A one-sided p^+n silicon diode has doping concentrations of...

A one-sided $\mathrm{p}^{+} \mathrm{n}$ silicon diode has doping concentrations of $N_{a}=5 \times 10^{17} \mathrm{~cm}^{-s}$ and $N_{d}=8 \times 10^{15} \mathrm{~cm}^{-3}$. The minority carrier lifetimes are $\tau_{n 0}=10^{-7} \mathrm{~s}$ and $\tau_{p 0}=8 \times$ $10^{-8} \mathrm{~s}$. The cross-sectional area is $A=2 \times 10^{-4} \mathrm{~cm}^{2}$. Calculate the $(a)$ reverse-biased saturation current, and $(b)$ the forward-bias current at (i) $V_{a}=0.45 \mathrm{~V}$, (ii) $V_{a}=0.55 \mathrm{~V}$, and (iii) $V_{a}=0.65 \mathrm{~V}$.

Transcript

00:01 Hello students in this question we are given a one -sided p .n.

00:05 Silicon diode has dropping concentration.

00:08 So we are given the value of n .a.

00:10 Is equal to 5 multiplied by 10 to the par 17 per centimeter cube.

00:15 And we are given nd which is 8 multiplied 10 to 5 .15 per centimeter cube.

00:22 And the minority carrier lifetimes are tau and knot.

00:29 This is tau n0 is equal to 10 to the by minus 7 seconds and tau p .0 is 8 multiplied 10 to the minus 8 seconds and the cross -sectional area a is equal to 2 multiplied 10 to minus 4 cm square so we have to calculate in the a part the reverse biased current reverse via saturation current now for calculating the reverse by saturation current that is i is which is equal to a multiplied by j is co -solution area j is the current density so for first we find out the current density because the area is giving we founded current density which is e n i square multiplied by 1 over n d root of d d p over tau p plus 1 over n a root of n a root of d n d n over tau n this is the formula for current density so we calculate j over here so we have e as 1 .6 multiply by 10 to the by minus 19 and we have n i over here which is 8 multiply 10 to 15 per centipote a over 1 over we have n d here we have n i which is 1 .5 multiplied by 10 to 10 per centimeter host square we have 1 over n d which is 8 multiplied by 10 to the bar 15 root of dp which is 10 over to be which is 8 multiplied by 10 to the by minus 8 plus we have 1 over n a which is 5 multiplied 10 to the by 17 root of d n which is 25 over we have 1 multiplied by 10 to the minus 7 so, here this comes, the value of j comes out to be 5 .15 multiplied by 10 to the 0 .11 ampere per centimeter square.

02:46 Now, is is j multiplied by 8, j is 5 .15 multiplied by 10 to minus 11 multiplied by a is 2 multiplied by 10 to 2 minus 4...

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