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Physics 116A NotesFall 2004David E. PellettDraft v.0.9• Notes Copyright 2004 David E. Pellett unless stated otherwise.• References:– Text for course:Fundamentals of Electrical Engineering, second edition, by LeonardS. Bobrow, published by Oxford University Press (1996)– Others as noted1Physics 116A, BJT Amplifiers: Outline• Small signal AC analysis (review)• A simple small signal AC BJT model and alternatives• Emitter follower small signal AC analysis– Voltage gain– Input impedance– Output impedance• Common emitter amplifier small signal AC analysis– Voltage gain– Effect of emitter resistor, RE2Simple BJT AmplifiersSimple one-BJT amplifier analysis:• Specify amplifier configuration (one BJT node is shared (common) be-tween the input and output circuit)– Common collector (emitter follower)– Common emitter– Common base• Specify BJT operating point (quiescent point or “Q point”)– BJT biasing• Small signal AC analysis of resulting circuit– Linearized small signal AC models of BJT– Small signal AC equivalent circuit– Use to calculate desired quantities such as Av, Rin.3Small Signal AC AnalysisAt Q point, VBQ= 1.7 V, IBQ= 20 µA, IEQ= 2.0 mA, VEQ= 1.0 V, etc.R1= 40 kΩ.• Introduce signals, vior ii, which are assumed to be small, first-order linearvariations about the Q point– E.g., VB= VBQ+ vb, where vbis the small signal AC base voltage.– Like X + ∆X in the calculus ⇒ f(X0+ ∆X) ≈ f(X0) +dfdX|X0∆X– Keeping just first-order terms allows us to “linearize” non-linear circuitelements like diodes and transistors– The Q-point DC values (including supply voltages) are ignored in thesmall signal model: keep variations from Q-point only.4Small Signal AC Analysis: refor BJT5A Simple Small Signal AC BJT Model• Note that vbecan be positive or negative (if negative, the currents areopposite).• The same small signal AC model suffices for npn or pnp transistors (justuse |IE| in finding re).• The BJT transconductance is defined as gm≡ ic/vbe≈ 1/re6Alternative Small Signal AC BJT Models7Small Signal AC Equivalent Circuit• We use superposition to separate the small signal AC circuit responsedue to vinfrom the DC Q point values.• If a voltage at som e point does not change due to vin(e.g., VCC), it willbe zero in the small signal equivalent circuit. Thus, R1and the BJTcollector connect to ground.• The BJT is replaced by a linear small signal model.8Small Signal AC Analysis: AvCalculate the voltage gain, Av= vout/vin• Since C1, C2→ ∞ they are short circuits for AC, so they are replaced withwires: vin= vband vout= ve• vb= iere+ ieRE, ve= ieRE, so Av= ve/vb= RE/(re+ RE).• In this case, Av= 500 Ω/(13 Ω + 500 Ω) = 0 .975.(Note that Avis not very sensitive to rehere since REis large.)9Small Signal AC Analysis: RinCalculate the input (AC) resistance, Rin= vin/iin.First calculate the resistance “looking into the BJT base,” Rb≡ vb/ib:vb= ie(re+ RE)ie= ib+ ic= ib+ hfeib= (hfe+ 1)ibRb= vb/ib= (hfe+ 1)(re+ RE) = (101)(513 Ω) = 52 kΩ.|| R0s ⇒ Rin−1= R1−1+ R2−1+ Rb−1⇒ Rin= 6.0 kΩ.10Small Signal AC Analysis: RoutRout= v/i for a s ignal applied a t the output (think Th´evenin).We also connect a signal source to the input (with vs= 0) since the sourceimpedance affects the answer. Let R = Rs||R1||R2. Then vb= ibR. Now findthe resistance looking into the emitter, Re≡ ve/ie: ve= iere+ vb= iere+ ibR =iere+ ieR/(hfe+ 1) ⇒ Re= re+ R/(hfe+ 1). Rout= Re||RE= ReRE/(Re+ RE).Put in numbers: if Rs= 600 Ω, Re= 18.5 Ω and Rout= 18 Ω. Suitable fordriving a low impedance load.11Common Emitter Amplifier: Q Point• Add RCin the collector circuit and take the output from the collector toget a CE amplifier.• Check that BJT in active region with this choice of RC. The base andemitter circuits are the same as before, so assume VBQ= 1.7 V, IBQ= 20 µA,IEQ= 2.0 mA, VEQ= 1.0 V. Find VCand VCE: VC= VCC− ICRC.IC= IE− IB≈ 2.0 mA. ⇒ VC= 10 V − 2.0 mA × 2.5 kΩ = 5 V.VCE= 5 V − 1 V = 4 V(> .2 V).• The BJT is in the active region with VCQ= 5 V.12CE Small Signal AC Analysis: AvAv= vout/vinvout= −icRCvin= ie(re+ RE)ic/ie= hfe/(hfe+ 1)Av= −hfeRC/((hfe+ 1)(re+ RE)) ≈ −RC/(re+ RE).• In this case, Av= −4.8 and the output resistance, Rout= RC= 2.5 kΩ.13Effect of REand Em itter Bypass Capacitor• REstabilizes amplifier operation and improves linearity for large signalsat the expense of gain.– Characteristic of negative feedback• You can bypass all or part of REfor AC with a large capacitor (CE||RE)– Keeps Q point sta bility since DC characteristics are the same– If REfully bypassed, gain is maximum (Av= −RC/re≈ −190) butexpect nonlinearity and gain variations for different BJ


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UCD PHY 116A - LECTURE NOTES

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