00:01
Hello students, here this is common emitter configuration where r pi and i b is the base current and here it is grounded voltage in v in is given.
00:19
The sinusoidal voltage is 0 .5 sin omega frequency given 10 kilohertz so that it can be written as 0 .5 sin 2 pi 10 k t.
00:35
Now we know that voltage in is given by from ohm's law r pi into i b then v out can be written as minus r c into beta into i b.
01:01
From this we can obtain the voltage gain that is the ratio of output to input voltage minus r c beta i b divided by r pi i b.
01:18
I b get cancelled minus r c beta divided by r pi.
01:25
Here r pi is nothing but beta divided by g m therefore voltage gain will become minus g m into r c and g m value is given by current i c divided by voltage t.
01:48
Then we can write minus g m into r c must be greater than 10 which is given in the equation.
02:00
Let us consider current i c 1 milliampere and register r c value 0 .5 kilo ohm.
02:09
Then by substituting here for g m that is 1 by 26 into r c value 0 .5 kilo which gives the value 19 .23 and it is greater than as it satisfies the condition we can write condition is true and also base voltage minus base emitter voltage is zero.
02:44
From this base voltage will be equal to 0 .7 volt.
02:50
Next we will move on to check it is working in active mode or not.
02:57
This circuit is considered for dc analysis...