Why does a LED burn out with 2x AA (or AAA) in parallel series, without resistor, but not with a single CR2032?
Both are practically the same voltage, right?
The led is a yellow from the Arduino Starter Kit.
Electronic – LED burnt out with 2x AA (or AAA) in series
batteriesled
Best Answer
A battery isn't just a simple source of voltage. Every battery has an internal resistance. This is what limits the amount of current the battery is capable of providing.
In a CR2032 button cell that internal resistance is so high the current is limited to a few tens of milliamps.
Conversely, with AA or AAA batteries the internal resistance is considerably lower. They are capable of providing many amps.
Running an LED from a single CR2032 is light running it from a current-limited supply. Running it from AA batteries is like running it from a non current-limited supply. A big difference.
To illustrate, consider these two circuits with rough ball-park figures:
simulate this circuit – Schematic created using CircuitLab
Assume the LED has a 2V forward voltage.
In the left hand circuit the current through the LED would be \$\frac{3-2}{100}=0.01A\$, whereas the right hand circuit would be \$\frac{3-2}{1}=1A\$
10mA is well within specification for the LED, whereas 1A is more than enough to make the LED go up in smoke.