00:01
Hello students, in this question, j -5th in voltage divider bias configuration is given, input parameters are given.
00:09
First, we have to calculate the quotient gate source voltage that is vgsq.
00:18
This can be calculated from the formula vgsq is equal to minus of vp minus idss into the source resistance.
00:29
Now, plucking the values given minus 10 volt into 11 milliampere into 1 .2 kiloohms.
00:41
So, these values are given, taken from the given data, which is minus 10 volt minus 13 .2 volt.
00:50
This gives the quotient gate source voltage vgsq is equal to minus 23 .2 volt.
00:59
Next, we have to calculate the transconductance gm.
01:06
Transconductance gm is equal to 2 into square root of idss into ido.
01:28
So, gm equal to 2 into square root of 11 milliampere into 8 milliampere and this gives 2 into square root of 0 .088 amperes, which is 2 into 0 .297 ampere and we get the value of transconductance equal to 0 .594 amperes.
02:00
Next is we have to calculate the voltage gain apl.
02:14
So, this is calculated by minus transconductance gm into r, the parallel resistance rd parallel to rl here and this will be equal to minus 0 .594 into 3 kiloohm is parallel to your rls 4 .7 kiloohm and solving this, we get 1 .68 kiloohm into minus 0 .594.
02:53
This gives a voltage gain of minus 0 .9979.
03:04
Fourth one, we have to calculate the input thermal resistance, which is rthi.
03:25
So, this is given by rsig, which is parallel to rg1 into 1 by transconductance.
03:40
Now, plugging in the values here, 2 kiloohm is parallel to 46 kiloohm plus here plus will be plus 1 by 0 .594 and this gives 1 .94 kiloohm plus 1 .68 ohms and if we are adding this, we get the input thermal resistance is equal to 1 .94 kiloohms.
04:16
Since it is a very small value, we can take it approximately as 1 .94 kiloohm...