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The current voltage relation of diode is given by $I=(e^{1000V/T} -1)\;mA$, where the applied voltage $V$ is in volts and the temperature $T$ is in degree Kelvin. If a student makes an error measuring $ \mp 0.01\;V$ while measuring the current of $5\; mA$ at $300\; K$, what will be the error in the value of current in $mA$ ?
$0.02$
$0.5$
$0.05$
$0.2$
Solution
$5 = {e^{\log \frac{V}{T}}} – 1$
$\Rightarrow \quad \mathrm{e}^{1000} \frac{\mathrm{V}}{\mathrm{T}}=6$ …………….. $(1)$
Again, $I=e^{1000 \frac{V}{T}}-1$
$\frac{\mathrm{dI}}{\mathrm{dV}}=\mathrm{e}^{\frac{1000 \mathrm{v}}{\mathrm{T}}} \frac{1000}{\mathrm{T}}$
${\rm{dI}} = \frac{{1000}}{{\rm{T}}}{{\rm{e}}^{\frac{{1000}}{{\rm{T}}}{\rm{V}}}}{\rm{dV}}$
Using $(1)$
$\Delta \mathrm{I}=\frac{1000}{\mathrm{T}} \times 6 \times 0.01=\frac{60}{\mathrm{T}}=\frac{60}{300}=0.2 \mathrm{\,mA}$
