(a)The reverse−saturation current of a gallium arsenide pn junction diode is I S = 10 − 22 A . Determine the diode current for diode voltages of 0.8, 1.0, 1.2, −0.02, −0.2, and −2 V. (b) Repeat part (a) for I S = 5 × 10 − 24 A .

Question

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Answer 1

Textbook Question

Chapter 1, Problem 1.35P

(a)The reverse−saturation current of a gallium arsenide pn junction diode is I S = 10 − 22 A . Determine the diode current for diode voltages of 0.8, 1.0, 1.2, −0.02, −0.2, and −2 V. (b) Repeat part (a) for I S = 5 × 10 − 24 A .

(a)

Expert Solution

To determine

Diode current for the given saturation current and diode voltages.

Answer to Problem 1.35P

For VD=0.8V , ID1=2.3063 nA

For VD=1.0V , ID2=5.054 μA

For VD=1.2V , ID3=11.0754 mA

For VD=−0.02V , ID4=−5.3663×10−23 A

For VD=−0.2V , ID5=−9.995×10−23 A

For VD=−0.02V , ID6=−1×10−22A

Explanation of Solution

Given:

Thermal voltage is VT=0.026 V .

Reverse-saturation current is IS=10−22A .

Different values of diode voltagesare

VD=0.8V .

VD=1.0V .

VD=1.2V .

VD=−0.02V .

VD=−0.2V .

VD=−2V .

Concept Used:

For forward diode voltage, the diode current is

ID=ISexp(VDVT)

For reverse diode voltage, the diode current is

ID=IS[exp( V D V T)−1]

Calculation:

For VD=0.8V ,

ID1=ISexp( V D V T )=( 10 −22)exp{0.80.026}=2.3063 nA

For VD=1V ,

ID2=ISexp( V D V T )=( 10 −22)exp{10.026}=5.054 μA

For VD=1.2V ,

ID3=ISexp( V D V T )=( 10 −22)exp{1.20.026}=11.0754 mA

For reverse voltage, VD=−0.02V .

ID4=IS[exp( V D V T )−1]=( 10 −22)[exp( −0.02 0.026)−1]=−5.3663×10−23 A

For reverse voltage, VD=−0.2V .

ID5=IS[exp( V D V T )−1]=( 10 −22)[exp( −0.2 0.026)−1]=−9.995×10−23 A

For reverse voltage, VD=−2V .

ID6=IS[exp( V D V T )−1]=( 10 −22)[exp( −2 0.026)−1]=−1×10−22 A

(b)

Expert Solution

To determine

Diode current for the given saturation current and diode voltages.

Answer to Problem 1.35P

For VD=0.8V , ID1=115.313 pA

For VD=1.0V , ID2=0.2527 μA

For VD=1.2V , ID3=0.55377 mA

For VD=−0.02V , ID4=−2.6832×10−24 A

For VD=−0.2V , ID5=−4.997×10−24 A

For VD=−2V , ID6=−5×10−24 A

Explanation of Solution

Given:

Thermal voltage is VT=0.026 V .

Reverse-saturation current is IS=5×10−24A .

Different values of diode voltagesare

VD=0.8V .

VD=1.0V .

VD=1.2V .

VD=−0.02V .

VD=−0.2V .

VD=−2V .

Concept Used:

For forward diode voltage, the diode current is

ID=ISexp(VDVT)

For reverse diode voltage, the diode current is

ID=IS[exp( V D V T)−1]

Calculation:

For VD=0.8V ,

ID1=ISexp( V D V T )=(5× 10 −24)exp{0.80.026}=115.313 pA

For VD=1V

ID2=ISexp( V D V T )=(5× 10 −24)exp{10.026}=0.2527 μA

For VD=1.2V

ID3=ISexp( V D V T )=(5× 10 −24)exp{1.20.026}=0.55377 mA

For reverse voltage, VD=−0.02V

ID4=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −0.02 0.026)−1]=−2.6832×10−24 A

For reverse voltage, VD=−0.2V

ID5=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −0.2 0.026)−1]=−4.997×10−24 A

For reverse voltage, VD=−2V

ID6=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −2 0.026)−1]=−5×10−24 A

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Answer 2

Textbook Question

Chapter 1, Problem 1.35P

(a)The reverse−saturation current of a gallium arsenide pn junction diode is I S = 10 − 22 A . Determine the diode current for diode voltages of 0.8, 1.0, 1.2, −0.02, −0.2, and −2 V. (b) Repeat part (a) for I S = 5 × 10 − 24 A .

(a)

Expert Solution

To determine

Diode current for the given saturation current and diode voltages.

Answer to Problem 1.35P

For VD=0.8V , ID1=2.3063 nA

For VD=1.0V , ID2=5.054 μA

For VD=1.2V , ID3=11.0754 mA

For VD=−0.02V , ID4=−5.3663×10−23 A

For VD=−0.2V , ID5=−9.995×10−23 A

For VD=−0.02V , ID6=−1×10−22A

Explanation of Solution

Given:

Thermal voltage is VT=0.026 V .

Reverse-saturation current is IS=10−22A .

Different values of diode voltagesare

VD=0.8V .

VD=1.0V .

VD=1.2V .

VD=−0.02V .

VD=−0.2V .

VD=−2V .

Concept Used:

For forward diode voltage, the diode current is

ID=ISexp(VDVT)

For reverse diode voltage, the diode current is

ID=IS[exp( V D V T)−1]

Calculation:

For VD=0.8V ,

ID1=ISexp( V D V T )=( 10 −22)exp{0.80.026}=2.3063 nA

For VD=1V ,

ID2=ISexp( V D V T )=( 10 −22)exp{10.026}=5.054 μA

For VD=1.2V ,

ID3=ISexp( V D V T )=( 10 −22)exp{1.20.026}=11.0754 mA

For reverse voltage, VD=−0.02V .

ID4=IS[exp( V D V T )−1]=( 10 −22)[exp( −0.02 0.026)−1]=−5.3663×10−23 A

For reverse voltage, VD=−0.2V .

ID5=IS[exp( V D V T )−1]=( 10 −22)[exp( −0.2 0.026)−1]=−9.995×10−23 A

For reverse voltage, VD=−2V .

ID6=IS[exp( V D V T )−1]=( 10 −22)[exp( −2 0.026)−1]=−1×10−22 A

(b)

Expert Solution

To determine

Diode current for the given saturation current and diode voltages.

Answer to Problem 1.35P

For VD=0.8V , ID1=115.313 pA

For VD=1.0V , ID2=0.2527 μA

For VD=1.2V , ID3=0.55377 mA

For VD=−0.02V , ID4=−2.6832×10−24 A

For VD=−0.2V , ID5=−4.997×10−24 A

For VD=−2V , ID6=−5×10−24 A

Explanation of Solution

Given:

Thermal voltage is VT=0.026 V .

Reverse-saturation current is IS=5×10−24A .

Different values of diode voltagesare

VD=0.8V .

VD=1.0V .

VD=1.2V .

VD=−0.02V .

VD=−0.2V .

VD=−2V .

Concept Used:

For forward diode voltage, the diode current is

ID=ISexp(VDVT)

For reverse diode voltage, the diode current is

ID=IS[exp( V D V T)−1]

Calculation:

For VD=0.8V ,

ID1=ISexp( V D V T )=(5× 10 −24)exp{0.80.026}=115.313 pA

For VD=1V

ID2=ISexp( V D V T )=(5× 10 −24)exp{10.026}=0.2527 μA

For VD=1.2V

ID3=ISexp( V D V T )=(5× 10 −24)exp{1.20.026}=0.55377 mA

For reverse voltage, VD=−0.02V

ID4=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −0.02 0.026)−1]=−2.6832×10−24 A

For reverse voltage, VD=−0.2V

ID5=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −0.2 0.026)−1]=−4.997×10−24 A

For reverse voltage, VD=−2V

ID6=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −2 0.026)−1]=−5×10−24 A

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Answer 3

Textbook Question

Chapter 1, Problem 1.35P

(a)The reverse−saturation current of a gallium arsenide pn junction diode is I S = 10 − 22 A . Determine the diode current for diode voltages of 0.8, 1.0, 1.2, −0.02, −0.2, and −2 V. (b) Repeat part (a) for I S = 5 × 10 − 24 A .

(a)

Expert Solution

To determine

Diode current for the given saturation current and diode voltages.

Answer to Problem 1.35P

For VD=0.8V , ID1=2.3063 nA

For VD=1.0V , ID2=5.054 μA

For VD=1.2V , ID3=11.0754 mA

For VD=−0.02V , ID4=−5.3663×10−23 A

For VD=−0.2V , ID5=−9.995×10−23 A

For VD=−0.02V , ID6=−1×10−22A

Explanation of Solution

Given:

Thermal voltage is VT=0.026 V .

Reverse-saturation current is IS=10−22A .

Different values of diode voltagesare

VD=0.8V .

VD=1.0V .

VD=1.2V .

VD=−0.02V .

VD=−0.2V .

VD=−2V .

Concept Used:

For forward diode voltage, the diode current is

ID=ISexp(VDVT)

For reverse diode voltage, the diode current is

ID=IS[exp( V D V T)−1]

Calculation:

For VD=0.8V ,

ID1=ISexp( V D V T )=( 10 −22)exp{0.80.026}=2.3063 nA

For VD=1V ,

ID2=ISexp( V D V T )=( 10 −22)exp{10.026}=5.054 μA

For VD=1.2V ,

ID3=ISexp( V D V T )=( 10 −22)exp{1.20.026}=11.0754 mA

For reverse voltage, VD=−0.02V .

ID4=IS[exp( V D V T )−1]=( 10 −22)[exp( −0.02 0.026)−1]=−5.3663×10−23 A

For reverse voltage, VD=−0.2V .

ID5=IS[exp( V D V T )−1]=( 10 −22)[exp( −0.2 0.026)−1]=−9.995×10−23 A

For reverse voltage, VD=−2V .

ID6=IS[exp( V D V T )−1]=( 10 −22)[exp( −2 0.026)−1]=−1×10−22 A

(b)

Expert Solution

To determine

Diode current for the given saturation current and diode voltages.

Answer to Problem 1.35P

For VD=0.8V , ID1=115.313 pA

For VD=1.0V , ID2=0.2527 μA

For VD=1.2V , ID3=0.55377 mA

For VD=−0.02V , ID4=−2.6832×10−24 A

For VD=−0.2V , ID5=−4.997×10−24 A

For VD=−2V , ID6=−5×10−24 A

Explanation of Solution

Given:

Thermal voltage is VT=0.026 V .

Reverse-saturation current is IS=5×10−24A .

Different values of diode voltagesare

VD=0.8V .

VD=1.0V .

VD=1.2V .

VD=−0.02V .

VD=−0.2V .

VD=−2V .

Concept Used:

For forward diode voltage, the diode current is

ID=ISexp(VDVT)

For reverse diode voltage, the diode current is

ID=IS[exp( V D V T)−1]

Calculation:

For VD=0.8V ,

ID1=ISexp( V D V T )=(5× 10 −24)exp{0.80.026}=115.313 pA

For VD=1V

ID2=ISexp( V D V T )=(5× 10 −24)exp{10.026}=0.2527 μA

For VD=1.2V

ID3=ISexp( V D V T )=(5× 10 −24)exp{1.20.026}=0.55377 mA

For reverse voltage, VD=−0.02V

ID4=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −0.02 0.026)−1]=−2.6832×10−24 A

For reverse voltage, VD=−0.2V

ID5=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −0.2 0.026)−1]=−4.997×10−24 A

For reverse voltage, VD=−2V

ID6=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −2 0.026)−1]=−5×10−24 A

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Answer 4

Textbook Question

Chapter 1, Problem 1.35P

(a)The reverse−saturation current of a gallium arsenide pn junction diode is I S = 10 − 22 A . Determine the diode current for diode voltages of 0.8, 1.0, 1.2, −0.02, −0.2, and −2 V. (b) Repeat part (a) for I S = 5 × 10 − 24 A .

(a)

Expert Solution

To determine

Diode current for the given saturation current and diode voltages.

Answer to Problem 1.35P

For VD=0.8V , ID1=2.3063 nA

For VD=1.0V , ID2=5.054 μA

For VD=1.2V , ID3=11.0754 mA

For VD=−0.02V , ID4=−5.3663×10−23 A

For VD=−0.2V , ID5=−9.995×10−23 A

For VD=−0.02V , ID6=−1×10−22A

Explanation of Solution

Given:

Thermal voltage is VT=0.026 V .

Reverse-saturation current is IS=10−22A .

Different values of diode voltagesare

VD=0.8V .

VD=1.0V .

VD=1.2V .

VD=−0.02V .

VD=−0.2V .

VD=−2V .

Concept Used:

For forward diode voltage, the diode current is

ID=ISexp(VDVT)

For reverse diode voltage, the diode current is

ID=IS[exp( V D V T)−1]

Calculation:

For VD=0.8V ,

ID1=ISexp( V D V T )=( 10 −22)exp{0.80.026}=2.3063 nA

For VD=1V ,

ID2=ISexp( V D V T )=( 10 −22)exp{10.026}=5.054 μA

For VD=1.2V ,

ID3=ISexp( V D V T )=( 10 −22)exp{1.20.026}=11.0754 mA

For reverse voltage, VD=−0.02V .

ID4=IS[exp( V D V T )−1]=( 10 −22)[exp( −0.02 0.026)−1]=−5.3663×10−23 A

For reverse voltage, VD=−0.2V .

ID5=IS[exp( V D V T )−1]=( 10 −22)[exp( −0.2 0.026)−1]=−9.995×10−23 A

For reverse voltage, VD=−2V .

ID6=IS[exp( V D V T )−1]=( 10 −22)[exp( −2 0.026)−1]=−1×10−22 A

(b)

Expert Solution

To determine

Diode current for the given saturation current and diode voltages.

Answer to Problem 1.35P

For VD=0.8V , ID1=115.313 pA

For VD=1.0V , ID2=0.2527 μA

For VD=1.2V , ID3=0.55377 mA

For VD=−0.02V , ID4=−2.6832×10−24 A

For VD=−0.2V , ID5=−4.997×10−24 A

For VD=−2V , ID6=−5×10−24 A

Explanation of Solution

Given:

Thermal voltage is VT=0.026 V .

Reverse-saturation current is IS=5×10−24A .

Different values of diode voltagesare

VD=0.8V .

VD=1.0V .

VD=1.2V .

VD=−0.02V .

VD=−0.2V .

VD=−2V .

Concept Used:

For forward diode voltage, the diode current is

ID=ISexp(VDVT)

For reverse diode voltage, the diode current is

ID=IS[exp( V D V T)−1]

Calculation:

For VD=0.8V ,

ID1=ISexp( V D V T )=(5× 10 −24)exp{0.80.026}=115.313 pA

For VD=1V

ID2=ISexp( V D V T )=(5× 10 −24)exp{10.026}=0.2527 μA

For VD=1.2V

ID3=ISexp( V D V T )=(5× 10 −24)exp{1.20.026}=0.55377 mA

For reverse voltage, VD=−0.02V

ID4=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −0.02 0.026)−1]=−2.6832×10−24 A

For reverse voltage, VD=−0.2V

ID5=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −0.2 0.026)−1]=−4.997×10−24 A

For reverse voltage, VD=−2V

ID6=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −2 0.026)−1]=−5×10−24 A

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

Diode current for the given saturation current and diode voltages.

Answer to Problem 1.35P

For VD=0.8V , ID1=2.3063 nA

For VD=1.0V , ID2=5.054 μA

For VD=1.2V , ID3=11.0754 mA

For VD=−0.02V , ID4=−5.3663×10−23 A

For VD=−0.2V , ID5=−9.995×10−23 A

For VD=−0.02V , ID6=−1×10−22A

Explanation of Solution

Given:

Thermal voltage is VT=0.026 V .

Reverse-saturation current is IS=10−22A .

Different values of diode voltagesare

VD=0.8V .

VD=1.0V .

VD=1.2V .

VD=−0.02V .

VD=−0.2V .

VD=−2V .

Concept Used:

For forward diode voltage, the diode current is

ID=ISexp(VDVT)

For reverse diode voltage, the diode current is

ID=IS[exp( V D V T)−1]

Calculation:

For VD=0.8V ,

ID1=ISexp( V D V T )=( 10 −22)exp{0.80.026}=2.3063 nA

For VD=1V ,

ID2=ISexp( V D V T )=( 10 −22)exp{10.026}=5.054 μA

For VD=1.2V ,

ID3=ISexp( V D V T )=( 10 −22)exp{1.20.026}=11.0754 mA

For reverse voltage, VD=−0.02V .

ID4=IS[exp( V D V T )−1]=( 10 −22)[exp( −0.02 0.026)−1]=−5.3663×10−23 A

For reverse voltage, VD=−0.2V .

ID5=IS[exp( V D V T )−1]=( 10 −22)[exp( −0.2 0.026)−1]=−9.995×10−23 A

For reverse voltage, VD=−2V .

ID6=IS[exp( V D V T )−1]=( 10 −22)[exp( −2 0.026)−1]=−1×10−22 A

(b)

Expert Solution

To determine

Diode current for the given saturation current and diode voltages.

Answer to Problem 1.35P

For VD=0.8V , ID1=115.313 pA

For VD=1.0V , ID2=0.2527 μA

For VD=1.2V , ID3=0.55377 mA

For VD=−0.02V , ID4=−2.6832×10−24 A

For VD=−0.2V , ID5=−4.997×10−24 A

For VD=−2V , ID6=−5×10−24 A

Explanation of Solution

Given:

Thermal voltage is VT=0.026 V .

Reverse-saturation current is IS=5×10−24A .

Different values of diode voltagesare

VD=0.8V .

VD=1.0V .

VD=1.2V .

VD=−0.02V .

VD=−0.2V .

VD=−2V .

Concept Used:

For forward diode voltage, the diode current is

ID=ISexp(VDVT)

For reverse diode voltage, the diode current is

ID=IS[exp( V D V T)−1]

Calculation:

For VD=0.8V ,

ID1=ISexp( V D V T )=(5× 10 −24)exp{0.80.026}=115.313 pA

For VD=1V

ID2=ISexp( V D V T )=(5× 10 −24)exp{10.026}=0.2527 μA

For VD=1.2V

ID3=ISexp( V D V T )=(5× 10 −24)exp{1.20.026}=0.55377 mA

For reverse voltage, VD=−0.02V

ID4=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −0.02 0.026)−1]=−2.6832×10−24 A

For reverse voltage, VD=−0.2V

ID5=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −0.2 0.026)−1]=−4.997×10−24 A

For reverse voltage, VD=−2V

ID6=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −2 0.026)−1]=−5×10−24 A

Answer 8

(a)

Expert Solution

To determine

Diode current for the given saturation current and diode voltages.

Answer to Problem 1.35P

For VD=0.8V , ID1=2.3063 nA

For VD=1.0V , ID2=5.054 μA

For VD=1.2V , ID3=11.0754 mA

For VD=−0.02V , ID4=−5.3663×10−23 A

For VD=−0.2V , ID5=−9.995×10−23 A

For VD=−0.02V , ID6=−1×10−22A

Explanation of Solution

Given:

Thermal voltage is VT=0.026 V .

Reverse-saturation current is IS=10−22A .

Different values of diode voltagesare

VD=0.8V .

VD=1.0V .

VD=1.2V .

VD=−0.02V .

VD=−0.2V .

VD=−2V .

Concept Used:

For forward diode voltage, the diode current is

ID=ISexp(VDVT)

For reverse diode voltage, the diode current is

ID=IS[exp( V D V T)−1]

Calculation:

For VD=0.8V ,

ID1=ISexp( V D V T )=( 10 −22)exp{0.80.026}=2.3063 nA

For VD=1V ,

ID2=ISexp( V D V T )=( 10 −22)exp{10.026}=5.054 μA

For VD=1.2V ,

ID3=ISexp( V D V T )=( 10 −22)exp{1.20.026}=11.0754 mA

For reverse voltage, VD=−0.02V .

ID4=IS[exp( V D V T )−1]=( 10 −22)[exp( −0.02 0.026)−1]=−5.3663×10−23 A

For reverse voltage, VD=−0.2V .

ID5=IS[exp( V D V T )−1]=( 10 −22)[exp( −0.2 0.026)−1]=−9.995×10−23 A

For reverse voltage, VD=−2V .

ID6=IS[exp( V D V T )−1]=( 10 −22)[exp( −2 0.026)−1]=−1×10−22 A

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Diode current for the given saturation current and diode voltages.

Answer 13

Answer to Problem 1.35P

For VD=0.8V , ID1=2.3063 nA

For VD=1.0V , ID2=5.054 μA

For VD=1.2V , ID3=11.0754 mA

For VD=−0.02V , ID4=−5.3663×10−23 A

For VD=−0.2V , ID5=−9.995×10−23 A

For VD=−0.02V , ID6=−1×10−22A

Answer 14

Explanation of Solution

Given:

Thermal voltage is VT=0.026 V .

Reverse-saturation current is IS=10−22A .

Different values of diode voltagesare

VD=0.8V .

VD=1.0V .

VD=1.2V .

VD=−0.02V .

VD=−0.2V .

VD=−2V .

Concept Used:

For forward diode voltage, the diode current is

ID=ISexp(VDVT)

For reverse diode voltage, the diode current is

ID=IS[exp( V D V T)−1]

Calculation:

For VD=0.8V ,

ID1=ISexp( V D V T )=( 10 −22)exp{0.80.026}=2.3063 nA

For VD=1V ,

ID2=ISexp( V D V T )=( 10 −22)exp{10.026}=5.054 μA

For VD=1.2V ,

ID3=ISexp( V D V T )=( 10 −22)exp{1.20.026}=11.0754 mA

For reverse voltage, VD=−0.02V .

ID4=IS[exp( V D V T )−1]=( 10 −22)[exp( −0.02 0.026)−1]=−5.3663×10−23 A

For reverse voltage, VD=−0.2V .

ID5=IS[exp( V D V T )−1]=( 10 −22)[exp( −0.2 0.026)−1]=−9.995×10−23 A

For reverse voltage, VD=−2V .

ID6=IS[exp( V D V T )−1]=( 10 −22)[exp( −2 0.026)−1]=−1×10−22 A

Answer 15

(b)

Expert Solution

To determine

Diode current for the given saturation current and diode voltages.

Answer to Problem 1.35P

For VD=0.8V , ID1=115.313 pA

For VD=1.0V , ID2=0.2527 μA

For VD=1.2V , ID3=0.55377 mA

For VD=−0.02V , ID4=−2.6832×10−24 A

For VD=−0.2V , ID5=−4.997×10−24 A

For VD=−2V , ID6=−5×10−24 A

Explanation of Solution

Given:

Thermal voltage is VT=0.026 V .

Reverse-saturation current is IS=5×10−24A .

Different values of diode voltagesare

VD=0.8V .

VD=1.0V .

VD=1.2V .

VD=−0.02V .

VD=−0.2V .

VD=−2V .

Concept Used:

For forward diode voltage, the diode current is

ID=ISexp(VDVT)

For reverse diode voltage, the diode current is

ID=IS[exp( V D V T)−1]

Calculation:

For VD=0.8V ,

ID1=ISexp( V D V T )=(5× 10 −24)exp{0.80.026}=115.313 pA

For VD=1V

ID2=ISexp( V D V T )=(5× 10 −24)exp{10.026}=0.2527 μA

For VD=1.2V

ID3=ISexp( V D V T )=(5× 10 −24)exp{1.20.026}=0.55377 mA

For reverse voltage, VD=−0.02V

ID4=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −0.02 0.026)−1]=−2.6832×10−24 A

For reverse voltage, VD=−0.2V

ID5=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −0.2 0.026)−1]=−4.997×10−24 A

For reverse voltage, VD=−2V

ID6=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −2 0.026)−1]=−5×10−24 A

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

Diode current for the given saturation current and diode voltages.

Answer 20

Answer to Problem 1.35P

For VD=0.8V , ID1=115.313 pA

For VD=1.0V , ID2=0.2527 μA

For VD=1.2V , ID3=0.55377 mA

For VD=−0.02V , ID4=−2.6832×10−24 A

For VD=−0.2V , ID5=−4.997×10−24 A

For VD=−2V , ID6=−5×10−24 A

Answer 21

Explanation of Solution

Given:

Thermal voltage is VT=0.026 V .

Reverse-saturation current is IS=5×10−24A .

Different values of diode voltagesare

VD=0.8V .

VD=1.0V .

VD=1.2V .

VD=−0.02V .

VD=−0.2V .

VD=−2V .

Concept Used:

For forward diode voltage, the diode current is

ID=ISexp(VDVT)

For reverse diode voltage, the diode current is

ID=IS[exp( V D V T)−1]

Calculation:

For VD=0.8V ,

ID1=ISexp( V D V T )=(5× 10 −24)exp{0.80.026}=115.313 pA

For VD=1V

ID2=ISexp( V D V T )=(5× 10 −24)exp{10.026}=0.2527 μA

For VD=1.2V

ID3=ISexp( V D V T )=(5× 10 −24)exp{1.20.026}=0.55377 mA

For reverse voltage, VD=−0.02V

ID4=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −0.02 0.026)−1]=−2.6832×10−24 A

For reverse voltage, VD=−0.2V

ID5=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −0.2 0.026)−1]=−4.997×10−24 A

For reverse voltage, VD=−2V

ID6=IS[exp( V D V T )−1]=(5× 10 −24)[exp( −2 0.026)−1]=−5×10−24 A

Answer 22

Ch. 1 - Calculate the intrinsic carrier concentration in...Ch. 1 - (a) Calculate the majority and minority carrier...Ch. 1 - Consider ntype GaAs at T=300K doped to a...Ch. 1 - Consider silicon at T=300K . Assume the hole...Ch. 1 - Determine the intrinsic carrier concentration in...Ch. 1 - (a) Consider silicon at T=300K . Assume that...Ch. 1 - Using the results of TYU1.2, determine the drift...Ch. 1 - The electron and hole diffusion coefficients in...Ch. 1 - A sample of silicon at T=300K is doped to...Ch. 1 - (a) Calculate Vbi for a GaAs pn junction at T=300K...

(a)The reverse−saturation current of a gallium arsenide pn junction diode is I S = 10 − 22 A . Determine the diode current for diode voltages of 0.8, 1.0, 1.2, −0.02, −0.2, and −2 V. (b) Repeat part (a) for I S = 5 × 10 − 24 A .

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Answer to Problem 1.35P

Explanation of Solution

Answer to Problem 1.35P

Explanation of Solution

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