Electric field in diode I read that the electric field in a diode aids the flow of minority carriers (drift current). However, why is the eletric field direction simply represented with a single arrow? Shouldn't it look like the diagram below? Based on the lines of force, shouldn't minority carriers be repelled away?


 A: The electric field inside the diode will be neither homogeneous like the single arrow suggests nor will it look like the field between two point charges in vacuum, like in the second image. Instead, it will be a rather complex superposition of the outer electric field and the fields of the charges inside the semiconductors. However, most of the time when talking about diodes, these details are not of importance and everything is neglected except the fact that there is an electric field which, on average, points in the direction of the one arrow (or against it, depending on the outer electric field, see below).
Now without any further explanation, your first image is a little confusing, because there are two electric fields and two voltages involved, the difference between which is not really made clear: Firstly, there is the electric field between the charges in the depletion region, which, in your image, points from right to left. Secondly, there is an outer field which can point

*

*from right to left, too. In this case, the diode's resistance is high, because the negative (left) side is flooded with negative charges, which repel each other.

*from left to right. In this case, the negative charges on the left side of the depletion region are pulled away to the left and current can flow.

In both cases, the analoguous happens on the right side.
I think in the image, $V_0$ is the outer voltage which creates the outer electric field. It's polarity is chosen so that the diode will have low resistance. On the other hand, $E$ is the electric field of the charges in the depletion region as long as they are not pulled away by the outer field.
