I have always been told that LEDs need a resistor in series when powered from a battery.
If this is true, how is it possible to power an LED from a coin cell battery without a resistor? I have tried this, and it works without burning out the LED.
Electronic – Why can a coin cell power an LED without a resistor
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Related Solutions
Coin cell batteries generally have a voltage of 3V. Your Gemma takes in 5V. Even though your LEDs are powering on, your Gemma doesn't have enough voltage to actually run.
The voltage regulator doesn't take whatever input you give it and make the proper output, it needs either a lower OR a higher voltage. In this case, and this is the most often case, you'll need a higher voltage than what your voltage regulator will output
I would caution you regarding putting 2 CR2032 cells together because this will result in 6V. If the Gemma has a regulator, then that will be safe but your leds might not have the proper resistor for a 6V source.
This should work: -
Rset and Vset determine the current that flows through the load resistor: -
\$I_{LOAD} = \dfrac{V_{SET}}{R_{SET}}\$
However, the op-amp needs to be able to work down to 1.8 volts or lower and will need power connections (not shown above) to V and ground. It also must have rail-to-rail inputs and also preferably outputs.
If Vset is made to be 0.15 volts and Iset is required to be 15mA then Rset is 10 ohms.
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Best Answer
No battery is an ideal voltage source. This is easy to demonstrate: measure the voltage of a coin cell battery with a voltmeter. For your typical lithium coin cell, it should be around 3V. Now connect the LED to it, and measure the voltage again. The voltage should be around the forward voltage of the LED.
This works because real batteries have some internal resistance. A fresh coin cell might be approximately modelled by adding a series resistor. With an LED connected, it might look something like this:
simulate this circuit – Schematic created using CircuitLab
It's the internal resistance of the battery, R1 here, which limits the current. Just like a discrete resistor, this internal resistance drops a voltage proportional to the current through it according to Ohm's law. It also makes the battery warm.
The internal resistance also increases significantly as the battery discharges, and this is why a fresh alkaline battery might measure 1.65V, whereas a dead one might still measure 1.4V, but is unable to provide enough current to work as intended.
There's another way to demonstrate a battery's internal resistance: short a coin cell's terminals with a wire and note that it does not explode. The internal resistance of the battery limits the current through the wire (which has nearly zero resistance), and thus, limits the rate at which electrical energy is converted to heat.
Now try the same experiment with a lithium ion battery (which has a very low internal resistance) and a heavy gauge wire (which also has a very low resistance) and you will create a sizable explosion, or at least a fire.
On second thought, don't try that second experiment.