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# Course materials - V out = 0 K s s s4 = V O V S = V O V S =...

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5.9.3 Part I Low Frequency Response Lecture 4 Part I Wed. Jan. 20, 2010

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2 AC - Coupled CE Amplifier We design to have gain at mid-band. C in , C E , C out are ac short-circuits |A mid | R C || R L r e = V T / I E =
3 Suppose C in out = infinity (Very Large). Then, C E determines low frequency response. When C E reactance is large enough, C E is an open-circuit. |A mid | R C || R L r e + R E = Resistance in Collector circuit Resistance in Emitter circuit = Transfer Function V O V S = K * s + s1 / s + s2 s = jω ω = 2 π f Let s infinity V O V S = K = {(R C || R L ) / r e } Let s 0 V O V S = K * s1 / s2 = {(R C || R L ) / (r e + R E )} We can solve circuit for s 1 2

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4 Suppose C E out = infinity C in determines frequency response At f = 0 C in is an ac open-circuit O = 0 V O V S = V O V S = V O V S = K * s / s + s4 s = jω ω = 2 π f As s infinity K = (R C || R L ) / r e As s 0 0 We can solve circuit for s 4
5 V O V S Suppose C E in = infinity C out determines frequency response At f = 0 Hz C out = ac open-circuit

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Unformatted text preview: V out = 0 K * s / s + s4 = V O V S = V O V S = As s infinity K = (R C || R L ) / r e As s We can solve circuit for s 6 6 Method of Short-Circuit Time Constants We do not need to solve for the poles & zeros. Our objective is to select, C in , C out & C E to obtain a desired frequency response usually characterized by the low half-power frequency. Bode Plot 7 The Amplifier Transfer Function A (s) = V O (s) V i (s) DC Amplifier Capacitively Coupled Amplifier 8 ω L Lower Half - Power Frequency ω L = 2 π f L Can be estimated using Method Of Short - Circuit Time Constants ω H Upper Half - Power Frequency ω H = 2 π f H Can be estimated using Method Of Open Circuit Time Constants Gray & Searle “Electronic Principles” John Wiley 1969...
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Course materials - V out = 0 K s s s4 = V O V S = V O V S =...

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