chap10 - CHAPTER 10 112.1 (a) Common base. (b) Common...

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Unformatted text preview: CHAPTER 10 112.1 (a) Common base. (b) Common emitter. 7 (c) Saturation, active, inverted, and cutoff. (d) The buried layer serves as a low~resistance path between the active collector region of the BJT and the top-side collector contact. (6) NAB >> NDB > NAC '(t) W in both cases. (g) The width of the base is less than, typically much less than, the minority carrier diffusion length in the base. (h) The narrow nature of the base couples the current flow across the E-B and GB junctions, a prerequisite for transistor action. (i) The emitter efficiency specifies the fraction of the emitter current that is associated with carrier injection from the emitter into the base. (i) The base transport factor is the fraction of the minority carriers injected into the base that successfully diffuse across the quasineutral width of the base and enter the collector. . 10—1 LQQ (a) For the given doping concentrations, one computes EF— E = ~0.459eV, 0.298eV, and —0.239eV respectively in the emitter, base, and collector. Also, with N A}; >> NDB, the E—B depletion width will lie almost exclusively in the base. Likewise, the majority of the GB depletion width will lie in the collector. The diagram produced by the BJTWEband program is displayed below. P+ N . I I I I I I I I I l r P i. I I (b) 10~2 (C) AVCE =7 (I/Q)[(Ei—EF)emitter‘ (Ei‘EFkollectorl = (kT/CI)[1H(NAE/ni) — 1n(NAdng)] or AVCE= (kT/q)1n(NAE/NAC) = (0.0259)1n(5><1017/1014) = 0.221 v ((1) As noted in the text (Ed. 103), W e" WB“anB"xnCB 1/2 <14 , “2 “EB E [M550 VbiEB] = [(2)(11.8)(8.85X10 )(0.757)] : 9.94X10_50m NDB (1.6X10‘19X1015) [21(580 NAC Jill (2X11.8)(8.85X10‘14)(1014)(0.537) 1’2 xncg = *“mmvbiCB = w an3 NAc+NDB (1.6x10—19)(1015)(1.1x10”) = 2.52 x 10-5 cm and therefore W = 3 x 10-4 — 9.94 x 10-5 — 2.52 x 10-5 = 1.75 x 10—4 em = 1.75 film The emitter—base and collector—base built~in voltages (ngEB and Vbicg) were deduced from the EF — E values computed in pan (a). (e) -19 15 -5 lame—B) = qNDanEB z W 2 1,52 x 194 Wm K380 (11.8)(8.85X10‘14) ~19 15 -5 engage—B) = W“ xnCB : (1‘6X10 X10 )(Z'SZXIO ) = 3.86 x 103 V/cm K580 (11.8)(8.85X10‘14) 10—3 11):}; (a) For the given doping concentrations, one computes EF— E = 0.477eV, ~0.358eV, and 0.298eV respectively in the emitter, base, and collector. Also, with NDE >> NAB, the E-B depletion width will lie almost exclusively in the base. Likewise, the majority of the OB depletion width will lie in the collector. The diagram produced by the BJT_Eband program is displayed below. (b) 10—4 (C) » AVCE = (1/51)[(EF“Ei)collector“ (EF-Ei)emiuer] = (kT/q)[1n(NDC/ni) - 1n(N DIE/’10] 01' AVCE= (kT/q)1n(NDc/NDE) = (0.0259) 111(1015/1018) = —0.179 V (d) Analogous to Eq.(10.3) in the text, W = WB"xpEB“xpCB 1 2 [21cm MEET/2 : [(2)01.8)(8.85><10‘14)(0.835)] ’2 = 330x105 cm X EB ._ p qNAB (1.6x10-19x1016) x - [ZKseo NDC Wear/2 _ (2)(11.8)(8.85X10‘14)(1015x0656) 1’2 B — ~———_ -— —————————————————————-—-—-——————— 9C qNAB NDC+NAB 1 (1.6x10-19)(1016)(1.1x10“) = 8.82X 10—6 cm and therefore W = 2 x104 — 3.30 x 10-5 — 8.82 x 10-6 = 1.58 x 10*4 em = 1.58 pm The emitter—base and collector-base built-in voltages (Vb-[EB and VbiCB) were deduced from the E1: — E values computed in part (a). “*quAB _~ (1.6X10‘19)(1016)(3.30><10‘5) B w W (11.8)(885X10‘14) 2 5.06 X 104 V/em -19 16 6 [ElmaXCC-B) = W“ xpCB = (“X”) X10 )(g'gzxm ) = 1.35 x 104 V/cm K880 (11.8)(8.85><10-14) 10—5 105 (b) 10~6 M The energy band diagram for a typically doped Si npn transistor under equilibrium conditions was sketched in Fig. E10.1(a). Under active mode biasing in the npn transistor VBE > 0 and VBC < 0. Appropriately modifying the Fig. E10.1(a) diagram to account for the applied biases, we conclude Following the usual procedures in interpreting the energy band diagram to deduce the electrostatic variables, we conclude V 10—7 (Arrows Show the direction of carrier and/or current flow.) 10—8 19:3; ‘ ‘I-—u—-u-~----- 10«9 m : Leg 2 0.98 mA : (a) wr IE? lmA 0.9800 (b) 7 - [5" — ————————1mA = 0.9901 ” [EPHEB ‘ 1mA+0.01mA (C) 1E [1243+IEn = 1mA+0.01mA = 1.01 mA 5‘ 0 1CD +1'Cn = 0.98mA + 0.1,uA = 0.9801 mA. {3 = IE—Ic = 1.01mA—0.9801mA = 29.9 uA (d) adc = 705T = 0.9703 = 00c = 0.9703 z 27 fidc l—adc 14.9703 3', (e) As given by Eq. (10.12), 1080 =ICn = O‘IflA Likewise, Eq. (10.17) states [CEO _ 0.1,uA I = - = .7A CEO lmadc 1—0.9703 33” (f) The Icp increase while I Ep remains fixed indicates that the base transport factor has been improved. An increase in Ctr in turn leads to an increase in adc = 341T and therefore to an increase in fidc. (g) An increase in [En while 15;, remains fixed indicates that the emitter efficiency has been degraded. A decrease in y in turn leads to a decrease in (me 2 WT and therefore to a decrease in fidc. 10« 10 10.19 (a) ocr ~ ~— — IE" = 100% =09901 IEn+1Ep 100,uA+1,uA ' (b) 3’ = (c) [E = IEn+IEp = 100;LA+1}1A = 1011113. IC = ICn+Icp = 99yA+0.1,uA = 99.1;1A [B = IE—IC = IOWA—99.1w = 1.91m ((1) (Ida = year = 0.9802 2 vac = 0.9302 :4. fidc 1—0416, 1419802 95 (e) Analogous to Eq. (10.12), ICBO =1Cp = 0-1flA Likewise, analogous to Eq. (10.17), 1.0%C 1—09802 '05“ [CEO I (t) The [Cu increase while I En remains fixed indicates that the base transport factor has been improved“ An increase in oer in turn leads to an increase in adc = WT and therefore to an increase in fidc. (g) An increase in 13;, while [En remains fixed indicates that the emitter efficiency has been degraded. A decrease in y in turn leads to a decrease in adc = my and therefore to a decrease in fidco 10-11 1.11 As pictured below, there will indeed be some minority canier holes in the base that wander into the OB depletion region and thereby contribute to ICBO. However, because the base is very narrow, the quasineutral region generation that sustains the hole current is expected to be small, and the hole current itself is therefore expected to be negligible compared to [CI]. Quantitatively, employing an analysis similar to that in Exercise 6.4, and where the B and C subscripts refer to parameters in the base and collector, respectively. Since NDB > N Ac and W/LB << 1, and assuming Dch ~ DB/LB, we again conclude , << ICn. 10-12 ...
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chap10 - CHAPTER 10 112.1 (a) Common base. (b) Common...

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