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# Lect28 7.5.2 Part 1 MOSDiffAmpWithAcitveLoad - EE 311...

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Unformatted text preview: EE 311 Lecture 28 EE 311 Lecture 28 March 22, 2010 March 22, 2010 7.5.2 MOS Differential Amplifier 7.5.2 MOS Differential Amplifier With Active Load With Active Load Zubair Rehman Q1 i d1 = g m1 v d / 2 Q2 i d2 = - g m2 v d / 2 Q3 i d3 = i d1 = g m1 v d / 2 Q4 i d4 = i d3 = g m1 v d / 2 i o = G m v i v i = v d i o = (g m2 v d / 2) + (g m1 v d / 2) = g m v d G m = g m Circuit For Determining R Circuit For Determining R o o v d = 0 & V DD is an ac short & I is an ac open-circuit (1) i enters d 2 & leaves s2 i = g m2 v gs2 + i r02 (2) i enters s1 (3) i leaves d1 i = -[g m1 v gs1 + i r01 ] (4) i enters d3 of diode corrected Q3 (i = g m3 v gs3 + i r03 ) diode connected transistor presents resistance r 03 || (1/g m3 ) = 1 / g m3 (1) i enters drain of Q4 because v gs4 = v gs3 (Mirror) i does not include current through r 04 Eq. 6.101 R o for a CG with R s between s & gnd. R = r + [1 + (g + g )r ]R Correction Q4 is a CS V x = v x R o = r o + [1 + (g m + g mb )r ]R s Eq. 6.101 for a CG output resistance For Q2 add subscript 2 R 02 = r 02 + [1 + (g m2 + g mb2 )r 02 ]R s2 Sedra-Smith either ignores g mb or includes it in the g m term R s2 = R INQ1 = R INCG...
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Lect28 7.5.2 Part 1 MOSDiffAmpWithAcitveLoad - EE 311...

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