1.8 The diode equation

The reason for calling the proportionality constant $I_{\mathrm{sat}}$ will become obvious when we consider reverse bias. Let us now make $V_a$ negative instead of positive. The applied electric field now adds in the same direction to the built-in field. This means the barrier will increase instead of decrease, and so we have what is shown in . Note that we have marked the barrier height as $q(V_{\mathrm{bi}}-V_a)$ as before. It is just that now, $V_a$ is negative, and so the barrier is bigger.

Remember, the electrons fall off exponentially as we move up in energy, so it does not take much of a shift of the bands beforethere are essentially no electrons on the n-side with enough energy to get over the barrier. This isreflected in the diode equation where, if we let $V_a$ be a negative number, $e^{\frac{q{V}_a}{kT}}$ very quickly goes to zero and we are left with

Digital volt meters (DVM's) use this characteristic for their"diode check" function. What they do is, when the "red" or positive lead is connected to the p-side (anode, or arrow in thediagram) and the "black" or negative lead is connected to the n-side (cathode, or bar in the diagram) of a diode, the meterattempts to pass (usually) 1 mA of current through the diode. If the 1 mA of current is allowed to flow, the meter thenindicates the amount of forward voltage developed across the diode. If it reads something like 0.673 volts, then you can bepretty sure the diode is OK. Reverse the leads, and the diode is reverse biased, and the meter should read "OL" (overload) orsomething like that to indicate that no current is flowing.

The diode equation is usually approximated by two somewhat simpler equations, depending upon whether the diode is forward orreverse biased:

Now let's connect this "ideal diode equation" to the real world. One thing you might ask yourself is "How could I check to see ifan actual diode follows the equation given here ?" As we said, $I_{\mathrm{sat}}$ is a very small current, and so trying to do the reverse test is probably not going to be successful.What is usually done is to measure the diode current (and forward voltage) over several orders of magnitude of current.

If we take the natural log of both sides of the second piece of , we find:

Well, I went into the lab, grabbed a real diode and made some measurements. is a plot of the natural log of the current as a function of voltage from 0.05 to 0.70volts. Included with this plot, is a linear curve fit to the data which is plotted as a dotted line. The linear fit goesthrough the data points quite nicely, so the current is surely an exponential function of the applied voltage! From theexpression for the best fit, which is printed above the graph, we see that $\ln I_{\mathrm{sat}}=-19.68$ . That means that $I_{\mathrm{sat}}=e^{-19.68}=2.89E-9$ amps, which is indeed a very small current. Look at the slope however. Its supposed to be 40, and yet it turns outto be slightly more than 20! This comes about because of some complex details of exactly what happens to the electronsand holes when they cross the junction. In what is called the diffusion dominated situation electrons and holes are injected across the junction, after which they diffuse awayfrom the junction, and also recombine, until eventually they are all gone. This is shown schematically in . The other regime is called recombination dominated and here, the majority of the current is made up of the electrons and holes recombiningdirectly with each other at the junction. This is shown in . For recombination dominated diode behavior, it turns out that the current is given by