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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 Basics: Outline• Bipolar Junction Transistor (BJT) fabrication (diagram)• BJT regions of operation• BJT active region behavior and Ebers-Moll model• Overview of simple BJT amplifiers• Q-point analysis• Small signal AC analysis• A simple small signal AC BJT model2BJT Regions of Operation• VBE≡ VB− VE, VCE≡ VC− VE, voltage polarities are for npnDiode Biases BC Reverse BC ForwardBE Reverse Cutoff RegionVBE< 0.5 V• BJT acts like openswitchReverse Active Regionor Inverse Region• Usually to be avoidedBE Forward Active RegionVBE∼ 0.7 VVCE> 0.2 V• BJT acts like dependentcurrent sourceSaturation RegionVBE' 0.8 V• BJT acts like closedswitch3BJT Active Region OperationThe BE diode is forward biased, base is very thin, BC diode reverse biased• Consider NPN case (for PNP, substitute ”holes” for ”electrons”)• IEdetermined by VBEthrough diode equation• Electrons from emitter diffuse across thin base into BC depletion region• Here,~E causes them to drift on through BC depletion region• Geometry such that most electrons from emitter make it to collector:IC= αIE, α ≈ 0.99• ICroughly independent of VCEso collector acts as current source• IB= IE− IC= IE(1 − α) ≈ 0.01IE• Define hF E≡ IC/IB= αIE/(1 − α)IE= α/(1 − α) ≈ 100– Thus BJT acts as dependent current source: IC= hF EIB– Current amplifier with large current gain4BJT Active Region IEvs. VBE• BJT regions – cutoff: VBE< 0.5 V; active: VBE∼ 0.7 V; saturation:VBE' 0.8 V05101520253035404550IE vs. VBE for 2N2222A npn BJT (SPICE simulation)VBE (mV)IE (mA)0 100 200 300 400 500 600 700 800• VCE= 10 V• I vs V characteristic of a forward-biased diode5BJT Active Region ICvs. VCE• BJT in active region for VCE> 0.2 V a nd IB> 0.0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0-0.50.00.51.01.52.02.53.03.5IC vs VCE for 2N2222A npn BJT (SPICE simulation) IC (mA)VCE (V)IB = 8 µAIB = 0 µAIB = 4 µAIB = 12 µAIB = 16 µA• Looks like current source in active region: IC= hFEIB, hFE≈ 160• Slight slope due to narrowing of base as BC depletion region grows6Simple BJT Active Region Model7Simple 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.8Emitter Follower DC Analysis I(a) VB:VB= VBE+ VE= 0.7 V + IEREVB= 0.7 V + 2.0 mA × 500 Ω = 1.7 V(b) IB:IE= IC+ IB= hF EIB+ IB= (hF E+ 1)IBIB= IE/(hF E+ 1) = 2.0 mA/101 = 20 µA9Emitter Follower DC Analysis IIVB= 1.7 V, IB= 20 µA. (c) I1:I1= IB+ I2= 20 µA + VB/R2I1= 20 µA + 1.7 V/8100 Ω = 0.23 mA(d) R1:R1= (VCC− VB)/I1= (10 V − 1.7 V)/0.23 mA = 40 kΩ10Small 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 t he calculus ⇒ f(X0+ ∆X) ≈ f(X0) +dfdX|X0∆X– Keeping just first-order terms allows us to “linearize” non-linear circuitelements like diodes and transistors11Small Signal AC Analysis: refor BJT12A Simple Small Signal AC BJT Model• Note that reis a small signal AC resistance. It is a linear approximationto the diode curve near the Q point. Also, note that the Q point VBEQ=0.7 V does not appear in the small signal AC model since it is a constantvoltage.• We will consider alternate and more detailed models later as needed.• Next: apply to find voltage gain of emitter


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UCD PHY 116A - Lecture Notes

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