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I am experimenting with a simple circuit. I have an Arduino board, a proto-board and an IR sensor (that uses electricity from the Arduino board). I use a 9V battery to feed the Arduino.

My problem is that when the voltage in the battery drops to about 6V, the sensor and the Arduino board don't work anymore. I suspect that is because the voltage used for the logic (since the sensor is fed from the Arduino board (3.3V pin)), is not enough for both components.

I'd like to be able to use an electronic component to stabilize the voltage coming out from the battery while preserving the same voltage for the board and the sensor, 'till the battery dies. Would a capacitor do that for me? If not, what should I use instead?

Thanks.

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2 Answers 2

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A battery's voltage naturally drops as it discharges. To get a stable voltage out of it, you want a voltage regulator. Voltage regulators come in two flavors, linear and switching. If you want the output voltage of the regulator to ever be higher than the battery voltage, you need a switching regulator. You'll also get higher efficiency that way, so your battery should last longer.

Now, in your specific case, you really need to look at the input voltage spec for your Arduino board. If it says it can handle 6-9 volts, or something like that, then an external regulator's not going to help. The board is already doing the best it can. To make it last longer, you'd need to parallel more batteries.

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    \$\begingroup\$ Specifically you want a boost regulator if you want it to be higher than the battery voltage. \$\endgroup\$
    – NickHalden
    Commented Jan 20, 2014 at 21:43
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Bypassing the onboard linear regulator and using an external switching one that is more efficient can help BUT it comes with increased danger for the board in case you mess up the switching circuit.

Arduino boards have the option to connect a 5v voltage directly in the output side of the onboard regulator, refer to the manual of your specific board.

A buck & boost regulator would help get the most of a battery that start with a voltage higher than the arduino needs and gradually drops lower than the needed arduino voltage.

Refer to Recommendations for converting 2~7V to 5V? for actual solutions.

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