A soldering gun is typically 150 watts; a woodburning kit, like for writing your name on your baseball bat, is 25 watts. Compound parabolic concentrators, if pointed correctly at the sun, can theoretically achieve 11000 suns of concentration without refraction (1/sin²(0.54°)), and the solar constant is about 1kW/m². At 11MW/m², 150 watts is 14 mm² and 25 watts is 2.3 mm², and a piece of dark metal placed at that point could conduct the heat to a smaller point if there’s a heatsink there.
Suppose we want a CPC with a 2.3mm² absorber area. That’s a 1.5-millimeter square. What does it look like?
We could make it square rather than round, which should make it easier to fabricate. Its eventual opening would be 25W of 1kW/m², which is a square 15.8 centimeters on a side, but it’s quite long indeed: to achieve its theoretical ideal performance, that 15.8 centimeters actually subtends 0.54° as seen from the absorber, which is to say it’s 16.8 meters long.
However, most of that length reflects very little light. If we’re willing to accept the reflected image of the sun only filling up, say, the outer 45 degrees of our viewing angle, which should reduce the power received only by a factor of 2 or so, then we should be able to truncate the CPC at a much more reasonable height.
...but how do I calculate that height? I mean I guess I could plot points and solve it numerically...
...maybe actually a CHC or something similar would be a better way to achieve such a large concentration?
What about imaging optics, like a magnifying glass? Can you solder with a magnifying glass? A magnifying glass of 15.8 centimeters square probably can’t focus light any closer than about 15.8 centimeters focal length (aperture f/1). At that distance 0.54° gives you a 1.49-millimeter-wide image of the sun, covering 2.2 square millimeters. So yeah, a short-focal-length magnifying glass would work for that. Typical focal lengths are longer, but they wouldn’t have to be.