how can i know the output power of a LED?

Thread Starter

fragrance2008

Joined Nov 18, 2009
22
The datasheet is the best way. Without it it takes some specialized equipment, or an educated guess.
I couldn't find the characteristic curves on the datasheet. It only gives out the corresponding output power at 10mA forward current. I need to know the output power in some other cases, then I can know the power receivered by a corresponding photodiode. could you give me some hints?
 

Thread Starter

fragrance2008

Joined Nov 18, 2009
22
That, and give us the specs you do have, including part numbers and where you're buying them from.
the transmitter I'm going to use is: SFH756v from Avago Technology;
and the receiver is: SFH250v.
I haven't bought them yet. I was just thinking about to choose the two devices and I am not sure if my choice is right or not.
 
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Papabravo

Joined Feb 24, 2006
21,094
The simple answer to your original question is that since you are converting electrical energy to light and some heat there will be some loss in the process. As a starting point I would hypothesize that this process was 80% efficient. Then I would compute the input power as the product of the forward voltage and the forward current. If you take 80% of that number that would be your first approximation. Now you have to scan the datasheet or find a way to measure the luminous intensity for a given power input to confirm or refute the 80% efficiency hypothesis. Rinse and repeat until you are satisfied.

The radiometric unit for light output is the lumen. It is the amount of light produced by a 1 candela point source into a solid angle of 1 steradian. The scaling factor between watts, a unit of power, and lumens is complicated by the dependence of the frequency(wavelength) content of the light. More info can be found at:

http://en.wikipedia.org/wiki/Photometry_(optics)
http://en.wikipedia.org/wiki/Luminous_flux
http://en.wikipedia.org/wiki/Power_(physics)

Is that what you were looking for?
 
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Thread Starter

fragrance2008

Joined Nov 18, 2009
22
What range do you want?
the signal I want to transmit is an AC signal around 1mV. but when this signal comes to the LED, the current through the LED may be around 25mA. But these values are only some testing ones, I have to modify the circuits to get better performance. That's why I need the output power as a function of the current, not only a few values.
 

Papabravo

Joined Feb 24, 2006
21,094
the signal I want to transmit is an AC signal around 1mV. but when this signal comes to the LED, the current through the LED may be around 25mA. But these values are only some testing ones, I have to modify the circuits to get better performance. That's why I need the output power as a function of the current, not only a few values.
Do you understand the difference between electrical power and optical power and why this quantity may be difficult to obtain?

Certainly an AC signal of 1mV (RMS or P2P?) will not light up any LED that I am familiar with. Even if you put that signal on top of a DC offset it is doubtful that you could detect the difference at the other end. An LED is basically a digital device not well suited for the transmission of analog signals.
 

Papabravo

Joined Feb 24, 2006
21,094
These devices are fiber optic transmitter/receiver pairs. The stated value of output power delivered to the fiber is between 100 and 200 μWatts at 660 nm with a fairly narrow bandwidth. You should be able to go back to the wikipedia article and find the equivalent in lumens. This efficiency is just over 1% if we assume Vf=1.85V @ 10 mA. I guess my earlier estimate was off by a considerable margin. Oh well.

I repeat, they are completely unsuitable for the transmission of analog signals.
 

Thread Starter

fragrance2008

Joined Nov 18, 2009
22
The simple answer to your original question is that since you are converting electrical energy to light and some heat there will be some loss in the process. As a starting point I would hypothesize that this process was 80% efficient. Then I would compute the input power as the product of the forward voltage and the forward current. If you take 80% of that number that would be your first approximation. Now you have to scan the datasheet or find a way to measure the luminous intensity for a given power input to confirm or refute the 80% efficiency hypothesis. Rinse and repeat until you are satisfied.

The radiometric unit for light output is the lumen. It is the amount of light produced by a 1 candela point source into a solid angle of 1 steradian. The scaling factor between watts, a unit of power, and lumens is complicated by the dependence of the frequency(wavelength) content of the light. More info can be found at:

http://en.wikipedia.org/wiki/Photometry_(optics)
http://en.wikipedia.org/wiki/Luminous_flux
http://en.wikipedia.org/wiki/Power_(physics)

Is that what you were looking for?
Yes, that guides me much. Thank you so much. Now the problem is that I could not find luminous intensity as a function of input power on the data sheet. I am so confused. I downloaded the datasheet on internet and I also checked on other datasheets, they were the same. I could not find a datasheet which contains the luminous intensity information. I don't know what's wrong. Could you give me some hints again?
 

Thread Starter

fragrance2008

Joined Nov 18, 2009
22
These devices are fiber optic transmitter/receiver pairs. The stated value of output power delivered to the fiber is between 100 and 200 μWatts at 660 nm with a fairly narrow bandwidth. You should be able to go back to the wikipedia article and find the equivalent in lumens. This efficiency is just over 1% if we assume Vf=1.85V @ 10 mA. I guess my earlier estimate was off by a considerable margin. Oh well.

I repeat, they are completely unsuitable for the transmission of analog signals.
"between 100 and 200 μWatts at 660 nm",
how do you get these values? On the datasheet, it only says 200uW@10mA.
"they are completely unsuitable for the transmission of analog",
Do you mean, the SFH756v and SFH250v not suitable for analog signals? My superviser told me they would like to detect the variation in DC source, and asked me to do some simulation about small AC signal to imitate that variation. I don't know which devices are suitable for transmit the AC signal. Could you give me some hints?
 

Papabravo

Joined Feb 24, 2006
21,094
"between 100 and 200 μWatts at 660 nm",
how do you get these values? On the datasheet, it only says 200uW@10mA.
"they are completely unsuitable for the transmission of analog",
Do you mean, the SFH756v and SFH250v not suitable for analog signals? My superviser told me they would like to detect the variation in DC source, and asked me to do some simulation about small AC signal to imitate that variation. I don't know which devices are suitable for transmit the AC signal. Could you give me some hints?
On the datasheet from Infineon was 200 μWatt @ 10 mA number and below that was the notation >100. In terms of specmanship I interpret this to mean that the mean of the normal distribution of values occurs at 200 μWatts and that the -3σ point occurs at 100 μWatts. No maximum value is specified. In the absence of better information I'd be tempted to infer that the +3σ point would be at 300 μWatts.

Fiber optic transmitters and receivers are inherently digital devices. What your supervisor wants is completely irrelevant. What he or you can actually demonstrate is the point. I don not believe that you can detect meaningful changes in intensity of a fiber optic transmitter with a fiber optic detector to recreate an analog signal waveform.

I don't think you could do it with an opto-isolator where the light source and the receiver are in the same package.

If you're looking for an optical method to pass analog signals I gotta say I don't think there is one, but I could be wrong. It should be possible to sample the analog singal, convert the sampes to a digital word, pass that down the fiber, recover the data at the other end, and use a DAC to recreate the original signal. I don't that was what he had in mind.
 
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