Correction to the last part of the graph : For VX > 1V , Vx will be the drain , So VGS will be 1V and the mos will be on . Under this condition , for VX >=1V MOS will be in saturation. *For VX=1V IX=0 , VDS=0 here. *For VX < 1V , Vx wil be the source , under this condition , as rightly pointed out by you, we are at the edge of saturation. So for VX
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Hi Praveen , thank you for pointing out. There is a correction needed in the last part of the graph. *For VX > 1V , Vx will be the drain , So VGS will be 1V and the mos will be on . Under this condition , for VX >=1V MOS will be in saturation. *For VX=1V IX=0 , VDS=0 here. *For VX < 1V , Vx wil be the source , under this condition , as rightly pointed out by you, we are at the edge of saturation. So for VX
Hi. I am not use if i understand your question correctly. For VDS > 0 and Vgs-Vth =0(saturation) . There will be current . You can also look at it from the I-V curve perspective . Thanks
Correction to the last part of the graph :
For VX > 1V , Vx will be the drain , So VGS will be 1V and the mos will be on . Under this condition , for VX >=1V MOS will be in saturation.
*For VX=1V IX=0 , VDS=0 here.
*For VX < 1V , Vx wil be the source , under this condition , as rightly pointed out by you, we are at the edge of saturation. So for VX
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.
This link has a discount coupon valid till 2021 30th October. Happy learning 😃
For Vx>1V --------> Ix = 0.
For Vx
Hi Praveen , thank you for pointing out. There is a correction needed in the last part of the graph.
*For VX > 1V , Vx will be the drain , So VGS will be 1V and the mos will be on . Under this condition , for VX >=1V MOS will be in saturation.
*For VX=1V IX=0 , VDS=0 here.
*For VX < 1V , Vx wil be the source , under this condition , as rightly pointed out by you, we are at the edge of saturation. So for VX
@@vlsiorg For Vx>1, the MOS is in saturation, but (Vgs-Vt) = 0, so by the square law, Ix should also be 0
Hey when you make regarding this for correction in last paragraph??
If Vx is really high, we agree that the VDS is high and the MOS would be at saturation. However, VGS-VTH = (2V-1V)-1V = 0V, hence, shouldn't Ix= 0A?
Hi.
I am not use if i understand your question correctly.
For VDS > 0 and Vgs-Vth =0(saturation) . There will be current .
You can also look at it from the I-V curve perspective .
Thanks
Nice
This plot is incorrect.
Check once..
Re-check please.
Hi vishal , i have written the correction in the comments. I will soon make a update on this in my next video. Thank you 😃
@@vlsiorg since the gate to source voltage is not more than threshold, will there be any current. That's my doubt