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<m:math><m:mrow><m:mn>0.50</m:mn><m:mtext> V</m:mtext></m:mrow></m:math>

Answer to Problem 54P

Explanation of Solution

The resistance for a 0.50 V<m:math><m:mrow><m:mn>0.50</m:mn><m:mtext> </m:mtext><m:mtext>V</m:mtext></m:mrow></m:math> forward bias and the corresponding current.

Answer to Problem 54P

Explanation of Solution

The AC resistance for a 0.50 V<m:math><m:mrow><m:mn>0.50</m:mn><m:mtext> </m:mtext><m:mtext>V</m:mtext></m:mrow></m:math> forward bias voltage.

Answer to Problem 54P

Explanation of Solution

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Chapter 38 Solutions

The resistance for reverse bias of $0.50 V$

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

The AC resistance for a $0.50 V$ forward bias voltage.