07_OPAMP2

07_OPAMP2 - EE6326 Analog IC Design Fall 2008 Topic 6....

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1 1 WLCHEN EE6326 Analog IC Design – Fall 2008 Topic 6. Operational Amplifier (OPAMP) - 2 Wenliang Chen, Ph. D. Instructor
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2 2 WLCHEN Outline • 2-stage/7-transistor OP AMPs Without Compensation Miller Compensation Design of a Two-Stage Amplifier Step-by-step procedure
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3 3 WLCHEN Review of Bode Plots (1) Transfer Functions: A(s) = s A(s) = 1/s A(s) = 1/(1+s/p) (LHP pole @ p) |A(s)| 20dB/decade 0dB s 90 o 0 ph(A(s))
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4 4 WLCHEN Review of Bode Plots (2) A(s) = 1+s/z (LHP zero @ z) Both magnitude and phase increase A(s) = 1-s/z (RHP zero @ z) Magnitude increases but phase decreases!
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5 5 WLCHEN Phase Margin (PM) & Gain Margin (GM) fu |A0|(dB) |A(f)| |A0|-3dB fp1 fp2 GM Ang(A(f)) O dB 0 deg -90 deg -180 deg PM
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6 6 WLCHEN Differential Pair with Active Current Mirror In contrast to previous fully differential configuration, this topology contains two signal paths with different transfer functions. Mirror Pole at X = gm3/Cx Cx includes C GS3 , C GS4 ,C BD3 ,C BD1 , Miller effect of C GD1 and C GD4 Output Pole = 1 o4 o2 )C r || (r 1
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7 7 WLCHEN Differential Pair with Active Current Mirror Mirror Pole at X: Output Pole: 1 o4 o2 p1 )C r || (r 1 ω = Zero: The circuit consists of a “slow path” (M1, M3, and M4) and a “fast path” (M2) ) S/ ω )(1 S/ ω (1 ) S/ ω (2 A 1) S/ ω 1 1 ( S/ ω 1 A Vin Vout ) S/ ω )(1 S/ ω (1 A S/ ω 1 A Vin Vout p2 p1 p2 0 p2 p1 0 p2 p1 0 p1 0 + + + = + + + = + + + + = X m3 p2 C g ω = Dominant Pole Zero: X m3 p2 z C 2g 2 ω ω = =
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8 8 WLCHEN Input Common Mode Range for different types of differential pairs (Important basic concept) 1. Tie both inputs together. 2. Sweep the input voltage and find the input voltage range in which all transistors are saturated. 3. Device has high gain & output impedance in saturation region th1 OD3 th3 DD in,cm th1 od1 od0 V ) V (V V V V V V + + < < + + th1 OD3 th3 DD in,cm th1 od1 od0 V ) V (V V V V V V + + < < + + Vout Vb M1 M2 Vout2 Vin,cm Vb M1 M2 Vout1 M3 M4 M3 M4 Vin,cm M0 Vout2 Vb M1 M2 Vout1 M3 M4 Vin,cm M0 M0 th1 od3 DD in,cm th1 od1 od0 V V V V V V V + < < + + VDD VSS VDD VSS VDD VSS
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9 9 WLCHEN Two-Stage Amplifier Without Compensation Fig. 2 is the signal representation of Fig. 1 –g m1 (fig. 2) = g m1,2 (fig. 1) mL (fig. 2) = g mL (fig. 1) –r 1 , C 1 (fig. 2) = equivalent output resistance (r o2 //r o4 ), capacitance (C dtot,M4 +C dtot,M2 +C gs,ML ) at V 1 (fig. 1) L , C L (fig. 2) = output resistance (r oL //r ob3 ), capacitance (C L ) at V o (fig. 1) X m3 p2 z C 2g 2 ω ω = = Fig. 1 Fig. 2 1 o4 o2 p1 )C r || (r 1 ω = X m3 p2 C g ω = L ob3 oL p3 )C r || (r 1 ω = (high freq) (high freq)
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10 10 WLCHEN Two-Stage Amplifier Without Compensation A(f) arg A(f) ω p1 ω p3 ω u ω p2 ω z Phase margin is significantly less than 45deg. This amplifier must be compensated!
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11 11 WLCHEN Miller Effect Miller’s Theorem: If the circuit of (a) can be converted to (b), then and , where Av=Vy/Vx. ) A Z/(1 Z v 1 = ) A Z/(1 Z 1 v 2 =
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12 12 WLCHEN Proof The current flowing through Z from X to Y is equal to (Vx-Vy)/Z. For the two circuits to be equivalent, the same current must flow through Z 1 , thus, similarly, 1 X Y X Z V Z V V = V X Y 1 A 1 Z V V 1 Z Z = = 1 V Y X 2 A 1 Z V V 1 Z Z = =
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13 13 WLCHEN Example ) A (1 SC 1 A 1 SC 1 Z A) (1 SC 1 A 1 SC 1 Z SC 1 Z 1 - F 1 - F 2 F F
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