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So many people have asked questions related to mine already, but mine is not quite out there. So my question is I have a 3.7V Lithium Ion battery and I need to power Arduino Uno, which requires a minimum of 7V. I found some voltage doubling circuits, either they are using an external component(like mosfets, etc) or the circuit is for AC.

Is there any simple voltage doubling or any another easier way using resistors and capacitors?

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  • \$\begingroup\$ How much current does the Arduino Uno Require? \$\endgroup\$ – Suirnder Apr 17 '13 at 14:08
  • \$\begingroup\$ Thanks Suirnder, it would be around 100 to 150 milliamps \$\endgroup\$ – Vikyboss Apr 17 '13 at 14:21
  • \$\begingroup\$ Note that you can run the ATMEGA328p chip itself from 3.7 volts, though you may need to use a slower clock. It may be possible to modify an Uno to support this, or you might want to look at a minimum circuit rather than a full Arduino. Power consumption will depend on the program - battery devices often use sleep mode to keep the average low. \$\endgroup\$ – Chris Stratton Apr 17 '13 at 14:23
  • \$\begingroup\$ I would like to work just with the microcontroller but I have a shield that needs to be plugged into Arduino Uno. That is the only reason why I needed to double it. Thanks Chris! \$\endgroup\$ – Vikyboss Apr 17 '13 at 14:28
  • \$\begingroup\$ What you are really asking is how to make 7V at 150mA from a 3.7V lithium battery, having nothing really to do with the arduino. \$\endgroup\$ – Olin Lathrop Apr 17 '13 at 14:28
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As noted by others, there is no simple way to boost a DC voltage, without an IC.

First, a short note on how the Arduino works. The Uno can use a Vin barrel jack that goes to a NCP1117 5v regulator, that needs at least 7v In for proper regulation. It can also use the VIN header pin for the same regulator. Or it can be powered via it's USB connector, where the USB 5V is almost directly connected, through a fuse and a switch (so that you don't break your arduino or computer, if usb and the Vin jack is used at the same time).

That said, you can simply use a Regulated 5v through the USB jack, instead of needing 7V, which would waste 2V * Current in wasted heat.

So you have a 3.7v Lipo battery. You will need a Boost Regulator to bring that up to a regulated 5V. You can get individual ICs, and the passive parts they need (caps, inductor, etc), or you can get ready made parts. 5V boost Regulators for batteries, with USB out, lockout protection (to prevent the LIPO from being drained so much it dies), and constant regulation are common to find. You can even get Arduino LIPO shields, or Lipo Charging and regulating boards. These can charge your Lipo from a usb port (and/or solar panels). And frankly, at about 10 dollars, is probably easier than designing and getting pcbs made. Boost regulator modules are cheaper, 4~6 bucks, and since you already have a lipo charger, it might be a better fit.

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  • \$\begingroup\$ I recently got a Lithium Ion USB power pack for around 6 bucks. It's a german name brand and rated at 2500mAh. Would the circuit out of one of those work with any lith-ion, especially a smaller one? \$\endgroup\$ – Killroy Aug 11 '17 at 8:39
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"Simple voltage doubling or any another easier way using resistors and capacitors" = NO

Resistors take energy and convert it to heat and capacitors on there own (or in conjunction with resistors) cannot convert 3.7V to anything significantly higher.

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  • \$\begingroup\$ From the following website I found that one can double AC using simple resistors and capacitors: electronics-projects-and-circuits.com/… \$\endgroup\$ – Vikyboss Apr 17 '13 at 14:26
  • \$\begingroup\$ @Vikyboss these use diodes and an AC source \$\endgroup\$ – Andy aka Apr 17 '13 at 14:30
  • \$\begingroup\$ Sorry my bad. But still I can use Diodes too but mine is a DC source. Will this work for DC? \$\endgroup\$ – Vikyboss Apr 17 '13 at 15:01
  • \$\begingroup\$ Read what it says in your link - that gives you the answer \$\endgroup\$ – Andy aka Apr 17 '13 at 15:10
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There is something called a charge pump that can boost a voltage, but at 150 mA out that is not applicable here. The capacitors required would need to be quite large and would need to tolerate significant ripple current. While that is all theoretically possible, it is not a practical solution.

What you want is called a bosst converter. That is a switching power supply that uses a inductor to make a higher output voltage than the input voltage.

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As pointed out by Chris, you just can't use resistors and capacitors. You can use LM2663M configured as a voltage doubler. As you may know, a single fully charged Lithium Ion battery cell will put out 4.2V. If you are going to need a charging circuit, LTC1734 works great.

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