1
\$\begingroup\$

I have a project that involves detecting blue light (400nm-490nm) from Cherenkov radiation to find the power that is being generated by the reactor. I understand the photodiode will give me a leakage current that varies directly with light intensity, but how do I calculate power? Does it involve using the input voltage from a battery that my photodiode will use? The resistance in the photodiode circuit?

\$\endgroup\$
1
\$\begingroup\$

First, you consult the photodiode data sheet to find the quantum efficiency of the PD at the wavelength of interest.

Then, you measure the PD current. Using the standard relationship, you convert this to electrons/second. (1 amp equals 1 coulomb/sec).

Now, knowing the electrons/second impinging on the PD, you use the QE to determine the number of photons hitting the PD.

Finally, since you know the photon flux, and you know the wavelength of the photons, you can calculate the power.

If you are looking for total power, you take the PD active area and calculate the total area illuminated by the Cerenkov radiation, calculate the ratio of the two, and apply this to the PD power.

\$\endgroup\$
  • \$\begingroup\$ We're actually using voltage from the photodiode to calculate the power. \$\endgroup\$ – Macuser Nov 19 '17 at 20:51
  • \$\begingroup\$ @Macuser - Probably not. Pnotodiodes are intrinsically current devices. I suspect you mean that you are using one in photovoltaic mode, but this does not measure the photodiode voltage, despite the name. In fact, the PD voltage is held to zero for such circuits. \$\endgroup\$ – WhatRoughBeast Nov 20 '17 at 0:20
  • \$\begingroup\$ We're definitely using voltage output to find the power through a linear relationship. We've got 3 of them wired in parallel at the moment. We tested them by holding them up to the sun and got voltage readings. \$\endgroup\$ – Macuser Nov 20 '17 at 3:51
  • \$\begingroup\$ @Macuser - Then (if I understand you correctly) you are using them wrong. Please edit your question to show exactly what you are doing. I suspect that you are seeing the PD output current times the meter input resistance - that is, you are using the meter as a resistive load, and effectively measuring the voltage developed across that. Be aware that meter resistances are often not closely controlled, so doing it this way may give somewhat inaccurate results. You really ought to be using a well-characterized circuit to generate the voltage from the PD, then use the meter to measure the voltage. \$\endgroup\$ – WhatRoughBeast Nov 21 '17 at 2:05
  • \$\begingroup\$ Ah, so you're saying I should have some sort of op amp in my photodiode circuit. The reason we're using voltage is because the current (dark current) is way too small for the multimeter's sensitive to detect. \$\endgroup\$ – Macuser Nov 21 '17 at 12:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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