00:01
Hello students, this question is on diode.
00:03
Here we have ideal diode equation.
00:12
It is given by expression.
00:14
We have id is equal to is exponential vd divided by n vt minus 1.
00:26
Here id it is diode current, is it is reverse saturation current, vd it is diode voltage, n it is given as equal to 1 and vt it is equal to kt by q.
00:43
Here k it is boltzmann's constant, t it is temperature and q it is charge.
00:48
So here we are given vd it is equal to 0 .6 volt.
00:55
We have id it is equal to 0 .1 milliampere and we have here n it is equal to 1.
01:03
So we can write here 0 .1 into 10 to the power minus 3 amperes, id it is equal to is into exponential vd that is 0 .6 volt divided by 1 into vt.
01:21
Vt it is given as 0 .026 volt into 0 .026 minus 1.
01:30
Here we get 0 .1 into 10 to the power minus 3 amperes.
01:35
It is equal to is into 1 .0523988 into 10 to the power 10.
01:45
From here we get reverse saturation current it is equal to 0 .1 into 10 to the power minus 3 amperes divided by 1 .0523988 into 10 to the power 10.
02:03
This comes out to be equal to 0 .095 into 10 to the power minus 13 amperes.
02:09
Here we get reverse saturation current it is equal to 9 .5 into 10 to the power minus 15 amperes.
02:20
Now we are given diode voltage it is equal to 0 .65 volt.
02:25
Here we have id it is equal to 9 .5 into 10 to the power minus 15 amperes into exponential 0 .65 volt divided by 1 into 0 .026 volt minus 1.
02:43
Here we get id it is equal to 0 .000684 amperes or this comes out to be equal to 0 .684 into 10 to the power minus 3 amperes or we can write id it is equal to 0 .684 milli amperes.
03:06
Here for vd it is equal to 0 .70 volt.
03:10
We have id equal to 9 .5 into 10 to the power minus 15 amperes into exponential of 0 .70 volt divided by 1 into 0 .026 volt minus 1.
03:28
Here we get id it is equal to 0 .00468 amperes so this is equal to 4 .68 into 10 to the power minus 3 amperes...