0
$\begingroup$

I am trying to find the minimum power that can be detected by a photodetector for which I know its noise-equivalent power (NEP).

The standard formulas are:

$$\mathrm{SNR} = \frac{\langle i_s^2(t)\rangle}{\langle i_\mathrm{th}^2\rangle},$$

where $\langle i_s^2(t)\rangle$ is the mean-square signal current and $\langle i_\mathrm{th}^2\rangle$ is the mean-square current due to the photodetector's thermal noise, which is the dominant noise at low light intensity. We can express the thermal noise:

$$\langle i_\mathrm{th}^2\rangle = \tilde{\rho}_{\mathrm{pd}}^2\cdot B \cdot \left(\mathrm{NEP}\right)^2,$$

where $\tilde{\rho}_{\mathrm{pd}}$ is the photodetector responsivity (in units A/W), $\mathrm{NEP}$ is the photodetector's noise equivalent power and $B$ is the detection bandwidth.

So far, so good. My problem comes with $\langle i_s^2(t)\rangle$. I now consider the detection of an optical signal that is the noise coming from an erbium-doped amplifier (EDFA), having a certain optical power spectral density $S$ (expressed in W/Hz). Let's consider $S$ independent of frequency to simplify. I want to know how much I can attenuate this optical signal(that is actually noise) before I am no longer able to detect it with my detector (SNR <1). I am allowed any bandwidth I want at detection. I ignore here the effect of photons(shot noise). According to the formula for the SNR, the smaller the bandwidth $B$, the higher the SNR, so in principle for any input power, I can always reduce sufficiently my bandwidth to detect the signal.

But if we reduce the bandwidth, intuitively I think that we will filter the electrical power spectral density and decrease $\langle i_s^2(t)\rangle$ as well. So it looks that the bandwidth will have no effect. This is my intuition, is it correct? and if yes, how to compute $\langle i_s^2(t)\rangle$ in function of the bandwidth ?

EDIT: According to "The Photon", I am right and $\langle i_s^2(t)\rangle$ is proportional to the bandwidth. Then I have an additional, related question: I detect such a signal with an optical spectrum analyzer (OSA) and I have a choice of sensitivity levels (normal, high-sens1, high-sens2,...). I am guessing that these higher sensitivities decrease the detection bandwidth (e.g. by averaging). Thus, when I detect my EDFA noise with the OSA, I would expect that when I choose a higher sensitivity, both the signal and the noise should go down. However, when I do it, only the noise goes down but the signal remains at the same level (I keep the resolution bandwidth at the same value). How is that possible?

$\endgroup$
1
$\begingroup$

According to the formula for the SNR, the smaller the bandwidth $B$, the higher the SNR, so in principle for any input power, I can always reduce sufficiently my bandwidth to detect the signal.

This is true if changing the bandwidth reduces $\langle i_{th}^2(t)\rangle$ more than it reduces $\langle i_s^2(t)\rangle$. This isn't an uncommon scenario in communications systems, or if you want to detect a optical source modulated with a single sinusoidal tone.

But you are trying to detect a white noise source (because you assumed $S$ is independent of frequency).

In this case when you reduce your receiver bandwidth, you will reduce your "signal" by just as much as you reduce your receiver's noise.

But if we reduce the bandwidth, intuitively I think that we will filter the electrical power spectral density and decrease $\langle i_s^2(t)\rangle$ as well. So it looks that the bandwidth will have no effect.

Agreed.

how to compute $\langle i_s^2(t)\rangle$ in function of the bandwidth ?

Because you have a white noise source, $\langle i_s^2(t)\rangle$ is proportional to bandwidth.

$\endgroup$

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.