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Is there a mathematical relationship between the optical power transmitted by a light source (LED) and the optical power received by the photo-diode and its transformation into electrical power.

Lets assume we transmit with 20mw optical power, and we receive 10mw in optical power at the photo-diode. What is this 10mw equivalent to in terms of electrical power.

I was intimidated with this question because I read a paper saying:

X denotes the electrical signal-to-noise ratio (which is proportional to the square of the received optical power Pr)

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  • $\begingroup$ Can you provide the citation for the paper? I would guess that the electrical signal is proportional to the optical power. For the signal to noise ratio to go like the power squared then the noise must be inversely proportional to the power. The following link says that generally the noise improves with power (rp-photonics.com/signal_to_noise_ratio.html) which is suggestive of that type of relationship. $\endgroup$ – LasersMatter Sep 7 '16 at 13:53
  • $\begingroup$ I would try looking at the section in these notes on SNR and thermal noise. course.ee.ust.hk/elec509/notes/…. $\endgroup$ – LasersMatter Sep 7 '16 at 14:02
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The photodetectors used in optical communications are square law detectors. The electrical power they output is proportional to the square of the optical power input.

This is because they generally generate one charge carrier for each photon received. Thus the electrical current is proportional to the optical power. And the electrical power, being proportional to the square of current, is proportional to the square of the optical power.

There are some variations. In any detector, not every incident photon generates a carrier, so the quantum efficiency must be taken into account. In a APD or photomultiplier, the generated carrier triggers an electrical gain mechanism, resulting in an effective response greater than one carrier per photon.

Lets assume we transmit with 20mw optical power, and we receive 10mw in optical power at the photo-diode. What is this 10mw equivalent to in terms of electrical power.

You need to multiply by the responsivity of the detector to get the photocurrent.

$$I = P_o R$$

where $P_o$ is the incident optical power.

The responsivity is given by

$$R = \frac{\eta q}{h \nu}$$

where $\eta$ is the quantum efficiency of the device, $q$ is the electron charge, $h$ is Planck's constant, and $\nu$ is the frequency of the incident light. Generally, $\eta$ will be a frequency-dependent parameter.

To get the electrical power generated, you need to multiply the photocurrent by the potential difference across the detector.

Generally, the electrical power is not a primary concern when analyzing receivers in telecommunications (although it is fundamental to the definition of signal-to-noise ratio). However, solar cells work on the same principle, and there electrical power output is of course the primary goal.

X denotes the electrical signal-to-noise ratio (which is proportional to the square of the received optical power Pr)

This is true when the dominant source of noise in the receiver is from the receiver amplifier (so the noise term is constant and only the signal term varies), which is a very common, but not universal, situation.

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  • $\begingroup$ can we infer the optical power if we know the Tx current that was used to drive the light source? $\endgroup$ – Kristof Pal Sep 9 '16 at 10:27
  • $\begingroup$ @KristofPal No. For example, if we're talking about LEDs, the efficiency vary depending on the temperature of the device. $\endgroup$ – The Photon Sep 9 '16 at 15:14
  • $\begingroup$ why the electrical SNR == optical SNR^2. Is this related to the fact that the path loss coefficient of visible light being 4 instead of 2? $\endgroup$ – Kristof Pal Sep 12 '16 at 13:37
  • $\begingroup$ @KristofPal, There is no such rule, since generally the receiver noise dominates the electrical SNR and it isn't present in the optical signal, so there's no reason there should be a connection between electrical SNR and optical SNR. If some other noise source dominates, or you want to relate the receiver noise back to an equivalent optical noise, then the rule would hold, for the reasons outlined in my answer. $\endgroup$ – The Photon Sep 12 '16 at 15:38
  • $\begingroup$ the responsivity of the PD is given over wavelength. For an LED emitting white, what would be the responsivity? The average responsivity across all wavelengths? $\endgroup$ – Kristof Pal Sep 13 '16 at 11:01

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