WHY is drift current through a PN junction independent of applied bias voltages?

1. May 2, 2007

leright

I have looked all over in many textbooks and I cannot find a decent explanation of this. Most textbooks simply say that it is independent of applied bias because all of the minority carriers generated by thermal energy within a diffusion length of the depletion region will essentially get swept across the junction due to the field, regardless of the applied bias and the strength of the electric field. So, the drift current is essentially limited by the thermal generation rate of minority carriers in the neutral regions, and not by the field strength...however, it seems obvious that the field strength will affect the speed at which minority carriers drift, which will in turn affect the drift current, even if the number of charges that drift is the same. So, how do you explain this?? No textbook mentions that the drift speed changes depending on the bias.

I considered that even though the speed at which the carreirs drift changes depending on the bias, the width of the depletion region is also changing, which could compensate, but when I work out the drift current I always end up with a bias dependence. I figure the drift current is of course J = qnv, where q is the change of a hole (or electron). I also assume that n, the number of minority carriers generated within a diffusion length of the depletion region is the same regardless of bias. So, the only thing that could change with respect the changes in bias is the speed, v, which is W/tdrift, but when I figure the velocity I always get a bias dependence. So, I am a bit lost as to why drift is independent of bias.

Also, when the diode equation is derived only diffusion current is considered...but they call the reverse current Io in the resulting IV characteristics the 'reverse generation current', but reverse gen current is a DRIFT effect! So, how do you get drift characteristics out a derivation that only considers diffusion! This doesn't make much sense to me at all. I think the -Io term isn't really the reverse generation current....

Maybe someone can explain this nonsense to me.

2. May 2, 2007

marcusl

The equation J=nqv doesn't properly treat the time dependence of this situation. As you point out, the rate of thermal carrier creation dn/dt = const, but of course qdn/dt=J. Another way of looking at it is this: no matter what the velocity, the number of charges arriving at the contact per unit time (per unit area) is constant, and that is the definition of current density, so J=const independent of bias.

3. May 2, 2007

leright

hmmm....well I feel stupid now. :p I realize now that if you're only producing, say, 5 carriers per second then moving them a lot faster due to reverse bias doesn't really affect much, since the speed at which the charges can get to the contact is still limited by the generation rate of 5 carriers per second....now, if you have a practically unlimited carrier supply then the speed at which the carriers are transferred has an effect.

So, the point here is the minority carriers are generated much slower than they are swept due to the field, and you're limited by the generation rate. If the generation rate were higher than the speed at which they are swept due to the field then the electric field strength would have an effect. ok, I'm convinced.

Now, explain the IV characteristic derivation.

edit: on second thought, I guess this sorta explains the derivation....the excess minority carrier concentration is used to calculate the diffusion current. the excess minority carrier concentration is simply the total minority carrier concentration minus the equilibrium minority carrier concentration....it is the "minus the equilibrium carrier concentratioin" part that gives rise to the reverse generation current in the IV Characteristics, and the equilibrium carrier concentration is the thermally generated minority carriers.

Interesting. This is quite relieving.

Last edited: May 2, 2007