# PN Junction Diodes

 The Equation Derivations from the Ideal Let's Draw! Related Topics

### The Ideal Diode Equation:

Diodes should be familiar to us by now.  We use them when we need current to flow in only one direction.  The question is, just how much current is there?  Of course, we can just give you the equation to figure it out, but what good will that do?  Instead let's derive it!  You should know by now that when we derive something, we must first make a few assumptions.  These are similar to those we make when we use the MCDEs.  As a matter of fact, we will be using the MCDEs to derive the ideal diode equation. Next, we have to recognize that we have to consider three regions (instead of just one like we're used to), the quasineutral p-region, the quasineutral n-region, and the depletion region.  Are you wondering what quasineutral is? Take a detour here.  The quasineutral p-region is from the edge of the depletion region, which we will call -xp, to the edge of the diode, which we assume is an infinite distance away from -xp.  The quasineutral n-region is from the edge of the depletion region, which we will call xn, to the edge of the diode, which we assume is an infinite distance away from xnIn the quasineutral regions there is no electric field.  This allows us to use the MCDEs to find the current densities in these regions.

In order to solve the MCDE for the quasineutral regions we must first determine the boundaries and the boundary conditions.  (Hint:  You were just given the boundaries for solving the MCDEs in these regions.)  We assume the edges of the diode are an infinite distance away from any actions taking place in the depletion region.  This means that there isn't any variation of carrier concentrations as we get to the edges of the diode: At the edges of the depletion region, -xp and xn, equilibrium conditions do not prevail so we must use the "law of the junction".

 The Law of the Junction: To find the boundary conditions at -xp and xn we use the law of the junction and solve for the minority carrier in each region to obtain:  Using the assumptions we made, the MCDE and J in the quasineutral regions simplify to:

 On the n-side: x >= xn  On the p-side: x <= -xp  Next, using the boundary conditions, we solve the MCDEs for each quasineutral region.  You can do the math and come up with:  Then, to find the current densities of the quasineutral regions, we take the derivatives of Dnp and Dpn and plug them into the equations for JP and JN.  Next, we evaluate JP and JN at the depletion region edges, -xp and xn respectively, to obtain the current density in each region.  By adding them together, we obtain the current density in the depletion region.

 Derivatives of Dnp and Dpn:  JP and JNat the depletion region edges:  The current density in the depletion region: Did you already forget what we're deriving?  Don't worry, we're almost done.  We have solved for the current densities in the quasineutral region to obtain the current density in the depletion region, but what we're looking for is current through the diode.  If you recall, current is charge crossing an area, therefore we multiply (you can do this) the current density (J) by the area (A) to obtain the ideal diode equation (emphasis on ideal): 