GATE Papers >> ECE >> 2019 >> Question No 59

Question No. 59 ECE | GATE 2019

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.

$ \lbrack k=1.38\times10^{-23}\;\mathrm{JK}^{-1},\;h=6.625\times10^{-34}\;\mathrm J-\mathrm s,\;q=1.602\times10^{-19}\mathrm C\rbrack $


Answer : 34.00 to 38.00


Solution of Question No 59 of GATE 2019 ECE Paper

Given

$ \mathrm\eta $ = 1

T = 300 K

$ {\mathrm I}_{\mathrm f}={\mathrm I}_0\left(\mathrm e^{-\frac{\mathrm V}{\mathrm{ηVT}}}-1\right) $

From I-V Characteristic the equation will be

$ \begin{array}{l}-0.75\;{\mathrm I}_0={\mathrm I}_{\mathrm o}\left(\mathrm e^\frac{-{\mathrm V}_{\mathrm R}}{{\mathrm V}_{\mathrm T}}-1\right)\\0.25=\mathrm e^{-\frac{{\mathrm V}_{\mathrm R}}{25\mathrm m}}\\\ln\;0.25=-\frac{{\mathrm V}_{\mathrm R}}{25\mathrm m}\\{\mathrm V}_{\mathrm R}=-25\mathrm m\times\ln\;0.25\\{\mathrm V}_{\mathrm R}=34.65\;\mathrm{mV}\end{array} $

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