PN Junction Diodes

A Brief Description Definitions The Depletion Region Varying VA Diode Currents
The Equation Derivations from the Ideal Let's Draw! Related Topics

Diode Currents:

Remember when drift current was first introduced?  What about diffusion current?  No?  Well then, let's refresh your memory!  Drift current is electrons and holes responding to an applied electric field.  Holes move in the direction of the electric field while electrons move opposite the electric field.  This occurs as long as there are carriers available.  Diffusion current is holes and electrons moving from areas of high concentration, where they are the majority carrier, to areas of low concentration, where they become minority carriers.  This occurs until they are uniformly distributed.  Unlike drift, diffusion takes place without an electric field being present.  Back then you thought, "When will I use this?" Well, here's your chance!

Now we're talking about pn junction diodes.  The band diagram that models a pn junction has band bending at the metallurgical junction.  Remember that band bending is caused by an applied electric field and that non-uniform doping can cause an electric field.  In a pn junction the electric field is produced when a p-type and an n-type semiconductor are brought together; therefore it is present even when it is in equilibrium.  This means both drift current and diffusion current are present in equilibrium but the total current density J = 0, now how could that be?

Keep in mind that the p-side of the diode contains many holes and few electrons and that the n-side has many electrons and few holes.  When we draw the band diagram for a pn junction, the p-side is always at a higher level (potential), do you know why?  (Hint:  It has to do with the Fermi level.)  This means there is a potential hill the majority carriers must climb to get to the other side, n-side electrons to get to the p-side and p-side holes to get to the n-side.  Do you recognize this activity?  It's diffusion.

This potential hill has a totally different meaning to the electrons on the p-side and holes on the n-side, otherwise known as the minority carriers.  They don't see a hill, they see a slide!  Remember electrons flow against the electric field and holes flow with it, therefore carriers near the depletion region get caught in the electric field and quickly drift to the other side.  Do you recognize this activity?  We call it drift.

Applying a potential to the diode affects the current flowing through the diode.  The potential hill varies with the applied voltage VA.  As discussed earlier, if VA = 0 the diode is in equilibrium, if VA > 0 it is forward biased, and if VA < 0 it is reverse biased.  Earlier we said that drift current and diffusion current are present even though J = 0 in equilibrium.  This is because for every electron that diffuses from the n-side to the p-side there is an electron that drifts from the p-side to the n-side.  The same goes for holes.  The two currents balance each other and the total current density J = 0.

When the diode is forward biased, VA > 0, the potential hill is still present, but is is less steep.  A smaller potential hill means electrons from the n-side and holes from the p-side need less energy to climb the hill, therefore diffusion current increases.  This current is positive, electrons flowing opposite the electric field produce a positive current and holes flowing with the electric field also produce a positive current.  Under forward biasing, the applied voltage needs only to be increased by tenths of a volt to see a significant change in current.  As the voltage is increased slightly, the current increases exponentially because of the increasing number of carriers that have enough energy to cross the junction.  Remember, majority carrier concentrations are around the magnitude of 1017/cm3, so if even one third cross the junction it is still a lot.

When the diode is reverse biased, VA < 0, the potential hill gets steeper.  If the majority carriers had difficulty climbing the potential hill when the diode is in equilibrium, what do you think happens when the hill is even steeper?  If you thought, "They won't be able to get to the other side!", then you're right!  By setting VA just a few tenths of a volt less than zero, the potential hill gets steep enought that the majority carriers do not have enough energy to climb it and make it to the other side.  This causes the diffusion current to become negiligible.  Now you're thinking "But why does the ID vs. VA curve show a negative current when VA = 0?"  Well, we said that diffusion current goes to zero, we haven't mentioned what happens with drift current yet.

Do you think VA affects drift current?  If you thought "No!", you're right!  As we mentioned before, the potential hill affects drift  in a totally different way than diffusion.  It doesn't matter how steep or flat the potential hill is, as long as there is an electric field and available carriers (ionized dopants don't count) drift will remain the same.  The current produced is negative, electrons are flowing with the electric field producing a negative current and holes are flowing against the electric field also producing a negative current.  The reverse bias current is expected to be small, unlike forward bias current, because it depends on the minority carrier drift.

When the diode is forward biased drift current is present, but because diffusion current grows exponentially, it dominates.  The total current flowing through the depletion region under forward biasing is made up of mostly majority carrier diffusion.  When the diode is reverse biased diffusion is negligible, but drift remains constant.  The total current flowing through the depletion region under reverse biasing is made up of mostly of minority carrier drift.  


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