3.4 Mos regimes  (Page 2/2)

At the drain end of the channel, when $V_{\mathrm{ds}}$ just equals $V_{\mathrm{dsat}}$ , the difference between the gate voltage and the channel voltage, $V_{\mathrm{gc}}(L)$ is just equal to $V_T$ , the threshold voltage. Any further increase in $V_{\mathrm{ds}}$ and the difference between the gate and the channel ( in the channel region just near the drain ) will drop below the threshold voltage. This means that when $V_{\mathrm{ds}}$ gets bigger than $V_{\mathrm{dsat}}$ , the channel just near the drain region disappears! We no longer have sufficient voltage betweenthe gate and the channel region to maintain an inversion layer, so we simply revert to a depletion condition. This is called pinch off , as seen in .

What happens to the drain current when we hit pinch off? It looks like it might go to zero, but that is not the rightanswer! Although there is no active channel in the pinch-off region, there is still silicon - it just happens to be depletedof all free carriers. There is an electric field, going from the drain to the channel, and any electrons which move along thechannel to the pinch-off region are sucked across by the field, and enter the drain. This is just like the current that flows inthe reverse saturation condition of a diode. There are no free carriers in the depletion region of the diode, yet $I_{\mathrm{sat}}$ does flow across the junction region.

Under pinch-off conditions, further increases in $V_{\mathrm{ds}}$ , does not result in more drain current. You can think of the pinched-off channel as a resistor,with a voltage of $V_{\mathrm{dsat}}$ across it. When $V_{\mathrm{ds}}$ gets bigger than $V_{\mathrm{dsat}}$ , the excess voltage appears across the pinch-off region, and the voltage across the channel remainsfixed at $V_{\mathrm{dsat}}$ . If the channel keeps the same charge, and has the same voltage across it, then the currentthrough the channel (and into the drain) will remain fixed, at a value we will call $I_{\mathrm{dsat}}$ .

There is one other figure which sometimes helps in seeing what is going on. We will plot potential energy for an electron, as ittraverses across the channel. Since the source is at zero potential and the drain is at $V_{\mathrm{ds}}$ , an electron will loose potential energy as it flows from the source to the drain. shows some examples for various values of $V_{\mathrm{ds}}$ :

For the first two drain voltages, $V_{\mathrm{ds1}}$ and $V_{\mathrm{ds2}}$ , we are below pinch-off, and so the voltage drop across $R_{\mathrm{channel}}$ increases as $V_{\mathrm{ds}}$ increases, and hence, so does $I_d$ . At $V_{\mathrm{dsat}}$ , we have just reached pinch-off, and we are starting to see the "high field" depletion region beginto develop. Since electric field is just the derivative of the potential, the slope of curves in gives you an idea of how big the electric field will be. For furtherincreases in $V_{\mathrm{ds}}$ , $V_{\mathrm{ds4}}$ and $V_{\mathrm{ds5}}$ all of the additional voltage just shows up as a high field drop at the end of the channel. Thevoltage drop across the conducting part of the channel stays fixed (more or less) at $V_{\mathrm{dsat}}$ and so the drain current stays more or less fixed at $I_{\mathrm{dsat}}$ .

Substituting the expression for $V_{\mathrm{dsat}}$ into the expression for $I_d$ we can get an expression for $I_{\mathrm{dsat}}$

What this means for is that when $V_{\mathrm{ds}}$ gets to $V_{\mathrm{dsat}}$ , we simply hold $I_d$ fixed from then on, with a value of $I_{\mathrm{dsat}}$ . For different values of $V_g$ , the gate voltage, we are going to have a different $I_d-V_{\mathrm{ds}}$ curve, and so once again, we end up with a family of "characteristic curves" for the MOSFET. These are shown in .

This also gives us a fairly easy way in which to "sketch" a set of characteristic curves for a given device. Suppose we have aMOS field effect transistor which has a threshold voltage of 2 volts, a width of 10 microns, and a channel length of 1 micron,an oxide thickness of 150 angstroms, and a surface mobility of $400\frac{c}{V\sec }$ . using ${\epsilon }_{\mathrm{ox}}=3.310^{-13}\frac{F}{\mathrm{cm}}$ , we get a value of $2.210^{-7}\frac{F}{c}$ for $c_{\mathrm{ox}}$ . This then makes k have a value of