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An abrupt silicon pn junction at T = 300 K has impurity doping concentrations of Na = 5 X 1016 cm-3 and Na = 1015 cm-3. Calculate (a) Vbi, (b) W at (i) VR = 0 and (ii) VR = 5 V, and (c) |Emax | at (i) VR = 0 and (ii) VR = 5. 7.16 7.17 Consider the pn junction described in Problem 7.10 for T = 300 K. The cross-

An abrupt silicon pn junction at T = 300 K has impurity doping concentrations of Na = 5 X 1016 cm-3 and Na = 1015 cm-3. Calculate (a) Vbi, (b) W at (i) VR = 0 and (ii) VR = 5 V, and (c) |Emax | at (i) VR = 0 and (ii) VR = 5. 7.16 7.17 Consider the pn junction described in Problem 7.10 for T = 300 K. The cross-

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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7.15 Examine the electric field versus distance through a uniformly doped silicon pn junction at T = 300 K as a function of doping concentrations. Assume zero applied bias. Sketch the electric field versus distance through the space charge region and calculate |Emax| for: (a) Na = 1017 cm-3 and 1014 < Na< 107 cm 3 and (b) Na = 1014 cm 3 and 1014 < Na< 1017 cm 3. (c) What can be said about the results for Na 100 Na or Na 100 Na? AV Section 7.3 Reverse Applied Bias 7.16 An abrupt silicon pn junction at T = 300 K has impurity doping concentrations of Na = 5 X 10'6 cm 3 and Na= 1015 cm-3. Calculate (a) Vi, (b) W at (i) VR = 0 and (ii) VR = 5 V, and (c) |Emax | at (i) VR = 0 and (ii) VR = 5. 7.17 Consider the pn junction described in Problem 7.10 for T = 300 K. The cross- sectional area of the junction is 2 X 10-4 cm? and the applied reverse-biased voltage is VR = 2.5 V. Calculate (a) Vi, (b) Xn, Xp, W, (c) |Emaxl, and (d) the junction capacitance. An ideal one-sided silicon pn junction at T = 300 K is uniformly doped on both sides of the metallurgical junction. It is found that the doping relation is Na = 80 Na and the built-in potential barrier is Vi = 0.740 V. A reverse-biased voltage of VR = 10 V is applied. Determine (a) Na, Na; (b) Xp, Xn; (C) |Emax |; and (d) C. A silicon n*p junction is biased at VR = 5 V. (a) Determine the change in built-in potential barrier if the doping concentration in the p region increases by a factor of 3. (b) Determine the ratio of junction capacitance when the acceptor doping is 3Na com- pared to that when the acceptor doping is N.. (c) Why does the junction capacitance increase when the doping concentration increases? 7.18 %3D 7.19 7.20 (a) The peak electric field in a reverse-biased silicon pn junction is [Emax| = 3 X 10 V/cm. The doping concentrations are Na = 4 X 1015 cm 3 and Na = 4 X 1017 cm-3. Find the magnitude of the reverse-biased voltage. (b) Repeat part (a) for Na = 4 × 1016 cm 3 and %3D 1Y 1017 cm 3
Transcribed Image Text:7.15 Examine the electric field versus distance through a uniformly doped silicon pn junction at T = 300 K as a function of doping concentrations. Assume zero applied bias. Sketch the electric field versus distance through the space charge region and calculate |Emax| for: (a) Na = 1017 cm-3 and 1014 < Na< 107 cm 3 and (b) Na = 1014 cm 3 and 1014 < Na< 1017 cm 3. (c) What can be said about the results for Na 100 Na or Na 100 Na? AV Section 7.3 Reverse Applied Bias 7.16 An abrupt silicon pn junction at T = 300 K has impurity doping concentrations of Na = 5 X 10'6 cm 3 and Na= 1015 cm-3. Calculate (a) Vi, (b) W at (i) VR = 0 and (ii) VR = 5 V, and (c) |Emax | at (i) VR = 0 and (ii) VR = 5. 7.17 Consider the pn junction described in Problem 7.10 for T = 300 K. The cross- sectional area of the junction is 2 X 10-4 cm? and the applied reverse-biased voltage is VR = 2.5 V. Calculate (a) Vi, (b) Xn, Xp, W, (c) |Emaxl, and (d) the junction capacitance. An ideal one-sided silicon pn junction at T = 300 K is uniformly doped on both sides of the metallurgical junction. It is found that the doping relation is Na = 80 Na and the built-in potential barrier is Vi = 0.740 V. A reverse-biased voltage of VR = 10 V is applied. Determine (a) Na, Na; (b) Xp, Xn; (C) |Emax |; and (d) C. A silicon n*p junction is biased at VR = 5 V. (a) Determine the change in built-in potential barrier if the doping concentration in the p region increases by a factor of 3. (b) Determine the ratio of junction capacitance when the acceptor doping is 3Na com- pared to that when the acceptor doping is N.. (c) Why does the junction capacitance increase when the doping concentration increases? 7.18 %3D 7.19 7.20 (a) The peak electric field in a reverse-biased silicon pn junction is [Emax| = 3 X 10 V/cm. The doping concentrations are Na = 4 X 1015 cm 3 and Na = 4 X 1017 cm-3. Find the magnitude of the reverse-biased voltage. (b) Repeat part (a) for Na = 4 × 1016 cm 3 and %3D 1Y 1017 cm 3
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