L13 Part 1 TwoStageAmp_1

L13 Part 1 TwoStageAmp_1 - EE 311 EEO 311 Lect 13 6.11...

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EE 311 & EEO 311 Lect. 13 6.11 Useful Transistor Pairings: Two - Stage Amplifier EE 353 Exp. 4 Ref 1: 5.7 Single-Stage Amp Eq. For BJT Slides 6 & 7 Ref. 2: L11 Part 2 R10 & R20 Formulas Zubair Rehman
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2 Direct Coupled CE - CC V CC = 18 V Reference Sedra - Smith 5.5 Biasing in BJT Amplifier Circuit For Q1 Transform V CC , R11 & R12 to V B & R B DC Bias Circuit For Q1 1 1 11 || 12 12 11 12 * 1 1 1 1 1 1 1 1 + = + = + + = = + = 1 FE E E B E E BE B B B B CC B h I I I I R V I R V R R R R R R V V β
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3 From 2N3904 Data Sheet At I C = 10 mA V BE = 0.72 V h FE = I C / I B β At I C = 10 mA: min h FE = 100 max h FE = 300 Keep R B * [I E1 / ( β+129] < 0.2 * R E1 I E1 e.q. R B * [I E1 / ( β+129] = 1 V R B = ( β+129 * (1 V / I E1 ) maxR B = [(h FE,min + 1) * 1 V] / I E1 = (101*1 V) / 10 mA ≈ 10 k Why use minimum value of h FE ? V B = 1 V + 0.72 V + 6 V = 7.72 V 1 1 1 E E BE E B B I R V I R V + + 1 + = β Divide V CC equally across R C1 , V CE & R E1 I C1 R C1 = V CE1 = I E1 R E = V CC /3 = 18 V / 3 = 6 V V BE is in the range 0.6 to 0.8 V Set I E = 10 mA R E1 = 6 V / 10 mA = 0.6 k
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4 Check on I E1 for max h FE = 300 I E1 = (V B - V BE ) / {[R B / ( β+129] + R E1 } Let β = max h FE = 300 Assume V BE1 = 0.72 V (unchanged) I E1 = (7.72 - 0.72) / [(10 k / 301) + 0.6 k] = 7 V / 0.633 = 11.1 mA
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