I need to make something that reads brightness value of an LED from voltage output of photodiode.
I will be using the following parts:


  • Forward voltage Vf (typ.): 20.8 V at forward current If = 700 mA
  • light emitting surface: 2.1 mm x 3.2 mm
  • Luminance: 22 x 10^6 cd/m^2
  • optical efficiency: 65 lm/W at 350 mA

Hamamatsu Si-PIN diode S6036 datasheet

  • Reverse voltage Vr: 12 V
  • lens: phi 7 mm
  • sensitivity: 0.56 A/W
  • short circuit current: 30 uA
  • dark current: 0.1 nA

I will see the voltage output as drawn in this figure

And voltage input range of my ADC equipment is within 2 volts peak to peak and +/- 3 volts.

I need to know which resistor I need to use, or whether I need put amplifier or attenuator, in order to adjust output voltage level from photodiode towards ADC input.

To figure out it, here's my calculation:

sensing area of photodiode: 3.14 * (0.007 / 2) ^2 = 3.848451e-5 m^2 # 0.007: lens phi
candela value into sensing area: 22e6 * area = 846.65922 cd
candela-Watt conversion: 1 cd = 18.3988 mW over a complete sphere
watt value: 15.57751 W
current output from photodiode: watt * sensitivity = 8.7234 A
assume that + 2 volts is the maximum acceptable, 2 V = 8.7234 A * R
thus, I should use resistor with R = 0.2292684 ohm

But, I think 0.23 ohm is absurdly small.

Since I'm not familiar with electrical engineering, circuitry and optical, I have totally no idea whether my calculation is correct or not, and whether I use the terms in the equations with right interpretation.

Another problem is, I think current output from photodiode towards ADC does matter. If it is too high or low, ADC will be fried or breakdown due to too much current drain. I'm not sure that it is correct or not. If it is wrong, please fix the statement. Back to the question, anyway, is it correct that the current matters for ADC?

FYI, ADC that I will use is on LFRX daughterboard (schematics) for USRP.

  • 1
    \$\begingroup\$ I didn't check all the numbers... (I hate candela's) But not all the light from the LED will make it into the Photodiode. There is at least a solid angle correction depending on the distance between the two. It's also hard to know the real area with the lens... does all the light into the lens get focused onto the photodiode? I've used just bare photodiodes for this. Measure the PD current with the uA (current) scale of your DMM. Then you can pick your resistor. \$\endgroup\$ Commented Feb 11, 2015 at 16:57
  • \$\begingroup\$ To expand on George Herold's comment. Let's think about this - your calculation assumes that all of the LED light is focused on the photodiode. And what would be the use of an invisible LED? I think you're better off making a convenient light path to tap off a very small fraction of the LED light, then illuminating a photodiode and measuring the response. You're overthinking the problem. \$\endgroup\$ Commented Feb 11, 2015 at 17:47

1 Answer 1


I think current output from photodiode towards ADC does matter. If it is too high or low, ADC will be fried or breakdown due to too much current drain.

If you're doing a purely resistive conversion from PD current to voltage, then the PD current flows through the resistor, not into the ADC.


simulate this circuit – Schematic created using CircuitLab

The input to the ADC typically has high impedance (at least kohms), so long as the input voltage doesn't exceed the power supply rails of the ADC device. So not much current will flow into the ADC inputs. If you find you need a load resistor over 1 kohm to obtain the voltage you need, then you might want to re-check this assumption or consider using a trans-resistance amplifier instead of just a resistor for current-to-voltage conversion.

  • \$\begingroup\$ Different ADC's are different (read the specs), but I know that most PICs' internal 8-12 bit SAR can handle up to 10k source impedance. I'm not sure how they measure that, but if I were to guess, I'd say it has something to do with the RC time-constant when combined with the internal sample-and-hold capacitor, the settling time (before converting), and the voltage difference between successive output codes. \$\endgroup\$
    – AaronD
    Commented Feb 11, 2015 at 17:33
  • \$\begingroup\$ @AaronD, yes, but other types might have input impedance as low as 10 kohms. That's why I advise OP to check his specs if his load resistor gets in the range of 1 kohm. \$\endgroup\$
    – The Photon
    Commented Feb 11, 2015 at 17:43
  • 1
    \$\begingroup\$ Just to be safe, I like to read the specs anyway, regardless of any rules of thumb. I might have an oddball for whatever reason (maybe someone else loved it and chose it for me, or some other aspect was too good to for me pass up), and it might mean the difference between a direct measurement or a buffer amp that requires board space and possibly some non-trivial effort to design and lay out. I like to solve those kinds of problems before the first run. :) \$\endgroup\$
    – AaronD
    Commented Feb 11, 2015 at 17:59
  • \$\begingroup\$ Well, yes. OP should read the specs anyway. If he wants 12-bit resolution, he doesn't want 1 kohm source impedance even with a 100 kohm ADC input, for example. \$\endgroup\$
    – The Photon
    Commented Feb 11, 2015 at 18:01
  • \$\begingroup\$ As far as overvoltage to the ADC as long as the PD bias voltage is at or below the ADC maximum, then it will be fine. \$\endgroup\$ Commented Feb 11, 2015 at 18:36

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