Welcome to the boards, tony.
You might think that eventually a heat issue might arise from the excess voltage & current but I have placed a plastic crystal (clear or colored) around some of the LEDs to contain the excess heat while testing and they did not burn out either.
Any thoughts?
You've got it--heat is the enemy here.
When you limit current, you're limiting the power dissipated in the LED (calculated easily: V * I). In typical 5mm LEDs, the heat has only one path to follow, and that's down the electrical leads. So for example if you use a piece of copper-clad board with large areas of metal on the surface, that can help conduct some of that heat away from the junction of the LED, which is the part that gets damaged. If your LED leads are in contact with the button cells, they are heatsinking for you as well.
The reason you haven't run into a catastrophe is because you're using batteries. Batteries may say "3V" on the label, but if you measure with a meter, you'll find you're not getting 6V from the two button cells under any load (without a load, you may well measure 6V or close to it, but start drawing current and the voltage sags noticeably).
Another way to look at that situation is that the button cells have
internal resistance. It's as if you have an ideal 3V voltage source in series with a resistor, but the resistance is an effect of the physical construction of the button cells. If you had a well-regulated 6V voltage supply (well-regulated means that the output voltage doesn't change much over a wide range of load currents), you'd quickly release the magic smoke from your LEDs without a limiting resistor.
So to get back to the first question:
But, practically, when do those limits actually come into play?
Every current path has
some resistance. What you're looking to do is ensure there's enough resistance in the circuit to avoid damage to your LEDs. As noted above, the damage comes from heat.
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