Question

Does anybody help me with Q5 ? Here, VS=80V, R=1.5kΩ, and the transistor parameter f=2mA/V^3. The amplifier is biased with VIN=2V.

Answer 1

Vout=Vs-(f*(Vin)^2)*R

-3*f*(Vgs)^2*RL

Answer 2

I did it.. but not sure about the answer ..

Answer 3

because I don't know if I've applied it correctly .. when I replace this value f=2mA/V^3
do i need to use 2*10^-3 ?

Answer 4

yes...as it is in mili ampere......

Answer 5

man.. I'll solve Q5 , but I'd like your help wether it's correct or not when I finish this. can you help me this way ?

Answer 6

kk...let me solve it.....nd then i'll tl u the values

Answer 7

Vout is 56

Answer 8

what is the value given for RL

Answer 9

hey .. it's weird.. there is not a RL given LOL
it's Q5 Nurali .. thanks for helping me

Answer 10

Here, VS=80V, R=1.5kΩ, and the transistor parameter f=2mAV3. The amplifier is biased with VIN=2V.

Answer 11

sorryy.....got confused....nd the second part is -24

Answer 12

thanks you very much Kavi.. but the second part is not -24 .. got confused about this question

Answer 13

Nurali.. do you know how Can I solve Rth in Q6 ?

Answer 14

@Logan2 .....calculate it by using the formula....its -36.....m sure

Answer 15

are you sure now ?

Answer 16

it's my last chance :/

Answer 17

see the formula is -3*f*(Vgs)^2*R......Vgs is same as Vin......so it comes out to be...-36 only...i have calculated my value using the same.....nd it was correct....

Answer 18

THAAAANKS GUY !! :D

Answer 19

@logan2: Rl is same as resistor R.