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L14 Part 1 6.4.4 Miller'sTheorom_1

L14 Part 1 6.4.4 Miller'sTheorom_1 - EE 311 EEO 311 Lecture...

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EE 311 & EEO 311 Lecture 14 Part 1 6.4.4 Miller’s Theorem Zubair Rehman
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2 I 1 = Y(V 1 - V 2 ) I 1 = Y 1 V 1 I 2 = Y(V 2 - V 1 ) I 2 = Y 2 V 2 We must be able to relate V 2 & V 1 V 2 = KV 1 & V 1 = 1 / K V 2 I 1 = Y(V 1 - KV 1 ) I 2 = Y(V 2 - 1 / K V 2 ) I 1 = V 1 [Y(1-K)] I 2 = V 2 [Y(1- 1 / K )] Y 1 = Y(1-K) Y 2 = Y(1- 1 / K ) Figure 6.15 The Miller equivalent circuit. (Sedra/Smith)
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4 R s Thevenin Equivalent Resistance Includes Bias Resistances & Output Resistance of previous stage. V s Thevenin Equivalent Source R L Thevenin Equivalent Load Includes Bias Resistance & Input Resistance of the next stage. From Sedra/Smith 4 th  edition
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5 Note: Figure 6.20 in Sedra/Smith 5 th ed.
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6 Note: Figure 6.25 in Sedra/Smith 5 th ed.
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7 V S Thevenin Equivalent Source Voltage R S Thevenin Equivalent Resistance Includes Bias Resistors, Output Resistance of Previous Stage R L Thevenin Equivalent Resistance Includes Bias Resistors, Input Resistance of Next Stage R L ’ = R L || r O where r O = (V A / I C ) BJT CE Amplifier V TH = V s * [r π / (R s + r x + r π )] = Open-Circuit Voltage R TH = r π || (r x + R s ) = Resistance with V s = 0
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