EE 435 Lect 6 Spring 2010

EE 435 Lect 6 Spring 2010 - EE 435 Lecture 6: Current...

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1 EE 435 Lecture 6: Current Mirrors Signal Swing
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2 Performance with Common-Mode Input Consider tail-current bias amplifier V BB OUTC V c V I BIAS /2 C L V DD Common-Mode Half-Circuit C V X V OUTC V ( ) OUTC 1 2 M1 1 1 x C1 X X1 M 11 O U T C 1 sC+G +G +G = G = + G - G = G V V V Solving, we obtain V OUTC =0 thus A C =0 Review from last lecture:
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3 Performance with Common-Mode Input Consider tail-voltage bias amplifier Common-Mode Half-Circuit () OUTC 1 2 M1 1 C1 sC+G +G +G = 0 = V V Solving, we obtain OUTC V c C V OUTC V OUTC M1 C 2 -G =A = sC+G +G This circuit has a rather large common-mode gain and will not reject common-mode signals Review from last lecture:
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4 Applications of Quarter-Circuit Concept to Op Amp Design consider initially the basic single-ended amplifier Quarter Circuit Review from last lecture:
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5 Single-stage single-input low- gain op amp Basic Structure Practical Implementation Quarter Circuit Counterpart Circuit Review from last lecture:
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6 Single-stage low-gain differential op amp Need a CMFB circuit to establish V b1 3 1 1 2 O O L m g g sC g ) s ( A + + = O3 O1 m1 O g g 2 g A + = L m1 2C g GB = What are the number of degrees of freedom? Natural Parameters: 35 1 135 ,,, B1 B3 V, V WW W LL L ⎧⎫ ⎨⎬ ⎩⎭ (assume V DD , C L fixed) Constraints: I D5 ± 2I D3 Net Degrees of Freedom: 4 Practical Parameters: { } EB1 EB3 EB5 V , V , P Review from last lecture:
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7 Expressions valid for both tail-current and tail-voltage op amp So which one should be used? V DD V B1 M 1 M 2 V B2 M 3 M 4 V IN V IN C L M 5 C L V OUT V OUT Common-mode input range large for tail current bias Improved rejection of common-mode signals for tail current bias Extra design degree of freedom for tail current bias Improved output signal swing for tail voltage bias (will show later) Review from last lecture:
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8 Slew Rate Definition: The slew rate of an amplifier is the maximum rate of change that can occur at an output node SR is a nonlinear large-signal characteristic Input is over-driven hard (some devices in amplifier usually leave normal operating region) Magnitude of SR + and SR - usually same and called SR (else SR + and SR - must be given) Review from last lecture:
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9 Slew Rate With step input on V IN + , all tail current (I T ) will go to M 1 thus turning off M 2 thus current through M 4 which is ½ of I T will go to load capacitor C L The I-V characteristics of any capacitor is dV I=C dt Substituting I=I T /2, V=V OUT + and C=C L obtain a voltage ramp at the output thus + + OUT T LD D L dV I P SR dt 2C V 2C == = Review from last lecture:
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10 Slew Rate It can be similarly shown that putting a negative step on the input steer all current to M 2 thus the current to the capacitor C L will be I T minus the current from M 2 which is still I T /2. This will cause a negative ramp voltage on V OUT + of value + - OUT T LD D L dV I P SR dt 2C V 2C == = Since the magnitude of SR + and SR - are the same, obtain a single SR for the amplifier of value DD L P SR V2 C = Review from last lecture:
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11 11 Reference Op Amp single-ended output DD B1 1 2 B2 3 4 IN
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EE 435 Lect 6 Spring 2010 - EE 435 Lecture 6: Current...

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