage v CM

8. Operation with common mode input T he two gate terminals are connected to a voltage V CM called common mode voltage. So V G1 = V G2 = V CM The drain currents are, Voltage at sources, will be

9. Neglecting the channel length modulation and using the relation between V GS and I D is,(at saturation) Where, W=width of the channel L=length of the channel V GS =gain to source voltage V t =threshold voltage K n ’ = µ n C ox

10. Overdrive Voltage Substituting for I D we get, The equation can be expressed in terms of overdrive voltage as, V OV = V GS -V T . overdrive voltage is defined as V GS - V T when Q 1 and Q 2 each carry a current of I /2. Thus In terms of V OV ,

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, Q 1 and Q 2 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.

12. Input common mode range The highest value of V CM ensures that Q 1 and Q 2 remains in saturation region. The lowest value of V CM is determined by presence of sufficient voltage V CS across current source I for its proper operation. This is the range of V CM over which the differential pair works properly.

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 /V 2 ), V t =0.7V,I=0.2mA,R D =5KΩ and neglect channel length modulation. a)Find V ov and V GS for each transistor. b)For V cm =0 find Vs,i D1, i D2, V D1 and V D2. c)For V cm =+1V. d)For V cm =-1V. e)What is the highest value of V cm for which Q 1 and Q 2 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 V s and hence for V cm ?

14. GIVEN: V DD =V SS =2.5V, Kn ’(W/L)=3( mA /V 2 ) , V t =0.7V , I=0.2mA, R D =5K Ω

15. SOLUTION: V OV = = =0.26V 1) V S1 = V S2 = V cm - V GS =0-0.96=-0.96V 2) I D1 =I D2 =I/2=0.1mA 3) V D1 =V D2 =V DD -(I/2)*R D =+2.5-(0.1*2.5)=2.25V c) If V cm =+1 1)V S1 = V S2 = V cm - V GS =1-0.96 =0.04V 2) I D1 =I D2 =I/2=0.1mA 3) V D1 =V D2 =V DD -(I/2)*R D =+2.5-(0.1*2.5)=2.25V.

16. Contd … d) If V cm =-1V 1)V S1 = V S2 = V cm - V GS =-1-0.96 =-1.96V 2) I D1 =I D2 =I/2=0.1mA 3) V D1 =V D2 =V DD -(I/2)*R D =+2.5-(0.1*2.5)=2.25V. e)V CMAX = V t +V DD -(I/2)*R D = 0.7+2.5-(0.1*2.5)=+2.95V. f)V CMIN = -V SS + V CS + V t +V OV =-2.5+0.3+0.7+0.26 = -1.24V V SMIN = V CMIN -V GS = -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 V id is applied to gate of Q 1 and gate of Q 2 is grounded. Applying KVL, V id = V GS1 - V GS2 we know that, V d1 = V dd - i d1 R D V d2 = V dd - i d2 R D case(i) V id is positive V GS1 > V GS2 i d1 > i d2 V d1 < V d2 Hence, V d2 - V d1 is positive.

20. case(ii) V id is negative V GS1 < V GS2 i d1 < i d2 V d1 > V d2 Hence, V d2 - V d1 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 Q 1 , V G2 is reduced to V t , at which point V S = - V t , i d1 = I. I = k n ’ ( by simplyfication , V GS1 = V t + But , V OV = hence, V GS1 = V t + V OV   Where, k n ’-process transconductance parameter which is the product of electron mobility( and oxide capacitance ( ). where V OV is the overdrive voltage corresponds to the drain current of I/2.  

22. Thus the value of V id at which the entire bias current I is steered into Q 1 is, V idmax = V GS1 +V S = V t + V OV - V t V idmax = V OV (i) V id > V OV i d1 remains equal to I V GS1 remains V t + V OV V S rises correspondingly( thus keeping Q 2 off) (ii) V id ≥ - V OV Q 1 turns off, Q 2 conducts the entire bias current I. Thus the current I can be steered from one transistor to other by varying V id in the range , - V OV ≤ V id ≤ V OV 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