0
\$\begingroup\$

Datasheets:

Will post below since I can't post more than two links.

LED circuit schematic:

LED Circuit Schematic

Photodiode circuit schematic:

PD circuit schematic

Details:

The resistor value for the LED circuit (top right) is 24 ohms rated at 5W. The resistor value for the photodiode circuit is 33 ohms. The op amp is an LM358.

Problem:

I've hooked up these schematics to an Arduino uno and have been pulsing the LED at a frequency of 10 kHz. For some reason, the photodiode isn't reacting when I point the LED towards it. I've tried several different resistor values and can't get the photodiode to react, unfortunately. I've read several different threads and haven't found the information I need. I can post the code, if needed, but I doubt that's the culprit considering I've tried the same code on a different photodiode.

\$\endgroup\$
6
  • \$\begingroup\$ Datasheets: LED: lmsnt.com/datasheets/Standard%20chip/Lms43LED-CG/… Photodiode: eoc-inc.com/LED%20Microsensor/lms43pd-05-cg.pdf \$\endgroup\$
    – Amaury
    Commented Jul 11, 2017 at 17:40
  • \$\begingroup\$ If you want a positive voltage output from your receiver circuit, reverse the orientation of the photodiode. \$\endgroup\$
    – The Photon
    Commented Jul 11, 2017 at 17:49
  • \$\begingroup\$ Also, 33 ohms for the transimpedance gain of the receiver is very low. Are you sure you don't want a higher value? Have you done the math to figure what amplitude output you should expect from your receiver? Megohm values aren't uncommon here (depending on desired bandwidth). \$\endgroup\$
    – The Photon
    Commented Jul 11, 2017 at 17:51
  • \$\begingroup\$ I've tried reversing the orientation of the photodiode, and still no luck. And yes, I've looked into the math behind the resistor value, but I may be doing it wrong. I want the output to be able to be detected on an Arduino, so anywhere on that 0-5V reference would work. I've also tried various magnitudes for the photodiode, from 30 to 1M, and still nothing. I'm willing to try specific values if you can suggest any. Thanks for the help. \$\endgroup\$
    – Amaury
    Commented Jul 11, 2017 at 17:56
  • \$\begingroup\$ Do you have a reason to use 4 um here? Because the dark current of your PD is massive compared to what you could get at near-IR wavelengths. \$\endgroup\$
    – The Photon
    Commented Jul 11, 2017 at 18:18

1 Answer 1

0
\$\begingroup\$

Assuming you fix the obvious problems with the circuit mentioned in comments, you will still have some difficulties.

Let's say you are coupling 10% of the light from your LED into your photodiode. There's no quantum efficiency spec on the photodiode, so just for giggles we'll assume it's 100%. Then the expected photocurrent is

$$\frac{ P_{src} \eta \lambda e}{hc}$$

where \$P_{src}\$ is the source power, \$\eta\$ is the transfer efficiency, \$e\$ is the fundamental charge, and \$\frac{hc}{\lambda}\$ is the photon energy at your operating wavelength. Plugging in numbers, I get,

$$\frac {(10 {\rm \mu W})(0.1)(1.6\times 10^{-19} {\rm C})}{(0.3 {\rm eV})(1.6\times 10^{-19}{\rm J/eV})}$$

or about 30 uA.

The dark current of your photodiode is specified as 8 - 25 mA.

This means you have a signal to noise ratio of not much more than 0.001.

You will probably need to do some fairly careful design to make this system work. Consider these changes:

  • Increase your source power dramatically

  • Use a lock-in amplifier to separate your signal (10 kHz) from the background noise of your receiver

  • Use well-designed optics to optimize the optical coupling from the source to the receiver. Optics for 4 um wavelength will need to be chosen specifically for this wavelength.

  • Shield your receiver carefully from any background light

  • Cool your receiver to reduce dark current

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.