Discussion Forum

The current-voltage relation of diode is given by  $I=(e^{1000V/T}-1)mA$, where the applied voltage V is in volt and the temperature T is in kelvin. If a student makes an error measuring the current of 5 mA at 300K, What will be the error in the value of current in m A?

Answer:

Explanation:

Given,   $I=(e^{1000V/T}-1)mA$ 

 $dV=\pm0.01V$

$T=300K\Rightarrow mA$

So,       $I= e^{1000V/T}-1$

$I+1= e^{1000V/T}$
Taking log onboth sides , we get

       log(I+1)= $\frac{1000V}{T}$

On differentiating,    $\frac{dI}{I+1}=\frac{1000}{T}dV$

$dI= \frac{1000}{T}\times (I+1)dV$

$\Rightarrow dI= \frac{1000}{300}\times (5+1)0.01=0.2mA$

So, error in the value of current is 0.2 mA