00:01
Hi, in this question first let us consider jfet.
00:04
Now we have to calculate the drain current which is given by idss divided by 2 into vp into so 2vp divided by vgs plus 5v.
00:15
So we have this is so 16 divided by 2 into minus 8 so magnitude of minus 8 divided by so bar 0v plus 5.
00:27
So we have the drain current is equal to 16 milli ampere.
00:31
So here so here we have vgs is 0.
00:36
So it is biased at pin off voltage and drain current is equal to the saturation current.
00:43
So now we have bjt.
00:46
So for second case we have bjt where we have the collector current is given by it is beta into ib which is the base current.
00:55
So we have first let us have base current which is vcc minus vbe divided by rb.
01:03
So this is equal to 25 minus 0 .7 divided by 4 which is equal to so 0 .675 milli ampere.
01:12
So minus so from this we have ic is equal to 85 into minus 0 .675 divided by so or ic will be equal to minus 57 .375 milli ampere.
01:27
So bjt is forward because collector current is negative.
01:31
In the next part we have to calculate the transconductance so small signal transconductance so which is equal to 2 into root of idss into magnitude of vp divided by vgs...