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

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(D) Given the equation: $I = (e^{1000V/T} - 1) 	ext{ mA}$.At $I = 5 	ext{ mA}$,we have $5 = e^{1000V/T} - 1$,which implies $e^{1000V/T} = 6$.Differe

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$0.2$

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(D) Given the equation: $I = (e^{1000V/T} - 1) 	ext{ mA}$.At $I = 5 	ext{ mA}$,we have $5 = e^{1000V/T} - 1$,which implies $e^{1000V/T} = 6$.Differentiating the current $I$ with respect to voltage $V$:$\frac{dI}{dV} = e^{1000V/T} \cdot \frac{1000}{T}$.Substituting the value of $e^{1000V/T} = 6$ and $T = 300 	ext{ K}$:$\frac{dI}{dV} = 6 \cdot \frac{1000}{300} = 6 \cdot \frac{10}{3} = 20 	ext{ mA/V}$.The error in current $\Delta I$ is given by $\Delta I = \left| \frac{dI}{dV} ight| \cdot \Delta V$.Given $\Delta V = 0.01 	ext{ V}$,we get:$\Delta I = 20 \cdot 0.01 = 0.2 	ext{ mA}$.

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