Lecture 11: Single-Stage BJT Amplifiers

Lecture 11:Single-Stage BJT AmplifiersGu-Yeon WeiDivision of Engineering and Applied SciencesHarvard Universityguyeon@eecs.harvard.eduWei1

Overview WeiReading– S&S: Chapter 4.11Supplemental ReadingBackground– Given our understanding of small-signal models and analysis, wecontinue our investigation of BJT circuits by looking at single-stageBJT amplifiers of various configurations. While there is a lot moredetail we can discuss in terms of BJTs and their large signalbehavior, we will stop with the discussion of BJT amplifiers andcontinue next lecture with MOSFETs which we will focus on for therest of the semester.– For more information on BJT circuits, read S&S 4.12 15ES154 - Lecture 112

Single-Stage Amplifier Configurations WeiThere are three basic configurations for single-stage BJTamplifiers:– Common-Emitter– Common-Base– Common-CollectorLet’s look at these amplifier configurations and their small-signaloperationES154 - Lecture 113

Common-Emitter Amplifier First, assume Re 0 (this is not re, but an explicit resistor)The BJT is biased with a current source (with high outputimpedance) and a capacitor connects the emitter to ground.– Cap provides an AC short at the emitter for small time-varyingsignals but is an open circuit for DC signalsCan redraw the circuit with an equivalent circuit that replaces theBJT with its hybrid-π modelRsBCgmvπvsvπrπrοRC voEvπrπ vs Rs rπvo g m vπ (RC ro )vo g m (RC ro )vπβ (RC ro )vo vo vπr π g m (RC ro ) vs vπ vsRs rπRs rπWeiES154 - Lecture 11Rin rπRo ro RC4

CE Amp with Emitter Degeneration Now, assume Re 0. First, find Ri – voltage applied to the base is across re and Revb ie (re Re )– base current isib – and let’s us find RiRi ie1 βvb (β 1)(re Re )ib– this tells us that adding Re increases the inputresistanceRi (w/ Re )R 1 e 1 g m ReRi (w/o Re )re– Can design the desired Ri by setting ReWeiES154 - Lecture 115

To determine the voltage gain, first find the gain from the base tothe collector (ignore ro b/c it complicates the analysis considerably)vo αie RCvb ie (re Re ) RCvo αRC vb re Re re Re–NOTE: Voltage gain between base and collector is equal toratio of total resistance in the collector to the total resistance inthe emitter.To find the total gain, RCβRCvo vo vbRi vs vb vb Ri Rs re ReRs (β 1)(re Re )Characteristics with Re :– gain is less with but less dependent on β– input resistance is higher– allows higher input signal voltage WeiES154 - Lecture 116

Common-Base Amplifier This time, ground the base and drive the input signal into the emitter through acoupling capacitor (only passes ac signals) Model the small signal approximation with a T-model– current source is an AC open and CC is an AC shortWeiES154 - Lecture 117

First, we can see that To find the gain, solve for vovo αie RC Ri reie vsRs reA voαRC vs Rs reThe output impedance is just R RoCCB amp characteristics:– voltage gain has little dependence on β– gain depends critically on Rs– is non-inverting– most commonly used as a unity-gain current amplifier or currentbuffer and not as a voltage amplifier: accepts an input signalcurrent with low input resistance and delivers a nearly equalcurrent with impedance– most significant advantage is its excellent frequency responseWeiES154 - Lecture 118

Common-Collector Amplifier (Emitter Follower) The last basic configuration is to tie the collector to a fixed voltage, drive aninput signal into the base and observe the output at the emitter Also called an emitter follower since the emitter follows the input signalUsed for connecting a source with a large Rs to a load with low resistanceWeiES154 - Lecture 119

Redraw the circuit to have ro in parallel with RL– now, find RiRi (β 1)[re (ro RL )]– when re RL ro Ri (β 1)RL– notice the amplifier has large input resistanceFind the gain with two voltage dividers(β 1)[re (ro RL )]vb vs Rs (β 1)[re (ro RL )]ro RLvo vb re ro RL(β 1)(ro RL )ro RLvoAv Rsvs Rs (β 1)[re (ro RL )] re (ro RL )β 1– gain is less than unity, but close (to unity) since βis large and re is smallWeiES154 - Lecture 1110

High-Frequency Model of BJTs When we looked at pn junctions, we found that the depletion regions can bemodeled as capacitances. Let’s see how these capacitances affect theperformance of BJTs.There are two depletion regions and therefore two capacitors we need to includeinto our small-signal model. Also, a base resistance rx is added because ofcurrent crowding at high frequencies (second-order effect in BJTs).BrxCµCgmvπrπvπCπroE– Cπ and Cµ are bias dependent values which can be found (approximated)from the DC bias conditions of the circuitWeiES154 - Lecture 1111

fT Transistor performance is often presented in terms of fT which is a value thatcorresponds to its short-circuit current gain bandwidth product (or unity-gainbandwidth). Let’s see how to solve for it BIbrxIc (gm-sCµ)VπCµCgmvπrπvπCπroE Find Ic/Ibg m sC µIc I b 1 rπ s (Cπ C µ )Vπ I b (rπ Cπ C µ )– at frequencies over which the model is valid, gm ωCµIcg m rπβ0 I b 1 s Cπ Cµ rπ 1 s Cπ Cµ rπ()()β0 is the low-frequency value of β and current gain rolls off with a single polegmg 2π1at ωβ fT mωβ ωT β 0ω β (Cπ Cµ )rπCπ CµCπ CµWeiES154 - Lecture 1112

BJT amplifiers of various configurations. While there is a lot more detail we can discuss in terms of BJTs and their large signal behavior, we will stop with the discussion of BJT amplifiers and continue next lecture with MOSFETs which we will focus on for the rest of the semester. – For more information on BJT circuits, read S&S 4.12 15