CLASS 3&4 BJT currents, parameters and circuit configurations
I E =I Ep +I En I C =I Cp +I Cn I B =I BB +I En -I Cn I BB =I Ep -I Cp I E = I B + I C I En = current produced by the electrons injected from B to E I Cn = current from the electrons thermally generated near the edge of the C-B junction that drifted from C to B. I B =I E -I C =I Ep +I En -I Cp -I Cn =I Ep -I Cp +I En -I Cn =I BB +I En -I Cn 2
An important BJT parameter is the common-base (CB) current gain, α o. α o = I Cp / I E = I Cp / (I Ep + I En ) = I Cp I Ep / [I Ep (I Ep + I En )] =[I Ep /(I Ep + I En )][I Cp / I Ep ] α o = γα T Emitter efficiency, γ = I Ep / (I Ep + I En ) γ = I Ep / I E Base transport factor, α T = I Cp / I Ep Since α o = γ α T and I En << I Ep, then I Ep I E. Hence, γ 1. I Cp I Ep. Thus, α T 1. Consequently, α o 1 3
I C = I Cp + I Cn As α T = I Cp / I Ep, then I C = α T I Ep + I Cn Since α o = γα T and γ = I Ep / I E : I C = (α o / γ ) γ I E + I Cn I C = α o I E + I Cn I E = I B + I C I Cn can be determined by measuring the current flowing across the B-C junction when E is an open-circuit. I E = 0. The value of I Cn under this condition is known as I CBO. I CBO represents the leakage current between C and B when E-B is open circuited. Collector current for the CB configuration is represented by the expression: I C = α o I E + I CBO 4
The conventional current flow is always in the opposite direction as the flow of electron. The conventional current flow is always in the same direction as the flow of holes. The flow of holes is always opposite with the flow of electrons. The general equation that relates the emitter, collector and base currents is: I E =I B +I C 5
Holes are injected from E to B when the E-B junction is fb. Holes oeswill then diffuse across B and reach the B-C junction. n noe where: p no = density of the minority carriers under equilibrium condition. P(0)= p e =n i2 /N B N B = donor density in B. /q = temperature equivalent voltage The existence of the density gradient of holes in B shows that the holes injected from E will diffuse across B to the edge of the B-C depletion region before they are swept into C by the electric field across B-C. 6
P(0)=p n no e If the E-B junction is fb, the minority carrier density at the edge of the E-B depletion region (at x=0) is increased beyond its equilibrium value by a factor of : e P n (W) = 0 Under the rb condition, the minority carrier density at the edge of the B-C depletion region (x = W) is 0. If the B is very narrow (i.e. W/L p << 1): n no ( ) P(x)=p e 1-xW = n ( ) P(0)1-xW 7
n no ( qv ) ( ) P(x)=p e 1-xW This expression is close to the real minority carrier distribution in B. The assumption that the minority carrier distribution in B is linear simplifies the derivation of the I-V characteristic. 8
n (x= x )=n e E E Eo C C Co -q V CB n (x=x )=n e =0 where n Eo and n Co are the electron densities under equilibrium condition for the E and C, respectively. ( x + x ) L E Eo Eo E E n (x)=n +n e 1 e for x x E n C(x)=nCo ncoe for x x C ( x xc ) LC 9
Transistor currents in the active mode of operation The hole current, I Ep, injected from E at x=0 is proportional to the p gradient of the minority carrier density. I dp Ep =A qd p dx where n no I Ep p no n (x) x=0 ( ) P(x)=p e 1-xW qad p W e The hole current collected by C at x=w is ( qv ) dpn (x) qadp pno I Cp =A qdp e dx x=w W I W Ep =I Cp for <<1(i (i.e. when B is L narrow) p 10
I En is produced by the flow of electrons from B to E. dn I E En = A qde dx x= x ( qv ) E qad En = Eo e 1 L E L E is the diffusion length of the electron in the E. D E is the diffusion i constant tfor the electron in E. 11
I Cn is produced by the flow of electrons from C to B. I dnc =A qd Cn C dx x=x C qad n = C Co L C L C is the diffusion length of the electron in the C. D C is the diffusion constant for the electron in C. 12
I I I E = Ep + En ( ) ( ) qv qad qv En + Eo L E qadppno = e e 1 W I I I C = Cp + Cn qad ppno = + W I I I B = E C qad n L E Eo E e e 1 qadcnco L C qadcnco L C 13
The current in each terminal (E,B and C) is determined mostly by the minority carrier distribution in B. I C is independent of V BC as long as the B-C junction is rb. If it is assumed that there is no recombination in B, I EP =I CP. Hence, I BB =I Ep -I Cp =0 I B =I BB +I En I Cn =I En I Cn 14
QUESTION The p + -n-p transistor has 10 19,10 17 and 5x 10 15 cm -3 impurity density in each E, B and C, respectively. The lifetime is 10-8,10-7 and 10-6 s. Assume that the cross-sectionsection area, A = 0.0505 mm 2 and the E-B junction is fb by a 0.6 V. Determine the common-base (CB) current gain, α o. Other device parameters are D E =1cm 2 /s, D B =10cm 2 /s, D C =2cm 2 /s, intrinsic electron-hole pair density = 9.65x10 9 cm -3 and W = 0.5 μm. 15
o I Cp IE E = Ep + En α = I I I ( ) ( ) qv D qv EnEo + L E Dppno =qa e e 1 W I Cp = qad p W p no e 16
D p = diffusion constant of hole in B = 10 cm 2 /s p no = hole minority carriers in B during equilibrium p no = n i2 /N B = (9.65x10 9 ) 2 /10 17 = 931.225 cm -3 D E = electron diffusion coefficient in E = 1 cm 2 /s n Eo = electron minority carrier in E during equilibrium n E n Eo = n i2 /N E = (9.65x10 9 ) 2 /10 19 = 9.3122 cm -3 L E = electron diffusion length in E -23 ο Boltzmann constant k = 1.381 10 J/ K -5 ο k = 8.620 10 ev/ K qadppno I Cp = I Ep = e W E E 2-8 -4 = D τ = 1 cm /s 10 s = 10 cm -19-2 2 2-3 1.6x10 C 0.05x10 cm 10 cm /s 931.225cm = -4 0.5x10 cm e 17
qad En I Eo En = e 1 L -14 10 I Cp = IEp 1.49x10 x1.1505x10 A -4 = 1.7142x10 A E -19-2 2 2-3 1.6x10 C 0.05x10 cm 1 cm /s 9.3122cm 10 = 1.1505x10 1-4 10 cm -8 = 8.5709x10 A -4 ICp 1.7142x10 o 0.9995 I -4 E α = = = 1.715x10 Emitter efficiency, γ = I Ep / (I Ep + I En ) = I Ep / I E In the case of narrow Base, I Ep = I Cp Thus, γ = α o. 18
γ= I Ep I E qadppno e = W Dp ( qv ) ( qv ) p no Dn E Eo qa e + e 1 W L E Dp p no = W Dp Dn = p no W + 1 D n 1+ D p L E Eo LE E Eo W p no E 2 2 n n n 1 Eo =,p =, γ= N N D N W i i no E B 1 E B + Dp NE LE 19
γ= 1 DE N 1 B W + D N L p E E To increase the emitter efficiency, γ, N B /N E has to be low. This indicates that E has to be doped higher than B. This is the reason why E is represented by p + for the p + -n-p transistor. To increase γ, the width of the B, W, should be small as compared to the diffusion length of the electrons in the E. 20
BJT CIRCUIT CU CONFIGURATIONS ONS 3 basic configurations: 1. Common Emitter (CE) 2. Common Collector (CC) 3. Common Base (CB) All transistor circuits, no matter how complex they are, are based on either one or combinations of 2 or all of these configurations. 21
Common Emitter (CE) E is the common point for both the input and output signals. Input signal is applied to B and output is at C. E is AC ground. Common Collector (CC) C is the common point for both the input and output signals. Input signal is applied to B and output is at E. C is AC ground. Common Base (CB) B is the common point for both the input and output signals. Input signal is applied to E and output is at C. B is AC ground. 22
23