Bipolar transport in a simple illuminated semiconductor bar

In summary, the delta n and delta p always accompany each other, but the current produced is zero when the material is externally biased.
  • #1
zhanghe
44
2
TL;DR Summary
light-conducivity and the practical movement of electron and hole.
I feel quite confused for a few days, when I apply the bipolar transport equation into a voltage-applied semicondutor material (e.g. p-type c-Si bar, or a resistor) which just have some light-generated electron-hole pairs by a pulse of photon at somewhere on the bar. In terms of bipolar transport theory, the delta n and delta p should go together along the bar in the direction as the so-called minority carrier should go (as the e.g. above, the electron for a p-type c-Si) . However, it seems that there will be no net light current, because delta n and delta p always accompany each other on the 1-D bar. On the other hand, we always use the delta n + delta p the sum to give a light electrical current (i.e. light conductivity).
Please give me some hints, there must be somewhere wrong.
 
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  • #2
This is why one needs a PN junction to produce photocurrent!
 
  • #3
There should also be a photocurrent on a p-type (or n-type) material without PN junction, the so-called the light-conductivity for material. :confused:
 
  • #4
zhanghe said:
TL;DR Summary: light-conducivity and the practical movement of electron and hole.

In terms of bipolar transport theory, the delta n and delta p should go together along the bar in the direction as the so-called minority carrier should go (as the e.g. above, the electron for a p-type c-Si)
But the material must be externally biased in order to produce current. Why do you think they are equal??
 
  • #5
When there is a current in a material externally biased, could you help me to analyze the composition of the current, the density of two carriers and their movement direct? I thought about it for a while last night and attach a file below, that could explain my confusion easily.
1679959298409.jpeg
 
  • #6
I believe the the conductivity comes mostly from the majority carriers whose number in the conduction band is promoted by the light, hence the photoresponse. I do not understand your sketch but I believe the "bipolar transport" is secondary.
I am not expert in this field so invite comment!!
 
  • #7
zhanghe said:
TL;DR Summary: light-conducivity and the practical movement of electron and hole.

I feel quite confused for a few days, when I apply the bipolar transport equation into a voltage-applied semicondutor material (e.g. p-type c-Si bar, or a resistor) which just have some light-generated electron-hole pairs by a pulse of photon at somewhere on the bar. In terms of bipolar transport theory, the delta n and delta p should go together along the bar in the direction as the so-called minority carrier should go (as the e.g. above, the electron for a p-type c-Si) . However, it seems that there will be no net light current, because delta n and delta p always accompany each other on the 1-D bar. On the other hand, we always use the delta n + delta p the sum to give a light electrical current (i.e. light conductivity).
Please give me some hints, there must be somewhere wrong.
Maybe, you will find some information on photoconductivity at
https://www.eeeguide.com/photoconductivity-definition-working-and-its-applications/
 
  • #8
Thanks guys. I could withdraw this "problem" for now. Finally, I found my mistake, the bipolar transport, ie. the movement of the delta n is totally different the electrical current. the bipolar is more like a kind of forms, but when you look into the current, you have to go to the movement of the n and p. Anyway, thanks your guys so much.
 
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