yes Ids is negative but Isd is positive as here the current is flowing from source to drain. That's also the reason, why K' was considered as negative in Ids equation
I don't understand the gamma value... for nmos Vth was increasing for increasing VSB and thus gamma was positive... shouldn't by the same logic, in PMOS Vth decrease further and gamma be a positive value?
For NMOS , the vsb should have been negative so that the net threshold of the transistor comes down , However in PMOS the vsb should be positive as my (surface potential chi s ) is negative , so if the value needs to be net positive the pmos gamma needs to be negative
by vth we mean effort by Vgs for surface inversion. For nmos Vth (effort for inversion of surface and magnitude of Vth(as Vth is positive) ) increase with increase in Vsb. Surface potential ( psi s) and Vto were positive. Therefore | psi s+ Vsb| was greater than | psi s |. Positive gamma will give Vth= Vto+(gamma)*(positive value) hence Vth> Vto i.e threshold potential increases. For pmos Vth(effort for inversion not magnitude of Vth( as Vth is -ve) ) decrease with increase in Vsb. Surface potential ( psi s) and threshold voltage are negative. Therefore | psi s+ Vsb| is smaller than | psi s |. Negative gamma will give Vth =Vto+(gamma)*( negative term). So we have Vth= Vto+(some positive term). Assume earlier Vth=Vto was -1V and after applying Vsb it becomes -0.8. That is Vth becomes less negative. So inversion occurs at lower -ve value of Vgs. Hope it helps.
@09:28 saturation equation should be divided by 2
nope. since Vds/2 is inside along with the Vgs-Vth. you may deduce that to get /2
@@bharadwaj767 bro the saturation current not the linear one. Saturation current should be divided by 2.
So, we should substitute actual values right?
I think yes
Sir, is Id negative because PMOS current direction is opposite of NMOS?
yes Ids is negative but Isd is positive as here the current is flowing from source to drain. That's also the reason, why K' was considered as negative in Ids equation
I don't understand the gamma value... for nmos Vth was increasing for increasing VSB and thus gamma was positive... shouldn't by the same logic, in PMOS Vth decrease further and gamma be a positive value?
For NMOS , the vsb should have been negative so that the net threshold of the transistor comes down , However in PMOS the vsb should be positive as my (surface potential chi s ) is negative , so if the value needs to be net positive the pmos gamma needs to be negative
by vth we mean effort by Vgs for surface inversion.
For nmos Vth (effort for inversion of surface and magnitude of Vth(as Vth is positive) ) increase with increase in Vsb.
Surface potential ( psi s) and Vto were positive. Therefore | psi s+ Vsb| was greater than | psi s |.
Positive gamma will give Vth= Vto+(gamma)*(positive value) hence Vth> Vto i.e threshold potential increases.
For pmos Vth(effort for inversion not magnitude of Vth( as Vth is -ve) ) decrease with increase in Vsb.
Surface potential ( psi s) and threshold voltage are negative.
Therefore | psi s+ Vsb| is smaller than | psi s |. Negative gamma will give Vth =Vto+(gamma)*( negative term). So we have Vth= Vto+(some positive term). Assume earlier Vth=Vto was -1V and after applying Vsb it becomes -0.8. That is Vth becomes less negative. So inversion occurs at lower -ve value of Vgs.
Hope it helps.
niceeeeeeeeee