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.

2 Differential Signal: - Measured between 2 nodes that have equal and opposite excursions around a fixed potential. - 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.

3 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. 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.

4 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). H. Aboushady University of Paris VI.

5 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.

6 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. 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.

7 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. 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.

8 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.

9 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.

10 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. 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.