0
\$\begingroup\$

I have a problem with circuit, for some reason I always have 4.8 volt in output.

Like a:

  • 4.8 V - when I put finger on photodiode
  • 4.83 V - in room light
  • 8.27 V - when I light it with IR laser

What am I doing wrong?

Here is my circuit:

Schematic

I'm using LM358, by the way when I'm trying to do same with LM258 it not working.

I also tried use it with single supply, without negative voltage, it is not working at all in this case.

I also own these amplifiers, which one will be better: LM258,AD8544, MCP6022? I read that rail-to-rail is better in this case, but why? I want to use output with Arduino, to measure amount of reflected IR light in long range 150-200 meters. What I can improve here?

\$\endgroup\$
7
  • \$\begingroup\$ A 9 volt battery with a resistive centre tap produces +4.5 volts and -4.5 volts with respect to GND and not +9 volts and -9 volts. This circuit should run from a single supply. \$\endgroup\$
    – Andy aka
    Commented Dec 13, 2022 at 20:53
  • \$\begingroup\$ The problem might be that the photodiode is inappropriate for the amount of light you're sensing. Indoor light irradiance is ~10μW/cm2 and the "reverse light current" chart for the BPW34 starts at that level. \$\endgroup\$
    – vir
    Commented Dec 13, 2022 at 21:01
  • \$\begingroup\$ Yuri, are you hoping to detect reflected IR over a range of 150-200 m? Meaning a total trip of 300-400 m, out and back? Is it the case that your emitter (the light source) and your detector (your PD) are near each other and that the reflection is taking place a long way away from them. Is that the situation, in very broad terms? \$\endgroup\$
    – jonk
    Commented Dec 14, 2022 at 0:21
  • \$\begingroup\$ @Andyaka yea, it my mistake on schematic. \$\endgroup\$ Commented Dec 14, 2022 at 7:39
  • \$\begingroup\$ @jonk yes, they are near each other. I have 808nm photodiode 300mW, i taked it of from green laser, as i know green laser with 100mW of power have beam with 1km range, so in this case, when i have shorten range & more powerfull laser, it should work, or do i miss something? Im also thinking if PIN photodiode will not work here, to use avalanche photodiode, they are used in measurement tools in big ranges, like a 100+m \$\endgroup\$ Commented Dec 14, 2022 at 7:42

1 Answer 1

2
\$\begingroup\$

As Andy says you can (and should) run this from a single supply and lose those 100k resistors and virtually useless (in the application of bypassing the supply rails) 9pF caps.

You have several problems from what I can tell.

  1. You are measuring the voltage relative to the negative terminal of the battery rather than your "ground".

  2. The "ground" is varying in voltage from roughly the middle of the +/- supplies to close to the positive supply, depending on the PD current. If you want to split a rail with resistors the voltage change for the current should be small. In this case, you may have 50 or 100uA flowing from the PD with a very bright light source. That will produce very little voltage across the 1K resistor (that's the desirable output voltage) and a lot of change in the 50kΩ source resistance of your 'ground'.

  3. The 1kΩ feedback resistor is far too low in value for normal light levels.

I suggest you try it with a single 9V supply. Measure relative to the battery negative terminal and swap the 1kΩ resistor for one of those 100kΩ resistors. Keep the 9pF across the 100kΩ feedback resistor.

That should give you some volts of change when exposed to a strong light source.

\$\endgroup\$
3
  • \$\begingroup\$ it gives me 0.04v in room light & 8.2v when i enable laser. Thank you. What about another amplifier, will it be maybe better to use? or it doesnt matter in my case? \$\endgroup\$ Commented Dec 14, 2022 at 7:43
  • \$\begingroup\$ 8.2V probably represents saturation so if you intend to measure (rather than detect) the laser light then a somewhat lower feedback resistor value may be required. You could use a different (better) op-amp but in this case the benefits would not be great (at least from what you've said). Possible advantages might include speed (especially when the circuit is optimized), noise level, getting a bit closer to the positive rail on the output, and accuracy. Those advantages, which may or may not matter to your application come with downsides generally. Cost almost for sure. Also... 1/2 \$\endgroup\$ Commented Dec 14, 2022 at 15:55
  • \$\begingroup\$ 2/2 very high speed op-amps such as GBW 1-2GHz for sure, and even ones with 1/10 that gain bandwidth at times, are a bit finicky and prone to oscillate under some conditions. \$\endgroup\$ Commented Dec 14, 2022 at 15:57

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.