GATE 2019 EC – Question 59
In an ideal $pn$ junction with an ideality factor of 1 at T=300 K, the magnitude of the reverse-bias voltage required to reach 75% of its reverse saturation current, rounded off to 2 decimal places, is ________ mV.
[$k=1.38\times10^{-23}\ \text{JK}^{-1}$, $h=6.625\times10^{-34}$ J-s, $q=1.602\times10^{-19}$ C]
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Correct answer: 35.62 to 35.98
Explanation
The reverse current is $I=I_s\left(1-e^{-V_R/V_T}\right)$. Setting it equal to $0.75I_s$ gives $e^{-V_R/V_T}=0.25$, so $V_R=V_T\ln4=25.85\text{ mV}\times1.386=35.8$ mV.