Why photodiode is operated in reverse bias?

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Photodiodes are operated in reverse bias to widen the depletion region, allowing for more efficient collection of electron-hole pairs generated by incoming light. In reverse bias, only the charges generated within the depletion region contribute to the current, enhancing sensitivity. Forward bias reduces the depletion layer, decreasing sensitivity and increasing the forward current, which overshadows the photocurrent. While photodiodes can operate without bias, their efficiency is significantly lower compared to reverse bias operation. Overall, reverse bias is crucial for maximizing the performance of photodiodes in light detection applications.
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
 
I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question. Ex. 1 - A moving object impacts a static object at a constant velocity. Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact. Assuming the mass of the objects is the same and the velocity at the moment of impact...

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