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Chapter 26, Problem 46AP

(a)

To determine

A spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=280K,300K,and320K .

(a)

Expert Solution
Check Mark

Answer to Problem 46AP

A spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=280K is,

ΔV

(Volts)

I

(Amperes)

R=ΔVI (ohms)
0.4 0.015932 25.1067
0.405 0.019602 20.66116
0.41 0.024117 17.00046
0.415 0.029673 13.98578
0.42 0.036508 11.50433
0.425 0.044918 9.461686
0.43 0.055264 7.780834
0.435 0.067995 6.397529
0.44 0.083657 5.259572
0.445 0.102927 4.323453
0.45 0.126637 3.553464
0.455 0.155807 2.92028
0.46 0.191697 2.39962
0.465 0.235855 1.97155
0.47 0.290184 1.619662
0.475 0.357027 1.330432
0.48 0.439268 1.092727
0.485 0.540454 0.897394
0.495 0.818117 0.605048
0.5 1.006569 0.496737
0.505 1.238432 0.407774
0.51 1.523704 0.334711
0.515 1.874688 0.274712
0.52 2.306521 0.225448
0.525 2.837827 0.185001
0.53 3.491518 0.151796
0.535 4.295787 0.124541
0.54 5.285319 0.10217
0.545 6.502788 0.08381
0.55 8.000701 0.068744
0.555 9.843657 0.056381
0.56 12.11114 0.046238
0.565 14.90093 0.037917
0.57 18.33335 0.031091
0.575 22.55642 0.025492
0.58 27.75228 0.020899
0.585 34.145 0.017133
0.59 42.01028 0.014044
0.595 51.68732 0.011512
0.6 63.59346 0.009435

A spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=300K is,

ΔV

(Volts)

I

(Amperes)

R=ΔVI (ohms)
0.4 0.005274 75.84672
0.405 0.0064 63.28565
0.41 0.007766 52.79679
0.415 0.009423 44.03979
0.42 0.011435 36.72991
0.425 0.013876 30.62901
0.43 0.016838 25.53795
0.435 0.020432 21.29022
0.44 0.024793 17.74668
0.445 0.030086 14.79101
0.45 0.036508 12.32605
0.455 0.044301 10.27061
0.46 0.053758 8.556892
0.465 0.065233 7.128278
0.47 0.079158 5.937492
0.475 0.096055 4.945067
0.48 0.11656 4.118066
0.485 0.141441 3.428998
0.495 0.20827 2.376718
0.5 0.252728 1.978408
0.505 0.306677 1.646686
0.51 0.372141 1.370449
0.515 0.451579 1.140443
0.52 0.547974 0.948949
0.525 0.664947 0.789537
0.53 0.806888 0.656844
0.535 0.979129 0.546404
0.54 1.188137 0.454493
0.545 1.44176 0.37801
0.55 1.749522 0.314372
0.555 2.122981 0.261425
0.56 2.576159 0.217378
0.565 3.126073 0.180738
0.57 3.793374 0.150262
0.575 4.603119 0.124915
0.58 5.585715 0.103836
0.585 6.778058 0.086308
0.59 8.224923 0.071733
0.595 9.98064 0.059615
0.6 12.11114 0.049541

A spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=320K is,

ΔV

(Volts)

I

(Amperes)

R=ΔVI (ohms)
0.4 0.002004 199.5582
0.405 0.002403 168.5349
0.41 0.002881 142.3127
0.415 0.003454 120.1526
0.42 0.004141 101.4283
0.425 0.004964 85.60991
0.43 0.005952 72.24847
0.435 0.007135 60.96416
0.44 0.008554 51.43551
0.445 0.010256 43.39059
0.45 0.012295 36.59933
0.455 0.014741 30.86719
0.46 0.017672 26.02967
0.465 0.021187 21.9477
0.47 0.0254 18.50372
0.475 0.030452 15.59839
0.48 0.036508 13.14778
0.485 0.043769 11.08098
0.495 0.062909 7.868498
0.5 0.07542 6.629515
0.505 0.09042 5.585066
0.51 0.108402 4.704703
0.515 0.129961 3.962729
0.52 0.155807 3.337456
0.525 0.186794 2.810585
0.53 0.223943 2.366674
0.535 0.26848 1.992698
0.54 0.321875 1.67767
0.545 0.385889 1.412324
0.55 0.462633 1.188846
0.555 0.554641 1.000647
0.56 0.664947 0.842173
0.565 0.79719 0.70874
0.57 0.955733 0.596401
0.575 1.145807 0.50183
0.58 1.373682 0.422223
0.585 1.646877 0.355218
0.59 1.974404 0.298824
0.595 2.367069 0.251366
0.6 2.837827 0.211429

Explanation of Solution

Given information: Th first symbol i.e. Euler’s number is e , the second symbol i.e. magnitude of electron charge is e , Boltzmann’s constant is kB , the absolute temperature is T , the value of current across a semiconductor diode temperature 0K is 1.00nA .

It is given that the expression for the current-voltage characteristic curve for a semiconductor diode as a function of temperature T is,

I=I0(eeΔVkBT1) (1)

Here,

I is the current across a semiconductor diode temperature TK .

I0 is the current across a semiconductor diode temperature 0K .

e is the first symbol i.e.Euler’s number.

e is the second symbol i.e. magnitude of electron charge.

kB is the Boltzmann’s constant.

ΔV is the voltage across the diode.

T is the absolute temperature.

Formula to calculate the resistance across the diode is,

R=ΔVI (2)

Here,

R is the resistance across the diode.

The value of magnitude of electron charge is 1.602×1019C .

The value of Boltzmann’s constant is 1.38×1023J/K .

The value of voltage across the diode varies from 0.400V to 0.600V in increments of 0.005V .

From equation (1), formula to calculate the current across a semiconductor diode temperature 280K is,

I1=I0(eeΔV1kBT1) (3)

Here,

I1 is the current across a semiconductor diode temperature 280K .

ΔV1 is the initial voltage across the diode for temperature 280K .

Substitute 1.602×1019C for e , 1.38×1023J/K for kB , 0.400V for ΔV1 , 280K for T in equation (3) to find I1 ,

I1=(1.00nA×1A109nA)(e(1.602×1019C)(0.400V)(1.38×1023J/K)(280K)1)=0.15932A

Thus, the current across a semiconductor diode temperature 280K is 0.15932A .

From equation (2), formula to calculate the resistance across the diode is,

R1=ΔV1I1 (4)

Here,

R1 is the resistance across the diode.

Substitute 0.15932A for I1 , 0.400V for ΔV1 in equation (4) to find R1 ,

R1=0.400V0.15932A=25.1067Ω

Thus, the resistance across the diode is 25.1067Ω .

As the value of voltage across the diode varies from 0.400V to 0.600V in increments of 0.005V , the values for the current and resistance are calculated by same procedure as above.

Thus, a spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=280K is,

ΔV

(Volts)

I

(Amperes)

R=ΔVI (ohms)
0.4 0.015932 25.1067
0.405 0.019602 20.66116
0.41 0.024117 17.00046
0.415 0.029673 13.98578
0.42 0.036508 11.50433
0.425 0.044918 9.461686
0.43 0.055264 7.780834
0.435 0.067995 6.397529
0.44 0.083657 5.259572
0.445 0.102927 4.323453
0.45 0.126637 3.553464
0.455 0.155807 2.92028
0.46 0.191697 2.39962
0.465 0.235855 1.97155
0.47 0.290184 1.619662
0.475 0.357027 1.330432
0.48 0.439268 1.092727
0.485 0.540454 0.897394
0.495 0.818117 0.605048
0.5 1.006569 0.496737
0.505 1.238432 0.407774
0.51 1.523704 0.334711
0.515 1.874688 0.274712
0.52 2.306521 0.225448
0.525 2.837827 0.185001
0.53 3.491518 0.151796
0.535 4.295787 0.124541
0.54 5.285319 0.10217
0.545 6.502788 0.08381
0.55 8.000701 0.068744
0.555 9.843657 0.056381
0.56 12.11114 0.046238
0.565 14.90093 0.037917
0.57 18.33335 0.031091
0.575 22.55642 0.025492
0.58 27.75228 0.020899
0.585 34.145 0.017133
0.59 42.01028 0.014044
0.595 51.68732 0.011512
0.6 63.59346 0.009435

From equation (1), formula to calculate the current across a semiconductor diode temperature 300K is,

I1=I0(eeΔV1kBT1) (5)

Here,

I1 is the current across a semiconductor diode temperature 300K .

ΔV1 is the voltage across the diode for temperature 300K .

Substitute 1.602×1019C for e , 1.38×1023J/K for kB , 0.400V for ΔV1 , 300K for T in equation (5) to find I1 ,

I1=(1.00nA×1A109nA)(e(1.602×1019C)(0.400V)(1.38×1023J/K)(300K)1)=0.005274A

Thus, the current across a semiconductor diode temperature 300K is 0.005274A .

From equation (2), formula to calculate the resistance across the diode is,

R1=ΔV1I1 (6)

Here,

R1 is the resistance across the diode.

Substitute 0.005274A for I1 , 0.400V for ΔV1 in equation (6) to find R1 ,

R1=0.400V0.005274A=75.84672Ω

Thus, the resistance across the diode is 75.84672Ω .

As the value of voltage across the diode varies from 0.400V to 0.600V in increments of 0.005V , the values for the current and resistance are calculated by same procedure as above.

Thus, a spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=300K is,

ΔV

(Volts)

I

(Amperes)

R=ΔVI (ohms)
0.4 0.005274 75.84672
0.405 0.0064 63.28565
0.41 0.007766 52.79679
0.415 0.009423 44.03979
0.42 0.011435 36.72991
0.425 0.013876 30.62901
0.43 0.016838 25.53795
0.435 0.020432 21.29022
0.44 0.024793 17.74668
0.445 0.030086 14.79101
0.45 0.036508 12.32605
0.455 0.044301 10.27061
0.46 0.053758 8.556892
0.465 0.065233 7.128278
0.47 0.079158 5.937492
0.475 0.096055 4.945067
0.48 0.11656 4.118066
0.485 0.141441 3.428998
0.495 0.20827 2.376718
0.5 0.252728 1.978408
0.505 0.306677 1.646686
0.51 0.372141 1.370449
0.515 0.451579 1.140443
0.52 0.547974 0.948949
0.525 0.664947 0.789537
0.53 0.806888 0.656844
0.535 0.979129 0.546404
0.54 1.188137 0.454493
0.545 1.44176 0.37801
0.55 1.749522 0.314372
0.555 2.122981 0.261425
0.56 2.576159 0.217378
0.565 3.126073 0.180738
0.57 3.793374 0.150262
0.575 4.603119 0.124915
0.58 5.585715 0.103836
0.585 6.778058 0.086308
0.59 8.224923 0.071733
0.595 9.98064 0.059615
0.6 12.11114 0.049541

From equation (1), formula to calculate the current across a semiconductor diode temperature 320K is,

I1=I0(eeΔV1kBT1) (7)

Here,

I1 is the current across a semiconductor diode temperature 320K .

ΔV1 is the voltage across the diode for temperature 320K .

Substitute 1.602×1019C for e , 1.38×1023J/K for kB , 0.400V for ΔV , 320K for T in equation (7) to find I1 ,

I1=(1.00nA×1A109nA)(e(1.602×1019C)(0.400V)(1.38×1023J/K)(320K)1)=0.002004A

Thus, the current across a semiconductor diode temperature 320K is 0.002004A .

From equation (2), formula to calculate the resistance across the diode is,

R1=ΔV1I1 (8)

Here,

R1 is the resistance across the diode.

Substitute 0.002004A for I1 , 0.400V for ΔV1 in equation (8) to find R1 ,

R1=0.400V0.002004A=199.5582Ω

Thus, the resistance across the diode is 199.5582Ω .

As the value of voltage across the diode varies from 0.400V to 0.600V in increments of 0.005V , the values for the current and resistance are calculated by same procedure as above.

Thus, a spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=320K is,

ΔV

(Volts)

I

(Amperes)

R=ΔVI (ohms)
0.4 0.002004 199.5582
0.405 0.002403 168.5349
0.41 0.002881 142.3127
0.415 0.003454 120.1526
0.42 0.004141 101.4283
0.425 0.004964 85.60991
0.43 0.005952 72.24847
0.435 0.007135 60.96416
0.44 0.008554 51.43551
0.445 0.010256 43.39059
0.45 0.012295 36.59933
0.455 0.014741 30.86719
0.46 0.017672 26.02967
0.465 0.021187 21.9477
0.47 0.0254 18.50372
0.475 0.030452 15.59839
0.48 0.036508 13.14778
0.485 0.043769 11.08098
0.495 0.062909 7.868498
0.5 0.07542 6.629515
0.505 0.09042 5.585066
0.51 0.108402 4.704703
0.515 0.129961 3.962729
0.52 0.155807 3.337456
0.525 0.186794 2.810585
0.53 0.223943 2.366674
0.535 0.26848 1.992698
0.54 0.321875 1.67767
0.545 0.385889 1.412324
0.55 0.462633 1.188846
0.555 0.554641 1.000647
0.56 0.664947 0.842173
0.565 0.79719 0.70874
0.57 0.955733 0.596401
0.575 1.145807 0.50183
0.58 1.373682 0.422223
0.585 1.646877 0.355218
0.59 1.974404 0.298824
0.595 2.367069 0.251366
0.6 2.837827 0.211429

Conclusion:

Therefore, a spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=280K is,

ΔV

(Volts)

I

(Amperes)

R=ΔVI (ohms)
0.4 0.015932 25.1067
0.405 0.019602 20.66116
0.41 0.024117 17.00046
0.415 0.029673 13.98578
0.42 0.036508 11.50433
0.425 0.044918 9.461686
0.43 0.055264 7.780834
0.435 0.067995 6.397529
0.44 0.083657 5.259572
0.445 0.102927 4.323453
0.45 0.126637 3.553464
0.455 0.155807 2.92028
0.46 0.191697 2.39962
0.465 0.235855 1.97155
0.47 0.290184 1.619662
0.475 0.357027 1.330432
0.48 0.439268 1.092727
0.485 0.540454 0.897394
0.495 0.818117 0.605048
0.5 1.006569 0.496737
0.505 1.238432 0.407774
0.51 1.523704 0.334711
0.515 1.874688 0.274712
0.52 2.306521 0.225448
0.525 2.837827 0.185001
0.53 3.491518 0.151796
0.535 4.295787 0.124541
0.54 5.285319 0.10217
0.545 6.502788 0.08381
0.55 8.000701 0.068744
0.555 9.843657 0.056381
0.56 12.11114 0.046238
0.565 14.90093 0.037917
0.57 18.33335 0.031091
0.575 22.55642 0.025492
0.58 27.75228 0.020899
0.585 34.145 0.017133
0.59 42.01028 0.014044
0.595 51.68732 0.011512
0.6 63.59346 0.009435

A spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=300K is,

ΔV

(Volts)

I

(Amperes)

R=ΔVI (ohms)
0.4 0.005274 75.84672
0.405 0.0064 63.28565
0.41 0.007766 52.79679
0.415 0.009423 44.03979
0.42 0.011435 36.72991
0.425 0.013876 30.62901
0.43 0.016838 25.53795
0.435 0.020432 21.29022
0.44 0.024793 17.74668
0.445 0.030086 14.79101
0.45 0.036508 12.32605
0.455 0.044301 10.27061
0.46 0.053758 8.556892
0.465 0.065233 7.128278
0.47 0.079158 5.937492
0.475 0.096055 4.945067
0.48 0.11656 4.118066
0.485 0.141441 3.428998
0.495 0.20827 2.376718
0.5 0.252728 1.978408
0.505 0.306677 1.646686
0.51 0.372141 1.370449
0.515 0.451579 1.140443
0.52 0.547974 0.948949
0.525 0.664947 0.789537
0.53 0.806888 0.656844
0.535 0.979129 0.546404
0.54 1.188137 0.454493
0.545 1.44176 0.37801
0.55 1.749522 0.314372
0.555 2.122981 0.261425
0.56 2.576159 0.217378
0.565 3.126073 0.180738
0.57 3.793374 0.150262
0.575 4.603119 0.124915
0.58 5.585715 0.103836
0.585 6.778058 0.086308
0.59 8.224923 0.071733
0.595 9.98064 0.059615
0.6 12.11114 0.049541

A spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=320K is,

ΔV

(Volts)

I

(Amperes)

R=ΔVI (ohms)
0.4 0.002004 199.5582
0.405 0.002403 168.5349
0.41 0.002881 142.3127
0.415 0.003454 120.1526
0.42 0.004141 101.4283
0.425 0.004964 85.60991
0.43 0.005952 72.24847
0.435 0.007135 60.96416
0.44 0.008554 51.43551
0.445 0.010256 43.39059
0.45 0.012295 36.59933
0.455 0.014741 30.86719
0.46 0.017672 26.02967
0.465 0.021187 21.9477
0.47 0.0254 18.50372
0.475 0.030452 15.59839
0.48 0.036508 13.14778
0.485 0.043769 11.08098
0.495 0.062909 7.868498
0.5 0.07542 6.629515
0.505 0.09042 5.585066
0.51 0.108402 4.704703
0.515 0.129961 3.962729
0.52 0.155807 3.337456
0.525 0.186794 2.810585
0.53 0.223943 2.366674
0.535 0.26848 1.992698
0.54 0.321875 1.67767
0.545 0.385889 1.412324
0.55 0.462633 1.188846
0.555 0.554641 1.000647
0.56 0.664947 0.842173
0.565 0.79719 0.70874
0.57 0.955733 0.596401
0.575 1.145807 0.50183
0.58 1.373682 0.422223
0.585 1.646877 0.355218
0.59 1.974404 0.298824
0.595 2.367069 0.251366
0.6 2.837827 0.211429

(b)

To determine

To draw: The graph for R versus ΔV for T=280K,300K,and320K .

(b)

Expert Solution
Check Mark

Answer to Problem 46AP

The graph for R versus ΔV for T=280K is,

Bundle: Physics For Scientists And Engineers With Modern Physics, Loose-leaf Version, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Single-term, Chapter 26, Problem 46AP , additional homework tip  1

The graph for R versus ΔV for T=300K is,

Bundle: Physics For Scientists And Engineers With Modern Physics, Loose-leaf Version, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Single-term, Chapter 26, Problem 46AP , additional homework tip  2

The graph for R versus ΔV for T=320K is,

Bundle: Physics For Scientists And Engineers With Modern Physics, Loose-leaf Version, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Single-term, Chapter 26, Problem 46AP , additional homework tip  3

Explanation of Solution

Given information: The first symbol i.e. Euler’s number is e , the second symbol i.e. magnitude of electron charge is e , Boltzmann’s constant is kB , the absolute temperature is T , the value of current across a semiconductor diode temperature 0K is 1.00nA .

The different values of the R and ΔV is given in spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=280K .

Thus, the graph for R versus ΔV for T=280K is,

Bundle: Physics For Scientists And Engineers With Modern Physics, Loose-leaf Version, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Single-term, Chapter 26, Problem 46AP , additional homework tip  4

The different values of the R and ΔV is given in spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=300K .

Thus, the graph for R versus ΔV for T=300K is,

Bundle: Physics For Scientists And Engineers With Modern Physics, Loose-leaf Version, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Single-term, Chapter 26, Problem 46AP , additional homework tip  5

The different values of the R and ΔV is given in spreadsheet for I and R=ΔVI for ΔV=0.400Vto0.600V in increments of 0.005V for T=320K .

The graph for R versus ΔV for T=320K is,

Bundle: Physics For Scientists And Engineers With Modern Physics, Loose-leaf Version, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Single-term, Chapter 26, Problem 46AP , additional homework tip  6

Conclusion:

Therefore, the graph for R versus ΔV for T=280K is,

Bundle: Physics For Scientists And Engineers With Modern Physics, Loose-leaf Version, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Single-term, Chapter 26, Problem 46AP , additional homework tip  7

Therefore, the graph for R versus ΔV for T=300K is,

Bundle: Physics For Scientists And Engineers With Modern Physics, Loose-leaf Version, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Single-term, Chapter 26, Problem 46AP , additional homework tip  8

Therefore, the graph for R versus ΔV for T=320K is,

Bundle: Physics For Scientists And Engineers With Modern Physics, Loose-leaf Version, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Single-term, Chapter 26, Problem 46AP , additional homework tip  9

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The current–voltage characteristic curve for a semiconductor diode as a function of temperature T is given by                                  I = I0(eeΔV/kBT - 1)Here the first symbol e represents Euler’s number, the base of natural logarithms. The second e is the magnitude of the electron charge, the kB stands for Boltzmann’s constant, and T is the absolute temperature. (a) Set up a spreadsheet to calculate I and R = ΔV/I for ΔV = 0.400 V to 0.600 Vin increments of 0.005 V. Assume I0 = 1.00 nA. (b) Plot R versus ΔV for T = 280 K, 300 K, and 320 K.
In the classic model for electrical conduction, electron mobility u in a metal wire is defined as where v, is the drift velocity of the electrons and E is the electric field in the metal. This parameter is a measure of how freely electrons can travel through a metal when an electric field is applied, such as when the ends of a metal wire are connected to a potential difference. (a) Show that where a is the conductivity of the metal, n is the number density of the charge carriers, and q is the charge on each carrier. (Submit a file with a maximum size of 1 MB.) Choose File no file selected This answer has not been graded yet. (b) B = AVH is often written as AVH - R IB where R, is the Hall coefficient of the metal. Show that the following is true. (Submit a file with a maximum size of 1 MB.) Hyo -n (Choose File no file selected This answer has not been graded yet (c) The table below shows some measured values of Hall coefficients for some elemental materials. What is an average number…
The current I is measured through a sample of an ohmic material as a voltage V is applied. (a) What is the current when the voltage is doubled to 2V (assume the change in temperature of the material is negligible)? (b) What is the voltage applied is the current measured is 0.2I (assume the change in temperature of the material is negligible)? What will happen to the current if the material if the voltage remains constant, but the temperature of the material increases significantly?

Chapter 26 Solutions

Bundle: Physics For Scientists And Engineers With Modern Physics, Loose-leaf Version, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Single-term

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Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning