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I've found some Do-it-yourself guides to making a 4-AA battery gadget charger, which are basically four AA batteries connected to a cut up usb cable:

http://robotification.com/2007/09/15/diy-usb-travel-charger/

http://www.instructables.com/id/Build-a-4-x-AA-USB-Altoids-Battery/

Such a device uses the AA batteries to recharge your gadget.

This looks like a useful project, but I was wondering, is this safe to use long-term, for gadgets like smartphones/android devices/mp3 players/tablets/etc.? Assuming it is connected with the right polarity, and lower-voltage rechargeable batteries are always used, it should always be around 5 volts, which is standard for all usb, but isn't there supposed to be a circuit to prevent overcharging? Is the worst case here simply not charging, or bricking the device?

Also, what would keep this from reversing? If the AAs went dead, would plugging in your phone actually drain the phone's battery to the rechargeable AA's?

Update -

A 5v regulator with red battery connected to input, red usb connected to output, both blacks connected to ground, doesn't seem to be charging when the battery is dead. I'm wondering why this recommends connecting green and white to ground. Wouldn't this make voltage return through -, d+, and d-, which is not recommended?

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The use of 4 x AA Alkaline would usually be safe BUT does exceed the USB spec and damage may occur in some cases. I have seen IC's in this role with max operating voltages of 5.5V (which is ludicrous) - you'd hope designers had more sense, but it can't be guaranteed.

While some devices may use converters between charge input and battery, many don't (probably most). A LiIon battery has max charging voltage of 4.2V so a 5V nominal USB input will usually meet this need with enough headroom for a linear regulator.

An Alkaline cell can be nearly 1.6V when fully charged - about 1.55V is common or 6.2V for 4, and up to 6.4V may be seen. There is not much energy in this initial high voltage "tail" and voltage falls to 1.5V or below very quickly.

So, you should be safe, but YMMV, alas.

A solution would be to use an LDO (low dropout voltage) regulator OR a clamp regulator which takes the peak energy out of the battery or a series diode to drop 0.4 to 0.8V (Schotky / Silicon).

  • LDO is best solution but you want as little drop as possible.

  • Clamp to drain peak battery voltage is unusual but viable. A zener could be used but is too inexact. An eg TL431 clamp regulator in a TO92 or other largish package (to get OK dissipation capability) ould do. A TL431 plus a transistor would be safer.

  • Series diode is cheap and easy but prevents full battery use. Say minimum usable battery voltage is 4.6V (may be higher). At 1.15V/cell there is still some battery capacity left. Adding a Schottky diode increases minimum battery voltage to 4.6 + 0.4 = 5V or 1.25 V/cell. Some capacity wasted. At the top end a 0.4V drop diode results in Vbattmax of say (1.55V x 4 - 0.4) = 5.8V or 1.45V/cell."Almost certainly safe".

Using NimH works but is more marginal at bottom end and safer at top end. At 4.6V, V per cell is 1.15V where NimH still has modest energy left. At top end Vmax = say 1.35V, maybe 1.4V for short periods at start. 4 x 1.4V = 5.6V. Very probably safe.

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  • \$\begingroup\$ Good info, thanks! Do you know if the white, green, AND black are supposed to be connected to ground, as described on that linked page in updated question? \$\endgroup\$ – NoBugs Aug 9 '11 at 0:59
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    \$\begingroup\$ Instructables advice is bad. Signal wires should not be used as shown for power feed. // Whether it chargs depends on what you are charging. Some phones etc try to stop you using non standard chargers. Some try harder than others. See Minty Boost page which (from memory) talks about compatible phones. Note extra contact used by some phones to detect chargers. What device are you trying to charge? \$\endgroup\$ – Russell McMahon Aug 9 '11 at 2:12
  • \$\begingroup\$ Apparently the d+ and d- need to be soldered together, as described here? bec-systems.com/site/800/how-do-modern-usb-chargers-work \$\endgroup\$ – NoBugs Aug 10 '11 at 5:32
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    \$\begingroup\$ Yes. Connecting d+ & d- signals charge system that a peripheral charger is in use. It's not obvious to me that connecting these two to ground is either required or wise. It may work OK, but ... ? \$\endgroup\$ – Russell McMahon Aug 10 '11 at 16:14
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The charging intelligence is in the phone that's being charged (or even its battery). When you charge via USB it sees just a fixed 5V, current limited at 500mA, so on that side there's no control over the charging.
The only thing that might make you frown is that the 4 AAs don't give the 5V a USB port would. Most chargers can work with voltages to at least 6V, so you should be safe.

Your phone's battery won't drain to the discharged AAs; they're not directly connected, or simply over a resistor. There's a DC-DC converter in between, which may actually charge the phone's battery from a voltage which is lower than the battery's.

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  • \$\begingroup\$ If the four batteries are NiMH batteries with 1.25 volts, would it be the same as a usb port? \$\endgroup\$ – NoBugs Jul 24 '11 at 7:38
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    \$\begingroup\$ @NoBugs - Yes, but the capacity of your phone's battery will be likely higher than the batteries, esp. if they're AAAs, and then the voltage will begin to fall long before the phone is fully charged. \$\endgroup\$ – stevenvh Jul 24 '11 at 8:00
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Take a look at the Mintyboost from Adafriut, it has a proper regulator that allows it to output a constant 5V and drain the battery completely.

Here's a writeup on its build process from Lady Ada:
http://learn.adafruit.com/minty-boost

It's using a MAX756, which only works down to 0.7V, but that's more than low enough to call 2 AA batteries completely drained.

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    \$\begingroup\$ a link would be nice. \$\endgroup\$ – stevenvh Jul 24 '11 at 9:36
  • \$\begingroup\$ @stevenvh: Here's a link for you. To everyone who was involved in this (there are a few deleted comments here), try to keep it civil - LMGTFY links aren't a good use of other people's time, and don't take things personally. \$\endgroup\$ – Kevin Vermeer Jul 24 '11 at 19:46
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The protection is the in equipment that is being charged.. Take a look at this IC (STC4054 http://www.st.com/stonline/books/pdf/docs/12666.pdf ) from ST. This is a common charger IC that guides the charging process.

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