GT ECE 3050 - Thévenin Base Circuit

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Thévenin Base CircuitAlthough the base is not an output terminal, the Thév enin equivalen t circuit seen lookinginto the base is useful in calculating the base cu rrent. It consists of a v olta ge sou rce vb(oc)in series with a resistor ribfromthebasenodetosignalground. Fig. 1(a)showstheBJTsym bol with a Thévenin source connected to its emitter. Fig. 1(b) shows the T model forcalculating the open-circuit base vo ltage. B ecau se ib=0, it follo w s that i0e=0.Thusthereis no drop across rxand reso that vb(oc)is giv e n byvb(oc)= ve= vter0+ RtcRte+ r0+ Rtc(1)Figure 1: (a) BJT with Thev enin source connected to the emitter. (b) T model for calculatin gvb(oc).The next step is to solve for the resistance seen looking into the base. It ca n be calculatedb y setting vte=0and connecting a test current source itto the base. It is given b y rib= vb/it.Fig. 2(a) shows the T circuit for calcu latin g vb, where the current source βithas beendivided into identical series sources with their common node grounded to simplify use ofsuperposition. By superposition of itand the two βitsources, we can writevb= itrx+(it+ βit)[re+ Rtek (r0+ Rtc)] − βitRtcRteRtc+ r0+ Rte(2)This can be solved for ribto obtainrib=vbit= rx+(1+β)[re+ Rtek (r0+ Rtc)] −βRtcRteRtc+ r0+ Rte= rx+(1+β) re+ Rte(1 + β) r0+ Rtcr0++Rte+ Rtc= rx+ rπ+ Rte(1 + β) r0+ Rtcr0++Rte+ Rtc(3)The Thévenin base circuit is shown in Fig. 2(b).1Figure 2: (a) Circuit for calculating vb. (b) Thév en in base


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