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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.

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  • \$\begingroup\$ I suspect that you connected the AA cells in series, to get 3 volts, which would damage a yellow LED. A parallel connection would provide 1.5 volts, which shouldn't light a yellow LED. \$\endgroup\$ Mar 14, 2015 at 16:05
  • \$\begingroup\$ Yup, series. Edited. \$\endgroup\$ Mar 14, 2015 at 23:18
  • \$\begingroup\$ Use of the word "burn" alone is unfortunate, since it can mean "glow" or "stopped glowing due to failure". As in, "The light continues to burn", or "After five years, the light burned out". Better to say "burned out", if that's what happened. \$\endgroup\$
    – gbarry
    Mar 14, 2015 at 23:50
  • \$\begingroup\$ Ok, got it. Lost in translation! \$\endgroup\$ Mar 15, 2015 at 23:02

2 Answers 2

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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:

schematic

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.

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  • \$\begingroup\$ Besides voltage, do I only have to mind the current specification for LEDs? Or also other components like motors, ics, etc? \$\endgroup\$ Mar 15, 2015 at 23:11
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    \$\begingroup\$ LEDs fall into a fairly small category of non-linear current driven devices along with other diodes and bipolar junction transistors. With things like motors you just need to worry if you can provide enough current for them. The same with ICs. They will have an amount of current they will draw from the source, so the source has to be able to supply at least that amount. \$\endgroup\$
    – Majenko
    Mar 16, 2015 at 0:04
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CR2032 batteries have a lot of internal resistance so the current is limited. It's in the tens of ohms at least.

A fresh AA cell can supply amperes, enough to easily burn out a small LED. They still have internal resistance, just much, much less.

The above is the quick TL;DR explanation. To really understand it accurately you'd have to look at the LED voltage vs. current curve and at battery polarization effects which also particularly affect the CR2032.

https://www.dmcinfo.com/Portals/0/Blog%20Files/High%20pulse%20drain%20impact%20on%20CR2032%20coin%20cell%20battery%20capacity.pdf

According to this graph, the internal resistance starts out at about 10 ohms with 30mA draw and increases in seconds to about 15 ohms (polarization effect). Higher current draw will hasten and increase the polarization effect- the battery can only supply so much current because of the electrochemical reaction taking place. A fresh AA alkaline cell has an internal resistance of perhaps 0.2 or 0.3 ohm.

http://rohmfs.rohm.com/en/products/databook/datasheet/opto/led/chip_mono/sml-p11_p-eco.pdf

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