3 ms·
Fixed voltage drop on an LED is too much of a simplification for this application. A better mental model is to imagine a piecewise linear approximation of curre
by sobriquet9 5y ago
Fixed voltage drop on an LED is too much of a simplification for this application. A better mental model is to imagine a piecewise linear approximation of current–voltage characteristic aka I–V curve.
I-V curve of a resistor is a straight line that goes through (0,0) point, with slope inversely proportional to resistance. The curve for LED has a bend somewhere between 1.5V and 2V, where its slope changes from low to high.
- compiler-guy 5y agoAnd yet it is exactly how many hobbyists approach the topic, and quite successfully. Not that he does hint at these curves when he talks about how running it at lower current will make it less bright.
- sobriquet9 5y agoMy problem with the way the topic is being presented here is not just that the mental model is too simple, but that it is so much simplified that it becomes internally inconsistent. Resistance is voltage over current, here's LED, it does not have resistance. Or maybe it has more than one resistance, depending on voltage. Battery is a voltage source, wait, no it's not, it has internal resistance.
- compiler-guy 5y agoEveryone reading that article knows it is heavily simplified.