I think at the heart of the MOSFET and related MISFET-type FET's, including those based on GaAS, lies the metal-insulator-semiconductor structure, which, in some ways, is essentially a capacitor. What I hadn't realized until today was the critical role that the chemical potential (or Fermi surface) of the metal plays in the whole thing.
These problems should help us understand that role. Once we understand it, maybe it won't seem so important or mysterious, but i think that until we do there may be a cloud of confusion hanging over whatever we do regarding metal-insulator-semiconductor structures.
So I would recommend doing this problem (really 3 problems) before any of the others assigned this week.
1. Consider a metal-insulator-semiconductor sandwich. Suppose the semiconductor is p-doped, and, for definiteness, suppose the band gap is 1 eV and $\mu - E_v$ = 0.2 eV far to the right (far from the metal) (moderate doping).
Sketch the band edges, the space-charge density, electric field, etc., in the semiconductor as a function of z (distance from the insulator):
a) for the case where the chemical potential of the metal is matched to that of the semiconductor.
b) for the case where the chemical potential of the metal is (or would like to be*) somewhat higher than that of the semiconductor.
c) for the case where the chemical potential of the metal is (would like to be*) somewhat lower than that of the semiconductor.
* (The problem statement is a little confusing, since the chemical potential should be independent of z, so maybe someone can clarify that here via comments. Maybe someone would like to outline their understanding and approach.)
PS. Even though there is no charge in the insulator, there is probably an electric field, in some cases, so it might be important to include that as well in lining things up.
PPS. This draws on exactly the same skills and principles we developed in studying p-n junctions.
Subscribe to:
Post Comments (Atom)
hey everyone, i have a suggestion: how about if we put all comments related to homework 6 on this post (the most recent one)? i just commented on the first homework 6 post, but then realized that it is waaay down there, past that beast of a diamond structure post- it might be easier if we just keep all hw6 discussion in one place so everyone's questions get answered.
ReplyDeletethat said, here's my first question on this homework:
ReplyDeletethis might be a dumb question, but where exactly is the gate voltage applied? i am picturing a battery with one terminal connected to the metal gate, and the other terminal connected to the semiconductor, but i just want to make sure that's right...
When the chemical potential of the metal is/would like to be different from that of the semiconductor, does that imply a voltage across them? That's at least a parameter we would expect to vary with z.
ReplyDeleteI think one can think of the voltage as applied between the gate and the "channel", that is, the semiconductor. So the voltage is applied across the insulator (the oxide). The voltage drop occurs across the insulator/oxide, i think, and that is also where any chemical potential/Fermi energy dependence on z would occur. I think that within the metal gate, there is no voltage drop or z dependence.
ReplyDeleteNote, however, that band-bending can occur without any applied voltage. Like in b), band-bending is needed to keep $\mu$ constant (independent of z) and it is associated with depletion and "inversion" which lead to "space charge", somewhat like in a p-n junction.
ReplyDelete(The same principles apply, so this is a good test of how well we understand those principles.)
Okay. Thanks for the clarification.
ReplyDeleteso the z dependence comes from the applied voltage which shifts the chemical potential in the semiconductor with respect to the metal. This will determine whether the fet is on or off. But then the bending of the bands "cancels" this as zack posted above?? i am a little confused with this relationship.
ReplyDeleteno, i don't think the band-bending cancels that. Not at all.
ReplyDeleteWhat's $E_i$ in the images posted below? I figure it has something to do with the intrinsic carriers, but I'm not sure exactly what.
ReplyDelete$E_i$ is a "marker" that indicates where $\mu$ would be if the material were undoped ("intrinsic"). (When $\mu$ is near $E_i$ the material is depleted, when it is well-above, n doped, when well below, p-doped.) (Technically it is the cross-over from p to n, though practically speaking it is more useful to view the middle region as depleted.
ReplyDeleteAnyone, feel free to elucidate this.
This comment has been removed by the author.
ReplyDeleteI think I'm missing some basic stuff, but one more thing: one of the problems says to assume that there are no carriers within the depletion region. This region extends from the O-S junction to some length W, like with p-n junctions. But doesn't making that assumption ignore the channel that gives us all the useful properties in the first place?
ReplyDeleteQuoting Zack: "...When $\mu$ is near $E_i$ the material is depleted..."
ReplyDeleteIs "intrinsic" or "intrinsic-like" a better word than "depleted" there? Depleted makes me think that there are very few free carriers, like in our depletion region from before.
But maybe that actually is the case. If it is, then would that mean that the point where $E_i$ crosses $\mu$ is the point where the 2-d electron gas (2DEG) starts, because that's where it starts to look n-doped? Or does that not form until $E_c$ is below $\mu$?
I thought the latter, but the diagram doesn't show this happening. Then again, it never claimed that there was a 2DEG present in that example.
intrinsic and depleted are not at all the same. Intrinsic means undoped.
ReplyDeleteOnly something that is first doped can be depleted. Depleted is not the same as undoped.
I made a mistake in my earlier post. The semiconductor is uniformly p-doped. The band bending can change the carrier concentrations in different regions but not the doping itself. so I should not have said:
ReplyDelete"when it is well-above, n doped, when well below, p-doped."
(What is a better way to express what i meant.)
I realize that they're not the same, that's why I asked what you meant. I thought you might have meant intrinsic-like because in an undoped (intrinsic) semiconductor, $E_i = /mu$ by definition, correct?
ReplyDeleteAnyway, so there are practically zero carriers where $E_i = /mu$, then, and not the intrinsic amount - 10^10 or whatever - hence the use of "depleted" instead of "intrinsic-like?" Or does depleted just mean that a doped semiconductor is acting like an intrinsic one at that point?
I'm probably still missing a lot, so I apologize if these questions are very basic or just don't make sense.
This comment has been removed by the author.
ReplyDeleteOops, didn't see your latest post. Did you mean to rephrase that by replacing "n-doped" with "electrons as the majority carrier" and "p-doped" with "holes as the majority carrier?"
ReplyDeleteyes. exactly.
ReplyDeleteAlso, in response to your earlier post, when you have doped something to 10^17, 10^10 is pretty much zero...
And, depleted might refer to anything below about 10^16 or so. (I am not really sure, it probably depends on what you are thinking about, but if it is rho(x), the "space charge" then depleted could apply to any region where the (majority) carrier concentration has dropped to like 30% or so of its initial value.
Sorry to change the subject, but the books discussion (pg 263 and 265) about pinning is confusing to me.
ReplyDeleteI thought I understood that the space charge increases very strongly with surface potential.
But then the band-bending is said to be pinned, or small, because of this same effect.
Why don't large increases in the space charge layer bend the bands appreciably?
Is this what is getting in my way of understanding why there isn't a depletion region involved in the accumulation scenario?
i am still having trouble on problem 2: has anyone been able to express n as a function of Vgate? and it seems to me that part c is asking the same thing as part b... what is the difference between these?
ReplyDeletehttp://www.ee.ui.ac.id/~astha/courses/ts/teksem/contents.htm
ReplyDeleteThis site is awesome. I didn't know where else to post it so it's going here now.
How does incorporating silicon into the AlGaAs layer produce negative charge at the interface?
ReplyDeleteI initially thought that it would induce a positive layer since it was like doping with a donor... but then I started thinking it wouldn't do anything.
I'm basing this off of
Al-> III
Ga-> III
As-> V
Si-> IV
So when the undoped crystal forms, galium and aluminium accept electrons from arsenic making neutral compounds.
When silicon is incorporated, it works its way into the crystal without accepting an electron from arsenic, and without giving one to galium or aluminium.
Referring to the same question (regarding finding the flatband voltage for the Metal/Si@AlGaAs/GaAs):
ReplyDeleteDo we need the width of the insulator to determine the capacitance between the metal and semiconductor? Is it correct to say that we want to counteract the energy involved in this capacitance?
So, on problem 8.7, I wasn't sure if what I was doing was correct. I used $W_m=\sqrt{\frac{2\epsilon_s \phi_s}{eN_a}}$, where $\phi_s = 2\phi_F$ and $\phi_F=\frac{kT}{e}\log{\frac{N_a}{n_i}}$. Doing this I got 420nm. Does this compare with what any of you got?
ReplyDeleteIs anyone else confused by the books problems? They ask for $V_T$, but give two different forms for it: $\Phi_{ms}-\frac{Q_i}{C_i}-\frac{Q_d}{C_i}+2d\phi_F$ and $-\frac{Q_d}{C_i}+2d\phi_F$. The latter the text says is the ideal. For 8.7 and 8.8, are we meant to use this formula, or the other one? Any comments? (Zack, some clarification here would be great.)
ReplyDeleteMike,
ReplyDeletefirst, I got 420nm too.
second, I used the first equation you wrote (the longer one)
mike: i got 407 nm for 8.7a, which is really close... we are probably off because i used 1.5e10 cm^-3 for n_i - i found that in the book, so i used it instead of 10^10. but yea, same method, same order of magnitude, so i think we are good there
ReplyDeleteso how are you guys calculating Φms? i just took a look at Streetman's Figure 8-14, and kind of guessed a value, but there must be an actual equation for it somewhere right?
ReplyDeleteI'm not sure about a formula for $\Phi_{ms}$. I guess we should use the table.
ReplyDeleteThanks for the answer comparisons by the way. I guess I'm on the right track now