I recently purchased the ODA-6WB-100M photodiode with preamp and I am having trouble getting it to work. I use two DC power supplies to supply positive and negative supply voltages and when I measure the output compared to ground, I do not get a good response. After several different tests, I have come to the conclusion that the photodiode works fine, but the amplifier does not. I believe it has to do with my method or circuit setup. Any tips?
Greetings,
Could you describe in some more detail what precisely it is that you’re applying in terms of power supplies and photo stimulus, what output you’re seeing, and how you came to the conclusion you mention?
This is a fairly simple device as such things go, but simplicity often invites silly mistakes; check your supplies and their connections carefully, giving attention to the possibilities like faulty test leads, incorrect reading of the pin diagram, forgetting to turn the power supply on, etc.
If all the connections and test equipment settings are proper yet a proper output is not forthcoming, the possibility of a damaged device is the next most likely explanation. It’s a static sensitive device, and application of supplies with incorrect polarity or excess voltage can cause damage also. Note that some power supply outputs can overshoot at turn-on to a significant degree.
There’s a rational explanation to be found somewhere, and in cases like this it’s often hiding in plain sight.
1 Like
My setup is as follows: I have the photodiode on a small breadboard and I apply positive and negative voltages to the V+ and V- pins (from top-view and case pin pointed left, the V+ pin is top right and the V- pin is bottom right). To create the positive and negative supplies, I used two DC power supplies. I connected the negative end of power supply #1 and the positive end of power supply #2 together to create the common reference node. I used the positive terminal of #1 as V+ and the negative terminal of #2 as V-. I connected the common reference node mentioned earlier into the GND pin of the photodiode. I left the CASE pin floating, and the Vout pin was connected to the multimeter (reference to common node). I also added capacitors in parallel with the supply inputs like the schematic in the data sheet.
For testing, I would apply my smart phone flashlight and observe the reading. I observed that without any power, the photodiode will display a voltage response, but the response is very small. However, with power, the output will jump to whatever V+ is and provide no output response to light. I have taken care to limit the supply voltages to within the suggested range. I used +5V and -5V for my tests.
I’d wager one beverage that this test was not conducted in a darkened room…
The built-in gain of the amplifier is likely high enough to drive the amplified output into saturation in response to the photosignal from ambient room lighting. Without power, the signal you’re seeing at the output pin is likely some unamplified version of the raw photodiode output.
See what happens at the output if you block all light from the sensor completely, perhaps with a piece of opaque tape over the window.
1 Like
Thank you for your help! I was conducting my tests in both a lighted and darkened room, however, the darkened room was not darkened enough. After taking some light insulation steps, my diode works perfectly.
I do have one more question. It appears that this device is very sensitive, to the point that it saturates very quickly even in a dark room. Are there any tips to prevent the device from saturating so easily?
To reduce the sensitivity I’d add a precision pinhole lens over the photodiode (I’m also an ancient photography nerd :-).
The reason for the $70.00 price of that photodiode assembly is the extreme sensitivity. Garden variety $0.50 photodiodes with a $0.50 Op-Amp will have lower sensitivity. So rather than waste an expensive high accuracy component I’d switch to cheaper parts rather than ty to limit the sensitivity of the high end part.
2 Likes