1.7 Forward biased

Now let's take a look at what happens when we apply an externalvoltage to this junction. First we need some conventions. We make connections to the device using contacts , which we show as cross-hatched blocks. These contacts allow thefree passage of current into and out of the device. Current usually flows through wires in the form of electrons, so it iseasy to imagine electrons flowing into or out of the n-region. In the p-region, when electrons flow out of the device into the wire, holes will flow into the p-region (so as to maintain continuity of current through the contact.) When electrons flowinto the p-region, they will recombine with holes, and so we have the net effect of holes flowing out of the p-region.

With the convention that a positive applied voltage means that the terminal connected to the p-region is positive with respect to the terminal connected to the n-region. This is easy to remember; "p is positive, n is negative". Let ustry to figure out what will happen when we apply a positive applied voltage $V_a$ . If $V_a$ is positive, then that means that the potential energy for electrons on the p-side must be lower than it was under the equilibrium condition. We reflect this on the band diagram by lowering the bands on the p-side from where they were originally. This is shown in .

As we can see from , when the p-region is lowered a couple of things happen. First of all, the Fermilevel (the dotted line) is no longer a flat line, but rather it bends upward in going from the p-region to the n-region. Theamount it bends (and hence the amount of shift of the bands) is just given by $q{V}_a$ , where the energy scale we are using for the band diagram is in electron-volts which, as we said before, is a common measure of potential energy when we aretalking about electronic materials. The other thing we can notice is that the electrons on the n-side and the holes on thep-side now "see" a lower potential energy barrier than they saw when no voltage was applied. In fact, it looks as if a lot ofelectrons now have sufficient energy such that they could move across from the n-region and flow into the p-region. Likewise,we would expect to see holes moving across from the p-region into the n-region.

This flow of carriers across the junction will result in a current flow across the junction. In order to see how thiscurrent will behave with applied voltage, we have to use a result from statistical thermodynamics concerning thedistribution of electrons in the conduction band, and holes in the valence band . We saw from our "cups" analogy, that theelectrons tend to fill in the lowest states first, with fewer and fewer of them as we go up in energy. For most situations, avery good description of just how the electrons are distributed in energy is given by a simple exponential decay. (This comesabout from a statistical analysis of electrons, which belong to a class of particles called Fermions . Fermions have the properties that they are:

• A)

  indistinguishable from one another

• B)

  obey the Pauli Exclusion Principle which says that two Fermions can not occupy the same exact state (energy and spin)

• C)

  remain at some fixed total number $N$ .)

If $n(E)$ tells us how many electrons there are with an energy greater than some value $E_c$ then $n(E)$ is given simply as:

If the energy $E$ is $E_c$ the energy level of the conduction band, then $n(E_c)=N_d$ , the density of electrons in the n-type material. As $E$ increases above $E_c$ , the density of electrons falls off exponentially, as depicted schematically in : Now let's go back to the unbiased junction.

Remember, as we said before, there are currents flowing across the junction, even if there is no bias. The current we haveshown as $I_f$ is due to those electrons which have an energy greater than the built-in potential. They are flowingfrom right to left, as shown by the open arrow, which, of course, gives a current flowing from left to right, as shown bythe solid arrows. Based on the current should be proportional to:

Now, what happens when we apply the bias? For the electrons over on the n-side, the barrier has been reduced from a heightof $q{V}_{\mathrm{bi}}$ to $q(V_{\mathrm{bi}}-V_a)$ and hence the forward current will be significantly increased.

We can take the effect of the holes, and the other unknowns about the proportionality, and bind them all intoone constant called $I_{\mathrm{sat}}$ , so that we write: