Guys - I am revisiting the basics of the p-n junction because of this thread. Please see here the fundamental question I have now:
Let's look at the diode's p-n junction. Now it has 0.7v barrier potential.
We connect it in forward bias.
With anything less than 0.7v from the power cell, no current flows because the electrons from the battery do not have sufficient energy to cross the junction, because they are repelled from the NEGATIVE IONS on the p-type side of the junction (which have stabilized in number to reach 0.7v, relative to the positive ions on the n-type side).
OK now we increase voltage, so that it exceeds 0.7v just a bit. Current starts to flow.
HOWEVER my question now is:
If we increase the voltage, more electrons come from the wire connected on the n-type side. So, there is a net increase of new free electrons coming across from the n-type. These ostensibly knock off more electrons from the n-type, but the holes there are quickly filled back up, because we have so many new electrons coming in.
OK, so here is the question:
With the p-type side getting bombarded with more electrons that have cross the depletion layer, we are creating even MORE negative ions out of the doped p-type material, right?
So shouldn't these negative ions INCREASE the REPULSION of the new negative charge carriers (now the majority carrier electrons) and this should adversely affect current?
I assume that as voltage increases tremendously, any such repulsion should be overwhelmed even if the entire junction now is filled with negative charge.
But from the moment beyond 0.7v barrier potential, to full current flow, should there not be a DECREASE/resistance in current flow, because of the vast number of negative ions (and almost zero positive ions)?