(a)
The resistance for small reverse-bias voltages.
(a)
Answer to Problem 54P
The resistance for small reverse-bias voltages is
Explanation of Solution
Given:
The saturation current is
The value of
Formula used:
The expression for Ohm’s Law is
Here,
The expression for current in semiconductor is,
Calculation:
For small reverse-bias voltages i.e
The expression for current in semiconductor is then reduced to,
The expression for Ohm’s Law is then derived as,
The resistance is further calculated as,
Conclusion:
Therefore, the resistance for small reverse-bias voltages is
(b)
The resistance for reverse bias of
(b)
Answer to Problem 54P
The resistance for reverse bias of
Explanation of Solution
Formula used:
The resistance for reverse bias of
Calculation:
Evaluating the term
The resistance for reverse bias of
Conclusion:
Therefore, the resistance for reverse bias of
(c)
The resistance for a
(c)
Answer to Problem 54P
The resistance for forward bias of
Explanation of Solution
Formula used:
The resistance for forward bias of
Calculation:
Evaluating the term
The resistance for reverse bias of
Current is calculated from Ohm’s Law,
Conclusion:
Therefore, the resistance for forward bias of
(d)
The AC resistance for a
(d)
Answer to Problem 54P
The AC resistance for forward bias of
Explanation of Solution
Formula used:
The resistance for forward bias of
The AC resistance is expressed as,
Calculation:
The AC resistance is calculated as,
The calculation is further simplified as
Conclusion:
Therefore, the AC resistance for forward bias of
Want to see more full solutions like this?
Chapter 38 Solutions
Physics for Scientists and Engineers
- Why does the horizontal Line in the graph in Figure 9.12 suddenly stop at the Fermi energy? Figure 9.12 (a) Density of state for a free electron gas; (b) probability that a state is occupied at T = 0 K; (c) density if occupied states at T = 0 k.arrow_forwardFor silicon at T = 500 K with donor density N_D = 5* 10^{13} cm^ and acceptor density N_A = 0 calculate the equilibrium hole concentration in cm^{-3}. In this problem, you can assume the bandgap energy and effective masses are independent of temperature and use the room temperature values for them. Values within 5% error will be considered correct.arrow_forwardConsider a p+-n Si diode with NA = 1018 cm-3 and ND = 1016 cm-3. The hole diffusion coefficient in the n-side is 10 cm2/s and carrier lifetime τp =10-7 s. The device area is 10-4 cm2. Calculate the reverse saturation current and the forward current at a forward bias of 0.7 V at 300 K.arrow_forward
- Modern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage LearningUniversity Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStaxGlencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-Hill