*3.29 A short-base diode is one where the widths of the p and n regions are much smaller than Ln and Lp, respectively. As a result, the excess minority-carrier distribution in each region is a straight line rather than the exponentials shown in Fig. 3.12. (a) For the short-base diode, sketch a figure corresponding to Fig. 3.12 and assume as in Fig. 3.12 that NA ND. (b) Following a derivation similar to that given in Section 3.5.2, show that if the widths of the p and n regions are denoted W p and Wn then I = Aqni2Wn −Dxp nND + Wp −DnxpNA eV/VT −1 and Qp = 1 2 Wn −xn2 D p Ip 1 2 W 2 n D p Ip , for Wn xn (c) Also, assuming Q Qp, I Ip, show that Cd = τ T VT I where τ T = 1 2 W 2 n D p (d) If a designer wishes to limit Cd to 8 pF at I = 1 mA, what should W n be? Assume Dp = 10 cm2/s.
3.28 For the p+n junction specified in Problem 3.22, find τp and calculate the excess minority-carrier charge and the value of the diffusion capacitance at I = 0.1 mA.
3.27 A pn junction operating in the forward-bias region with a current I of 1 mA is found to have a diffusion capacitance of 5 pF. What diffusion capacitance do you expect this junction to have at I = 0.1 mA? What is the mean transit time for this junction?
3.26 The junction capacitance Cj can be thought of as that of a parallel-plate capacitor and thus given by Cj = eA W Show that this approach leads to a formula identical to that obtained by combiningEqs.(3.43) and(3.45)[or equivalently, by combining Eqs. (3.47) and (3.48)].
3.25 For a particular junction for which Cj0 = 0.4 pF, V0 = 0.75 V, and m = 1/3, find Cj at reverse-bias voltages of 1 V and 10 V
3.23 A pn junction for which the breakdown voltage is 12 V has a rated (i.e., maximum allowable) power dissipation of 0.25 W. What continuous current in the breakdown region will raise the dissipation to half the rated value? If breakdown occursfor only 10ms in every 20ms, what average breakdown current is allowed?
3.22 A p+n junction is one in which the doping concentration in the p region is much greater than that in the n region. In such a junction, the forward current is mostly due to hole injection across the junction. Show that I I p = Aqni2 Dp Lp ND eV/VT −1 For the specific case in which ND = 1017/cm3, Dp = 10 cm2/s, Lp = 10 μm, and A = 104 μm2, find IS and the voltage V obtained when I = 1 mA. Assume operation at 300 K where n i = 1.5×1010/cm3.
3.21 Assuming that the temperature dependence of IS arises mostly because IS is proportional to ni2, use the expression for n i in Eq. (3.2) to determine the factor by which ni2 changes as T changes from 300 K to 305 K. This will be approximately the same factor by which IS changes for a 5°C rise in temperature. What is the factor?
3.20 Calculate I S and the current I for V = 750 mV for a pn junction for which NA = 1017/cm3, ND = 1016/cm3, A = 100 μm2, ni = 1.5 × 1010/cm3, Lp = 5 μm, Ln = 10 μm, D p = 10 cm2/s, and Dn = 18 cm2/s.