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 I B or voltage V BE, controls large current I c 1
2. How does it work? As we learned previously, for an asymmetric pn junction: I D = I Dn + I Dp electron & hole diffusion currents a = qan 2 i Dn N a (wp x p ) }{{} (e qvd/kt 1) contribution from (1) e diffusion in p-region b = qan 2 i Dp N } d (wn x {{ n ) } (e qvd/kt 1) contribution (2) from hole diffusion in n-region I D = a + b If N d N a, the contribution from hole (2) is much lower than that from electron (1) For BJT, the doping profiles are very asymmetric N de N ab N dc For example,n de = 10 20 cm 3, N ab =10 18 cm 3 N dc = 10 16 this is critical for BJT to function well. Why? 2
Minority carriers concentration under T.E. Minority carrier concentration under FAR: (broken linear y axis) (a) At contacts, equilibrium concentration (P neo,p nco ) (b) BE junction forward biased: minority carrier concentration increased by e qv BE/kT (c) Base-collector (BC) junction reverse biased: minority carrier concentration decreased by e qvbc/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 (I D ) are drift currents). 3
Electrons diffuse across Base V BC 0 {}}{ 0 dn pb n pbo (e qvbe/kt e qvbc/kt ) J nb = qd n = qdn dx 0 w B qd n n pbo e qv B/kT qd n n2 V BE = = i e V th w B wb N ab All electrons swept into C by large E field: (BC reverse biased) I c = J nb A E to make positive into c terminal; A E is the area of Emitter/Base interface = qa ED n n2 i e qvbe/kt =I S e qv B/kT N ab w B qa E D n n 2 i I 10 20 S = similar to I o in diode, remember on the order of 10 15 A N ab w B Holes diffuse across Emitter p ne p neo (e qv BE/kT 1) kt J pe = qd p = qd p V BE > dx O ( w E ) q I B = J PE A E = qa ED p n i 2 e qv BE/kT to make positive into B terminal N de w E Because E-B we have n-p from left to right, but we define positive direction left the current we calculate will be from p-n which will be right left. Now let us draw a flux picture. right, 4
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 V BC. What happened to the hole diffusion current due to the doping file in collection region in Fig. 5, I C = I S e qvbe/kt. It is too small to be counted. V BC is reverse biased, I D = I o (e qvbc/kt 1) I o 3. What about I E? I E = (I B + I C ) 4. Relationship between I B and I C. We see that both I B and I C e qv BE/kT I C D n /N ab w B D n N de w E β F = = = I B D p /N de w E D p N ab w B I C = I S e qv BE/kT ; I B = I S e qv BE /kt β F (a) We would like large β F. To make β F large, we need N de N ab, and w E >w B. D n in the background of N ab (b) D p in the background of N de = D n will be larger than D p, N ab N de plus e vs. hole (c) If we make pnp, then D p N ae w E β F = Dn N db w B The advantage of D n vs. D p will be gone 5
5. Why N ab N dc? When we apply negative V BC, the depletion layer will increase. Effectively, w B 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: A E = 10μm 1 μm = 10μm 2 = 10 (10 4 ) 2 cm 2 = 10 7 cm 2 D nb = 4cm 2 /s, D PE =1.3cm 2 /s I C D n N de w E β F = = I B D p N ab w B 4 10 20 0.2 = 13 5 10 18 =3 20 2 = 120 0.1 I S = qa E D n n 2 i 1.6 10 19 (C) 10 7 (cm 2 ) 4 (cm 2 /s) 10 20 (cm 6 ) = N w B 5 10 18 cm 3 0.1 10 4 ab cm I S = 1.07 10 21 A β F The e qv B/kT will make I c,i B into 10 6 A 1 μa. =1.28 10 19 A 6
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