@@coolwinder No, is not. The gain factor on a Jfet is its transconductance and it is dependent of the current in the drain. The gain of the first excercise explained is Vout = Vin*gm*R ... all this assuming it is not connected to any load in witch case the final expression will be Vout = Vin*gm*(R//Rl). One more thing, you might see gm as Yfs. Same thing, different name.
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![เบิ้ล ปทุมราช - สาวสอง ( มีมี่ของอ้าย ) ft. ตาต้า,ต้องเต [OFFICIAL MV]](http://i.ytimg.com/vi/mqlMjnP08XI/mqdefault.jpg)
You're wrong! The small signal voltage gain of that circuit is not equal to zero.
It's not wrong, check this video: th-cam.com/video/P0OjpBJsLOE/w-d-xo.html
It is correct. Unless the model is different then the one showed. Pls grasp whats your model.
@@coolwinder Totally nonsense!
@@coolwinder No, is not. The gain factor on a Jfet is its transconductance and it is dependent of the current in the drain. The gain of the first excercise explained is Vout = Vin*gm*R ... all this assuming it is not connected to any load in witch case the final expression will be Vout = Vin*gm*(R//Rl). One more thing, you might see gm as Yfs. Same thing, different name.
👍
I am New, thank you😊
vout2/vin=Av=GM*Rout=-gm1*ro...answer
vout1/vin=gm/1+gmRs..answer
vout/vin=GM*Rout=0..answer
This was so much fun! Pls continue :)
The circuit on the left is source degenerated so the effective gm will be gm/(1+gmR). And the gain will be gm effective *ro
@pragz92 my answer is the same as You
sir make more questions like this
I support ur channel.
Plz upload circuit design related videos for interviews
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