A4 · Semiconductor basics

The pn Junction: A Border That Builds Its Own Barrier

What happens when you join n-type and p-type? In this unit you follow, step by step, how the majority carriers cross the boundary and the fixed ions left behind raise a barrier that stops the diffusion by itself.

The path to junction equilibrium

The left side is p-type, where the blue hollow circles are holes, and the right side is n-type, where the gold dots are electrons. Step through the stages and a band emptied of carriers forms at the central boundary, the signs of the fixed ions inside it appear, and an arrow for the built-in electric field shows up across it. Step through to see when the net current stops.

Step through the stages with the buttons.

Equilibrium condition and barrier

Jdiff >> Jdrift

The concentration difference is huge. Holes try to cross to the right and electrons to the left, and there is no barrier yet to stop them.

Net diffusion is large

At contact, diffusion begins

The n side has an overwhelming number of electrons and the p side an overwhelming number of holes. The concentration difference across the boundary is so large that the moment they are joined, electrons pour across to the p side and holes to the n side. This is the diffusion you saw in A3.

Departure exposes the fixed ions

Even as carriers cross, the dopant atoms that supplied them are stuck in the lattice and cannot move. As you saw in A2, the n-side donors that gave up electrons remain as positive ions, and the p-side acceptors that took in electrons to make holes remain as negative ions. So the region near the boundary becomes a depletion layer, emptied of carriers and left with only fixed charge, and between the fixed charges on the two sides a built-in electric field points from the n side to the p side.

ObserveJdrift = Jdiff (Jnet = 0)

At equilibrium drift and diffusion cancel exactly.

ChooseVbi = (kTq) ln(?)

The built-in potential goes as the log of the doping product over ni squared.

Fill inNA xp = ?

The total fixed charge on each side of the depletion layer is equal.

On your ownVbi = (kTq) ln(?)

The heavier the doping, the higher the barrier.

The barrier stops diffusion: equilibrium

The built-in field pushes back carriers trying to cross. This drift grows in the direction opposite to diffusion until the two are equal in size and the net current becomes 0, which is the cancellation foreshadowed in A3. At that point the width of the depletion layer and the built-in potential barrier Vbi stop growing, and their size is set by the doping concentrations on the two sides. This boundary, which builds its own barrier to stop diffusion, is the core of the diode.

Back to the first screen

Right after contact the net diffusion was large. The more carriers crossed, the more fixed ions were exposed and the more the built-in field grew, and that barrier pushed the diffusion back through drift. Once the two were equal in size the net current was 0, and the depletion layer and the barrier stopped growing. Nothing was done from outside, yet the boundary built its own barrier and reached equilibrium. The fixed dopant ions of A2 and the drift-diffusion cancellation of A3 meet in one place to form the core of the diode.

In a pn junction the majority carriers cross the boundary by diffusion, and fixed dopant ions (positive on the n side, negative on the p side) are exposed where they left. This fixed charge sets up a carrier-free depletion layer and a built-in potential barrier Vbi. When the drift set up by the barrier exactly cancels the diffusion, the net current is 0 and the junction is at equilibrium, with Vbi = (kT/q) ln(NA ND / ni²).

What comes next

Now you hold the barrier in hand. In the next set, B, you apply an external voltage across this barrier. In forward bias the barrier drops and carriers pour across; in reverse bias the barrier rises and blocks the flow. This asymmetry, flowing one way only, is the diode's I-V curve and its rectifying action.