Why is the output from my photodiode not linearly related to laser power?

Question

I have some basic understanding of electronics but am having trouble optimizing our setup for laser intensity measurements.

Currently, we are using a Thorlabs photodiode PDA100A with 50Ω terminators connected to a Thorlabs 10kHz lowpass filter. The filter is then connected to our NI BNC-2110 connector (I would add more links to products but apparently I need 10 rep to post more than 2 links) on a analog input channel using differential terminal setup. We are reading the voltage using a NI PCI 6122 DAQ board. We are sampling at the max sample rate and using a trigger to determine which sample to read from the card.

The problem is, the relationship is non linear. We can double our laser power without doubling the voltage read on the DAQ card. The photodiode uses a transimpedance amplifier which we have set so that there is no gain. We are not reaching anywhere near the saturation limit of the photodiode - our measurements are more around 2-3V.

My question is at what point in our set up can you imagine nonlinearity taking over? I am skeptical of the low pass filter but that's partially because I don't know enough of the response voltages to understand how they would affect the data. Let me know if you have any suggestions.

Thanks in advance!

Answer 1

If you look at the data sheet, p. 9, you'll see that the maximum output current of the PDA100A is 100 mA. Since you have terminated the output with 50 ohms, 5 volts is all you can expect.

Answer 2

With 100mA limit at a peak response of 0.65A/W You can expect a max input of ~ 150 mW optical input.

Also you can expect efficacy of your laser to decline with increasing power due to more losses in bulk resistance or I^2 * ESR.

You can verify Pd vs If in your laser.

Take care in temperature rise which can also reduce efficacy of the emitter.

Answer 3

The PDA100A manual ( Download from menu on right side of page here ) says that there is an internal 50 Ohm series resistor which has a significant effect on Vout when low load resistances are used.

From page 3 here

• For low terminating resistors, <5kΩ or 1% error, an additional factor needs to be included in the above formula.

  As described above the output includes a 50Ω series resistor (RS). The output load creates a voltage divider with the 50Ω series resistor as follows:

  Scale Factor = RLOAD / (RLOAD + RS)
  Where RLOAD is the terminating resistor and RS = 50Ω.
  For a standard 50Ω terminator, the gain will be scaled by ½ as shown below:
  Scale Factor = 50Ω / (50Ω + 50Ω) = 0.5

  Output [V/W] = transimpedance gain [V/A] x ℜ(λ) [A/W] x Scale Factor