microelectronics by sedra and smith 8th edition chapter 3 Archives

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venkyelectrical

Asked: January 6, 2022In: microelectronics

*3.31 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 Wp and Wn then

*3.31 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 ...

venkyelectrical

Asked: January 6, 2022In: microelectronics

3.30 A pn junction operating in the forward-bias region with a current I of 0.4 mA is found to have a diffusion capacitance of 1 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? 0.25 pF; 64.8 ps

3.30 A pn junction operating in the forward-bias region with a current I of 0.4 mA is found to have a diffusion capacitance of 1 pF. What diffusion capacitance do you expect this junction to have at I = 0.1 ...

venkyelectrical

Asked: January 6, 2022In: microelectronics

3.29 The junction capacitance Cj can be thought of as that of a parallel-plate capacitor and thus given by Show that this approach leads to a formula identical to that obtained by combining Eqs. (3.42) and (3.44) [or equivalently, by combining Eqs. (3.44) and (3.45)].

3.29 The junction capacitance Cj can be thought of as that of a parallel-plate capacitor and thus given by

venkyelectrical

Asked: January 6, 2022In: microelectronics

3.28 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 3 V. 0.23 pF

venkyelectrical

Asked: January 6, 2022In: microelectronics

3.27 For the pn junction specified in Problem 3.14, find Cj0 and Cj at VR = 2V.

venkyelectrical

Asked: January 6, 2022In: microelectronics

3.26 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 occurs for only 10 ms in every 20 ms, what average breakdown current is allowed? 10.42 mA; 41.7 mA

3.26 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 ...

venkyelectrical

Asked: January 6, 2022In: microelectronics

3.25 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 For the specific case in which ND = 1016/cm3, Dp = 10 cm2/s, Lp = 10 µm, and A = 104 µm2, find IS and the voltage V obtained when I = 0.2 mA. Assume operation at 300 K where ni = 1.5 × 1010/cm3.

3.25 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. ...

venkyelectrical

Asked: January 6, 2022In: microelectronics

3.24 Assuming that the temperature dependence of IS arises mostly because IS is proportional to , use the expression for ni in Eq. (3.2) to determine the factor by which changes as T changes from 300 K to 310 K. This will be approximately the same factor by which IS changes for a 10°C rise in temperature. What is the factor?

3.24 Assuming that the temperature dependence of IS arises mostly because IS is proportional to , use the expression for ni in Eq. (3.2) to determine the factor by which changes as T changes from 300 K to 310 K. ...

venkyelectrical

Asked: January 6, 2022In: microelectronics

3.23 Calculate IS and the current I for V = 780 mV for a pn junction for which NA = 1017/cm3, ND = 1016/cm3, A = 20 µm2, ni = 1.5 × 1010/cm3, Lp = 5 µm, Ln = 10 µm, Dp = 10 cm2/s, and Dn = 18 cm2/s. 1.57 × 10−17 A; 1.88 mA

3.23 Calculate IS and the current I for V = 780 mV for a pn junction for which NA = 1017/cm3, ND = 1016/cm3, A = 20 µm2, ni = 1.5 × 1010/cm3, Lp = 5 µm, Ln = 10 ...

venkyelectrical

Asked: January 6, 2022In: microelectronics

3.22 In a forward-biased pn junction show that the ratio of the current component due to hole injection across the junction to the component due to electron injection is given by Evaluate this ratio for the case NA = 1018/cm3, ND = 1016/cm3, Lp = 5 µm, Ln = 10 µm, Dp = 10 cm2/s, and Dn = 20 cm2/s, and hence find Ip and In for the case in which the pn junction is conducting a forward current I = 100 µA.

3.22 In a forward-biased pn junction show that the ratio of the current component due to hole injection across the junction to the component due to electron injection is given by

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