sir why pMos to saturation region when given Ov as input according to Vds < vgs - vth condition is false , it should drive in linear region for high output 13:29 please some one tell me if I wrong
At 9.15 you said the upper MOSFET is pMos but I think it's actually a Nmos because the arrow from source to Gate is inside.. am I wrong or right ? Please reply as soon as possible thank you.
At switching threshold voltage( Vm= Vdd/2)both NMOS and PMOS will be ON. At that point Current will flow from Vdd to ground and as they are in series current will be same in both of them.
As in saturation MOS behaves as a current source ideally with resistance infinity...then why has sir considered the source to drain resistance as a finite resistance. Anyone please clarify.
Because , the input is dynamic and it`s fluctuates between 0(T/2) and 1(T/2) so for that half of the time Pmos will switch on and Nmos will cutt off and for another half time vice versa, So the Pmos will work in 3 rd quadrant for half of the time and the other half of the time Nmos will work as low resistance path to discharge the capacitor and current will be decreases at that moment in the 1st quadrant .
you speak very good english
Very good prof
sir why pMos to saturation region when given Ov as input according to Vds < vgs - vth condition is false , it should drive in linear region for high output 13:29 please some one tell me if I wrong
If change the source of pmos to gnd and nmos to vdd what will happen? I just want know answer
It will not be inverter it will act as voltage follower
buffer but bad one
best👌🏻
Pmos becomes on when Vsg > |Vtp|, so when the input is 0 Pmos becomes on.
yeah as vgs=0-vdd=-vdd < Vtp
At 9.15 you said the upper MOSFET is pMos but I think it's actually a Nmos because the arrow from source to Gate is inside.. am I wrong or right ? Please reply as soon as possible thank you.
upper one is pmos only, refer the first lecture of this series
You are wrong because for NMOS the arrow will be from gate to source
Why at 29:44 you said current flowing through them will be equal to each other? Isn't one always off?
At switching threshold voltage( Vm= Vdd/2)both NMOS and PMOS will be ON. At that point Current will flow from Vdd to ground and as they are in series current will be same in both of them.
As in saturation MOS behaves as a current source ideally with resistance infinity...then why has sir considered the source to drain resistance as a finite resistance. Anyone please clarify.
Watch previous lecture
Thats "ideally". In practical obviously there is nothing like infinite resistance. Watch previous videos, your doubt will be solved.
how i can download ppt which explained by sir
Mail me for ppt....premanand.patel@gmail.com, you will get it.
@@dhairyaclasses6156 i will mail you, please send it
@@zaidakhtar3093 mail me then you will get it soon.
i.e in transcripts
Why there is the need of shifting from second quadrant to first quadrant
Because if no shift on 2nd quad then that means I_ds_p = - I_ds_n ... the current for nMos flows in opposite direction to that of pMos
Because , the input is dynamic and it`s fluctuates between 0(T/2) and 1(T/2) so for that half of the time Pmos will switch on and Nmos will cutt off and for another half time vice versa, So the Pmos will work in 3 rd quadrant for half of the time and the other half of the time Nmos will work as low resistance path to discharge the capacitor and current will be decreases at that moment in the 1st quadrant .
@@nafizalam8097 no
it's because the x-axis changes from Vds to Vin. so you add 2.5
doubt at graphing part
What is threshold voltage
Threshold voltage is the minimum voltage required for forward biasing the source and body to conduct the charge carriers and switch onn the mos