Canon apparently hasn't ever herd of something called USB-charging. The factory AC adapter for this battery(NB-5L) output's 4.2V's at 700mA. Most standard USB chargers output 5V at only 500mA. Would it be feasible to step up 500mA to 700mA for this application? Because it's a Lithium battery I also need some sort of IC to protect from over-charging, I haven't worked much with micro controllers, what kind would I need? I also need to add a IC that can access the batteries communication terminal to light up the LED when the battery is fully charged.

This is one of my first projects of this complexity, and I'm not sure where I should start. Thank you for any help you can provide!

  • 1
    \$\begingroup\$ Can you give us some links to your product and charger? It may seem minor but sometimes it can answer questions you did not realize we would have. \$\endgroup\$
    – Kortuk
    Commented Nov 10, 2011 at 20:20
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    \$\begingroup\$ Well, here is the charger (Canon CB-2LX): shop.usa.canon.com/webapp/wcs/stores/servlet/… And I'm sorry if I didn't make it clear in my original post, I'm not actually doing anything with this charger, but rather making my own to charge the Canon NB-5L battery from a USB power source. \$\endgroup\$
    – Matt
    Commented Nov 10, 2011 at 20:41
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    \$\begingroup\$ Someone will give a full explanation of what you are trying to do, but what you are requesting here violates conservation of power. \$\endgroup\$
    – Kortuk
    Commented Nov 10, 2011 at 21:02
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    \$\begingroup\$ @Matt: If you're trying to convert 5 V|500 mA to 4.2 V|700 mA, you can't. That violates conservation of energy. You'd be creating energy from nothing. At 95% efficiency, you could only produce 565 mA at 4.2 V. V×I will always be lower at the output of a power converter. \$\endgroup\$
    – endolith
    Commented Nov 10, 2011 at 21:16
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    \$\begingroup\$ Why not just charge the battery at 500ma? \$\endgroup\$ Commented Nov 10, 2011 at 21:30

4 Answers 4


USB supplies 5V at maximum 500mA. Power-wise this is 5 x 0.5 = 2.5W.

Your charger supplies 4.2V at 700mA. Power wise this is 4.2 * 0.7 = 2.94W.

You cannot get more power out than goes in, so you cannot charge at 4.2V @ 700mA. To increase the current to 700mA you would need to lower the voltage below 4V, which is not really an option for Li-Ion charging. The maximum current you could achieve (assuming 100% conversion efficiency) would be 2.5W / 4.2V = 0.595A. In reality due to losses it would be more like 595mA * 0.9 = 535mA.
So you might as well skip the conversion and just use the 500mA directly.

At 500mA, it will just take a bit longer to charge.

Li-Ion charging can get quite complicated, but to get to started the MCP73831 is a very simple/cheap charge control IC from Microchip, who make quite a few decent charge control ICs of varying complexity. It is selectable between 100mA and 500mA, and has a pin for LED indicator.
The MCP73837 can switch between a DC apadtor and USB, has more control over charge current, and a thermistor input.
Read both datasheets thoroughly, they give plenty of good information and example schematics.


Regarding what @Kortuk said:

  • 4.2 V x 700 mA = 2940 mW (needed)
  • 5.0 V x 500 mA = 2500 mW (from USB)

So you can't get the power you want from USB (within spec), even if you had perfect conversion.


(Been sitting in drafts).

Technical questions about charging Lithium Ion batteries have, as their first answer "If you are asking this question you probably shouldn't do it" :-). There is an element of humor there, but also a lot of truth. Refernces to "some sort of IC to protect from overcharging" suggest that a modicum of learning is required. This can be great fun BUT is a necessity if you start producing your own LiIon chargers.

A number of people seem to have assumed that you wanted to up convert 500 mA at 5V to more current at less voltage - but I do not read it that way.
I assume that you are trying to use a USB port to charge and want as much current as possible - say 750 mA. The following addresesses that perspective.


A USB connection does not "output" current - it outputs Voltage, and current then flows depending on the load. This is an important and non trivial distinction even though it may not appear so.

The maximum current which CAN be drawn from a USB outlet depends on both its physical capabilities and the current which MAY or SHOULD be drawn depends both on the design of the port and the ability of your equipment to "negotiate" with that port.

So you have CAN, and two flavors of MAY.
Regarding "MAY" - Recent revisions to the USB standard allow up to 750 mA to be drawn from a USB port, but this capablity is not manadatory for the supplier to provide, and there is now a spec for chargers only, which can provide as much as 5 Amps !

This is subject to negotiation with the port

Here is a somewhat informal but practially useful discussion of using USB as a power source


The essentially trivial matter of drawing current from a nominally 5 Volt USB power supply is complicated by a large range of practical issues.These are summarised reasonably well in the following text. There is nothing technically hard here - just lots of little issues bundled together to help confuse.

See also this excellent related prior stack exchange discussion.

The following, relating to USB Power-Supply issues was copied directly from
The Wikipedia USB page

  • The USB 1.x and 2.0 specifications provide a 5 V supply on a single wire from which connected USB devices may draw power. The specification provides for
    no more than 5.25 V and
    no less than 4.75 V (5 V±5%) between the positive and negative bus power lines.

    • For USB 3.0, the voltage supplied by
      low-powered hub ports is 4.45–5.25 V.[36]

    • A unit load is defined as
      100 mA in USB 2.0, and 150 mA in USB 3.0.
      A device may draw a maximum of
      5 unit loads (500 mA) from a port in USB 2.0;
      6 unit loads (900 mA) in USB 3.0.

    • There are two types of deviceslow-power and high-power:

    A low-power device draws at most 1 unit load, with minimum operating voltage of
    4.4 V in USB 2.0, and
    4 V in USB 3.0.

    A high-power device draws the maximum number of unit loads permitted by the standard.
    Every device functions initially as low-power
    but the device may request high-power
    and will get it if the power is available on the providing bus.[37]

    • Some devices, such as high-speed external disk drives, require more than 500 mA of current[38]
      and therefore cannot be powered from one USB 2.0 port.
      Such devices usually come with Y-shaped cable that has two USB connectors to be plugged into a computer. With such a cable, a device can draw power from two USB ports simultaneously.[39]

    • A bus-powered hub initializes itself at 1 unit load and transitions to maximum unit loads after it completes hub configuration.
      Any device connected to the hub will draw 1 unit load regardless of the current draw of devices connected to other ports of the hub
      (i.e. one device connected on a four-port hub will draw only 1 unit load despite the fact that more unit loads are being supplied to the hub).[37]

    • A self-powered hub will supply maximum supported unit loads to any device connected to it.
      In addition, the VBUS will present 1 unit load upstream for communication if parts of the Hub are powered down.[37]

    • In Battery Charging Specification,[40]
      new powering modes are added to the USB specification.
      A host or hub Charging Downstream Port can supply
      a maximum of 1.5 A when communicating at low-bandwidth
      or full-bandwidth,
      a maximum of 900 mA when communicating at high-bandwidth,
      and as much current as the connector will safely handle when no communication is taking place;

    • USB 2.0 standard-A connectors are rated at 1.5 A by default.
      A Dedicated Charging Port can supply a maximum of 1.8 A of current at 5.25 V.
      A portable device can draw up to 1.8 A from a Dedicated Charging Port. *** The Dedicated Charging Port shorts the D+ and D- pins with a resistance of at most 200 Ω.
      The short disables data transfer, but allows devices to detect the Dedicated Charging Port and allows very simple, high current chargers to be manufactured.
      The increased current (faster, 9 W charging) happens if both the host/hub and devices implement the new charging specification.

*** DCP now seems to allow up to 5 Amps.

This would happily destroy any standard USB connector.

See Battery charging V1.2 spec and adopters agreement DECEMBER 7TH, 2010.

Dedicated charging port

ie no data - just charging MUST operate in dark grey area.
MAY operate in light grey area.

NB note that requirement for DCP is LESS stringent than CDP = Charging Diwnstream Port = data equipped downstream port,
which must maintain Voltage above 4.75V out to 1.5A.

enter image description here

Charging downstream port:

ie has data and charging capability.
MUST operate in dark grey area.
May operate in light grey area.
MUST exit light grey area via right hand side only - ie while providing between 4.75V and 5.25V current drain must increase through 1.5A.
After that any area in light grey is valid.

enter image description here


It's a 3.7v Lithium Ion battery, single cell by the looks of it. You can definitely build a circuit to charge a single cell li-ion battery from USB power.

Li-ion batteries require careful charging. Your supply would be a current limited voltage source. To reach a full charge you need two stages first you charge at max current till cell voltage threshold. At this point the battery is not fully charged, maybe 80%. You then charge at the max cell voltage till your current drops to threshold. Obviously quite complicated and you need temperature sensing for maximum safety.

As others have mentioned you'll only have 500mA from a computer USB 2.0 port. USB 3.0 supports up to 900mA but I'm not sure what conditions need to be met for that. Otherwise some USB charging wall warts can supply more current.

So basically you'll need to accept 500mA as you baseline. The lower current means slower charging but is at least going to be within spec for your battery.

There's a few chips out there that will make things easier to build a charge circuit. They manage the constant current and temperature monitoring themselves. Something like a BQ2000T.

If you'd like to learn all about the complexities of charging a li-ion battery (or any battery actually) check out this site.

  • \$\begingroup\$ I would actually call it a current-limited voltage-source for charging. You limit your current to a safe current until you reach around 90% charge and then it slowly trickle charges until you reach 100%. If you would like to know more about current limits on usb there is a question for that. \$\endgroup\$
    – Kortuk
    Commented Nov 10, 2011 at 23:52
  • 1
    \$\begingroup\$ Thanks Kortuk! I just realised I was confusing li-ion with nimh methods! \$\endgroup\$
    – snoopen
    Commented Nov 11, 2011 at 0:25

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