If you look closely at these pictures, you will notice something. As we remove more and more electrons and holes, we are starting to "uncover" the fixed charges associated with the donors and acceptors. We are making what is known as a depletion region, so named because it is depleted of mobile carriers (holes and electrons). The uncovered net charge in the depletion region is separated, with negative charge in the p-region, and positive charge in the n-region. What will such a charge separation give rise to? Why, an electric field! Of course! Which way will the field point? The electric field which arises from a separation of charges always goes from the positive charge, towards the negative charge. This is shown in Figure 1.

Figure 1: The pn-junction with the resultant built-in electric field

What effect will this field have on our device? It will have the tendency to push the holes back into the p-region and the electrons into the n-region. This is just what we need to counteract the recombination which has been going on, and hopefully bring it to a stop.

Now try to think through what effect this field could have on our energy band diagram. The band diagram is for electrons, so if an electron moves from the right hand side of the device (the n-region) towards the left hand side (the p-region), it will have to move through an electric field which is opposing its motion. This means it has do some work, or in other words, the potential energy for the electron must go up. We can show this on the band diagram by simply shifting the bands on the left hand side upward, to indicate that there is a shift in potential energy as electrons move from right to left across the junction.

Figure 2: Energy band diagram for a p-n junction at equilibrium

The shift of the bands, which is just the difference between the location of the Fermi level in the n-region and the Fermi level is the p-region, is called the built-in potential, Vbi Vbi . This built-in potential keeps the majority of holes in the p-region, and the electrons in the n-region. It provides a potential barrier, which prevents current flow across the junction. (On the band diagram we have to multiply the built-in potential Vbi Vbi by the charge of an electron, q q, so that we can represent the shift in energy in terms of electron volts, the unit of potential energy used in band diagrams.)

How big is Vbi Vbi ? This is not too hard to figure out. Let's look at Figure 2 a little more carefully. Remember, we know from this equation and this equation that since n= N d n N d in the n-region and p= N a p N a in the p-region, we can relate the distance of the Fermi level from Ec Ec and Ef Ef by

Look at Figure 2 and see if you can agree that

Where Nd Nd and Na Na are the doping densities in the n and p sinc respectively. Remember, k⁢T=1/40⁢eV=0.025⁢eV k T 140 eV 0.025 eV , E g =1.1⁢eV E g 1.1 eV and Nc Nc and Nv Nv are both ≈⁢1019 ≈ 10 19 . Thus, q⁢ V BI =1.1⁢eV−0.025⁢eV⁢ln1038 N d ⁢ N a q V BI 1.1 eV 0.025 eV 10 38 N d N a Here the qq in front of the VBI VBI and the ee in eVeV are both the charge of 1 electron and they cancel out making V BI =(1.1−0.025⁢ln1038 N d ⁢ N a )⁢volts V BI 1.1 0.025 10 38 N d N a volts Suppose both Nd Nd and Na Na are both about [10 to the 15th] - not uncommon values. How big would the built-in potential be in this case?

It turns out that we can actually derive some specific details about the depletion region if we make only a coupled of simplifying (and often justified) assumptions. In order to make the math easier, and also because many p-n junctions are built this way, we will consider what is known as a one sided junction. Figure 3 is a picture of such a beast: In this diode, one side is much more heavily doped than the other. In this particular example, the p-side is heavily doped, and the n-side is relatively lightly doped. We can not show the true picture here, because typically, the more heavily doped side will be doped several orders of magnitude greater than the lightly doped side. Typical values might be N a =1019 N a 10 19 and N d =1016 N d 10 16 . Regardless of how big the difference is however, there must be exactly the same amount of "uncovered" charge on both side of the junction. Why? Because each time a hole and electron recombine to form the depletion region, they each leave behind either a donor or an acceptor. A careful count of the exposed charge in Figure 3 shows that I was careful enough to draw my figure accurately for you. We do not need to have a one-sided diode to do the analysis that will follow, but the equations are easier to solve if we do.

Figure 3: An example of a one-sided diode

In order to proceed from here, the first thing we do is make a plot of the charge density ρ⁢x ρ x as we move through the junction. Naturally, in the bulk, since the holes and the acceptors (in the p-side), or the electrons and the donors (in the n-side) just equal one another, the net charge density is zero. In the depletion region, the charge density is - (−q)⁢ N a q N a on the p-side and (q)⁢ N d q N d on the donor side. (All the mobile carriers are gone, and we are left with just the charged acceptors or donors.) We will make the assumption that on the n-side, the depletion extends a distance − x n x n from the junction. On the p-side, the acceptor charge density is so large, that we will treat it is a δ-function, with essentially no width. The areas of the two boxes must be the same (equal amount of positive and negative charge) and hence, the tall thin box actually has a width of N d N a ⁢ x n N d N a x n , which, since Na Na is several orders of magnitude greater than Nd Nd, means that the tall box has a very very small width compared to the lower, wider one, which is q⁢ N d q N d tall, and has a width of xn xn.

Figure 4: Charge density as a function of position

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Last edited by Raymond Wagner on Aug 14, 2007 11:46 am -0500.