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4/1/2011Example A Small Signal Analysis of a BJT Amp1/10Example: A Small-SignalAnalysis of a BJTAmplifier15.0 VConsider the following BJTamplifier:RC 5 KvO (t ) VO vo (t )RB 5 K β 100vi (t ) RE 5 K5.8 VCOUS Let’s determine its small-signal, open-circuit voltage gain:Avo vo (t )vi (t )

4/1/2011Example A Small Signal Analysis of a BJT Amp2/10To do this, we must follow each of our five small-signal analysissteps!Step 1: Complete a D.C. Analysis15.0 VThe DC circuit that we must analyze is:ICRC 5 KVORB 5 Kβ 100IB RE 5 KIE5.8 V Note what we have done to the original circuit:1) We turned off the small-signal voltage source(vi (t ) 0 ), thus replacing it with a short circuit.2) We replaced the capacitor with an open circuit—its DCimpedance.

4/1/2011Example A Small Signal Analysis of a BJT Amp3/10 V15.0Now we proceed with the DC analysis.RC 5 KWe ASSUME that the BJT is in activemode, and thus ENFORCE the equalitiesVBE 0.7 V and IC β IB .VORB 5 KWe now begin toANALYZE the circuit bywriting the BaseEmitter Leg KVL: 5.8 V IB and thus:β 100IB5.8 5IB 0.7 5(β 1)IB 0Therefore:ICRE 5 KIE5.1 0.01 mA5 5(101)IC βIB 1.0 mAIE IB IC 1.01 mAQ: Since we know the DC bias currents, we have all theinformation we need to determine the small-signal parameters.Why don’t we proceed directly to step 2?

4/1/2011Example A Small Signal Analysis of a BJT Amp4/10A: Because we still need to CHECK our assumption! To do this,we must determine either VCE or VCB .15.0 VNote that the Collector voltage is:VC 15 IC RC 15 (1.0)5RC 5 K 10.0 VVOAnd the Emitter voltage is:VE IE RERB 5 K (1.01)55.8 V 5.05 VICβ 100IB Therefore, VCE is:RE 5 KVCE VC VE 10.0 5.05 4.95 VWe now can complete our CHECK:IC 1.0 mA 0VCE 4.95 V 0.7Time to move on to step 2!IE

4/1/2011Example A Small Signal Analysis of a BJT Amp5/10Step 2: Calculate the small-signal circuit parameters foreach BJT.If we use the Hybrid-Π model, we need to determine gm and rπ :gm 1.0 mAICmA 40VT0.025VVrπ VT0.025 V 2.5 KIB 0.01 mAIf we were to use the T-model we would likewise need todetermine the emitter resistance:re VT0.025 V 24.7 ΩIB1.01 mAThe Early voltage VA of this BJT is unknown, so we will neglectthe Early effect in our analysis.As such, we assume that the output resistance is infinite( ro ).

4/1/2011Example A Small Signal Analysis of a BJT Amp6/10Step 3: Carefully replace all BJTs with their small-signalcircuit model.15.0 VRC 5 KRB 5 KBC vi (t)vbevO (t ) VO vo (t )2.5 K40 vbe 5.8 VE RE 5 KCOUS

4/1/2011Example A Small Signal Analysis of a BJT Amp7/10Step 4: Set all D.C. sources to zero.RC 5 KRB 5 Kvo (t )BC vi (t)vbe2.5 K40 vbeERE 5 KWe likewise notice that the large capacitor (COUS) is anapproximate AC short, and thus we can further simplify theschematic by replacing it with a short circuit.

4/1/2011Example A Small Signal Analysis of a BJT Amp8/10RC 5 KRB 5 KBib vi (t)vo (t )Cic vbe2.5 Kie40 vbeEWe notice that one terminal of the small-signal voltage source,the emitter terminal, and one terminal of the collector resistorRC are all connected to ground—thus they are all collected toeach other!We can use this fact to simplify the small-signal schematic.

4/1/2011Example A Small Signal Analysis of a BJT AmpRB 5 KBibic C vi (t)9/10rπ vbe2.5 K40 vbevo (t )RC 5 K-ieEThe schematic above is the small-signal circuit of thisamplifier. We are ready to continue to step 5!Step 5: Analyze small-signal circuit.This is just a simple EECS 211 problem! The left side of thecircuit provides the voltage divider equation:vbe rπRB rπvi2.5vi5.0 2.5vi3a result that relates the input signal to the base-emittervoltage.

4/1/2011Example A Small Signal Analysis of a BJT AmpRB 5 KBC vi (t)10/10rπ vbe2.5 K40 vbevo (t )RC 5 K-EThe right side of the schematic allows us to determine theoutput voltage in terms of the base-emitter voltage:vo ic RC (gmvbe ) RC 40(5)vbe 200vbeCombining these two equations, we find:vo 200vbevi 2003 66.7 viThe open-circuit, small-signal voltage gain of this amplifier gainis therefore:Avo vo 66.7vi

We notice that one terminal of the small-signal voltage source, the emitter terminal, and one terminal of the collector resistor R C are all connected to ground—thus they are all collected to each other! We can use this fact to simplify the small-signal schematic. v i (t) be R B 5 K R C 5 K v o()t B C E b i i e i c v 25K. 40v _