Razavi Electronics 1, Lec 31, MOS Characteristics II

MOS Characteristics II
(for next series, search for Razavi Electronics 2 or longkong)

[aaronnowack]

I love the intro to these lectures, I get HYPED to learn about MOSFETs!!

[devanshrajoria]

Sir you are an excellent teacher. Thank you for all these lectures

[kalebhernandez]

I would like to thank you for uploading this admirable lectures. It helps to understand better the Razavi's book.

[karamany]

Exercise: we know that Vgs-Vth determines whether we are in saturation or triode region, so when we start increasing Vgs, the value Vgs-Vth will be very small such that Vds>Vgs-Vth and therefore the device is in saturation. As Vgs increases, the device will go into the triode region

[prakhargupta]

For those who are confused about the question posed by Razavi sir, if I re-phrase it...
"Set the VDS as a constant > 0 (however small you want). Now, when you turn on (or "start") the MOS device, you will observe that it "starts" in saturation mode. Prove this result."

[sds]

Question at 49:00 "Prove that the mos device always turns on in saturation if VDS > 0?"


Answer:


1. Consider at the beginning VGS=0, VDS=0.
2. Now if I increase Vds to 0.1 V and I am fixing it.
3. I am increasing Vgs slowly from 0 to Vth. At Vth my mos device is on (Vgs =Vth).
4. For saturation the condition is Vds > Vgs- Vth.
5. as Vgs = Vth implies Vgs-Vth=0.
6. hence Vds > Vgs-Vth; Vds=0.1>0.
So the mos device will turns on n the saturation.

[stefano.a]

Perhaps is better to clear the question at 49:40 : the thesis is that when Vds is FIXED, and we increase Vgs, the MOSFET, *TURNS ON* in saturation, that is: when Vgs reach Vth, then the MOSFET is in saturation (then after Vgs exceeds Vds + Vth the MOSFET pass into triod region)

[himanshuful]

Hello Sir.
You have explained very well about Saturation of Drain current mathematically. By changing the limit of V(x) i.e. channel.voltage from Vds to Vds, sat= Vgs-Vth which results in Id independence on Vds.
But I also want to know how it is constant at semiconductor level..

[mennatallah]

الاسطوره واحد بس رزافي التوب والباقي كنتلوب

[tanmoydutta]

Question : If charge density Q(Ch, den) decreases with distance x from the Source end to the Drain end, then that creates a gradient of charge. So, in this expression of I(D), we have just included mobility (meu) which characterizes this current as drift current. However, from the lectures of pn junction and its biasing, we learnt that when this kind of charge density gradient is present, it also gives rise to diffusion current. So, does the expression for I(D) here include this diffusion current? If yes, then where? If not, then how?

[rohithbhaskar]

Wowww. Heard the best lecture on MOSFET char...

[theminertom]

Shouldn't that exercise read "prove that the MOS device always turns on in sat if Vds > (Vgs-Vt)" instead of Vds > zero? If Vds = 0 then there may be a channel (if Vgs > Vt) but there will be no current flowing.

[bluebox]

I think maybe we have some confusion here. It should be " the device will initially be in the saturation region"

[yeohoonyoon]

I don't understand: why does the overdrive voltage, VGS - VTH, appear in the charge density equation? Doesn't this mean that charge is zero when VGS = VTH even though the transistor is turned on and current flows due to VDS > 0?

[nevilpooniwala]

Prove that mos always turns on in saturation region if Vds >0. Please tell me if my answer is correct-

To turn on MOSFET, Vgs >= Vth so we have Vgs - Vth = 0 just when the device turns on. But given Vds>0, thus, Vds > Vgs - Vth
Thus the device is in sat region.

[mutasemwahbeh]

Thank you sir, That is indeed an informative lecture ...

[veys]

Thank you very much, amazing as always.

[filankestayn]

God bless you sir, thank you very very much.

[sumanth]

After pinchoff voltage there won't be any electrons...so how come that current Ids(saturation one) flows

[dilshadadil]

All right for those you are scanning the comment box for the answer to the question Razavi sir asked towards the end of this video let me explain

The question is if Vgs>Vth (just when the device is turned on) then for any positive value of Vds, the current through the MOS will be in saturation region. Note that this is a case for when Vgs has just reached Vth and not when Vgs is much greater than Vth ( to be exact Vgs= Vth+ whatever Vds you have taken)

1. So if we look carefully we will observe that saturation region is achieved only when there is a pinch off anywhere between the source and the drain.
2. We must notice that the value of Vds decrease as we move towards the source from drain (since source is grounded) and becomes equal to zero at the source
3. Now it is given that Vgs>Vth, therefore if we analyze the voltage drop at or at the very proximity of the source we notice that Vgs-Vds>Vth (since Vds at source is zero and it is mentioned that Vgs>Vth ). therefore a channel starts to build and since Vgs<Vth+Vds (you have taken some arbitrary value of Vds) the channel will quench before the drain and hence a saturation current is drawn .

hope I was able to explain you the solution properly
thanks