6.012 Recitation 17: BJT-Basic Operation in FAR

Recitation 17 BJT: Basic Operation in FAR 6.012 Spring 2009 Recitation 17: BJT-Basic Operation in FAR BJT stands for Bipolar Junction Transistor 1. Can be thought of as two p-n junctions back to back, you can have pnp or npn. In analogy to MOSFET small current IB or voltage VBE , controls large current Ic 1 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 + IDp electron & hole diffusion currents Dn a = qAn2i × Na (wp − xp )    ×(eqVD /kT − 1) contribution from (1) e− diffusion in p-region Dp N (wn − xn )  d   b = qAn2i × ×(eqVD /kT − 1) contribution (2) from hole diffusion in n-region ID = a + b If Nd  Na , the contribution from hole (2) is much lower than that from electron (1) • For BJT, the doping profiles are very asymmetric NdE  NaB  NdC For example, NdE = 1020 cm−3 , NaB = 1018 cm−3 NdC = 1016 this is critical for BJT to function well. Why? 2 Recitation 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 by eqVBE /kT (c) Base-collector (BC) junction reverse biased: minority carrier concentration decreased by eqVBC /kT −→ 0 (d) No recombination in QNR =⇒ linear profile (e) These profiles result in diffusion currents (In BJT, the current we calculate are diffusion currents; in contrast, for MOSFET, the currents we calculate (ID ) are drift currents). 3 Recitation 17 BJT: Basic Operation in FAR Electrons diffuse across Base 6.012 Spring 2009 ∵VBC ≤0    :0  /kT BC npBo (eqVBE /kT −  eqV ) dnpB = qDn dx 0 − wB qV /kT B qDn npBo e qDn n2i VVBE = − =− e th wB wB · NaB JnB = qDn All electrons swept into C by large E field: (BC reverse biased) Ic = −JnB · AE to make positive into c terminal; AE is the area of Emitter/Base interface qAE Dn n2i qVBE /kT = e = IS eqVB /kT NaB wB qAE Dn n2i IS = similar to Io in diode, remember on the order of 10−15 − 10−20 A NaB wB Holes diffuse across Emitter JpE = −qDp pnE pnEo (eqVBE /kT − 1) kT
VBE > = −qDp q dx O( − wE ) IB = −JPE · AE = qAE Dp n2i qVBE /kT e to make positive into B terminal NdE wE Because 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. 4 Recitation 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 be majority carrier current (drift-diffusion); in emitter, this becomes minority carrier diffusion current, as we calculated). Look at Fig. 1, the base is so thin that all the electron current is directly swept to the collector, can not reach hole contact 2. Current in the collector terminal, is only the electron current, and do not depend on VBC . What happened to the hole diffusion current due to the doping file in collection region in Fig. 5, IC = IS · eqVBE /kT . It is too small to be counted. VBC is reverse biased, ID = Io (eqVBC /kT − 1)  −Io 3. What about IE ? IE = −(IB + IC ) 4. Relationship between IB and IC . We see that both IB and IC ∝ eqVBE /kT βF ∴ IC     IC Dn /NaB wB Dn NdE wE = = = Dp /NdE wE Dp NaB wB IB I S qVBE /kT = IS eqVBE /kT ; IB = e βF (a) We would like large βF . To make βF large, we need NdE  NaB , and wE > wB . Dn in the background of NaB (b) Dp in the background of NdE =⇒ Dn will be larger than Dp , ∵ NaB  NdE plus e vs. hole (c) If we make pnp, then  βF = Dp Dn  NaE NdB The advantage of Dn vs. Dp will be gone 5  wE wB  Recitation 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, wB will decrease. ( =⇒ 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 through =⇒ undesirable =⇒ high enough doping in base compared to collector Exercise See figure below: AE = 10 μm × 1 μm = 10μm2 = 10 × (10−4 )2 cm2 = 10−7 cm2 DnB = 4 cm2 /s, DPE = 1.3 cm2 /s     IC Dn NdE wE βF = = Dp NaB wB IB     20 4 10 0.2 = = 3 × 20 × 2 = 120 18 13 5 × 10 0.1 qAE Dn n2i 1.6 × 10−19 (C) × 10−7 (cm2 ) × 4 (cm2 /s) · 1020 (cm−6 ) IS = = = 1.28 × 10−19 A 5 × 1018 cm−3 × 0.1 × 10−4 cm NaB · wB IS = 1.07 × 10−21 A βF The eqVB /kT will make Ic , IB into 10−6 A  1 μA. 6