# Lect27 7.5.5 Part1 BJTDiffAmpWithActiveLoad - EE 311...

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Unformatted text preview: EE 311 & EEO 311 EE Lecture 26 Part 1 Lecture March 21, 2008 March March 20, 2009 7.5 BJT Differential Amplifier 7.5 With Active Load With Zubair Rehman BJT Differential Amplifier & Active Load BJT vd = differential input Half-Circuit Approach for Q1 & Q2 ic1 = gm * vd/2 ic2 = -gm * vd/2 ic3 = ic1 = gm * vd/2 ic4 = ic3 = gm * vd/2 (veb3 = veb4) iout = gm * vd/2 + gm * vd/2 = gmvd Open Circuit vout = ioutRo = gmvdRo Where Ro is the output resistance Ro = ? Intuitive Approach from 4th edition of Sedra-Smith Wrong Analysis Happily Leads to Correct Result We shall do correctly at a later time. RO = v x / i x ix = (vx / r02) + (vx/r04) = vx[1/r02 + 1/r04] = vx / (r02 || r04) Ro = r02 || r04 r02 = VA2/IC2 = VA2/(I/2) r04 = VA4/IC4 = VA4/(I/2) If VA4 = VA2 r04 = r02 Ro = r02 / 2 vout = gmvdRo = gmvd * ro/2 vo / vd = gmro / 2 gm = IC2 / VT = (I/2) / VT ro = VA / IC2 = VA/(I/2) vo / vd = gmro / 2 = 1/2 * (I/2)/VT * (VA)/(I/2) = 1/2 * (VA/VT) For VA = 50 V & VT = 25 mV vo / vd = 1/2 * (50 V/25 mV) = 1000 What if next stage has non-infinite RIN? Transconductance Amplifier ...
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## Lect27 7.5.5 Part1 BJTDiffAmpWithActiveLoad - EE 311...

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