The current-voltage relation of a diode is gi | vedclass

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(B) Given the relation: $I = e^{\frac{1000V}{T}} - 1$.Differentiating with respect to $V$:$\frac{dI}{dV} = \frac{1000}{T} e^{\frac{1000V}{T}}$.Thus,$d

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$0.2$

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(B) Given the relation: $I = e^{\frac{1000V}{T}} - 1$.Differentiating with respect to $V$:$\frac{dI}{dV} = \frac{1000}{T} e^{\frac{1000V}{T}}$.Thus,$dI = \left( \frac{1000}{T} e^{\frac{1000V}{T}} ight) dV$.Given $I = 5 \, mA$,$T = 300 \, K$,and $dV = 0.01 \, V$.From the original equation,$I + 1 = e^{\frac{1000V}{T}}$.Substituting $I = 5$,we get $e^{\frac{1000V}{T}} = 5 + 1 = 6$.Now,substitute these values into the expression for $dI$:$dI = \left( \frac{1000}{300} 	imes 6 ight) 	imes 0.01$.$dI = \left( \frac{10}{3} 	imes 6 ight) 	imes 0.01$.$dI = 20 	imes 0.01 = 0.2 \, mA$.

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