MOS Transistor 2.15 2000/08/04 2007/08/14 11:41:20.842 GMT-5 Bill Wilson wlw@madriver.net Bill Wilson wlw@madriver.net Liqun Wang liqun@rice.edu Elizabeth Gregory elizabeth.gregory@gmail.com Jeffrey M Silverman JSilverman@astro.berkeley.edu Gerard Wysocki gerardw@rice.edu MOS Transistor Introduction of MOS Transistor, especially the structure and some attributes of MOS Transistor. Now we can go back now to our initial structure, shown in the introduction to MOSFETs, only this time we will add an oxide layer, a gate structure, and another battery so that we can invert the region under the gate and connect the two n-regions together. Well also identify some names for parts of the structure, so we will know what we are talking about. For reasons which will be clear later, we call the n-region connected to the negative side of the battery the source, and the other one the drain. We will ground the source, and also the p-type substrate. We add two batteries, Vgs between the gate and the source, and Vds between the drain and the source.

Biasing a MOSFET transistor

It will be helpful if we also make another sketch, which gives us a perspective view of the device. For this we strip off the gate and oxide, but we will imagine that we have applied a voltage greater than VT to the gate, so there is a n-type region, called the channel which connects the two. We will assume that the channel region is L long and W wide, as shown in .

The inversion channel and its resistance

Next we want to take a look at a little section of channel, and find its resistance ⅆ R , when the little section is ⅆ x long. ⅆ R dx σ s W We have introduced a slightly different form for our resistance formula here. Normally, we would have a simple σ in the denominator, and an area A, for the cross-sectional area of the channel. It turns out to be very hard to figure out what that cross sectional area of the channel is however. The electrons which form the inversion layer crowd into a very thin sheet of surface charge which really has little or no thickness, or penetration into the substrate. If, on the other hand we consider a surface conductivity (units: simply mhos), σs , where σ s μ s Q chan then we will have an expression which we can evaluate. Here, μs is a surface mobility, with units of cm 2 V sec . We ran into μ in earlier chapters, when we were building our simple conduction model. It was the quantity which represented the proportionality between the average carrier velocity and the electric field. v μ E μ q τ m The surface mobility is a quantity which has to be measured for a given system, and is usually just a number which is given to you. Something around 300 cm 2 V sec is about right for silicon. Qchan is called the surface charge density or channel charge density and it has units of Coulombs cm 2 . This is like a sheet of charge, which is different from the bulk charge density, which has units of Coulombs cm 2 . Note that: cm 2 Volt sec Coulombs cm 2 Coul sec Volt I V mhos It turns out that it is pretty simple to get an expression for Qchan , the surface charge density in the channel. For any given gate voltage Vgs , we know that the charge density on the gate is given simply as: Q g c ox V gs However, until the gate voltage Vgs gets larger than VT we are not creating any mobile electrons under the gate, we are just building up a depletion region. We'll define QT as the charge on the gate necessary to get to threshold. Q T c ox V T . Any charge added to the gate above QT is matched by charge Qchan in the channel. Thus, it is easy to say: Q channel Q g Q T or Q chan c ox V g V T Thus, putting and into , we get: ⅆ R ⅆ x μ s c ox V gs V T W If you look back at , you will see that we have defined a current Id flowing into the drain. That current flows through the channel, and hence through our little incremental resistance ⅆ R , creating a voltage drop ⅆ V c across it, where Vc is the channel voltage. ⅆ V c x I d ⅆ R I d ⅆ x μ s c ox V gs V T W Let's move the denominator to the left, and integrate. We want to do our integral completely along the channel. The voltage on the channel V c x goes from 0 on the left to Vds on the right. At the same time, x is going from 0 to L. Thus our limits of integration will be 0 and Vds for the voltage integral ⅆ V c x and from 0 to L for the x integral ⅆ x . Vc 0 Vds μs cox Vgs VT W x 0 L Id Both integrals are pretty trivial. Let's swap the equation order, since we usually want Id as a function of applied voltages. I d L μ s c ox W V gs V T V ds We now simply divide both sides by L, and we end up with an expression for the drain current Id , in terms of the drain-source voltage, Vds , the gate voltage Vgs and some physical attributes of the MOS transistor. I d μ s c ox W L V gs V T V ds