00:01
Hi, in this question we have to design abrupt silicon pn plus diode.
00:07
So here we are given with a value of ebd as 130 volts, ifd as 22 milliampere, v is given as 0 .7 and t is given as 10 power minus 7 seconds.
00:20
Now we have the breakdown voltage is given by es into e square max divided by so divided by 2q so 2q into ebd.
00:36
So now we have so here we have the value of so here we have for silicon so critical field is given by so critical field is given by it is 3 into 10 power 5 so 10 power 5 whereas es value is given as 1 .0355 into 10 power minus 12.
01:00
So now we have the value of nd is given by it is es into e critical square divided by so 2q into vbd.
01:12
So on simplifying we have 1 .0355 into 10 power minus 12 into 3 into 10 power 5 square divided by so divided by 2 into 1 .6 into 10 power minus 19 into 130 volt.
01:27
So from this we have the value of nd is equal to 2 .24 into 10 power 15 centimeter cube inverse.
01:35
Then we have the expression for diode current is is into exponential v naught divided by vt.
01:42
So we have it is q into ni square into a divided by nd root of so hp divided by tp into vt exponential va divided by vt.
01:53
So now this is the expression for is.
01:56
So now so this is so this is the root value...