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.
Suppose, you increase Vgs so much that Vgs-Vth>Vds as you have fixed Vds, this condition does not apply. Is it for this reason that he said for a very high Vgs, the quadratic equation is not valid? Could you clear my doubt.
@@roshansharma5251 Mr. Razavi says turns on in saturation region. Not always in saturation region if you increase Vgs so much. Turns on means the point that Vgs begins to be higher than Vth.
@@ozgunyurutken9485 Wrong. Dr.Razavi says it turns on. If you look at the graph , it turns on when Vgs > Vth, and this is the point where saturation occurs/ given that Vds > 0.
@@roshansharma5251 That case is much beyond turning on......When you "just" turn on the MOS, V(GS) = V(TH). So, condition of saturation only applies till V(GS) becomes greater than V(DS) by atleast one threshold (V(TH)).
But I do have a doubt like after vth, Vgs-Vds keeps increasing for we keep Vds constant and increase Vgs so this difference will be more than Vth right in that case the MOS is in Triode region right
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."
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
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)
When we look at the ID (Drain current) vs VGS (Gate-Source voltage) characteristic: 1. For VGS < Vth (threshold voltage): - The device is in the cutoff/subthreshold region - Very little to no current flows 2. For VGS > Vth: - A channel forms - If VDS > 0: - There will be a potential difference between the drain and the source - This causes current to flow through the channel - The device enters saturation when VDS > (VGS - Vth) - In saturation, ID = k(VGS - Vth)²/2 Therefore, to prove the device "always turns on in saturation if VDS > 0": - Once VGS exceeds Vth, a channel forms - Any positive VDS will cause current to flow - As VDS increases, the device naturally enters saturation - The key is that VDS > 0 ensures there's a driving force for current flow, and the quadratic relationship between ID and (VGS - Vth) in saturation ensures the device is "on."
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..
I found this lecture very confusing because Dr. Razavi skipped over so many things. First, at pinch off, the channel length, L, gets smaller, but he just ignored that and kept using L for the whole device. It is fine if that is justified, but he needs to make some justification for why that was OK. Second, he completely ignored the mechanism of current flow in the pinch off region. With zero density of charge carrier, how does the current get from the drain to the channel? Finally, when he talks about the Id-Vgs curve and says it always turns on in saturation, that is obviously true, but it is confusing without the top half of that curve. As soon as the overdrive voltage exceeds Vds, the circuit will fall out of saturation. And at 5mv, that is going to happen really fast. I'm guessing we'll get to this in the next lecture, but it deserves a mention here so we are not left scratching our heads.
I agree, it would be nice if he would have mentioned this. But I think these lectures are made for first year bachelor students, not us who want to refresh and improve our knowledge. Its much new information for them anyway.
depletion region is formed beyond pinch off and the field present in that depletion region swift away the electrons as soon as they reach the pinchoff,so current flow is not effected. hope u get it.. read sedra smith to make ur concepts much clear
Well he did explain that while the L is smaller... it is almost L... so it is an approximation... the mechanism in which electrons flow there... it is a drift current caused because of the electric field. If you notice... you will get a PN junction in reverse. While PN junction in reverse does not conduct by diffusion, it lets a drift current flow
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
No, the question is: 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
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?
I think it is because Diffusing current is something that we talk about with regards to majority charge carriers but here we are dealing only with minority carriers I am not sure though...
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.
Hi, I think you are correct. Prove that the MOS device always turns on in saturation if Vds>0 means that initially when the MOSFET is turned on, Vgs starts from 0 and increases. Whereas Vth and Vds remains constant, so Vds+Vth>Vgs. After Vgs is fully increased, MOSFET will enter Ohmic region.
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.
It will form basically a reverse diode.. a PN junction... the carriers will flow because of an electric field. ( If you saw bipolar transistors, it is similar to a bipolar transistor, since when the channel gets "cut" ... you will get a sort of p-n-p or n-p-n structure
The current can still flow through the "substrate" even though the channel is pinched. The reason why it saturates is that there will be a region of higher resistance of size proportional to the Drain-Source voltage, and therefore the resistance of this region will be proportional to the same voltage. But as current is voltage/resistance, the dependence will cancel out and you'll get "constant" current. From Wiki (): Even though the conductive channel formed by gate-to-source voltage no longer connects source to drain during saturation mode, carriers are not blocked from flowing. Considering again an n-channel enhancement-mode device, a depletion region exists in the p-type body, surrounding the conductive channel and drain and source regions. The electrons which comprise the channel are free to move out of the channel through the depletion region if attracted to the drain by drain-to-source voltage. The depletion region is free of carriers and has a resistance similar to silicon. Any increase of the drain-to-source voltage will increase the distance from drain to the pinch-off point, increasing the resistance of the depletion region in proportion to the drain-to-source voltage applied. This proportional change causes the drain-to-source current to remain relatively fixed, independent of changes to the drain-to-source voltage, quite unlike its ohmic behavior in the linear mode of operation. Thus, in saturation mode, the FET behaves as a constant-current source rather than as a resistor, and can effectively be used as a voltage amplifier. In this case, the gate-to-source voltage determines the level of constant current through the channel. Also, from the MOSFET operation description, under saturation: Since the drain voltage is higher than the source voltage, the electrons spread out, and conduction is not through a narrow channel but through a broader, two- or three-dimensional current distribution extending away from the interface and deeper in the substrate. The onset of this region is also known as pinch-off to indicate the lack of channel region near the drain. Although the channel does not extend the full length of the device, the electric field between the drain and the channel is very high, and conduction continues. Thanks to Clabacchio who answered this.
Maybe it is related to the fact that I have in any case Vds>Vod when I turn on the device. I mean if I have Vds=5mV, as soon as I turn on the device (when Vgs=Vth) the overdrive is negligible. So I have for sure Vds>Vod(Condiction for the saturation).
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?
Yeah! There is charge density only when VGS exceeds VTH and not when VGS=VTH . At any point we take the voltage to be the difference of VGS and VTH. There is a current flow soon after VGS exceeds VTH and VD>0. If VD=0 , you have no current flow since Drain to source has the same potential!
Electrons at the pinch off point will experience an electric field which will drift the electron towards the drain region. The electric field is from the drain region(holes in n+) to the depletion region(negative charged ions). Also since the depletion region does not have holes so recombination is not possible. This means every electron from the pinch off will be drifted to the drain region.
@@GiorgiAptsiauriX it's like the current that is swept in the reverse biased PN junction (base-collector) in the case of BJT, is this analogy sensible?
"once the electrons reach the end of the channel, they experience the high electric field in the depletion region surrounding the drain junction and are rapidly swept to the drain terminal." From textbook
Just when the MOS device turns on , we have that Vgs= Vth => Vgs - Vth =0. Its given that Vds >0 => Vds > Vgs - Vth which means that when the MOS turns on, we are in the saturation region.
I think it is the voltage across the gate and source rather than the voltage across the gate and substrate. Some times v_substrate is not equal to v_source.
Vds+Vth >Vgs is the condition for saturation region.Lets say Vds+Vth is 4V. If u now increase Vgs (from Vth)to plot d graph thn it would be in saturation region till Vgs=4V
@@andrejosue98 That's what he mentioned "If we keep on increasing the Vgs then the characteristics will be different and we will ignore that at present".Check the video
@@revanthvarma2000 yeah exactly at the point above VTH its well and good but increasing VGS so Much that VGS-VTH becomes greater than VDS in that case it will enter triode region but still its not true becoz something else might happen!!
