Familiar with P-type semiconductors and N-type semiconductors, coupled with the basic ways to calculate their values for holes and electrons, we can now move onto the next subject: P-N junction diodes. Diodes are shown by the following figure, and are characterized by allowing current to flow through them only if the voltage drop is greater than their turn on voltage. The diode will resist all current flow when the voltage dropped across its terminals are below the turn on voltage. The voltage dropped by the diode can usually be characterized by approximately 0.7 V, i.e., Vanode-Vcathode = 0.7V. The 0.7V is an example of a turn on voltage. If the voltage was less than this value, there would be no current flow.

Now, we must take a look under the hood of the diode, with the equations we currently know, to understand what physics determine our specified behavior. When a PN junction diode is created, there is a point where the two semiconductor material meets, and allows for a flow of electrons. Eventually, a equilibrium state is reached, at which point there is a space in the P side that is negatively charged (due to its doping atoms accepting electrons from the N type semiconductor’s donors), and conversely a space in the N side is positively charged. Remembering basic physics, we know that a separation of charges will create an electric field pointing from the positive charges to the negative charges. Recall that E = -grad(V).

Noticing this orientation, and direction of the electric field, one could wonder, what side of the PN junction is the cathode and anode? (Answer, the voltage drop has to overcome the electric field direction, thus the P side is the anode, and the N side is the cathode)

Now, it is important to create a relationship between the concentration of the doped atoms in either semiconductor, and the voltage needed to overcome the electric field to turn the device on. All of these variables will also be used to determine the length of the space charged region in either semiconductor.

Note the direction of the electric field in the diagram of a PN junction, and recall E = -grad(V), thus as the electric field increases, the voltage will decrease. Using this principle we can estimate that:

Note: The equations used in this post are for ABRUPT junctions. The graphs for abrupt junctions look like this:

Using the equations relating to holes and electron density in the semiconductor materials, we can derive the final equation to find the amount of voltage needed to turn the diode on. NOTE: the holes concentration and electron concentration are different in the two semiconductors, meaning it does not hold true that the doping concentration of the p-type, multiplied by the doping concentration of the n-type will equal the intrinsic concentration squared. That is, the two semiconductors have two separate equations to find the concentrations of holes and electrons.

With a relationship describing the turn on voltage, now we must create a relationship with the doping concentrations, applied voltage, turn on voltage, and depletion region width.

Using the Poisson Equation, we can make a relationship between the electric field (V/cm), the charge density (C/cm^3) and the permittivity (F/cm)

Now we have established what the gradient of the electric field for both sides of the PN junction in the depletion region. The xn and xp are the boundaries of the depletion (or charged) region. To make a relationship between this equation and electric field, we must integrate this equation. Due to the fact that we are only concerned with the gradient in the x direction, we are only mentioning the x direction in respect to voltage. Recall a gradient is equal to partial derivative x (i) + partial derivative y (j)+ partial derivative z (k).

With the relationship with electric field found, we must integrate again to find the relationship of the doping concentrations and voltage. Due to electric field being equal to the -grad(V), we know that the turn on voltage has to be equal to the electric field integrated from xp to xn. To find this value, we should take our previously derived equations for electric field along the p and n ranges, integrating them to find their associated voltage value.

Integrating the N side we will find:

To solve for the charged region aka depletion region, all one must do is set the equation to x = 0, where both p-type and n-type will equal each other. Thus: