Chapter 27, Problem 80AP

(a)

To determine

A spreadsheet for $I$ and $R=\frac{\Delta V}{I}$ for $\Delta V=0.400\text{V}\text{to}\text{0}\text{.600}\text{V}$ in increments of $\text{0}\text{.005}\text{V}$ for $T=280\text{K,}300\text{K,}\text{and}320\text{K}$.

Answer to Problem 80AP

A spreadsheet for $I$ and $R=\frac{\Delta V}{I}$ for $\Delta V=0.400\text{V}\text{to}\text{0}\text{.600}\text{V}$ in increments of $\text{0}\text{.005}\text{V}$ for $T=280\text{K}$ is,

$\Delta V$ (Volts)	$I$ (Amperes)	$R=\frac{\Delta V}{I}$ (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 $R=\frac{\Delta V}{I}$ for $\Delta V=0.400\text{V}\text{to}\text{0}\text{.600}\text{V}$ in increments of $\text{0}\text{.005}\text{V}$ for $T=300\text{K}$ is,

$\Delta V$ (Volts)	$I$ (Amperes)	$R=\frac{\Delta V}{I}$ (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 $R=\frac{\Delta V}{I}$ for $\Delta V=0.400\text{V}\text{to}\text{0}\text{.600}\text{V}$ in increments of $\text{0}\text{.005}\text{V}$ for $T=320\text{K}$ is,

$\Delta V$ (Volts)	$I$ (Amperes)	$R=\frac{\Delta V}{I}$ (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 $k_{\text{B}}$, the absolute temperature is $T$, the value of current across a semiconductor diode temperature $0\text{K}$ is $1.00\text{nA}$.

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

$I=I_0\left(e^{\frac{e\Delta V}{k_{\text{B}}T}}-1\right)$ (1)

Here,

$I$ is the current across a semiconductor diode temperature $T\text{K}$.

$I_0$ is the current across a semiconductor diode temperature $0\text{K}$.

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

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

$k_{\text{B}}$ is the Boltzmann’s constant.

$\Delta V$ is the voltage across the diode.

$T$ is the absolute temperature.

Formula to calculate the resistance across the diode is,

$R=\frac{\Delta V}{I}$ (2)

Here,

$R$ is the resistance across the diode.

The value of magnitude of electron charge is $1.602\times {10}^{-19}\text{C}$.

The value of Boltzmann’s constant is $1.38\times {10}^{-23}\text{J}/\text{K}$.

The value of voltage across the diode varies from $0.400\text{V}$ to $\text{0}\text{.600}\text{V}$ in increments of $\text{0}\text{.005}\text{V}$.

From equation (1), formula to calculate the current across a semiconductor diode temperature $280\text{K}$ is,

$I_1=I_0\left(e^{\frac{e\Delta V_1}{k_{\text{B}}T}}-1\right)$ (3)

Here,

$I_1$ is the current across a semiconductor diode temperature $280\text{K}$.

$\Delta V_1$ is the initial voltage across the diode for temperature $280\text{K}$.

Substitute $1.602\times {10}^{-19}\text{C}$ for $e$, $1.38\times {10}^{-23}\text{J}/\text{K}$ for $k_{\text{B}}$, $0.400\text{V}$ for $\Delta V_1$, $280\text{K}$ for $T$ in equation (3) to find $I_1$,

$\begin{array}{c}{I}_1=\left(1.00\text{nA}\times \frac{1\text{A}}{{10}^9\text{nA}}\right)\left(e^{\frac{\left(1.602\times {10}^{-19}\text{C}\right)\left(0.400\text{V}\right)}{\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\left(280\text{K}\right)}}-1\right)\\ =0.15932\text{A}\end{array}$

Thus, the current across a semiconductor diode temperature $280\text{K}$ is $0.15932\text{A}$.

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

$R_1=\frac{\Delta V_1}{I_1}$ (4)

Here,

$R_1$ is the resistance across the diode.

Substitute $0.15932\text{A}$ for $I_1$, $0.400\text{V}$ for $\Delta V_1$ in equation (4) to find $R_1$,

$\begin{array}{c}{R}_1=\frac{0.400\text{V}}{0.15932\text{A}}\\ =25.1067\Omega \end{array}$

Thus, the resistance across the diode is $25.1067\Omega$.

As the value of voltage across the diode varies from $0.400\text{V}$ to $\text{0}\text{.600}\text{V}$ in increments of $\text{0}\text{.005}\text{V}$, the values for the current and resistance are calculated by same procedure as above.

Thus, a spreadsheet for $I$ and $R=\frac{\Delta V}{I}$ for $\Delta V=0.400\text{V}\text{to}\text{0}\text{.600}\text{V}$ in increments of $\text{0}\text{.005}\text{V}$ for $T=280\text{K}$ is,

$\Delta V$ (Volts)	$I$ (Amperes)	$R=\frac{\Delta V}{I}$ (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 $300\text{K}$ is,

${I^{\prime }}_1=I_0\left(e^{\frac{e\Delta {V^{\prime }}_1}{k_{\text{B}}T}}-1\right)$ (5)

Here,

${I^{\prime }}_1$ is the current across a semiconductor diode temperature $300\text{K}$.

$\Delta {V^{\prime }}_1$ is the voltage across the diode for temperature $300\text{K}$.

Substitute $1.602\times {10}^{-19}\text{C}$ for $e$, $1.38\times {10}^{-23}\text{J}/\text{K}$ for $k_{\text{B}}$, $0.400\text{V}$ for $\Delta {V^{\prime }}_1$, $300\text{K}$ for $T$ in equation (5) to find ${I^{\prime }}_1$,

$\begin{array}{c}{I^{\prime }}_1=\left(1.00\text{nA}\times \frac{1\text{A}}{{10}^9\text{nA}}\right)\left(e^{\frac{\left(1.602\times {10}^{-19}\text{C}\right)\left(0.400\text{V}\right)}{\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\left(300\text{K}\right)}}-1\right)\\ =0.005274\text{A}\end{array}$

Thus, the current across a semiconductor diode temperature $300\text{K}$ is $0.005274\text{A}$.

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

${R^{\prime }}_1=\frac{\Delta {V^{\prime }}_1}{{I^{\prime }}_1}$ (6)

Here,

${R^{\prime }}_1$ is the resistance across the diode.

Substitute $0.005274\text{A}$ for ${I^{\prime }}_1$, $0.400\text{V}$ for $\Delta {V^{\prime }}_1$ in equation (6) to find ${R^{\prime }}_1$,

$\begin{array}{c}{R^{\prime }}_1=\frac{0.400\text{V}}{0.005274\text{A}}\\ =75.84672\Omega \end{array}$

Thus, the resistance across the diode is $75.84672\Omega$.

As the value of voltage across the diode varies from $0.400\text{V}$ to $\text{0}\text{.600}\text{V}$ in increments of $\text{0}\text{.005}\text{V}$, the values for the current and resistance are calculated by same procedure as above.

Thus, a spreadsheet for $I$ and $R=\frac{\Delta V}{I}$ for $\Delta V=0.400\text{V}\text{to}\text{0}\text{.600}\text{V}$ in increments of $\text{0}\text{.005}\text{V}$ for $T=300\text{K}$ is,

$\Delta V$ (Volts)	$I$ (Amperes)	$R=\frac{\Delta V}{I}$ (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 $320\text{K}$ is,

${I^{″}}_1=I_0\left(e^{\frac{e\Delta {V^{″}}_1}{k_{\text{B}}T}}-1\right)$ (7)

Here,

${I^{″}}_1$ is the current across a semiconductor diode temperature $320\text{K}$.

$\Delta {V^{″}}_1$ is the voltage across the diode for temperature $320\text{K}$.

Substitute $1.602\times {10}^{-19}\text{C}$ for $e$, $1.38\times {10}^{-23}\text{J}/\text{K}$ for $k_{\text{B}}$, $0.400\text{V}$ for $\Delta V^{″}$, $320\text{K}$ for $T$ in equation (7) to find ${I^{″}}_1$,

$\begin{array}{c}{I^{″}}_1=\left(1.00\text{nA}\times \frac{1\text{A}}{{10}^9\text{nA}}\right)\left(e^{\frac{\left(1.602\times {10}^{-19}\text{C}\right)\left(0.400\text{V}\right)}{\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\left(320\text{K}\right)}}-1\right)\\ =0.002004\text{A}\end{array}$

Thus, the current across a semiconductor diode temperature $320\text{K}$ is $0.002004\text{A}$.

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

${R^{″}}_1=\frac{\Delta {V^{″}}_1}{{I^{″}}_1}$ (8)

Here,

${R^{″}}_1$ is the resistance across the diode.

Substitute $0.002004\text{A}$ for ${I^{″}}_1$, $0.400\text{V}$ for $\Delta {V^{″}}_1$ in equation (8) to find ${R^{″}}_1$,

$\begin{array}{c}{R^{″}}_1=\frac{0.400\text{V}}{0.002004\text{A}}\\ =199.5582\Omega \end{array}$

Thus, the resistance across the diode is $199.5582\Omega$.

As the value of voltage across the diode varies from $0.400\text{V}$ to $\text{0}\text{.600}\text{V}$ in increments of $\text{0}\text{.005}\text{V}$, the values for the current and resistance are calculated by same procedure as above.

Thus, a spreadsheet for $I$ and $R=\frac{\Delta V}{I}$ for $\Delta V=0.400\text{V}\text{to}\text{0}\text{.600}\text{V}$ in increments of $\text{0}\text{.005}\text{V}$ for $T=320\text{K}$ is,

$\Delta V$ (Volts)	$I$ (Amperes)	$R=\frac{\Delta V}{I}$ (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 $R=\frac{\Delta V}{I}$ for $\Delta V=0.400\text{V}\text{to}\text{0}\text{.600}\text{V}$ in increments of $\text{0}\text{.005}\text{V}$ for $T=280\text{K}$ is,

$\Delta V$ (Volts)	$I$ (Amperes)	$R=\frac{\Delta V}{I}$ (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 $R=\frac{\Delta V}{I}$ for $\Delta V=0.400\text{V}\text{to}\text{0}\text{.600}\text{V}$ in increments of $\text{0}\text{.005}\text{V}$ for $T=300\text{K}$ is,

$\Delta V$ (Volts)	$I$ (Amperes)	$R=\frac{\Delta V}{I}$ (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 $R=\frac{\Delta V}{I}$ for $\Delta V=0.400\text{V}\text{to}\text{0}\text{.600}\text{V}$ in increments of $\text{0}\text{.005}\text{V}$ for $T=320\text{K}$ is,

$\Delta V$ (Volts)	$I$ (Amperes)	$R=\frac{\Delta V}{I}$ (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 $\Delta V$ for $T=280\text{K,}300\text{K,}\text{and}320\text{K}$.

Answer to Problem 80AP

The graph for $R$ versus $\Delta V$ for $T=280\text{K}$ is,

The graph for $R$ versus $\Delta V$ for $T=300\text{K}$ is,

The graph for $R$ versus $\Delta V$ for $T=320\text{K}$ is,

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 $k_{\text{B}}$, the absolute temperature is $T$, the value of current across a semiconductor diode temperature $0\text{K}$ is $1.00\text{nA}$.

The different values of the $R$ and $\Delta V$ is given in spreadsheet for $I$ and $R=\frac{\Delta V}{I}$ for $\Delta V=0.400\text{V}\text{to}\text{0}\text{.600}\text{V}$ in increments of $\text{0}\text{.005}\text{V}$ for $T=280\text{K}$.

Thus, the graph for $R$ versus $\Delta V$ for $T=280\text{K}$ is,

The different values of the $R$ and $\Delta V$ is given in spreadsheet for $I$ and $R=\frac{\Delta V}{I}$ for $\Delta V=0.400\text{V}\text{to}\text{0}\text{.600}\text{V}$ in increments of $\text{0}\text{.005}\text{V}$ for $T=300\text{K}$.

Thus, the graph for $R$ versus $\Delta V$ for $T=300\text{K}$ is,

The different values of the $R$ and $\Delta V$ is given in spreadsheet for $I$ and $R=\frac{\Delta V}{I}$ for $\Delta V=0.400\text{V}\text{to}\text{0}\text{.600}\text{V}$ in increments of $\text{0}\text{.005}\text{V}$ for $T=320\text{K}$.

The graph for $R$ versus $\Delta V$ for $T=320\text{K}$ is,

Conclusion:

Therefore, the graph for $R$ versus $\Delta V$ for $T=280\text{K}$ is,

Therefore, the graph for $R$ versus $\Delta V$ for $T=300\text{K}$ is,

Therefore, the graph for $R$ versus $\Delta V$ for $T=320\text{K}$ is,