I read statements like ‘An LED has a fixed voltage drop regardless of the current running through it’, but looking at the following examples (using arbitrary numbers), can someone confirm that the following are true/correct -
If I had a 12V battery, 500 Ohm resistor and an LED with a forward voltage of 2V, the current in the circuit would be 20mA ((12 – 2)/500). The current-voltage curve of an LED shows a small increase in current would equal a tiny increase in forwarding voltage, so for example if I wanted the current to be 40mA, the LED would have a tiny increase in voltage drop?
If I had a 3V battery, no resistor and an LED with a forwarding voltage of 2V, what happens? KVL states that the total voltage drop of the circuit must equal the power source. So in simplistic terms, does the battery just pump out as much current until the LED is forwarding 3V, and the current of the circuit is whatever the current-voltage curve of the LED says is at 3V?
If both of the above are true, then is the statement regarding LEDs having a fixed voltage drop, is it more correct to say that an LED has roughly a fixed voltage drop when a similar amount of current is running through it?
…and finally, and this may never happen or even be possible, for example a 20V battery and just an LED, apart from the LED disintegrating, would it ever be possible for the power supply to not be able to supply enough current so that the LED can reach a voltage drop of 20V, and therefore breaking KVL, what happens in that situation?
I am fairly new to this stuff, and not looking for a too technical explanation, but really just a confirmation of the above questions, and if they are wrong a simple explanation as to why that is.