collinge@uvicctr.UUCP (Doug Collinge) (01/26/89)
OK, net, here's a question for you: I know what a real MOSFET looks like and I have a vague notion of how it works but how about this: 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? Next step: replace the semiconductor with a metal (Making a MOMFET!) - increasing the gate voltage to a sufficiently high level should eventually drive away all the free electrons rendering the channel an insulator. What field strength would this require? Lastly: for my own arcane reasons I am interested in actually making a transistor. I don't want to have to make ultrapure silicon - is there some other semiconductor (e.g., copper oxide) that would be easier to use? I don't care if it is a good transistor; gains of around 10 would be just fine. While we are at it: I read somewhere that when junction transistors were perfected research on point-contact transistors ceased and that they are still poorly understood. Is this true? If not, how do they work? -- Doug Collinge School of Music, University of Victoria, PO Box 1700, Victoria, B.C., Canada, V8W 2Y2 collinge@uvunix.BITNET decvax!uw-beaver!uvicctr!collinge ubc-vision!uvicctr!collinge __... ...__ _.. . ..._ . __... __. _. .._ ..._._
myers@hpfcdj.HP.COM (Bob Myers) (01/31/89)
>OK, net, here's a question for you: >I know what a real MOSFET looks like and I have a vague notion of how it >works but how about this: 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. >Next step: replace the semiconductor with a metal (Making a MOMFET!) - >increasing the gate voltage to a sufficiently high level should eventually >drive away all the free electrons rendering the channel an insulator. >What field strength would this require? Lots and lots and lots. I haven't done the calculations, but remember what makes a conductor a conductor, and what makes a semiconductor a semiconductor. Now, your next problem is to find an insulator that will stand up to the field required (see "Dielectric Breakdown") :-). Bob Myers KC0EW HP Graphics Tech. Div.| Opinions expressed here are not Ft. Collins, Colorado | those of my employer or any other {the known universe}!hplabs!hpfcla!myers | sentient life-form on this planet.
nusip@maccs.McMaster.CA (Mike Borza) (01/31/89)
In article <609@uvicctr.UUCP> collinge@uvicctr.UUCP (Doug Collinge) 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? Hmmm, interesting question. Presumably, you're proposing to etch the Si back from the opposite side that you've just metallized. In principle, you could do this, but in practice, you'd have a very hard time controlling the characteristics of the device. The problem is that leakage along the exposed face would be very high-- you'd have a quite nonlinear, very noisy resistor. It's highly unlikely you'd be able to "pinch" the channel off, hence you'd have (virtually) no transistor action. You don't need the junction with the substrate (sometimes called the "back gate") to have a MOSFET... this is similiar to the structure of semiconductor-on-insulator (SOI) transistors. > >Next step: replace the semiconductor with a metal (Making a MOMFET!) - >increasing the gate voltage to a sufficiently high level should eventually >drive away all the free electrons rendering the channel an insulator. >What field strength would this require? Now you're in trouble! You can fabricate useful devices in semiconductors because the mobile charge in specific volumes of space can be controlled quite accurately. Trying to fabricate a device in which you control the charge flowing laterally in a metal under the control of a "vertical" electric field presents many practical problems. First, the electric field rises very quickly as a function of distance in a metal as charge is forced away from its surface. This presents one of (at least) two practical problems. Either you need a layer of metal which is no more than a few interatomic distances thick (say about a hundred Angstroms), or an insulator which can withstand huge electric fields without breaking down electrically. In fact, the space-charge in a metal drops to its equilibrium value exponentially as a function of distance into the metal, so the second possibility is out of the question. Unfortunately, the atoms in extremely thin layers of metal have a tendency to move under the influence of applie electric field, eventually causing the failure of your device There are other practical macroscopic problems which would be observed, and undoubtedly some interesting quantum-mechanical ones in a layer this thin, as well. > >Lastly: for my own arcane reasons I am interested in actually making a >transistor. I don't want to have to make ultrapure silicon - is there >some other semiconductor (e.g., copper oxide) that would be easier to >use? I don't care if it is a good transistor; gains of around 10 would >be just fine. This goes back a ways, but if I remember this right, Lillenfield (or some similar name) applied for a patent around the turn of the century for a field controlled resistor using aluminum oxide. I'll try to dig up a reference to this, but undoubtedly someone else has this at the tip of their tongue (or fingers). > >While we are at it: I read somewhere that when junction transistors were >perfected research on point-contact transistors ceased and that they are >still poorly understood. Is this true? If not, how do they work? sorry, no time for that one now... >-- > Doug Collinge > collinge@uvunix.BITNET mike borza <nusip@maccs.uucp or antel!mike@maccs.uucp>
dbraun@cadavr.intel.com (Doug Braun ~) (02/01/89)
One of the Scientific American "Amateur Scientist" columns, from the early '70s, I think, described how to make a crude FET. It noted that people almost discovered FETS long before the Bell Labs transistor. Doug Braun Intel Corp CAD 408 765-4279 / decwrl \ | hplabs | -| oliveb |- !intelca!mipos3!cadev4!dbraun | amd | \ qantel /
kluksdah@enuxha.eas.asu.edu (Norman C. Kluksdahl) (02/03/89)
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