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W2011_221A_HW4_soln

W2011_221A_HW4_soln - Prob 6.31 Prob 6.32 Part C Under high...

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Unformatted text preview: Prob. 6.31 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 region. 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 high. ...
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