Path: utzoo!utgpu!jarvis.csri.toronto.edu!mailrus!ncar!noao!asuvax!enuxha!kluksdah
From: kluksdah@enuxha.eas.asu.edu (Norman C. Kluksdahl)
Newsgroups: sci.electronics
Subject: Re: How do FETs REALLY work?
Summary: Enhancement vs. Depletion MOSFET
Message-ID: <22@enuxha.eas.asu.edu>
Date: 2 Feb 89 23:32:06 GMT
References: <609@uvicctr.UUCP> <16750015@hpfcdj.HP.COM>
Organization: Arizona State Univ, Tempe
Lines: 40

In article <16750015@hpfcdj.HP.COM>, myers@hpfcdj.HP.COM (Bob Myers) writes:
> 
> >Say we get a piece of, say, N type silicon and
> >we oxidize it and stick on a gate.  Now we etch the Si so that it is very
> >thin under the gate.  Is this thing now a MOSFET even though there is no
> >junction nearby?  Does the MOSFET actually need the junction between the
> >substrate and the channel?
> 
> Yes, it is a FET; no, it does not need a "junction", since there isn't a
> junction there in the first place.  The SiO2 is simply an insulator -
> the generic name for this device is actually MISFET, for Metal-Insulator-
> Semiconductor FET.
> 
You have to be VERY careful in defining the operating mode of the proposed
FET.  If you want an enhancement MISFET (or MESFET for the GaAs lovers out
there), then you do not need anything except for an insulated gate.  The
applied gate potential creates a field, which causes inversion in a thin
(quantum sized) layer immediately under the oxide.  I.e., if you have a
p-type semiconductor, the conducting channel is forced into being n-type,
and makes contact between the n+ source and drain contacts.  Take away the
gate potential, and the channel reverts to p-type and cuts off current flow.

For a depletion MISFET, however, you must have some finite-sized region
through which the carriers normally flow.  This is typically accomplished
with using an n epitaxial layer on a p substrate or vice versa (or in a
diffused tub, but that's another story).  Until you apply a potential,
current flows.  It is the gate field that depletes the carriers and turns
off the transistor.  So if there is no epi-substrate junction( or finite
thickness to the semiconductor material) the current will merely be driven 
deeper into the semiconductor, until the gate insulator breaks down.  A
Metal-Insulator-Metal semiconductor would by necessity operate in 
depletion mode, and since the number of mobile electrons is more than 4
orders of magnitude (actually 5-6) more than the number in a semiconductor
channel, it's going to be damn tough to deplete.


Norman Kluksdahl              Arizona State University
            ..ncar!noao!asuvax!enuxha!kluksdah

standard disclaimer implied