Struggling with light calculations (incident power output)

In summary: A reasonable approximation is to use the following formula:1 watt/m^2 = 10 lux"So, multiplying the watts by the square meters will give you the lux.
  • #1
enjoyice
2
0
Hi guys, not sure if this is the right place. Its to do with an electronic project I've been working on. However please point me somewhere else if you know anywhere better suited! (It is quite a long post sorry)

Basically, I have created an LED array used for stimulating cells with light and I need to calculate the output power from the LEDs incident on a surface a certain distance away from the LEDs. I have grabbed some TSL235r photodiodes for this and just measure the frequency at the specific distance away and then convert that frequency to mW/m^2 and then to mw/mm^2 (which is what I need my results in).

The bit I am struggling with is the theoretical calculations. I know the LEDs strength (7200mcd, their luminous efficacy etc from the datasheet.) and I need to somehow convert this into a theoretical power at a surface a specific distance away and also calculate a theoretical required mcd to give a specific power at the surface.. (so two calculations to give me a 'spec' to search for). I've been looking online for a while but sites/guides seem to use different terms for the same thing (flux/lux/lumens/iridescence/intensity etc) and its very confusing trying to get my head around it.

It would be great if somebody could walk me through the steps for this so I actually understand what I am doing (if I don't understand it, how can I be confident in the results??)

Am I correct in thinking I can go from a required mW/mm^2 -> lux -> luminous flux -> required mcd?
And I can go from a given mcd (with angle, distance, luminous efficacy) -> luminous flux -> lux -> mW/mm^2?

I have tried these calculations but I just don't believe in the steps since some sources don't seem to allow you to convert from one to another yet others do. I also don't know what a 'sensible' number is and I was getting lux in the 6 figure range. Is that normal?

The required information:
Required power at surface: 1mW/mm^2 (so what's the required LED strength to achieve this). 7200mcd (so what is the theoretical power at the surface)

Viewing angle of LED: 23 degrees.
Diameter of surface: 11.5mm.
Distance from surface to LED: 20mm.
(Apex Angle: 33.67 degrees)
Wavelength of LED: 470nm.
Luminous Efficacy (nv): 78.

That should be enough to do the calculations I think .. but I am rather stuck and very confused. Thanks in advance! :smile:

Datasheet for LED: http://www.farnell.com/datasheets/573417.pdf
 
Science news on Phys.org
  • #2
After having a little look around I think I probably should have posted this in the homework help forum? (It's not homework but that forum seems to have more similar type questions)..

Very sorry!
 
  • #3

FAQ: Struggling with light calculations (incident power output)

1. What is the purpose of calculating incident power output for light?

Calculating the incident power output for light helps us understand the intensity of light that is being emitted from a source. This information is important for various applications such as determining the appropriate light levels for different environments or measuring the efficiency of light sources.

2. How do you calculate incident power output for light?

To calculate the incident power output, you need to know the incident power density, or the amount of light energy per unit area, and the surface area that the light is incident upon. The formula for incident power output is Power = Incident Power Density x Surface Area.

3. What are the common units for incident power output?

The most common units for incident power output are watts (W) or milliwatts (mW). The unit used will depend on the scale of the calculation and the specific application. Other units such as lumens (lm) or foot-candles (fc) may also be used, but these are not as commonly used for incident power output.

4. How does the angle of incidence affect incident power output?

The angle of incidence, or the angle at which light hits a surface, can affect the incident power output. If the angle of incidence is perpendicular to the surface, the full amount of light energy will be incident upon it. However, if the angle of incidence is not perpendicular, the amount of light energy that is incident upon the surface will be reduced and therefore, the incident power output will be lower.

5. What are the factors that can affect incident power output?

There are several factors that can affect incident power output, including the distance between the light source and the surface, the type of material the light is passing through, and the presence of any obstacles or reflections that may alter the amount of light reaching the surface. It is important to consider these factors when calculating incident power output for accurate results.

Similar threads

Replies
1
Views
1K
Replies
2
Views
4K
Replies
7
Views
3K
Replies
1
Views
2K
Replies
5
Views
4K
Back
Top