Spectral Irradiance Calculation

[Itay Segal]

TL;DR Summary

How to calculate spectral irradiance [µW/cm^2/nm] from irradiance [µW/cm^2]

Hi everybody,
I'm a mechanical engineer working on a phototherapy project, which includes choosing a LED light source that emits a spectral irradiance of 30 µW/cm^2/nm.

however, most of the LEDs datasheets giving you the lumens at best.

i was able to calculate the irradiance [µW/cm^2] from lumens, but i have no idea how to calculate the spectral irradiance [µW/cm^2/nm] from it.

it's important to mention that I'm aiming for a wavelength of 460nm, and trying to narrow the bandwidth as much as i can (to around 450-470nm). not sure how and if it impacts the calculation.

would appreciate the help!

 

[tech99]

According to Wiki you have to divide the irradiance by the wavelength in nm to obtain spectral irradiance.

 

[Itay Segal]

According to Wiki you have to divide the irradiance by the wavelength in nm to obtain spectral irradiance.

Thanka for the answer.

That what i was thinking and found on some websites, but in others they talked about an integral on the band of wavelengths (like from 450 to 470).
Any solid reference link / example solution out there?

 

[sophiecentaur]

Summary:: How to calculate spectral irradiance [µW/cm^2/nm] from irradiance [µW/cm^2]

but i have no idea how to calculate the spectral irradiance [µW/cm^2/nm] from it

It depends on the bandwidth of the filtering (or perhaps the sensitivity curve of the 'detector'). If you are given the lumens (for a domestic lamp), that will imply the total light flux over the visible range (assuming a black body curve and a specified temperature. The spectral irradiance would be related to the fraction of the visible range that your defining filter admits.
If they give you the 450-470nm value then the integral over that range should be divided by the bandwidth of the filter.(Assuming the spectrum is flat, of course; if it's not then you'd need to some compensation, using the spectral plot)
This is a noddy approach but you can improve on it if it's too simplistic.

 

[Itay Segal]

It depends on the bandwidth of the filtering (or perhaps the sensitivity curve of the 'detector'). If you are given the lumens (for a domestic lamp), that will imply the total light flux over the visible range (assuming a black body curve and a specified temperature. The spectral irradiance would be related to the fraction of the visible range that your defining filter admits.
If they give you the 450-470nm value then the integral over that range should be divided by the bandwidth of the filter.(Assuming the spectrum is flat, of course; if it's not then you'd need to some compensation, using the spectral plot)
This is a noddy approach but you can improve on it if it's too simplistic.

Thanks for the response.

this is the response graph of a measuring device I'm planning on using:

if my LED gives me 450-470nm wavelength range,
you mean i need to do an integral between 450 & 470, then divide it by the response bandwidth (50)?
and btw, is the integral on a constant of the irradiance? or is there some form of equation i need to be aware of?

 

[Itay Segal]

also, if this is my response curve, what is the divide value?

 

[sophiecentaur]

and btw, is the integral on a constant of the irradiance? or is there some form of equation i need to be aware of?

So the LED spectrum is flat, which makes things easy; the power flux getting to the detector is source power / (475-425) W/metre. For a straightforward band pass filter, the usual thing is to use the half power bandwidth as the equivalent rectangular filter. The parts of the curve outside the bandwidth are more or less equal to the parts of the curve inside the bandwidth. For a simple band pass filter, that's what's done in RF. The receive filter width between half power points is about (500-445) - or whatever you think is nearer.
As you have the curves, all you need to do is to apply the above, using the two bandwidths that you have; just simple proportions will give you a result that can be compared with other sources and other receive filters. I guess the only reason to get more complicated than that is if the results from your experiments seem to disagree with what other people get or just look 'plain wrong'.
In an idle moment you could digitise the data for both curves at n wavelengths (say 10 points) λ(n)) and calculate the sum of S_λ(n) . D_λ(n).
That result should be much the same as using the rectangular response assumption. OTOH you could just go along with usual practice - for the first time round at least.

 

[Itay Segal]

So the LED spectrum is flat, which makes things easy; the power flux getting to the detector is source power / (475-425) W/metre. For a straightforward band pass filter, the usual thing is to use the half power bandwidth as the equivalent rectangular filter. The parts of the curve outside the bandwidth are more or less equal to the parts of the curve inside the bandwidth. For a simple band pass filter, that's what's done in RF. The receive filter width between half power points is about (500-445) - or whatever you think is nearer.
As you have the curves, all you need to do is to apply the above, using the two bandwidths that you have; just simple proportions will give you a result that can be compared with other sources and other receive filters. I guess the only reason to get more complicated than that is if the results from your experiments seem to disagree with what other people get or just look 'plain wrong'.
In an idle moment you could digitise the data for both curves at n wavelengths (say 10 points) λ(n)) and calculate the sum of S_λ(n) . D_λ(n).
That result should be much the same as using the rectangular response assumption. OTOH you could just go along with usual practice - for the first time round at least.

Thanks!

i still did not fully understand the calculation needed.

lets say i have a LED with irradiance of 5 µW/cm^2 & wavelength of 450-470nm, and a measurement device with a response band width of 450-500nm.

if i understand it correctly:
to calculate the spectral irradiance, i should do an integral between 450-470 [receiving 5*(470-450)=100µW*nm/cm^2], then divide it by (500-450)=50nm? it gives me 2µW/cm^2 which cancels the "nm" unit.

where do i understand it wrong?

 

[sophiecentaur]

try drawing a diagram with both curves on the same axes. Your example seems to show a receive filter which admits all the source spectrum. How will you get less power through?

 

[Itay Segal]

try drawing a diagram with both curves on the same axes. Your example seems to show a receive filter which admits all the source spectrum. How will you get less power through?

Perhaps one of us misunderstood the other one.
We are talking about a measuring device, yes? Not a filter. I want to measure the spectral irradiance, and validate it with calculation.

 

[hutchphd]

What exactlly is fig4?? Percent transmission through what??

 

[Tom.G]

From: https://www.hkcalibrations.com.au/bili-meter

Measuring Irradiance​

The Bili-Meter is a spectroradiometer that measures the therapeutic irradiance (radiant power) of neonatal phototherapy lights. It measures the irradiance of the wavelengths from 425 to 475 nanometers (nm), the blue-green portion of the spectrum, which includes the principal action spectrum of bilirubin.

Units of Measurement​

The Bili-Meter measures irradiance in units of microwatts per square centimeter per nanometer (pW/cm2/nm). A nanometer is a inmsure of wavelength equal to one-billionth of a centimeter. The term "per nanomcter" indicates the average irradiance per nanometer across the spectral hand being measured, which is 50 nm wide. This makes it possible to compare average irradiance across spectral bands of different widths.

Cheers,
Tom

 

[Itay Segal]

What exactlly is fig4?? Percent transmission through what??

through a measuring device. it's a response graph.

Cheers,
Tom

Thanks,
i actually found it before and even presented the graph from there. it's one of the measuring devices i consider using.
from their description I'm still not sure of the exact calculation needed.
i would be happy if someone gives me an example with units and all.

 

[sophiecentaur]

Perhaps one of us misunderstood the other one.
We are talking about a measuring device, yes? Not a filter. I want to measure the spectral irradiance, and validate it with calculation.

Hence my request for a diagram and advice to draw graphs.
Areceiving device has an implied filter ( its sensitivity curve) followed by an implied broad band detector.

 

[Itay Segal]

Hence my request for a diagram and advice to draw graphs.
Areceiving device has an implied filter ( its sensitivity curve) followed by an implied broad band detector.

in that case i agree that the receive filter admits all of the source spectrum, but shouldn't it be that way? i do want to measure it, not to filter it. I'm not aiming to receive less power while measuring.

further more, I'm aiming to perform the measurements in a dark room, with only the blue LEDs turned on. i mean, I'm not even trying to filter daylight.

i guess that's why I'm not fully understand what your'e trying to say.

[and again - thanks for the help and detailed explanations! not my field of work usually :) ]

 

[sophiecentaur]

A diagram would have avoided the misunderstanding. I made assumptions . What’s the Bilimeter response about? I asked are you trying to improve on it in some way? Do tell.

 

[Itay Segal]

A diagram would have avoided the misunderstanding. I made assumptions . What’s the Bilimeter response about? I asked are you trying to improve on it in some way? Do tell.

well i need LEDs at a specific wavelength to treat jaundice, which requires spectral irradiance over 30 µW/cm^2/nm.

for choosing the right LED at the right distance, i want to be able to calculate the spectral irradiance before purchasing it.

after purchasing the LED, i will measure it, to double check the theory.

however, I'm not sure how to calculate it. i do know how to calculate the irradiance [µW/cm^2], but not the spectral irradiance [µW/cm^2/nm].

that's the fully story more or less.

 

[hutchphd]

well i need LEDs at a specific wavelength to treat jaundice, which requires spectral irradiance over 30 µW/cm^2/nm.

The manufacturer of the LED can probably provide you with a spectral output curve (spectral irradiance vs wavelength). To provide more than a particular irradiance at a particular wavelength the line needs to be above the target point. Thats all.
All LEDs of a particular color (not white ones!) rely on a particular bandgap and produce nearly identical spectral irradiance except for total integrated power output. So if necessary check other manufacturers for the graph and scale the output to match your target LED for which you have done the other conversions.
This is really not difficult, but it is confusing until you get used to it . The mix of photometric and radiometric units is a pain in the patoots.

 

[Itay Segal]

The manufacturer of the LED can probably provide you with a spectral output curve (spectral irradiance vs wavelength). To provide more than a particular irradiance at a particular wavelength the line needs to be above the target point. Thats all.
All LEDs of a particular color (not white ones!) rely on a particular bandgap and produce nearly identical spectral irradiance except for total integrated power output. So if necessary check other manufacturers for the graph and scale the output to match your target LED for which you have done the other conversions.
This is really not difficult, but it is confusing until you get used to it . The mix of photometric and radiometric units is a pain in the patoots.

those units sure are a pain to deal with *sigh*

it's hard to believe that LED manufacturers provides spectral irradiance vs wavelengths curves, or any spectral irradiance related data, as those are heavily related on the application (put aside the fact that i didn't find any of this data at any LED datasheet).
i mean yes, they rely on wavelength, beam angle and luminous flux (which are all provided by the manufacturer), but also rely on the distance from the light source and (as i recently learned) the receiving response spectrum.

 

[hutchphd]

For instance LUMILEDS. What LED are you looking at?

 

[Itay Segal]

For instance LUMILEDS. What LED are you looking at?

i didn't find anything regarding spectral irradiance in the datasheet you sent.
are you sure? any specific page number?

 

[Tom.G]

if i understand it correctly:
to calculate the spectral irradiance, i should do an integral between 450-470 [receiving 5*(470-450)=100µW*nm/cm^2], then divide it by (500-450)=50nm? it gives me 2µW/cm^2 which cancels the "nm" unit.

where do i understand it wrong?

The photosensor already does the integration for you. You put a bandpass filter in front of the photosensor to restrict the wavelengths you want to measure.

If the photosensor does not have a flat response over the desired wavelengths, you will have to account for that. Probably 10 points across your filter spectral width would be plenty.

If the filter passband is not flat, do the same correction as for the photosensor.

Then divide the corrected reading by the width of the bandpass filter, 50 in your case because the filter passband is 50nm wide. See the bolded sentence in post #12 above.

Be sure to account for the transmittance of the filter, it will reduce the amplitude to some extent.

I'll leave the unit conversions to others here.

Cheers,
Tom

 

[sophiecentaur]

The photosensor already does the integration for you. You put a bandpass filter in front of the photosensor to restrict the wavelengths you want to measure.

If the photosensor does not have a flat response over the desired wavelengths, you will have to account for that. Probably 10 points across your filter spectral width would be plenty.

If the filter passband is not flat, do the same correction as for the photosensor.

Then divide the corrected reading by the width of the bandpass filter, 50 in your case because the filter passband is 50nm wide. See the bolded sentence in post #12 above.

Be sure to account for the transmittance of the filter, it will reduce the amplitude to some extent.

I'll leave the unit conversions to others here.

Cheers,
Tom

What you are implying is that the OP needs to calibrate his / her system. That makes good sense except it does open a can of more worms to deal with (he's a Mechanical Engineer).
Something that needs to be addressed is the required accuracy of spectral irradiance. It may be useful to look for data on the sensor and work things backwards from that. Use a broad band source on the sensor and the spec of the sensor should tell you the incident flux. Replacing with the required filter should tell you the flux in its bandwidth. A broadband source would give a high signal level and a nice high sensor reading. There are plenty of reasonably priced filters for astrophotography with well defined wide passbands. Then the ratio of outputs would give the performance of the 'receive' filter / sensor.

 

[hutchphd]

i didn't find anything regarding spectral irradiance in the datasheet you sent.
are you sure? any specific page number?

Page 6 Fig 1a

If you don't mind possibly missing the flux by a factor of two then the above suggestions are perfectly adequate and correct. Just take the irradiance and divide it by the full width half max of the spectrum. For the blue LEDs FWHM=~30nm.

 

[Itay Segal]

Page 6 Fig 1a

If you don't mind possibly missing the flux by a factor of two then the above suggestions are perfectly adequate and correct. Just take the irradiance and divide it by the full width half max of the spectrum. For the blue LEDs FWHM=~30nm.

mm so if i get you correctly, i just calculate the irradiance [µW/cm^2] and divide it by the FWHM, which will give me the spectral irradiance [µW/cm^2/nm]?

and to define the FWHM, i should look for a normalized power vs wavelength graph?

 

[sophiecentaur]

a normalized power vs wavelength graph?

If the spectrum is vaguely flat until the edges, you can treat it as a rectangle between the half power points; it's what transmitter Engineers do whenever they can. Compare the answer you get that way with what you get if you turn it into a bar chart and work out the areas of the rectangles. I think you'll find the difference is much less than other possible errors.

 

[Itay Segal]

If the spectrum is vaguely flat until the edges, you can treat it as a rectangle between the half power points; it's what transmitter Engineers do whenever they can. Compare the answer you get that way with what you get if you turn it into a bar chart and work out the areas of the rectangles. I think you'll find the difference is much less than other possible errors.

thanks, i think i understand what to do now.
i will run some tests to see the similarities between the theory and the measurements.

thanks again everybody :)