The resistance for small reverse-bias voltages.

Question

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Answer 1

Question

Chapter 38, Problem 54P

(a)

To determine

The resistance for small reverse-bias voltages.

(a)

Expert Solution

Answer to Problem 54P

The resistance for small reverse-bias voltages is 25 MΩ .

Explanation of Solution

Given:

The saturation current is I0= 1.0 nA .

The value of kT=0.025 eV .

Formula used:

The expression for Ohm’s Law is

R=VbI

Here, R is the resistance, Vb is the voltage and I is the current in the circuit.

The expression for current in semiconductor is,

I=I0(ee V b/kT−1)

Calculation:

For small reverse-bias voltages i.e eVb<<kT

ee V b/kT−1≈1+eVbkT−1=eVbkT

The expression for current in semiconductor is then reduced to,

I=I0(e e V b / kT −1)=I0eVbkT

The expression for Ohm’s Law is then derived as,

R=VbI0 e V b kT=kTeI0

The resistance is further calculated as,

R=kTeI0=( 0.025 eV)( 1.6× 10 −19 J/ eV )( 1.6× 10 −19 C)( 1.0× 10 −9 A)=25 MΩ

Conclusion:

Therefore, the resistance for small reverse-bias voltages is 25 MΩ .

(b)

To determine

The resistance for reverse bias of 0.50 V

(b)

Expert Solution

Answer to Problem 54P

The resistance for reverse bias of 0.50 V is 5.0×108 Ω .

Explanation of Solution

Formula used:

The resistance for reverse bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

Calculation:

Evaluating the term eVbkT for Vb=−0.50 V

eVbkT=( 1.6× 10 −19 C)( −0.50 V)( 0.025 eV)( 1.6× 10 −19 J/ eV )=−19.8

The resistance for reverse bias of 0.50 V is calculated as,

R=VbI0( e e V b / kT −1)=−0.50 V( 1.0× 10 9 A)( e −19.8 −1)=5.0×108 Ω

Conclusion:

Therefore, the resistance for reverse bias of 0.50 V is 5.0×108 Ω .

(c)

To determine

The resistance for a 0.50 V forward bias and the corresponding current.

(c)

Expert Solution

Answer to Problem 54P

The resistance for forward bias of 0.50 V is 1.3 Ω and corresponding current is 0.38 A .

Explanation of Solution

Formula used:

The resistance for forward bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

Calculation:

Evaluating the term eVbkT for Vb=−0.50 V

eVbkT=( 1.6× 10 −19 C)( 0.50 V)( 0.025 eV)( 1.6× 10 −19 J/ eV )=19.8

The resistance for reverse bias of 0.50 V is calculated as,

R=VbI0( e e V b / kT −1)=0.50 V( 1.0× 10 9 A)( e 19.8 −1)=1.3 Ω

Current is calculated from Ohm’s Law,

I=VR=0.50 V1.3 Ω=0.38 A

Conclusion:

Therefore, the resistance for forward bias of 0.50 V is 1.3 Ω and corresponding current is 0.38 A .

(d)

To determine

The AC resistance for a 0.50 V forward bias voltage.

(d)

Expert Solution

Answer to Problem 54P

The AC resistance for forward bias of 0.50 V is 63 mΩ .

Explanation of Solution

Formula used:

The resistance for forward bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

The AC resistance is expressed as,

Rac=dVdI=( dI dV)−1

Calculation:

The AC resistance is calculated as,

Rac={d dV( I 0 ( e e V b / kT −1 ))}−1={ e I 0 kTe e V b / kT }−1=kTeI0e−e V b/kT=(25 MΩ)e−19.8

The calculation is further simplified as

Rac=25×106 Ω×2.5×10−9 Ω=63 mΩ

Conclusion:

Therefore, the AC resistance for forward bias of 0.50 V is 63 mΩ .

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Answer 2

Question

Chapter 38, Problem 54P

(a)

To determine

The resistance for small reverse-bias voltages.

(a)

Expert Solution

Answer to Problem 54P

The resistance for small reverse-bias voltages is 25 MΩ .

Explanation of Solution

Given:

The saturation current is I0= 1.0 nA .

The value of kT=0.025 eV .

Formula used:

The expression for Ohm’s Law is

R=VbI

Here, R is the resistance, Vb is the voltage and I is the current in the circuit.

The expression for current in semiconductor is,

I=I0(ee V b/kT−1)

Calculation:

For small reverse-bias voltages i.e eVb<<kT

ee V b/kT−1≈1+eVbkT−1=eVbkT

The expression for current in semiconductor is then reduced to,

I=I0(e e V b / kT −1)=I0eVbkT

The expression for Ohm’s Law is then derived as,

R=VbI0 e V b kT=kTeI0

The resistance is further calculated as,

R=kTeI0=( 0.025 eV)( 1.6× 10 −19 J/ eV )( 1.6× 10 −19 C)( 1.0× 10 −9 A)=25 MΩ

Conclusion:

Therefore, the resistance for small reverse-bias voltages is 25 MΩ .

(b)

To determine

The resistance for reverse bias of 0.50 V

(b)

Expert Solution

Answer to Problem 54P

The resistance for reverse bias of 0.50 V is 5.0×108 Ω .

Explanation of Solution

Formula used:

The resistance for reverse bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

Calculation:

Evaluating the term eVbkT for Vb=−0.50 V

eVbkT=( 1.6× 10 −19 C)( −0.50 V)( 0.025 eV)( 1.6× 10 −19 J/ eV )=−19.8

The resistance for reverse bias of 0.50 V is calculated as,

R=VbI0( e e V b / kT −1)=−0.50 V( 1.0× 10 9 A)( e −19.8 −1)=5.0×108 Ω

Conclusion:

Therefore, the resistance for reverse bias of 0.50 V is 5.0×108 Ω .

(c)

To determine

The resistance for a 0.50 V forward bias and the corresponding current.

(c)

Expert Solution

Answer to Problem 54P

The resistance for forward bias of 0.50 V is 1.3 Ω and corresponding current is 0.38 A .

Explanation of Solution

Formula used:

The resistance for forward bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

Calculation:

Evaluating the term eVbkT for Vb=−0.50 V

eVbkT=( 1.6× 10 −19 C)( 0.50 V)( 0.025 eV)( 1.6× 10 −19 J/ eV )=19.8

The resistance for reverse bias of 0.50 V is calculated as,

R=VbI0( e e V b / kT −1)=0.50 V( 1.0× 10 9 A)( e 19.8 −1)=1.3 Ω

Current is calculated from Ohm’s Law,

I=VR=0.50 V1.3 Ω=0.38 A

Conclusion:

Therefore, the resistance for forward bias of 0.50 V is 1.3 Ω and corresponding current is 0.38 A .

(d)

To determine

The AC resistance for a 0.50 V forward bias voltage.

(d)

Expert Solution

Answer to Problem 54P

The AC resistance for forward bias of 0.50 V is 63 mΩ .

Explanation of Solution

Formula used:

The resistance for forward bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

The AC resistance is expressed as,

Rac=dVdI=( dI dV)−1

Calculation:

The AC resistance is calculated as,

Rac={d dV( I 0 ( e e V b / kT −1 ))}−1={ e I 0 kTe e V b / kT }−1=kTeI0e−e V b/kT=(25 MΩ)e−19.8

The calculation is further simplified as

Rac=25×106 Ω×2.5×10−9 Ω=63 mΩ

Conclusion:

Therefore, the AC resistance for forward bias of 0.50 V is 63 mΩ .

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Answer 3

Question

Chapter 38, Problem 54P

(a)

To determine

The resistance for small reverse-bias voltages.

(a)

Expert Solution

Answer to Problem 54P

The resistance for small reverse-bias voltages is 25 MΩ .

Explanation of Solution

Given:

The saturation current is I0= 1.0 nA .

The value of kT=0.025 eV .

Formula used:

The expression for Ohm’s Law is

R=VbI

Here, R is the resistance, Vb is the voltage and I is the current in the circuit.

The expression for current in semiconductor is,

I=I0(ee V b/kT−1)

Calculation:

For small reverse-bias voltages i.e eVb<<kT

ee V b/kT−1≈1+eVbkT−1=eVbkT

The expression for current in semiconductor is then reduced to,

I=I0(e e V b / kT −1)=I0eVbkT

The expression for Ohm’s Law is then derived as,

R=VbI0 e V b kT=kTeI0

The resistance is further calculated as,

R=kTeI0=( 0.025 eV)( 1.6× 10 −19 J/ eV )( 1.6× 10 −19 C)( 1.0× 10 −9 A)=25 MΩ

Conclusion:

Therefore, the resistance for small reverse-bias voltages is 25 MΩ .

(b)

To determine

The resistance for reverse bias of 0.50 V

(b)

Expert Solution

Answer to Problem 54P

The resistance for reverse bias of 0.50 V is 5.0×108 Ω .

Explanation of Solution

Formula used:

The resistance for reverse bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

Calculation:

Evaluating the term eVbkT for Vb=−0.50 V

eVbkT=( 1.6× 10 −19 C)( −0.50 V)( 0.025 eV)( 1.6× 10 −19 J/ eV )=−19.8

The resistance for reverse bias of 0.50 V is calculated as,

R=VbI0( e e V b / kT −1)=−0.50 V( 1.0× 10 9 A)( e −19.8 −1)=5.0×108 Ω

Conclusion:

Therefore, the resistance for reverse bias of 0.50 V is 5.0×108 Ω .

(c)

To determine

The resistance for a 0.50 V forward bias and the corresponding current.

(c)

Expert Solution

Answer to Problem 54P

The resistance for forward bias of 0.50 V is 1.3 Ω and corresponding current is 0.38 A .

Explanation of Solution

Formula used:

The resistance for forward bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

Calculation:

Evaluating the term eVbkT for Vb=−0.50 V

eVbkT=( 1.6× 10 −19 C)( 0.50 V)( 0.025 eV)( 1.6× 10 −19 J/ eV )=19.8

The resistance for reverse bias of 0.50 V is calculated as,

R=VbI0( e e V b / kT −1)=0.50 V( 1.0× 10 9 A)( e 19.8 −1)=1.3 Ω

Current is calculated from Ohm’s Law,

I=VR=0.50 V1.3 Ω=0.38 A

Conclusion:

Therefore, the resistance for forward bias of 0.50 V is 1.3 Ω and corresponding current is 0.38 A .

(d)

To determine

The AC resistance for a 0.50 V forward bias voltage.

(d)

Expert Solution

Answer to Problem 54P

The AC resistance for forward bias of 0.50 V is 63 mΩ .

Explanation of Solution

Formula used:

The resistance for forward bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

The AC resistance is expressed as,

Rac=dVdI=( dI dV)−1

Calculation:

The AC resistance is calculated as,

Rac={d dV( I 0 ( e e V b / kT −1 ))}−1={ e I 0 kTe e V b / kT }−1=kTeI0e−e V b/kT=(25 MΩ)e−19.8

The calculation is further simplified as

Rac=25×106 Ω×2.5×10−9 Ω=63 mΩ

Conclusion:

Therefore, the AC resistance for forward bias of 0.50 V is 63 mΩ .

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Answer 4

Question

Chapter 38, Problem 54P

(a)

To determine

The resistance for small reverse-bias voltages.

(a)

Expert Solution

Answer to Problem 54P

The resistance for small reverse-bias voltages is 25 MΩ .

Explanation of Solution

Given:

The saturation current is I0= 1.0 nA .

The value of kT=0.025 eV .

Formula used:

The expression for Ohm’s Law is

R=VbI

Here, R is the resistance, Vb is the voltage and I is the current in the circuit.

The expression for current in semiconductor is,

I=I0(ee V b/kT−1)

Calculation:

For small reverse-bias voltages i.e eVb<<kT

ee V b/kT−1≈1+eVbkT−1=eVbkT

The expression for current in semiconductor is then reduced to,

I=I0(e e V b / kT −1)=I0eVbkT

The expression for Ohm’s Law is then derived as,

R=VbI0 e V b kT=kTeI0

The resistance is further calculated as,

R=kTeI0=( 0.025 eV)( 1.6× 10 −19 J/ eV )( 1.6× 10 −19 C)( 1.0× 10 −9 A)=25 MΩ

Conclusion:

Therefore, the resistance for small reverse-bias voltages is 25 MΩ .

(b)

To determine

The resistance for reverse bias of 0.50 V

(b)

Expert Solution

Answer to Problem 54P

The resistance for reverse bias of 0.50 V is 5.0×108 Ω .

Explanation of Solution

Formula used:

The resistance for reverse bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

Calculation:

Evaluating the term eVbkT for Vb=−0.50 V

eVbkT=( 1.6× 10 −19 C)( −0.50 V)( 0.025 eV)( 1.6× 10 −19 J/ eV )=−19.8

The resistance for reverse bias of 0.50 V is calculated as,

R=VbI0( e e V b / kT −1)=−0.50 V( 1.0× 10 9 A)( e −19.8 −1)=5.0×108 Ω

Conclusion:

Therefore, the resistance for reverse bias of 0.50 V is 5.0×108 Ω .

(c)

To determine

The resistance for a 0.50 V forward bias and the corresponding current.

(c)

Expert Solution

Answer to Problem 54P

The resistance for forward bias of 0.50 V is 1.3 Ω and corresponding current is 0.38 A .

Explanation of Solution

Formula used:

The resistance for forward bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

Calculation:

Evaluating the term eVbkT for Vb=−0.50 V

eVbkT=( 1.6× 10 −19 C)( 0.50 V)( 0.025 eV)( 1.6× 10 −19 J/ eV )=19.8

The resistance for reverse bias of 0.50 V is calculated as,

R=VbI0( e e V b / kT −1)=0.50 V( 1.0× 10 9 A)( e 19.8 −1)=1.3 Ω

Current is calculated from Ohm’s Law,

I=VR=0.50 V1.3 Ω=0.38 A

Conclusion:

Therefore, the resistance for forward bias of 0.50 V is 1.3 Ω and corresponding current is 0.38 A .

(d)

To determine

The AC resistance for a 0.50 V forward bias voltage.

(d)

Expert Solution

Answer to Problem 54P

The AC resistance for forward bias of 0.50 V is 63 mΩ .

Explanation of Solution

Formula used:

The resistance for forward bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

The AC resistance is expressed as,

Rac=dVdI=( dI dV)−1

Calculation:

The AC resistance is calculated as,

Rac={d dV( I 0 ( e e V b / kT −1 ))}−1={ e I 0 kTe e V b / kT }−1=kTeI0e−e V b/kT=(25 MΩ)e−19.8

The calculation is further simplified as

Rac=25×106 Ω×2.5×10−9 Ω=63 mΩ

Conclusion:

Therefore, the AC resistance for forward bias of 0.50 V is 63 mΩ .

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Answer 5

Chapter 38, Problem 54P

(a)

To determine

The resistance for small reverse-bias voltages.

(a)

Expert Solution

Answer to Problem 54P

The resistance for small reverse-bias voltages is 25 MΩ .

Explanation of Solution

Given:

The saturation current is I0= 1.0 nA .

The value of kT=0.025 eV .

Formula used:

The expression for Ohm’s Law is

R=VbI

Here, R is the resistance, Vb is the voltage and I is the current in the circuit.

The expression for current in semiconductor is,

I=I0(ee V b/kT−1)

Calculation:

For small reverse-bias voltages i.e eVb<<kT

ee V b/kT−1≈1+eVbkT−1=eVbkT

The expression for current in semiconductor is then reduced to,

I=I0(e e V b / kT −1)=I0eVbkT

The expression for Ohm’s Law is then derived as,

R=VbI0 e V b kT=kTeI0

The resistance is further calculated as,

R=kTeI0=( 0.025 eV)( 1.6× 10 −19 J/ eV )( 1.6× 10 −19 C)( 1.0× 10 −9 A)=25 MΩ

Conclusion:

Therefore, the resistance for small reverse-bias voltages is 25 MΩ .

(b)

To determine

The resistance for reverse bias of 0.50 V

(b)

Expert Solution

Answer to Problem 54P

The resistance for reverse bias of 0.50 V is 5.0×108 Ω .

Explanation of Solution

Formula used:

The resistance for reverse bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

Calculation:

Evaluating the term eVbkT for Vb=−0.50 V

eVbkT=( 1.6× 10 −19 C)( −0.50 V)( 0.025 eV)( 1.6× 10 −19 J/ eV )=−19.8

The resistance for reverse bias of 0.50 V is calculated as,

R=VbI0( e e V b / kT −1)=−0.50 V( 1.0× 10 9 A)( e −19.8 −1)=5.0×108 Ω

Conclusion:

Therefore, the resistance for reverse bias of 0.50 V is 5.0×108 Ω .

(c)

To determine

The resistance for a 0.50 V forward bias and the corresponding current.

(c)

Expert Solution

Answer to Problem 54P

The resistance for forward bias of 0.50 V is 1.3 Ω and corresponding current is 0.38 A .

Explanation of Solution

Formula used:

The resistance for forward bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

Calculation:

Evaluating the term eVbkT for Vb=−0.50 V

eVbkT=( 1.6× 10 −19 C)( 0.50 V)( 0.025 eV)( 1.6× 10 −19 J/ eV )=19.8

The resistance for reverse bias of 0.50 V is calculated as,

R=VbI0( e e V b / kT −1)=0.50 V( 1.0× 10 9 A)( e 19.8 −1)=1.3 Ω

Current is calculated from Ohm’s Law,

I=VR=0.50 V1.3 Ω=0.38 A

Conclusion:

Therefore, the resistance for forward bias of 0.50 V is 1.3 Ω and corresponding current is 0.38 A .

(d)

To determine

The AC resistance for a 0.50 V forward bias voltage.

(d)

Expert Solution

Answer to Problem 54P

The AC resistance for forward bias of 0.50 V is 63 mΩ .

Explanation of Solution

Formula used:

The resistance for forward bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

The AC resistance is expressed as,

Rac=dVdI=( dI dV)−1

Calculation:

The AC resistance is calculated as,

Rac={d dV( I 0 ( e e V b / kT −1 ))}−1={ e I 0 kTe e V b / kT }−1=kTeI0e−e V b/kT=(25 MΩ)e−19.8

The calculation is further simplified as

Rac=25×106 Ω×2.5×10−9 Ω=63 mΩ

Conclusion:

Therefore, the AC resistance for forward bias of 0.50 V is 63 mΩ .

Answer 6

Chapter 38, Problem 54P

(a)

To determine

The resistance for small reverse-bias voltages.

(a)

Expert Solution

Answer to Problem 54P

The resistance for small reverse-bias voltages is 25 MΩ .

Explanation of Solution

Given:

The saturation current is I0= 1.0 nA .

The value of kT=0.025 eV .

Formula used:

The expression for Ohm’s Law is

R=VbI

Here, R is the resistance, Vb is the voltage and I is the current in the circuit.

The expression for current in semiconductor is,

I=I0(ee V b/kT−1)

Calculation:

For small reverse-bias voltages i.e eVb<<kT

ee V b/kT−1≈1+eVbkT−1=eVbkT

The expression for current in semiconductor is then reduced to,

I=I0(e e V b / kT −1)=I0eVbkT

The expression for Ohm’s Law is then derived as,

R=VbI0 e V b kT=kTeI0

The resistance is further calculated as,

R=kTeI0=( 0.025 eV)( 1.6× 10 −19 J/ eV )( 1.6× 10 −19 C)( 1.0× 10 −9 A)=25 MΩ

Conclusion:

Therefore, the resistance for small reverse-bias voltages is 25 MΩ .

Answer 7

Chapter 38, Problem 54P

(a)

To determine

The resistance for small reverse-bias voltages.

Answer 8

(a)

Expert Solution

Answer to Problem 54P

The resistance for small reverse-bias voltages is 25 MΩ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The saturation current is I0= 1.0 nA .

The value of kT=0.025 eV .

Formula used:

The expression for Ohm’s Law is

R=VbI

Here, R is the resistance, Vb is the voltage and I is the current in the circuit.

The expression for current in semiconductor is,

I=I0(ee V b/kT−1)

Calculation:

For small reverse-bias voltages i.e eVb<<kT

ee V b/kT−1≈1+eVbkT−1=eVbkT

The expression for current in semiconductor is then reduced to,

I=I0(e e V b / kT −1)=I0eVbkT

The expression for Ohm’s Law is then derived as,

R=VbI0 e V b kT=kTeI0

The resistance is further calculated as,

R=kTeI0=( 0.025 eV)( 1.6× 10 −19 J/ eV )( 1.6× 10 −19 C)( 1.0× 10 −9 A)=25 MΩ

Conclusion:

Therefore, the resistance for small reverse-bias voltages is 25 MΩ .

Answer 13

(b)

To determine

The resistance for reverse bias of 0.50 V

(b)

Expert Solution

Answer to Problem 54P

The resistance for reverse bias of 0.50 V is 5.0×108 Ω .

Explanation of Solution

Formula used:

The resistance for reverse bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

Calculation:

Evaluating the term eVbkT for Vb=−0.50 V

eVbkT=( 1.6× 10 −19 C)( −0.50 V)( 0.025 eV)( 1.6× 10 −19 J/ eV )=−19.8

The resistance for reverse bias of 0.50 V is calculated as,

R=VbI0( e e V b / kT −1)=−0.50 V( 1.0× 10 9 A)( e −19.8 −1)=5.0×108 Ω

Conclusion:

Therefore, the resistance for reverse bias of 0.50 V is 5.0×108 Ω .

Answer 14

(b)

To determine

The resistance for reverse bias of 0.50 V

Answer 15

(b)

Expert Solution

Answer to Problem 54P

The resistance for reverse bias of 0.50 V is 5.0×108 Ω .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Formula used:

The resistance for reverse bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

Calculation:

Evaluating the term eVbkT for Vb=−0.50 V

eVbkT=( 1.6× 10 −19 C)( −0.50 V)( 0.025 eV)( 1.6× 10 −19 J/ eV )=−19.8

The resistance for reverse bias of 0.50 V is calculated as,

R=VbI0( e e V b / kT −1)=−0.50 V( 1.0× 10 9 A)( e −19.8 −1)=5.0×108 Ω

Conclusion:

Therefore, the resistance for reverse bias of 0.50 V is 5.0×108 Ω .

Answer 20

(c)

To determine

The resistance for a 0.50 V forward bias and the corresponding current.

(c)

Expert Solution

Answer to Problem 54P

The resistance for forward bias of 0.50 V is 1.3 Ω and corresponding current is 0.38 A .

Explanation of Solution

Formula used:

The resistance for forward bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

Calculation:

Evaluating the term eVbkT for Vb=−0.50 V

eVbkT=( 1.6× 10 −19 C)( 0.50 V)( 0.025 eV)( 1.6× 10 −19 J/ eV )=19.8

The resistance for reverse bias of 0.50 V is calculated as,

R=VbI0( e e V b / kT −1)=0.50 V( 1.0× 10 9 A)( e 19.8 −1)=1.3 Ω

Current is calculated from Ohm’s Law,

I=VR=0.50 V1.3 Ω=0.38 A

Conclusion:

Therefore, the resistance for forward bias of 0.50 V is 1.3 Ω and corresponding current is 0.38 A .

Answer 21

(c)

To determine

The resistance for a 0.50 V forward bias and the corresponding current.

Answer 22

(c)

Expert Solution

Answer to Problem 54P

The resistance for forward bias of 0.50 V is 1.3 Ω and corresponding current is 0.38 A .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Formula used:

The resistance for forward bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

Calculation:

Evaluating the term eVbkT for Vb=−0.50 V

eVbkT=( 1.6× 10 −19 C)( 0.50 V)( 0.025 eV)( 1.6× 10 −19 J/ eV )=19.8

The resistance for reverse bias of 0.50 V is calculated as,

R=VbI0( e e V b / kT −1)=0.50 V( 1.0× 10 9 A)( e 19.8 −1)=1.3 Ω

Current is calculated from Ohm’s Law,

I=VR=0.50 V1.3 Ω=0.38 A

Conclusion:

Therefore, the resistance for forward bias of 0.50 V is 1.3 Ω and corresponding current is 0.38 A .

Answer 27

(d)

To determine

The AC resistance for a 0.50 V forward bias voltage.

(d)

Expert Solution

Answer to Problem 54P

The AC resistance for forward bias of 0.50 V is 63 mΩ .

Explanation of Solution

Formula used:

The resistance for forward bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

The AC resistance is expressed as,

Rac=dVdI=( dI dV)−1

Calculation:

The AC resistance is calculated as,

Rac={d dV( I 0 ( e e V b / kT −1 ))}−1={ e I 0 kTe e V b / kT }−1=kTeI0e−e V b/kT=(25 MΩ)e−19.8

The calculation is further simplified as

Rac=25×106 Ω×2.5×10−9 Ω=63 mΩ

Conclusion:

Therefore, the AC resistance for forward bias of 0.50 V is 63 mΩ .

Answer 28

(d)

To determine

The AC resistance for a 0.50 V forward bias voltage.

Answer 29

(d)

Expert Solution

Answer to Problem 54P

The AC resistance for forward bias of 0.50 V is 63 mΩ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Formula used:

The resistance for forward bias of 0.50 V is,

R=VbI0(e e V b / kT −1)

The AC resistance is expressed as,

Rac=dVdI=( dI dV)−1

Calculation:

The AC resistance is calculated as,

Rac={d dV( I 0 ( e e V b / kT −1 ))}−1={ e I 0 kTe e V b / kT }−1=kTeI0e−e V b/kT=(25 MΩ)e−19.8

The calculation is further simplified as

Rac=25×106 Ω×2.5×10−9 Ω=63 mΩ

Conclusion:

Therefore, the AC resistance for forward bias of 0.50 V is 63 mΩ .

Answer 34

Ch. 38 - Prob. 1P Ch. 38 - Prob. 2P Ch. 38 - Prob. 3P Ch. 38 - Prob. 4P Ch. 38 - Prob. 5P Ch. 38 - Prob. 6P Ch. 38 - Prob. 7P Ch. 38 - Prob. 8P Ch. 38 - Prob. 9P Ch. 38 - Prob. 10P

The resistance for small reverse-bias voltages.

The resistance for small reverse-bias voltages.

Answer to Problem 54P

Explanation of Solution

The resistance for reverse bias of 0.50 V

Answer to Problem 54P

Explanation of Solution

The resistance for a 0.50 V forward bias and the corresponding current.

Answer to Problem 54P

Explanation of Solution

The AC resistance for a 0.50 V forward bias voltage.

Answer to Problem 54P

Explanation of Solution

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