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MIT 6 012 - BJT-Basic Operation in FAR

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Recitation 17 BJT: Basic Operation in FAR 6.012 Spring 2009 Recitation 17: BJT-Basic Operation in FARBJT stands for Bipolar Junction Transistor1. Can be thought of as two p-n junctions back to back, you can have pnp or npn.In analogy to MOSFET small current IBor voltage VBE, controls large current Ic1  Recitation 17 BJT: Basic Operation in FAR 6.012 Spring 2009 2. How does it work?• As we learned previously, for an asymmetric pn junction:ID= IDn+ IDpelectron & hole diffusion currentsa = qAn2i×Na(wDpn− xp)×(eqVD/kT− 1)contribution from (1) e−diffusion in p-regionb = qAn2i×Nd(wDnp− xn)×(eqVD/kT− 1)contribution (2) from hole diffusion in n-regionID= a + bIf Nd Na, the contribution from hole (2) is much lower than that from electron(1)• For BJT, the doping profiles are very asymmetricNdE NaB NdCFor example,NdE=1020cm−3, NaB=1018cm−3NdC=1016this is critical for BJT to function well. Why?2Recitation 17 BJT: Basic Operation in FAR 6.012 Spring 2009 Minority carriers concentration under T.E. • Minority carrier concentration under FAR: (broken linear y axis) (a) At contacts, equilibrium concentration (PnEO,PnCO)(b) BE junction forward biased: minority carrier concentration increased byeqVBE/kT(c) Base-collector (BC) junction reverse biased: minority carrier concentrationdecreased by eqVBC/kT−→ 0(d) No recombination in QNR = linear profile⇒(e) These profiles result in diffusion currents (In BJT, the current we calculateare diffusion currents; in contrast, for MOSFET, the currents we calculate(ID) are drift currents).3 Recitation 17 BJT: Basic Operation in FAR 6.012 Spring 2009 Electrons diffuse across Base∵VBC≤0:0JnB= qDndnpB= qDnnpBo(eqVBE/kT− eqVBC/kT)dx 0 − wBqDnnpBoeqVB/kTqDnni2VBE=eVth−wB= −wB· NaBAll electrons swept into C by large E field: (BC reverse biased)Ic= −JnB· AEto make positive into c terminal; AEis the area of Emitter/Base interface=qAEDnni2eqVBE/kT=ISeqVB/kT NaBwB qAEDnn2i− 10−20A IS= similar to Ioin diode, remember on the order of 10−15NaBwBHoles diffuse across EmitterpnEpnEo(eqVBE/kT1) kTJpE= −qDpdx= −qDpO(− wE)−VBE>qIB= −JPE· AE=qAEDpni2eqVBE/kTto make positive into B terminalNdEwEBecause E-B we have n-p from left to right, but we define positive direction left right,→the current we calculate will be from p-n which will be right left.→Now let us draw a flux picture.4Recitation 17 BJT: Basic Operation in FAR 6.012 Spring 2009 1. For current in the base terminal, it is only the hole current. (in base, this will bemajority carrier current (drift-diffusion); in emitter, this becomes minority carrierdiffusion current, as we calculated). Look at Fig. 1, the base is so thin that all theelectron current is directly swept to the collector, can not reach hole contact2. Current in the collector terminal, is only the electron current, and do not depend onVBC. What happened to the hole diffusion current due to the doping file in collectionregion in Fig. 5, IC= IS· eqVBE/kT.It is too small to be counted. VBCis reverse biased, ID= Io(eqVBC/kT− 1) −Io 3. What about IE?IE= −(IB+ IC)4. Relationship between IBand IC. We see that both IBand IC∝ eqVBE/kT ICDn/NaBwBDnNdEwEβF== =IBDp/NdEwEDpNaBwB∴ IC= ISeqVBE/kT;IB=ISeqVBE/kTβF(a) We would like large βF. To make βFlarge, we need NdE NaB, and wE>wB.Dnin the background of NaB(b) Dpin the background of NdE=⇒ Dnwill be larger than Dp, ∵ NaB NdEplus e vs. hole(c) If we make pnp, then DpNaEwEβF=DnNdBwBThe advantage of Dnvs. Dpwill be gone5Recitation 17 BJT: Basic Operation in FAR 6.012 Spring 2009 5. Why NaB NdC?When we apply negative VBC, the depletion layer will increase. Effectively, wBwilldecrease. ( = base width modulation similar to channel length modulation in⇒MOSFET case) = undesirable⇒In addition, if depletion layers in the base region touch each other from both sides(emitter → collector) =⇒ punch throughin base compared to collector=⇒ undesirable =⇒ high enough dopingExerciseSee figure below:AE =10μm × 1 μm =10μm2 =10× (10−4)2cm2 =10−7cm2 DnB=4cm2/s, DPE=1.3cm2/s   ICDnNdEwEβF== IBDpNaBwB    410200.2= 13 5 × 10180.1=3× 20 × 2 = 120qAEDnn21.6 × 10−19(C) × 10−7(cm2) × 4 (cm2/s) 1020(cm−6)IS=wBi=5 × 1018cm−3× 0.1 × 10−4cm·=1.28 × 10−19ANaB·IS=1.07 × 10−21AβFThe eqVB/kTwill make Ic,IBinto 10−6A  1 μA.6MIT OpenCourseWarehttp://ocw.mit.edu 6.012 Microelectronic Devices and CircuitsSpring 2009 For information about citing these materials or our Terms of Use, visit:


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MIT 6 012 - BJT-Basic Operation in FAR

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