Photodiode - Reverse Bias Operation

[tomizzo]

Hello,

I'm attempting to use a photodiode as a light intensity sensor. However, I have a question regarding the operation of a photodiode in a reverse biased position. The current going through the photodiode is dependent on the light received, and is not affected much by the reverse biased voltage (assuming you're applying reasonable voltages). However, wouldn't the current be dependent on the resistance that is in series with the photodiode? The datasheets I've been reading illustrate the relationship between the light received and the current produced, but there is no mentioning of any resistance. Does this mean that photodiode will act as a constant current source regardless of the resistance used in series with the photodiode?

I've attached an image of the circuit in question.

I appreciate any help!

 

[berkeman]

Pretty much. The voltage across that resistor R is dependent on the photocurrent produced by the diode. The main reason to put the negative voltage bias across a photodiode is to improve the sensing bandwidth. Do you know how the reverse bias improves the sensing bandwidth of the photodiode?

 

[tomizzo]

Thanks berkeman! I'm not sure if I know the answer to your question, but I'll give it a shot:

The photodiode will produce a current in the absence of light which is undesired. By reverse biasing the diode, the 'dark' current will be decreased, thus the current that will be produced will be more indicative of the presence of light and will be less skewed by the 'dark' current. Am I in the ballpark?

 

[berkeman]

Thanks berkeman! I'm not sure if I know the answer to your question, but I'll give it a shot:

The photodiode will produce a current in the absence of light which is undesired. By reverse biasing the diode, the 'dark' current will be decreased, thus the current that will be produced will be more indicative of the presence of light and will be less skewed by the 'dark' current. Am I in the ballpark?

Interesting, I didn't know that about the dark current. Do you have a reference for that so I can refresh my memory?

As you increase the reverse bias of any diode, what happens to the depletion region width? What effect does that have on changing the capacitance across the junction?

 

[meBigGuy]

The photo diode is modeled as a current source. Current sources have high output impedance (voltage changes make little change in current, so effective internal R must be large).

In addition there is the effect berkeman is referring to that impacts the response time to light (not the accuracy). It is very important and is fundamental to any light sensing application. It is usually enumerated in the datasheet tables.

I'm not sure why you think increasing the voltage would decrease the dark current.

 

[Jeff Rosenbury]

It's my understanding that lacking light a photodiode is just a diode. It should have a tiny reverse bias current (dark current) that is slightly voltage dependent if I remember my VI curve.

I think the dark current is temperature dependent as well. I think it's caused in part by thermal hole/electron creation. Other radiation sources also play a part.

Forward biasing it would usually swamp the photo effect in the forward current. This could be avoided by keeping any forward voltage well below the bias current knee.

 

[meBigGuy]

Jeff, there are very specific reasons why photo diodes are reverse biased. I can't think of any implementations where they would be forward biased in any way (zero bias photo-voltaic mode is common though). Do you have some examples for forward bias?

 

[Jeff Rosenbury]

Jeff, there are very specific reasons why photo diodes are reverse biased. I can't think of any implementations where they would be forward biased in any way (zero bias photo-voltaic mode is common though). Do you have some examples for forward bias?

I'm not aware of any applications. But technically it could be done.