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Unformatted text preview: Prob.
6.32 Part. C:
Under high injection conditions, electric field increases from left to
right in the sub-collector region (as shown in previous part).
Breakdown field in material A is less than that in material B. To
obtain a higher breakdown voltage, material A should be placed in a
low-field region and material B should be placed in a high-field
Hence, material A should be placed next to the base. Prob. 7.7
Identify which gives the best diode characteristics. ApE (a) IE = Ic , IB =• 0- Since V is
large, the collector is strongly
reverse biased, Ape = —PnSince Is = Ic, APE = —Ape
= pn from Eq.(7-34). The area
under Sp(xn) is zero. S? + «fn:<& (b) VCB = 0, thus Ape = 0. Notice
that this is the narrow-base
diode distribution. Sp 901 V Aps
ApE Ape ctApE 0 Wh (c) Since Ic = 0, Ape = aAps from
Eq.(7-34b). Wh (d) VEB = VCB = V. TAtw Ape = Ap£ Connection (b) gives the best diode since the stored charge is least and the current is
good. Connection (a) is not a good diode since the current is small and symmetrical
about V=0. Connections (c) and (d) are not good diodes because the stored charge is
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This note was uploaded on 12/29/2011 for the course ECE 221a taught by Professor Mishra,u during the Fall '08 term at UCSB.
- Fall '08