I am running a photodiode through an op-amp circuit (trans-impedance amplifier.) I have been told I should characterize this device. I'm assuming he means using the voltage I get at a given distance and link this with the intensity of light by use of an equation.

I moved the sensor back 5mm at a time taking readings etc., etc.

I plotted it on a graph of voltage against \$\frac {1}{\text{distance}^2}\$ giving a slope of 109.38V/cm^2

I plan on using this relationship:

$$I \propto x^{-2} $$


$$I=k \times x^2 $$

where $$I = \text{intensity} $$

Would \$\frac{k}{x^2}\$ not just be the same as the gradient of my graph?

I just can't seem to wrap my head around this and it's driving me up the wall.

Any help would be greatly appreciated.

I have attached the graph if it sheds any light on the problem:

enter image description here

This is the photodiode I'm using, detecting at a wavelength of 310nm.

enter image description here

I think I am getting there. This is the current equation I have. I used two equations and substituted them into each other to work out k.

enter image description here

I'm sorry I forgot to mention that I was using an LED for this. The power values are between 1mW and 2mW at I'm assuming the device is at 1mW as it's limited at 20mA of current at 5V.

Data sheet for the LED.

  • \$\begingroup\$ Please Edit your question to put that extra information directly in the question instead of in a comment. I'm assuming he means using the voltage Look up what a "transimpedance amplifier" does. does it amplify a voltage or a current? Also realize that a photon can liberate an electron. Flowing electrons means current. Do you still want to measure the voltage? The plot may show a voltage but that doesn't mean you have to measure the voltage as well. \$\endgroup\$ Commented Jun 22, 2021 at 14:03
  • \$\begingroup\$ @craigvenables: I put the formulas into mathjax format. Please make sure I got the meaning right. \$\endgroup\$
    – JRE
    Commented Jun 22, 2021 at 14:05
  • \$\begingroup\$ @JRE yes they are correct sorry i didn't know could do this, \$\endgroup\$ Commented Jun 22, 2021 at 14:09
  • 1
    \$\begingroup\$ @Bimpelrekkie I'm using a oscilloscope to measure the voltage from the output from the amplifier ill add the schematic into the question now. Sorry I'm new to amplifiers but i believe the opamp I am using converts current to voltage. i may be completely wrong haha \$\endgroup\$ Commented Jun 22, 2021 at 14:09
  • \$\begingroup\$ In your "relationship" does I stand for current or intensity? What is the relationship? What does this quantity relate to and in what way? \$\endgroup\$
    – The Photon
    Commented Jun 22, 2021 at 14:27

1 Answer 1


For the measured voltage \$v\$ your graph shows

$$ v = 109.4\,\text{V/cm}^2 \times \frac{1}{d^2} + 0.86\,\text{V} $$ The op-amp circuit translates the current through the photodiode into a voltage via the relationship $$ i_{\text{diode}} = 100\,\text{nA/V} \times v $$ Photodiodes produce a current based on the amount of power they receive. At low power levels this relationship is linear. The responsivity constant \$R\$ has units of amps per watt. Therefore the power received by the photodiode is:

$$ P_\text{diode} = \frac{1}{R}\times 100\,\text{nA/V} \times v $$ $$ = \frac{1}{R}\times 100\,\text{nA/V} \times \left( 109.4\,\text{V/cm}^2 \times \frac{1}{d^2} + 0.86\,\text{V} \right) $$ Intensity is power per unit area. In this experiment we are only measuring the total power received by the photodiode, so we can only talk about an average intensity which is \$P_\text{diode}\$ divided by the photodiode's surface area, \$\sigma_\text{diode}\$. This gives us:

$$ I_\text{avg} = \frac{P_\text{diode}}{\sigma_\text{diode}} = \frac{1}{\sigma_\text{diode}R}\times 100\,\text{nA/V} \times \left( 109.4\,\text{V/cm}^2 \times \frac{1}{d^2} + 0.86\,\text{V} \right) $$

You don't know the constants \$R\$ and \$\sigma_\text{diode}\$, but you can see from this formula that \$I_\text{avg}\$ is basically proportional to \$1/d^2\$.

  • \$\begingroup\$ Thank you so much fr some reason my comment didn't post earlier, this helps massively just need to understand it now. can i ask though where you got the i diode relationship from? \$\endgroup\$ Commented Jun 25, 2021 at 10:20
  • \$\begingroup\$ Which formula about the diode are you referring to? \$\endgroup\$
    – ErikR
    Commented Jun 25, 2021 at 10:33
  • \$\begingroup\$ this equation, idiode=100nA/V×v just want to read around to see how you got too this, for my own understanding \$\endgroup\$ Commented Jun 25, 2021 at 10:40
  • \$\begingroup\$ That's how the op-amp and 10M resistor work. The op-amp translates the photodiode current into a voltage via V = I*R or I = (1/R)*V. Since R = 10M ohms, 1/R = 1/(10M ohms) = 100 nA/V. \$\endgroup\$
    – ErikR
    Commented Jun 25, 2021 at 10:57
  • \$\begingroup\$ ohhhh yes that makes so much sense sorry, didn't make the connection thankyou \$\endgroup\$ Commented Jun 25, 2021 at 11:11

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.