# Estimating quantum efficiency of gated avalanche photodiode

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?

• 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. – jayann May 4 '14 at 17:21