GATE Electrical Engineering (EE) 2020 Solved Paper

Question : 57 of 65

Marks: +1, -0

A non-ideal diode is biased with a voltage of

-0.03 {V}

, and a diode current of

{I}_{1}

is measured. The thermal voltage is

26 {mV}

and the ideality factor for the diode is

15 \/ 13

. The voltage, in

{V}

, at which the measured current increases to

1.5 {I}_{1}

is closest to:

Solution:

{I}_{ {D}}= {I}_{ {S}}[ {e}^{ {V} {/} {nVT}}-1]

in the diode current equality for forward bios and reverse bias.

I_{1}=I_{3}[e^{-0.03 \/ \;{15}/{3} × 26 m V}-1]

2 ÷ 1 ⇒ 1.5[e^{-0.03 {/} 0.03}-1]=[e^{ {VD} {/} 0.03}-1]

⇒ {V}_{ {D}}=-0.0887 {V} ∼eq-0.09 {V}

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