Yes you have to, but the expression and the circuit topology take care of that. The expression Vds > Vgs - Vth can be written as: (Vd - Vs)>(Vg - Vs) - Vth, where Vd, Vg, and Vs mean the voltage of drain, gate and source to the ground respectively. As you can see the voltage Vs cancel out on both side, leaving only Vd > Vg - Vth. Plugging in the expression of Vd (Vdd-Id*Rd) and Vg (which is Vo) gives you the inequality on the whiteboard. Since Vd (instead of Vds) is directly measured from drain to ground, it automatically includes the voltage drop across resistor Rs.
actually we need to consider only vds>vgs-Vth but instead of vgs we are considering Vo (voltage between the gate and ground) so, similarly in place of Vds i.e. Vdd- (Id*Rd)- (Id*Rs) we are considering Vdd-(Id*Rd) i.e (Voltage between the drain and ground) so, Vdd-(Id*Rd) > Vo-Vth is the condition instead of Vdd- (Id*Rd)- (Id*Rs) > Vgs-Vth (PS: even I too got the same doubt initially by after seeing yours and someone elses doubt I thought and came to this conclusion)
dear sir, at 12:34 when we are checking whether device really in saturation or not, why gate voltage is taken only Vo , it should be Vo-IdRd becz Vgs is equal to that
That will be for Diode connected device but that circuit have constant current circuit model. You can see a battery connected between gate and source. What you are saying is for when gate and drain is shorted.
what u r talking about is for a Diode Connected load But this is a current source device for which current is coming out of an arbitrary node and going to ground which can be made by an NMOS device in which the gate and source terminals are biased
At 57:32 Rout = (1+ gmRs)ro + Rs. When ro= infinite, we are still left with Rout = Rs.............isn't it so Sir? Then why are you telling Rout will be infinite in your next video?
Excellent explanation with mind blowing applications which are required for a good circuit designer..
Awesomeeeeee Prof Razavi..Whole lots of love!
@13:00 shouldn't we consider the drop across the resistor Rs as the condition is Vds >= Vgs-Vth?
yeah... I too noticed that
Yes you have to, but the expression and the circuit topology take care of that.
The expression Vds > Vgs - Vth can be written as: (Vd - Vs)>(Vg - Vs) - Vth, where Vd, Vg, and Vs mean the voltage of drain, gate and source to the ground respectively.
As you can see the voltage Vs cancel out on both side, leaving only Vd > Vg - Vth. Plugging in the expression of Vd (Vdd-Id*Rd) and Vg (which is Vo) gives you the inequality on the whiteboard.
Since Vd (instead of Vds) is directly measured from drain to ground, it automatically includes the voltage drop across resistor Rs.
prof razavi is just exceptional. I wish I could have gotten into ucla just to study under him :(
Very nicely explained the concept!❤️
Kitna padhega brooo😄
Pdhna pdega bhai!😂😂
Wrna lag jaenge
Sir I like your youtube name 恐龙, thanks a lot for this videos, I'm struggling to prepare for my electronics exams right now.
in 12:28 sir you took the vds for saturation as vdd-IdRd but it should be vdd-IdRd-IdRs right?
Its correct.. because it is wrt ground not source.. Hope this make sense
Vds=Vo-IdRs and Vgs=Vdd-IdRd-IdRs, so IdRs at both side cancels each other.
@@azusay167 Ya but I think you mixed Vgs and Vds
actually we need to consider only vds>vgs-Vth
but instead of vgs we are considering Vo (voltage between the gate and ground)
so, similarly in place of Vds i.e. Vdd- (Id*Rd)- (Id*Rs) we are considering Vdd-(Id*Rd) i.e (Voltage between the drain and ground)
so, Vdd-(Id*Rd) > Vo-Vth
is the condition instead of
Vdd- (Id*Rd)- (Id*Rs) > Vgs-Vth
(PS: even I too got the same doubt initially by after seeing yours and someone elses doubt I thought and came to this conclusion)
wow....superb sir .............thanks a lot
Gain is lower but it's independent of process variations
Sir, V0= Vg + (drop on Rs), why did you consider Vg=V0 @13:33
yes i was wondering the same
I suppose that what you said is Vgs, but when explaining pinch-off at p277 it says that the edge only required Vg-Vth. And V0=Vg
V0 = VGS + drop across RS, since VS = drop across RS, V0 = VG
Gem of a lecture!
Correction, a Gem of a lecture "series" 😁
Love you sir 😍😍😍
dear sir, at 12:34 when we are checking whether device really in saturation or not, why gate voltage is taken only Vo , it should be Vo-IdRd becz Vgs is equal to that
Here Vo is over all Vgs . over all Vgs= vgs -I'd rs
You are almost right, it should be Vo - Id Rs as a total of Vgs
That is what i am talking about..!
Thanks a lot!
Thank You Very Much.
The last circuit you are teaching shouldn't its resistance be 1/gm rather than r02?
actually should be 1/gm2 || r02, i think?
That will be for Diode connected device but that circuit have constant current circuit model. You can see a battery connected between gate and source. What you are saying is for when gate and drain is shorted.
what u r talking about is for a Diode Connected load But this is a current source device for which current is coming out of an arbitrary node and going to ground which can be made by an NMOS device in which the gate and source terminals are biased
Hi, Long, where can we find the problem solving strategies for electronics I and II ?
You have sedra smith and Razavi's book
Ty
sir tu se great ho (indian teacher should teach like him)
Itna knowledgeable bhi hona chahiye
Correct 👍
Correct 👍
At 57:32 Rout = (1+ gmRs)ro + Rs. When ro= infinite, we are still left with Rout = Rs.............isn't it so Sir? Then why are you telling Rout will be infinite in your next video?
when ro is infinite, Rout = infinite + Rs which is obviously an infinite value
Razavi YYDS!
1:02:10 Please upload circuit theory I&II
why gm vanishes at 18:00 ??
He fixed it at 18:55
checked!
nice
21:00
shutup u are crying not me🤧🤧🤧
never been this happy