Studio 9 Review Operational Amplifier Stability Compensation Miller Effect Phase Margin Unity Gain Frequency Slew Rate Limiting Reading: Text sec 5.


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1 Studio 9 Review Operational Amplifier Stability Compensation Miller Effect Phase Margin Unity Gain Frequency Slew Rate Limiting Reading: Text sec 5.2 pp
2 Twostage opamp
3 Analysis Strategy Recognize subblocks Represent as cascade of simple stages
4 Total opamp model Input differential pair Common source stage
5 DC operating point MOSFET I D [µa] V eff M M M M M M M M
6 Small signal parameters MOSFET I D [µa] V eff g m [µa/v] r ds M M kΩ M M MΩ M kΩ M M M kΩ Note: l n = V 1 ; l p = V 1
7 Total opamp model: Low frequency gain Input differential pair Common source stage ( ) a v1 = g m1 r ds2 r ds4 a v1 = ( 208mA V) ( 800kW 1.43MW) a v1 =106 ( ) a v2 = g m2 r ds5 r ds8 a v2 = ( 285mA V) ( 400kW 715kW) a v2 = 73
8 Total opamp model with capacitances Gate of M5 Ê C g = (900mm)(10mm) 4.17E  4 F Ë m 2 C g = 3.74 pf ˆ Load: scope probe 10pF
9 Total opamp model with capacitances f p1 = f p1 = First stage pole 1 2p( r ds2 r ds4 )C g5 1 2p 800kW 1.43MW ( ) 3.74pF ( ) f p1 = 82kHz Second stage pole f p1 = f p1 = 1 2p( r ds5 r ds8 )C L 1 2p 400kW 715kW ( ) 10 pf ( ) f p1 = 61kHz
10 Open loop transfer function Product of individual stage transfer functions A( jw) = g m1 ( r ds2 r ds4 )g m5 ( r ds5 r ds8 ) 1 + jw2p( r ds2 r ds4 )C g5 ( )C L [ ][ 1 + jw2p r ds5 r ds8 ]
11 Twostage opamp: Simulation Schematic
12 DC Operating Point Simulation
13 Bode plot Magnitude, phase on log scales Pole: Root of denominator polynomial
14 Open loop Bode plot Product of terms : Sum on loglog plot
15 Open Loop Bode Plot Simulation
16 Stability example: Closed loop follower Negative feedback: Output connected to inverting input
17 Unity gain: Why bother? v out = Ê Á Ë R L R L + R S ˆ v in v out = v in No buffer: Voltage divider Signal reduced due to voltage drop across R S With buffer: No current required from source
18 Lab 9 Problem: Instability Oscillation superimposed on desired output Output for zero input Why? Need...
19 Controls: ES3011 in 20 minutes General framework A: Forward Gain b: Feedback Factor fraction of output fed back to input
20 Example: Opamp, Noninverting Gain A: Forward Gain Opamp open loop gain V out =A(V + V  ) Transfer function A(jw) b: Feedback Factor b = R 1 R 1 + R 2
21 Output v out = A v in bv out 3 Solve for v out /v in Closed Loop Gain ( ) v +  v  v out = Av in  Abv out ( 1 + Ab)v out = Av in v out v in = A 1 + Ab
22 Opamp with negative feedback If Ab >> 1 v out A = v in 1 + Ab ª A Ab Æ v out ª 1 v in b Closed loop gain determined only by b Advantage of negative feedback: Open loop gain A can be ugly (nonlinear, poorly controlled) as long as it's large!
23 Example: Opamp, Noninverting Gain b: Feedback Factor R 1 b = R 1 + R 2 Closed loop gain v out v in = R 1 + R 2 R 1 = 1 b
24 Reexamine closed loop transfer function Output with no input: infinite gain Infinite when 1+Ab = 0 Condition for oscillation: 1+Ab = 0 In general A, b functions of w v out v in = A 1 + Ab If there's a frequency w at which 1+Ab = 0: Oscillation at that frequency!
25 b = 1 Æ v out v in = A 1 + A Use A(jw), solve for 1+A = 0 No thanks! Example: follower A( jw) = g m1 ( r ds2 r ds4 )g m5 ( r ds5 r ds8 ) 1 + jw2p( r ds2 r ds4 )C g5 ( )C L [ ][ 1 + jw2p r ds5 r ds8 ]
26 Reexamine condition for oscillation 1+Ab = 0 Æ Ab = 1 Magnitude and phase condition: Ab = 1 AND Ab = 180 Easier to get from Bode plot
27 Look at original Ab for 2 stage opamp Find w at which Ab = 1; Check Ab 180? Trouble!
28 Simulation Ab for 2 stage opamp
29 Move one pole to lower frequency How? Compensation: Dominant Pole
30 Need to increase capacitance by 1000X: Compensation: Dominant Pole BAD! Die area cost
31 Miller Effect Impedance across inverting gain stage G Reduced by factor equal to (1+G)
32 Math for Miller effect i x = v x  (Gv x ) Z i x = v x ( 1 + G ) Z v x = Z in = i x Z 1 + G ( ) Impedance across inverting gain stage G Reduced by factor equal to (1+G)
33 Example: Impedance is capacitive Capacitance multiplied by (1+G) Z in = Z 1+ G ( ) Z = 1 sc Æ Z in = 1 s( G 4 )C 3 C eq Equivalent capacitance higher by factor 1+G Problem for high bandwidth amplifiers Opportunity for compensation...
34 Miller Compensation Need effect of large capacitance Use Miller effect to multiply small onchip capacitance to higher effective value Effect of large capacitance without die area cost of large capacitance
35 New schematic Add C C across 2nd stage
36 New transfer function
37 No oscillation! New step response
38 How stable is new transfer function? Phase margin = Phase lag at Ab = 1 minus (180 ) "Phase margin"
39 Dominant pole opamp model Simpler model with dominant pole from C C
40 Approximate dominant pole transfer function A(jw) ª g m1 ( r ds2 r ds4 )A 2 1+ jw( r ds2 r ds4 )A 2 C C [ ] A 2 = g m5 ( r ds5 r ds8 )
41 Depends only on Unity gain frequency Input stage transconductance g m Compensation capacitor C C A(jw) ª g ( m1 r ds2 r ds4 )A 2 w( r ds2 r ds4 )A 2 C C [ ] A(jw) =1 at w T w T ª g m1 C C
42 I= C dv/dt Slew rate Only limited current I BIAS available to charge, discharge C C
43 I= C dv/dt fi Slew rate dv dt I BIAS = CC
44 Summary Opamp: Stability Compensation Miller effect Phase Margin Unity gain frequency Slew Rate Limiting
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