3.7 Inverters and logic

As you already know, or will find out shortly, from taking a class in digital logic, logic circuits are primarily based upon acircuit called an inverter. An inverter simply takes a signal and gives you the opposite one. For instance, if a high voltage (a"one") is placed on the input of an inverter, it returns a low voltage (a "zero"). is a simple inverter based on a MOSFET transistor:

If $V_{\mathrm{in}}$ is zero, the MOSFET is turned off ( $V_{\mathrm{gs}}$ is $<(V_T)$ ) and so no current flows through the resistor, and $V_{\mathrm{out}}=V_{\mathrm{dd}}$ , a high. If $V_{\mathrm{in}}$ is high (and we assume that $V_T$ for the MOSFET is significantly less than $V_{\mathrm{in}}$ ) then the transistor is turned on, and if $R$ and $\frac{W}{L}$ are chosen so that enough current flows through $R$ to drop most of $V_{\mathrm{dd}}$ across it, then $V_{\mathrm{out}}$ will be low.

The way this is usually described is through a transfer function which tells us what the output voltage is as a function of the input voltage. Let's digress for just a minuteand see how such a function can be arrived at. Looking back at it should be easy to see that

We can re-write this as an equation for $I_d$ .

This is called a load-line equation. It says that $I_d$ varies linearly with $V_{\mathrm{ds}}$ (with a negative slope) and has a vertical off-set of $\frac{V_{\mathrm{dd}}}{R_d}$ . Let's suppose we have the MOSFET transistor for which we have already plotted the characteristic curves in a previous plot . We will let $V_{\mathrm{dd}}=5$ Volts, and let $R_d=1\mathrm{k\Omega }$ . From we can see that when $V_{\mathrm{ds}}=0$ , $I_d$ will be 5 mA, and when $V_{\mathrm{ds}}=V_{\mathrm{dd}}$ , $I_d$ will be 0. This then gives us a straight line on the characteristic curve plot which is calledthe load line . This is shown in . By looking back at the schematic for the inverter in we see that the same current $I_d$ flows through the load resistor, $R_d$ , and through the transistor. Thus, the correct value of current and voltage for the circuit for anygiven gate voltage is the simultaneous solution of the load line equation and the transistor behavior, which, of course, is justthe intersection of the load line with the appropriate characteristic curve. Thus it is a simple matter of drawingvertical lines down from each $V_{\mathrm{in}}$ curve or $V_{\mathrm{gs}}$ value down to the horizontal axis to find out what the appropriate $V_{\mathrm{dd}}$ or output voltage will be for the inverter. Assuming that $V_{\mathrm{in}}$ only goes up to 5 Volts, the resulting curve that we get look like . This is not a great transfer characteristic. $V_{\mathrm{in}}$ has to get fairly large before $V_{\mathrm{out}}$ starts to fall, and even with the full 5 Volt input, $V_{\mathrm{out}}$ is still greater than 1 Volt. Picking a transistor with a small $V_T$ and a bigger load resistor would give us a better response, but at least with this example you can seewhat is going on.

Based on this simple inverter circuit, we can build circuits which perform the NOR and NAND function.

Let's look at the inverter a little more closely . Usually, the load for the inverter will be the next stage of logic which, along with theassociated interconnect wiring, we can model as a simple capacitor. The value of the capacitance will vary, but it willbe on the order of $10^{-12}$ F.

When the input to the inverter switches instantaneously to a low value, current will stop flowing through the transistor, andinstead will start to charge up the load capacitance. The output voltage will follow the usual $\mathrm{RC}$ charging curve with a time constant given just by the product of $R$ times $C$ . If $C$ is $10^{-13}$ F, then to get a rise time of 1 ns we would have to make $R$ about $10^4\Omega$ .

As we shall see later, it is virtually impossible to make a 10 kΩresistor using integrated circuit techniques. Remember:

And thus, to get a really big resistance we need either a very tiny A (Too hard to achieve and control.), a really BIG L (Takesup too much room on the chip) or a huge $\rho$ (Again, very hard to control when you get to the very low doping densities that would berequired.)

Even if we could find a way to build such big integrated circuit resistors, there would still be a problem. The current flowingthrough the resistor when the MOSFET is on would be approximately