I would like to power my Arduino Yun board with the following battery:

6V battery

Arduino Yun has a VIN pin that represents the input voltage to the Arduino board. The official documentation says:

Unlike other Arduino boards, if you are going to provide power to the board through this pin (VIN), you must provide a regulated 5V.

What I have to make (exactly)?

Note: I have some diode, resistor and other basic component...

  1. Your best option is to get a separate 5V regulator (Linear or SMPS will work) to power the board from your battery. Digikey, Mouser or many other component distributors supply power regulators. A linear regulator would probably be the easiest to use. Generally there's an input pin, an output pin and a ground. You put the positive battery terminal on the input, the output pin to Vin and ground to ground on the Yun board and to the negative terminal of the battery. The regulator should be a 5V regulator obviously.

  2. If this is somehow too costly, or you're under strict time constraints, you could probably get away with putting 2 high current diodes in series with the battery to Vin. This would drop approximately 1.4 volts from your battery voltage and hopefully get you in the range of 5 to 5.5 volts. 5.5 is the maximum rated voltage for your Atmega chip on there. Keep in mind this is a hack and should be done with caution. I do not recommend this method unless you're willing to potentially fry your board.

  • \$\begingroup\$ Thanks. For now I have to use the second solution but how should I connect diodes (note: my diodes are 1N4007) to the battery? In what way and on what (the positive or negative) side of the battery I have to connect those? \$\endgroup\$ – user502052 Apr 17 '14 at 17:34
  • \$\begingroup\$ For starters if you want to try #2, take two diodes with their stripe pointing away from positive terminal and place them in series. Add a resistor (1k ohms maybe) from them to the negative terminal. Then measure the voltage over the resistor and make sure the voltage is between 5-5.5V. The setup should look like this: Bat+ > Diode| > Diode| > resistor > Bat- Where the arrow is a wire connection and the vertical bar represents where the stripe on the diode should be facing. \$\endgroup\$ – horta Apr 17 '14 at 17:42
  • \$\begingroup\$ Then once you've confirmed that ~5 volts is dropping across the 1k resistor, replace the 1K resistor with a connection to Vin and connect the negative terminal of the battery to ground on your board. \$\endgroup\$ – horta Apr 17 '14 at 17:43
  • \$\begingroup\$ You can add a switch to shunt 1 of 2 diodes when capacity reduces battery voltage <6V. Two 3V LEDs in series with say 150 Ohms can monitor battery voltage by brightness when on and only consume 10mA max@ 6.5V and then be very dim at 6V and just faint when battery is near 5.5V . Use high current power diodes for lowest drop. Or multiple 1N400x in parallel like 2S2P or 2S3P \$\endgroup\$ – user38637 Apr 17 '14 at 18:13

One needs to know the max load current and max voltage from a charger if used in situ.

Then decide how much you can afford, Beware of drop voltage vs current on many 3 pin LDO's may be excessive for >1A. This link filters out solutions for high current @ low drop. So it depends on your application.

This result offers good regulation with low drop when battery is @5.5V


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