Suppose, positive battery terminal is connected to P-region of a semiconductor and the negative battery terminal to the N-region. In that case the junction is said to be biased in the forward direction because it permits easy flow of current across the junction. This current flow may be explained in the following two ways :
1. As soon as battery connection is made, holes are repelled by the positive battery terminal and electrons by the negative battery terminal with the result that both the electrons and the holes are driven towards the junction where they recombine. This masse movement of electrons to the left and that of holes to the right of the junction constitutes a large current flow through the semiconductor. Obviously, the junction offers low resistance in the forward direction.
As free electrons move to the left, new free electrons are injected by the negative battery terminal into the N-region of the semiconductor. Thus, a flow of electrons is set up in the wire connected to the negative battery terminal. As holes are driven towards the junction, more holes are created in the P-region by the breakage of covalent bonds. These newly-created holes are driven towards the junction to keep up a continuous supply. But the electrons so produced are attracted to the left by the positive battery terminal from where they go to the negative terminal and finally to the N region of the crystal. Incidentally, it may be noted that though there is movement of both electrons and holes inside the crystal, only free electrons move in the external circuit i.e. in the battery-connecting wires.
2. Another way to explain current flow in forward direction is to say that forward bias of V volts lowers the barrier potential to (V - VB) which now allows more current to flow across the junction.
Incidentally, it may be noted that forward bias reduces the thickness of the depletion layer. Energy band diagram for forward bias can be seen this reduction, conduction electrons in N-region are able to cross over to P-region. After reaching there, each electron falls into a hole (path A) and becomes a valence electron. In this way, it is able to continue its journey towards the left end of the crystal.
