Why photodiode is operated in reverse bias?

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Discussion Overview

The discussion centers around the operation of photodiodes, specifically why they are typically operated in reverse bias rather than forward bias. Participants explore the implications of biasing on the depletion region, sensitivity, and current generation in photodiodes.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants suggest that operating a photodiode in reverse bias widens the depletion region, which allows for more efficient collection of electron-hole pairs generated by incident light.
  • Others argue that in forward bias, the depletion layer is narrower, potentially reducing the sensitivity of the photodiode.
  • It is noted that only electron-hole pairs generated in the depletion region or near it contribute to the current due to the presence of an electric field, while pairs generated outside this region do not.
  • Some participants question whether reverse current can be generated without biasing the diode, with one noting that there is some reverse current when light is shined on the photodiode even without bias.

Areas of Agreement / Disagreement

Participants express varying views on the effects of biasing on photodiode performance, with no consensus reached on the necessity of biasing for optimal operation.

Contextual Notes

Participants mention the relationship between biasing and the depletion region but do not fully resolve the implications of operating in different bias states or the conditions under which reverse current can be generated.

hanii
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why photodiode is operated in reverse bias??

can anyone please explain why photodiode is operated only in reverse bias?? can we operate it forward bias state also??
 
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Good question. I think it's because when reverse biased the depletion region becomes wider. Only electron-hole pairs that are generated in the depletion region (or within a diffusion length of the depletion region) will be swept across the junction and contribute to reverse current. So having a large depletion region is beneficial. You can also operate the PD with no bias but it will not be as efficient.

-Matt Leright
 


so if we operate it in forward bias...the depletion layer length would be less compared to that as in reverse biased condition...there by reducing its sensitivity . why only the depletion layer is exposed to light... not all the device..?
 


hanii said:
so if we operate it in forward bias...the depletion layer length would be less compared to that as in reverse biased condition...there by reducing its sensitivity . why only the depletion layer is exposed to light... not all the device..?

yes. Plus I think the forward bias current is much higher than the photocurrent generated.

The reason only ehps generated in the depletion region (or very close to the depletion region) contribute to current is because those are the ehps that are in an electric field. Remember the region where there is a slope in the band diagram there is an e-field. the flat regions there is no field so no current.
 


so biasing only helps to improve the reverse current ...?? if it is so...can we get the reverse current flow through the load and produce potential difference across it without biasing the diode...??
 


hanii said:
so biasing only helps to improve the reverse current ...?? if it is so...can we get the reverse current flow through the load and produce potential difference across it without biasing the diode...??

The PD will have some reverse current when light is shined on it with no bias, yes.

-Matt
 

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