1. May 17, 2007

### jjjtttggg

Im working on designing a non contact thermal instrument. Need to estimate how much current I'll get from the photodiode.

First background: I'm working from Planck's law and Stephan's law. The temperature of my target will be between 1250 and 1750K. From Stephan's law I've worked out this will give me a total radiated power of 5.67e-8 * T in W per sq meter of surface area. or 0.138 - .532W/M^2. Now I've built a spread sheet using Planck's law to get spectral distribution. I confirmed that the total under the curves sums to match Stephan's law, so I feel good about that so far. Next I multiplied by the sensitivity curve for my photodiode in A/W vs. wavelength. That gives my a photodiode current if every bit of of radiant energy from a given area of the target landed on the PD, which of course it won't.

Here's where I'm getting confused . . .
My detector will mask off all but a 10mm^2 area of the target and the photodiode active area itself is 0.785mm^2 and will sit 400mm from the target surface, which is flat. How do I work out what portion of the total radiant energy will land on my detector? Does the surface of a black body radiate equally in all directions? Do I assume the total power from the exposed section spreads evenly over a hemisphere of radius 400mm and take the area of my detector as a fraction of the area of that hemisphere? That's the best I can come up with, but I'm not sure.

Any thoughts?

2. May 17, 2007

### Dr Transport

Think about the solid angles involved.....

3. May 17, 2007

### jambaugh

Your analysis is "good enough" for the case you stated. It will work for source sufficiently distant from the receiver.

In general you must integrate over the surface of the source and over the surface of the detector's aperture:

$$\Phi = \sigma T^4 \int_{A_s}da \int_{A_{bb}}da' \frac{1}{2\pi r^2} (-\hat{r}\cdot\hat{n} \hat{r}\cdot\hat{n}')$$
Phi is the flux and r the vector between the source da' and detector da element (hatted r is normalized) and you must take into account the angles of the normals to these surface elements with the radial between them.

If the detector is much smaller than the source is also possible to ray-trace from the detector to the source. Imagine a hemisphere of any radius centered at the detector. Take the image of the source as seen from the detector and project it onto this hemisphere. Find the percentage of the total area.

Now if the whole of this hemisphere is coated by black body radiator then the situation is that of a black body cavity and your influx is just the Stephan's law with the aperture of the detector as the radiating area at the given source temperature. Call the influx of energy in this case I.

Work out the percentage of the hemisphere which looks like the black-body radiator and you have that percentage of the influx I described in the previous paragraph.

This "ray-tracing" method is generally quicker and easier than the full blown calculus problem.

You are essentially doing the reverse ray-tracing method using the assumption that the source area is "small enough".

Hmmm.... let's see:
Let's consider a source with circular area of radius r a distance d from the detector. In your case d=400mm and r=sqrt(10/pi)mm.

The angle from the axis through detector and center of the source circle is:
$$\theta = \tan^{-1}\left \frac{r}{d}\right)$$

The area of a spherical section of radius R is:
$$A_S = 2\pi R^2(1-\cos(\theta))$$
Recalling that:
$$\cos^2(\theta) = \frac{1}{1+\tan^2(\theta)}$$
we get:
$$A_S = 2\pi R^2\left(1-\sqrt{\frac{1}{1+r^2/d^2}}\right) = A_{\pi/2}\left(1-\sqrt{\frac{1}{1+r^2/d^2}}\right)$$
where $$A_{\pi/2}=2\pi R^2$$ is the area of a hemisphere.

Let's get this in terms of the area of the source:
$$w = 1-\sqrt{\frac{1}{1+A_{src}/\pi d^2}}$$

Let's call the ratio in the denominator of the radicand:
$$\epsilon = \frac{A_{src}}{\pi d^2}$$

$$w = 1-\sqrt{1 - \frac{\epsilon}{1+\epsilon}}$$
and for small epsilon we get:
$$w \approx 1-\sqrt{1 - \epsilon} \approx 1 - (1 - \frac{1}{2}\epsilon) = \frac{1}{2}\epsilon$$

With the number you give this gives:
$$w=0.5 \epsilon = \frac{A_{src}}{2\pi d^2} = 0.9947\times 10^{-5}$$

The influx of energy is then w times the area of your detectors aperture times the Stephan factor:
$$E = (7.80854\times 10^-6 mm)\times \sigma T^4$$

Note this is the same as the answer you get with your guess. The denominator of the approximated weighting factor I worked out is the area of a hemisphere of radius d. So this ray tracing wasn't even necessary (though instructive).

Now I assumed the source was a circle. It shouldn't matter if it is some other shape.

Regards,
James Baugh

4. May 18, 2007

### jjjtttggg

Thanks Jambaugh!!!

Fantastic answer! Thanks so much. A followup question if I may . . .

Stephan law gives total energy radiated from unit area of black body. My image of this is a simple sphere of unit surface area radiating in all directions. But in my case, I'm working with a small, effectively flat, spot on a much larger body. Clearly, a flat spot can radiate only into a hemisphere. I think it is correct to assume that the spot will radiate the same total energy into the hemisphere above it, that it would radiate in all directions if wrapped around a sphere of the same surface area. It doesn't quite "feel" right, but I can't figure out why. Is it?

Thanks again,
J