EE 435 Lect 14 Spring 2010

EE 435 Lect 14 Spring 2010 - EE 435 Lecture 14 Compensation...

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EE 435 Lecture 14 Compensation of Cascaded Amplifier Structures
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Basic Two-Stage Op Amp + + = o6 o5 m5 o4 o2 m1 o g g g g g g A + + = o6 05 m5 C o5 o1 1 g g g C g g p L m5 2 C g p = C m1 C g GB = By inspection .• • • Review from last lecture .•
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Small Signal Analysis of Two-Stage Miller Compensated Op Amp 2 V d 2 V d Differential Small Signal Equivalent Norton Equivalent One-Port Two-Port .• • • Review from last lecture .•
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Small Signal Analysis of Two-Stage Op Amp ( ) 2 C 2 mo oo L C OUT V sC V g g sC sC V = + + + ( ) OUT C d md od C 2 V sC V g g sC V = + + Differential Small Signal Equivalent } ( ) () [] od oo od L od oo C C mo L C 2 C mo md d OUT g g g C g g C C g s C C s sC g g V V + + + + + = ( ) od oo C mo L C 2 C mo md d OUT g g C sg C C s sC g g V V + + This simplifies to: Solving we obtain: .• • • Review from last lecture .•
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() ( ) od oo C m5 L C 2 C m5 md g g C sg C C s sC g g s A + + = m2 m1 md g g g = = o4 o2 od g g g + = o6 o5 oo g g g + = where Small Signal Analysis of Two-Stage Op Amp Differential Small Signal Equivalent Summary: .• • • Review from last lecture .•
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() ( ) od oo C m5 L C 2 C m5 md g g C sg C C s sC g g s A + + = Small Signal Analysis of Two-Stage Miller- Compensated Op Amp 1 0 12 s +1 z As A ss +1 +1 pp = ⎛⎞ ⎜⎟ ⎝⎠ Note this is of the form: This has two negative real-axis poles and one positive real-axis zero 1 2 1 .• • • Review from last lecture .•
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How does the Gain of the Two-Stage Miller-Compensated Op Amp Compare with Internal Compensated Op Amp? () 1 0 12 s +1 z As A ss +1 +1 pp = ⎛⎞ ⎜⎟ ⎝⎠ md m5 C 2 CL m 5C o oo d gg - s C As= sCC+sg C +g g md m5 2 Co o o d As sCC+sCg +g g ± 0 1 +1 +1 = 2 00 1 p 4 β A2 β A p >> Compensation criteria: Re j ω p 1 p 2 z 1
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This note was uploaded on 01/31/2012 for the course EE 345 taught by Professor Geiger during the Fall '11 term at Iowa State.

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EE 435 Lect 14 Spring 2010 - EE 435 Lecture 14 Compensation...

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