I have a photon counting system that uses a gated avalanche diode to detect single photons. The repetition frequency of the gates is $f_1$ and the temporal gate width is $\tau_1$ (so the duty cycle is $\tau_1 f_1$)

I want to get an estimate of the quantum efficiency $\eta_{\lambda}$, i.e., the probability of detecting single photons, for this diode at some wavelength $\lambda$. Note $\lambda$ is not anywhere close to the peak wavelength $\lambda_p$ of the diode, so $\eta_{\lambda}$ is expected to be small.

I have a pulsed laser system with repetition frequency $f_2 >> f_1$, pulse width $\tau_2 << \tau_1$, and center wavelength $\lambda$. I connect the output of this laser to the avalanche diode through a variable optical attenuator (VOA) already characterized at this wavelength before. I control the VOA setting so that the power impinging on the diode is $P$ and observe a probability of detection per gate $p_d$.

The main thing to note is that in this experiment the laser pulses are not synchronized to the detection gates because the laser and photon counting system do not have/cannot share the same clock. In this case:

Q1. Can I assume the diode essentially sees the laser light as a quasi-CW source? As in, is the power P more or less uniformly distributed throughout the train of gates?

Q2. If the answer to the above is yes, the mean photon number seen inside a gate should be $\mu_p = P\lambda\tau_1/\hbar c$. Can I then invert the equation $p_d = 1 - e^{-\mu_p \eta_{\lambda}}$ to obtain an estimate for $\eta_{\lambda}$ (ignoring dark counts here)?

Q3. If the above is correct, can it be said that the actual value -- found using a more precise method -- would be surely larger/smaller than the estimated one? As in, does the estimate provide lower/upper bound in this experiment?


1 Answer 1


I think it's fair to treat the incoming signal (from the diode's point of view) as a quasi-CW source with power level equal to the laser power divided by the laser's duty cycle. Now, all your measurements depend critically on your external calibration , i.e. believing that you know exactly what power level your VOA is putting out.

The one important factor that I think you've left out is that APDs have "dead time," which means that, after a photon is detetected and an avalanche pulse generated, the APD cannot generate another pulse for some small period of time. (think of it as 'recharging'). If you don't know that value, you may underestimate the quantum efficiency because some photons will be detected, i.e. generate an electron, but no pulse is produced. One way to determine that is to vary the input power and see if you ever get a pulse followed by a weaker pulse, and use that info (delay time and peaks) to estimate the dead time. If you're lucky, your APD gate width is smaller than the dead time, in which case this doesn't matter.

  • $\begingroup$ The APD is configured to apply a deadtime $\tau_d$ of a fixed no. of gates after detecting an avalanche. I could change the gating frequency, so I chose $f_1 < 1/\tau_d$. And I also pre-checked the attenuation of the VOA for a range of powers (as $\lambda$ was not the standard wavelength for that VOA) to get the offset. $\endgroup$
    – jayann
    May 4, 2014 at 17:21

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.