Single Ended and Differential Operation Basic Differential ...

1 Differential Amplifiers Single Ended and Differential Operation Basic Differential Pair Common-Mode Response Differential Pair with MOS loads Hassan Aboushady University of Paris VI. References B. Razavi, Design of Analog CMOS Integrated Circuits , McGraw-Hill, 2001. H. Aboushady University of Paris VI. 1. Differential Amplifiers Single Ended and Differential Operation Basic Differential Pair Common-Mode Response Differential Pair with MOS loads Hassan Aboushady University of Paris VI. Single Ended and Differential Operation Single Ended Signal: - Measured with respect to a fixed potential, usually ground. Differential Signal: - Measured between 2 nodes that have equal and opposite excursions around a fixed potential.

2 - The center potential is called Common Mode (CM). H. Aboushady University of Paris VI. 2. Rejection of Common Mode Noise Single Ended Signal: - Due to capacitive coupling, transitions on the clock line corrupt the signal on L1 . Differential Signal: - If the clock line is placed midway, the transitions disturb the Differential signals by equal amounts, leaving the difference intact. H. Aboushady University of Paris VI. Rejection of Power Supply Noise Maximum Output Swing: Maximum Output Swing: Vout max = VDD (VGS VTH ) VX max VY max = 2[VDD (VGS VTH )]. H. Aboushady University of Paris VI. 3. Differential Pair Differential circuit sensitive Differential pair minimal to the input CM level. dependence on input CM level.

3 If Vin1 Vin 2 I SS = I D1 + I D 2. I D1 I D 2. if Vin1 = Vin 2. g m1 g m 2 I SS. I D1 = I D 2 =. 2. I. Output CM = VDD RD SS. 2. H. Aboushady University of Paris VI. Differential Pair: Qualitative Analysis Vin1 << Vin 2 M1 OFF, M2 ON. I D 2 = I SS Vout1 = VDD. Vout 2 = VDD I SS RD. Vin1 = Vin 2 M1 ON, M2 ON. I I SS. I D1 = I D 2 = SS Vout1 = Vout 2 = VDD RD. 2 2. Vin1 >> Vin 2 M1 ON, M2 OFF. I D1 = I SS Vout1 = VDD I SS RD. Vout 2 = VDD. H. Aboushady University of Paris VI. 4. Differential Pair: Qualitative Analysis Maximum and minimum levels are well-defined and independent of the input CM: VDD and VDD - RD ISS. The small signal gain (the slope of Vout1-Vout2 vs Vin1-Vin2). is maximum for Vin1=Vin2 (equilibrium).

4 H. Aboushady University of Paris VI. Differential Pair: Common-Mode Behavior To study Common-Mode Vin1 = Vin 2 = Vin ,CM. For proper Operation : M3 in saturation Vin ,CM VGS 1 + (VGS 3 VTH 3 ). I. M1 & M2 in saturation Vin ,CM VDD RD SS + VTH. 2. H. Aboushady University of Paris VI. 5. Differential Pair: Output Voltage Swing For M1 & M2 in saturation: Vout VP Vin ,CM VP VTH. Vout Vin ,CM VTH. Output Voltage Swing: VDD Vout Vin ,CM VTH. To increase output swing, we choose a low Vin ,CM. H. Aboushady University of Paris VI. Differential Pair: Quantitative Analysis VP = Vin1 VGS1 = Vin 2 VGS 2. Vin1 Vin 2 = VGS1 VGS 2 1. Assuming M1 & M2 in saturation: 2I D. (VGS VTH ) 2 =. W. nCox L. From eq. 1 & 2 : 2I D. VGS = + VTH 2 I D1 2I D 2.

5 W 2 Vin1 Vin 2 = . nCox nCox W. nCox W. L. L L. Squaring the 2 sides, and since: I SS = I D1 + I D 2. 2. (Vin1 Vin 2 ) 2 = ( I SS 2 I D1 I D 2 ). W. nCox H. Aboushady L University of Paris VI. 6. Differential Pair: Quantitative Analysis Previous equation can be written: nCox W. (Vin1 Vin 2 ) 2 I SS = 2 I D1 I D 2. 2 L. Squaring the 2 sides, and since: 4 I D1 I D 2 = ( I D1 + I D 2 ) 2 ( I D1 I D 2 ) 2. 2. = I SS ( I D1 I D 2 ) 2. We arrive at: 2. 1 W W. ( I D1 I D 2 ) = nCox (Vin1 Vin 2 ) 4 + I SS nCox (Vin1 Vin 2 ) 2. 2. 4 L L. nCox W 4 I SS. I D1 I D 2 = (Vin1 Vin 2 ) (Vin1 Vin 2 ) 2. 2 L W 3. nCox L. H. Aboushady University of Paris VI. Differential Pair: Quantitative Analysis Let Vin = Vin1 Vin 2 and I D = I D1 I D 2.

6 Deriving eq. 3 with respect to Vin 4 I SS. 2 Vin2. I D nCox W nCoxW / L. Gm = =. Vin 2 L 4 I SS. Vin2 4. nCoxW / L. W. For Vin = 0 , Gm = nCox I SS. L. Since: Vout1 Vout 2 = VDD I D1 RD1 VDD I D 2 RD 2. Vout = I D RD Vout = Gm Vin RD. The small signal Vout W. Av = = Gm RD = n Cox I SS RD. Differential voltage gain: Vin L. H. Aboushady University of Paris VI. 7. Drain Currents and Overall Transconductance 2 I SS. Vin1 is when I D1 = I SS Vin1 = VGS1 VTH 1 Vin1 =. W. nCox nCox W 4 I SS L. I D1 I D 2 = (Vin1 Vin 2 ) (Vin1 Vin 2 ) 2. 2 L W 3. nCox L. 4 I SS. 2 Vin2. C W nCoxW / L. Gm = n ox 2 L 4 I SS. Vin2. nCoxW / L 4. H. Aboushady University of Paris VI. ID vs VD. Plot the input-output characteristics W. of a Differential pair as the device.

7 Width and the tail current vary: L. 2 I SS. Vin1 =. W. nCox L. I SS . H. Aboushady University of Paris VI. 8. Differential Pair: Small Signal Gain Vout W. Av = = Gm RD = n Cox I SS RD. Vin L. In equilibrium, we have I SS. I D1 = I D 2 =. 2. Av = g m RD Where gm is the transconductance of M1 & M2. H. Aboushady University of Paris VI. Calculating Small Signal Gain by Superposition Set Vin2 =0. M1 forms a common source stage with a degeneration resistance VX g R. Av = = m1 D. Vin1 1+ g m1 RS. Neglecting channel length modulation and body effect RS = 1 / g m 2. VX g m1 RD. = . Vin1 1 + g m1 / g m 2. VX RD. = . Vin1 1 1. + 5. g m1 g m 2. H. Aboushady University of Paris VI. 9. Calculating Small Signal Gain by Superposition Replacing M1 by its Th venin equivalent: VT = Vin1 RT = 1 / g m1 Rin 2 = 1 / g m 2.

8 VY RD. =. Vin1 1 1 6. +. g m1 g m 2. From eq. 5 & 6, we get: 2 RD. (V X VY ) due to V = V. in1 1 1 in1. +. H. Aboushady g m1 g m 2 University of Paris VI. Calculating Small Signal Gain by Superposition 2 RD. (V X VY ) due to V = V. in1 1 1 in1. +. g m1 g m 2. Since: g m1 = g m 2 = g m (V X VY ) due to V = g m RDVin1. in 1. Similarly we can say that: (V X VY ) due to V = g m RDVin 2. in 2. The small signal (V X VY ) total = g m RD. Differential voltage gain: Vin1 Vin 2. H. Aboushady University of Paris VI. 10. The concept of Half Circuit If a fully symmetric Differential pair senses Differential inputs then the concept of half circuit can be applied. A Differential change in the inputs Vin1 and Vin2 is absorbed by V1 and V2 leaving VP constant H.

9 Aboushady University of Paris VI. Application of The Half Circuit Concept Since VP experiences no change, node P can be considered ac ground and the circuit can be decomposed into two separate halves Two common source amplifiers: VX VY. = g m RD = g m RD. Vin1 Vin 2. VX VY. = g m RD. Vin1 Vin 2. H. Aboushady University of Paris VI. 11. The Half Circuit Concept : Example Taking into account the output resistance (channel length modulation). Two common source amplifiers: VX VY. = g m (RD // rO1 ) = g m (RD // rO 2 ). Vin1 Vin 2. VX VY. = g m (RD // rO ). Vin1 Vin 2. H. Aboushady University of Paris VI. Arbitrary Inputs to a Differential Pair Conversion of arbitrary inputs to Differential and common-mode components: Differential Common H.

10 Aboushady Mode University of Paris VI. 12. Arbitrary Inputs to a Differential Pair: Example Calculate VX and VY if Vin1 =Vin2 and =0. For Differential mode Operation : V X = g m (RD // rO1 ). (Vin1 Vin 2 ) For common mode Operation : 2 I D1 = I D 2 = I SS / 2. VY = g m (RD // rO 2 ). (Vin 2 Vin1 ) Assuming fully symmetric circuit and Ideal Current Source: 2 ID1 and ID2 independent of VCM ,in VX VY = g m (RD // rO ) (Vin1 Vin 2 ) VX and VY independent of VCM ,in The Differential pair circuit: Amplifies Vin1 Vin 2 , Eliminates the effect of VCM ,in H. Aboushady University of Paris VI. Common Mode Response: Non-Ideal Current Source Assuming fully symmetric circuit with finite output impedance current source, RSS : Equivalent circuit: Degenerated Common Source Vout RD / 2.