The current-voltage characteristic of an ideal p-n junction diode is given by i=i0(ee v / k T-1) where the drift current i 0 equal 10 μ A. Take the temperature T to be 300 K. Find the voltage V 0 for which ee V / K T=100. One can neglect the term 1 for voltages greater than this value

Question

The current-voltage characteristic of an ideal p-n junction diode is given by i=i0(ee v / k T-1) where the drift current i 0 equal 10 μ A. Take the temperature T to be 300 K. Find the voltage V 0 for which ee V / K T=100. One can neglect the term 1 for voltages greater than this value (A) 0.12 V (B) 0.5 V (C) 0.25 V (D) 1.5 V

Tardigrade Admin · Accepted Answer

Given ee v / k T=100 or (e v/k T)= ln 100 V=(k T ln 100/e) =(1.38 × 10-23 × 300 × ln 100/1.6 × 1019)=0.12 V

$0.12 V$

$0.5 V$

$0.25 V$

$1.5 V$

Given $e^{e v / k T} = 100$
or $\frac{e v}{k T} = ln 100$
$V = \frac{k T l n 100}{e}$
$= \frac{1.38 \times 1 0 ^{- 23} \times 300 \times l n 100}{1.6 \times 1 0 ^{19}} = 0.12 V$

Q. The current-voltage characteristic of an ideal p-n junction diode is given by i=i0​(eev/kT−1) where the drift current i0​ equal 10μA. Take the temperature T to be 300K. Find the voltage V0​ for which eeV/KT=100. One can neglect the term 1 for voltages greater than this value

0.12V

0.5V

0.25V

1.5V

Given eev/kT=100 or kTev​=ln100 V=ekTln100​ =1.6×10191.38×10−23×300×ln100​=0.12V

$0.12 V$

$0.5 V$

$0.25 V$

$1.5 V$

Given $e^{e v / k T} = 100$
or $\frac{e v}{k T} = ln 100$
$V = \frac{k T l n 100}{e}$
$= \frac{1.38 \times 1 0 ^{- 23} \times 300 \times l n 100}{1.6 \times 1 0 ^{19}} = 0.12 V$