EE130/230A Discussion 14 Peng Zheng 1 “Game Plan” for I-V Derivation • Solve the minority-carrier diffusion equation in each quasineutral region to obtain excess minority-carrier profiles – different set of boundary conditions for each region • Find minority-carrier diffusion currents at depletion region edges dnE E dx" x"0 I En qAD dnC C dx ' x ' 0 I Cn qAD dp B B dx x 0 I Ep qAD dp B B dx x W I Cp qAD • Add hole & electron components together terminal currents EE130/230A Fall 2013 Lecture 26, Slide 2 BJT Terminal Currents • We know: I En qA DE LE I Ep qA DB LB I Cp qA DB LB I Cn qA nE 0 (e qVEB / kT 1) pB0 p DC LC 1 B 0 sinh(W / LB ) nC 0 (e qVCB / kT • Therefore: qA I E qA IC DE LE nE 0 DB LB EE130/230A Fall 2013 DB LB qVCB / kT qVEB / kT 1 ( e 1 ) e 1 sinh(W / LB ) sinh(W / LB ) cosh(W / LB ) (e qVEB / kT 1) e qVCB / kT 1 1) pB 0 sinh(W / LBB )) (eqVEB / kT 1) cosh(W / L cosh(W / LB ) sinh(W / LB ) pB0 sinh(W1 / LB ) (eqVEB / kT 1) DC LC Lecture 26, Slide 3 DB LB pB 0 sinh(W1 / LB ) eqVCB / kT 1 nC 0 DLBB pB 0 cosh(W / LB ) sinh(W / LB ) e qVCB / kT 1 BJT with Narrow Base • In practice, we make W << LB to achieve high current gain. Then, since sinh for 1 cosh 1 2 for 1 2 we have: pB ( x) pB 0 (e qVEB / kT 1)1 Wx pB 0 (e qVCB / kT 1)Wx EE130/230A Fall 2013 Lecture 26, Slide 4 R. F. Pierret, Semiconductor Device Fundamentals, Fig. 11.2 Ebers-Moll Model increasing (npn) or VEC (pnp) C. C. Hu, Modern Semiconductor Devices for Integrated Circuits, Figure 8-2 The Ebers-Moll model is a large-signal equivalent circuit which describes both the active and saturation regions of BJT operation. • Use this model to calculate IB and IC given VBE and VBC EE130/230A Fall 2013 Lecture 26, Slide 5 qA I E qA DE LE IC DB LB nE 0 DB LB 1 B 0 sinh(W / LB ) p pB 0 sinh(W / LB ) (eqVEB / kT 1) cosh(W / LB ) (e qVEB / kT 1) DC LC nC 0 If only VEB is applied (VCB = 0): DB LB DB LB e 1 pB 0 sinh(W1 / LB ) eqVCB / kT 1 cosh(W / LB ) pB 0 sinh(W / LB ) qVCB / kT V EB V CB I E I F 0 ( e qVEB / kT 1) IB I C F I F 0 ( e qVEB / kT 1) I B 1 F I F 0 ( e qVEB / kT 1) If only VCB is applied (VEB = 0): : I C I R 0 (e qVCB / kT 1) I E R I R 0 (e qVCB / kT 1) I B I R 0 (1 R )(e EE130/230A Fall 2013 qVCB / kT 1) E B C IC aR : reverse common base gain aF : forward common base gain Reciprocity relationship: DB pB 0 F I F 0 R I R 0 qA LB sinh( W / LB ) Lecture 26, Slide 6 In the general case, both VEB and VCB are non-zero: I C F I F 0 (e qVEB / kT 1) I R 0 (e qVCB / kT 1) IC: C-B diode current + fraction of E-B diode current that makes it to the C-B junction I E I F 0 (e qVEB / kT 1) R I R 0 (e qVCB / kT 1) IE: E-B diode current + fraction of C-B diode current that makes it to the E-B junction Large-signal equivalent circuit for a pnp BJT R. F. Pierret, Semiconductor Device Fundamentals, Fig. 11.3 EE130/230A Fall 2013 Lecture 26, Slide 7 Summary: BJT Performance Requirements • High gain (bdc >> 1) One-sided emitter junction, so emitter efficiency g 1 • Emitter doped much more heavily than base (NE >> NB) Narrow base, so base transport factor T 1 • Quasi-neutral base width << minority-carrier diffusion length (W << LB) • IC determined only by IB (IC function of VCE,VCB) One-sided collector junction, so quasi-neutral base width W does not change drastically with changes in VCE (VCB) • Based doped more heavily than collector (NB > NC) (W = WB – xnEB – xnCB for PNP BJT) EE130/230A Fall 2013 Lecture 26, Slide 8 Questions regarding the MOSFET design project? Good luck to Quiz#6! 9