GATE Electronics and Communications (EC) 2019 Solved Paper

Question : 58 of 65

Marks: +1, -0

In an ideal pn junction with an ideality factor of 1 at

{T}=300 {Km}

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 × 10^{-23} {JK}^{-1}, {h}=6.625 × 10^{-34} {J}-

s, q=1.602 × 10^{-19} {C}]

Solution:

{I}_{ {r}}=0.75 {I}_{5}

∴

Forward current

= {I}_{ {D}}=-0.75 {Is}

∴ I_{s}( {e}^{ {vo} {/} {nvt}}-1)=-0.75 {I}_{ {s}}

Now Take

{n}=1

∴ e^{ {vo} {/} {v} {T}}=0.25

∴ {V}_{ {D}}= {V} {T} \ln (0.25)

∴ {V}_{ {R}}=- {V} {T} \ln (0.25)

=-\;{1.38 × 10^{-23} × 300}/{1.6 × 10^{-19}} ×-1.386

∴ {V}_{ {R}}=35.87 {mv}

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