Improving Photodiode Circuit Performance: Understanding Slow Rise Time Issues

AI Thread Summary
The discussion focuses on improving the rise time of a photodiode circuit that currently takes approximately 130ms to switch, which is significantly slower than the expected 200µs. The circuit includes a photodiode connected to a current-to-voltage op-amp and a voltage divider, with a 1M feedback resistor chosen to achieve a 5V output before stepping down to 3.3V. Participants suggest that reverse biasing the photodiode can reduce its capacitance and that lowering the feedback resistance will improve the circuit's speed. Additionally, they emphasize the importance of accounting for the op-amp's input capacitance and PCB parasitic capacitance in the RC time constant calculations. Implementing these changes could enhance the circuit's performance and reduce the rise time.
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Hi,

I have a photodiode circuit (attached), where a photodiode is hooked up to a current to voltage opamp connected to a voltage divider. This circuit works as intended: when there is light, Vo ~ 3.3V, and when there is no light, Vo ~ 0.3V. However, when I measured the rise time of this circuit, I found out that it takes ~130ms to switch, which is much slower than I anticipated. The photodiode's spec says that its rise and fall time is 200us, and although I am not in the same operating condition, I expect the delay to be in the same order of magnitude, give or take. Therefore, if any1 has any idea on this, please let me know.

Thanks.

[opamp] http://focus.ti.com/lit/ds/symlink/opa657.pdf
[photodiode] http://catalog.osram-os.com/catalogue/catalogue.do;jsessionid=D791D4FBE4D533287FAB36C83ADE1303?act=downloadFile&favOid=020000030000c448000100b6
 

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Is that 1M for the feedback resistor - that's rather large.
You haven't show the bias supply for the photodiode, are you just using the opamp input to bias it?
 
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Yes, that's 1M ohm. The reason why it's made large is because I wanted the output voltage BEFORE the voltage divider to be 5V (saturating the opamp) and then step it down to 3.3V AFTER the voltage divider. In fact, based on my calculation, the current generated by the photodiode is ~5uA. So to get 5V, I did, 5V = 5u*1M.

It should be ok even if the Rf is large because it will simply saturate the opamp and get 5V as mentioned before.

As for biasing, you are correct and I didn't bias the photodiode. I didn't realize you have to. I saw a proposed design from Sharp that used a simliar circuit so I thought it was fine. It's on page 3 Figure 6 of the link below

http://vorlon.case.edu/~flm/eecs245/Datasheets/Sharp%20photodevices.pdf
 
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To speed up your photodetector I --> V circuit, you need to reverse bias the diode (to lower its capacitance) and lower the feedback resistance (to speed up the associated RC time constant). No circuit with a 1Meg resistor in it is going to be fast. Don't forget to include the input capacitance of the opamp and the PCB parasitic trace capacitance in your RC calculations.
 
Thanks for your suggestion. I will try them out.
 
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