Lect27 7.5.5 Part2 BJTDiffAmpWithActiveLoad

Lect27 7.5.5 Part2 BJTDiffAmpWithActiveLoad - EE 311...

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Unformatted text preview: EE 311 & EEO 311 EE 311 & EEO 311 Lecture 26 Part 2 Lecture 26 Part 2 March 21, 2008 March 21, 2008 March 20, 2009 March 20, 2009 7.5 The BJT Differential Amplifier 7.5 The BJT Differential Amplifier With Active Load With Active Load 5 5 th th ed. Sedra-Smith ed. Sedra-Smith Zubair Rehman Very Similar To MOS Except For Finite r & Finite DC I E1 = I E2 = I o / 2 I C3 = I C1 I C4 = I C2 (except for effects of base currents) AC i c1 = g m1 v d / 2 i c2 = - g m2 v d / 2 i c3 = i c1 = g m1 v d / 2 i c4 = i c3 = g m1 v d / 2 i o = g m1 v d / 2 + g m1 v d / 2 i o = g m1 v d = G m v d Output Resistance R o V CC is an ac short-circuit: e3 & e4 are ac grounds v d /2 is an ac short-circuit b1 & b2 are ac grounds I o is an ac open-circuit (R SS = infinity) Q 3 is a diode-connected transistor: replace by r e3 in parallel with r 03 which can be ignored The r 03 can be ignored because V A >> ( + 1)V T 1. i enters c2 and leaves e2 2. i leaves e2 and enters e1 3. i leaves c1 and enters junction of c3, b3 & b4 so i = i c3 + i b3 + i b4 i c3 for large enough 4. i c4 (0) = i c3 because v be3 = v be4 5. Thus i c4 (0) = i also where i c4 (0) = g m v 4 and does not include i ro4 R...
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Lect27 7.5.5 Part2 BJTDiffAmpWithActiveLoad - EE 311...

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