I've finally managed to set up a more efficient portable charger that uses 8 AA alkaline batteries(actually not so efficient). Schematic:


simulate this circuit – Schematic created using CircuitLab

There are two switches in the schematic:CH1 and CH2.When all the alkaline batteries are fully charged,only CH1 is closed.That will make the regulator dissipate less power.After the batteries are discharged enough(until each one of the batteries reaches ~1,15V) so that togheter they output <7 volts,CH1 is opened and CH2 is closed to continue charging(until the top 2 batteries each reach 1,2V and the rest reach 0,8V(per battery),shutting down the voltage regulator).

I'm not really sure in what package is the regulator,but it has a heatsink attached and I attached another bigger improvised heatsink.It's not heating up too much.

I am sadly using batteries from different brands and I don't think they have the same age(I used some more than the others),but I am constantly keeping an eye out on them.

According to the datasheet,it should output 5V and 1A.

The role of this pack is to charge a dead phone battery for a bit so you can solve an emergency(call someone).

I already tested it on another phone(and it worked great) and on another device which showed some unexpected behaviour(the battery charge level flipping back and forth between fully discharged and a bit charged,but it had some problems before testing the circuit on it).

The original charger of the smartphone I wish to connect it to has an output of 5 V and 1,3 A

To conclude:
Can I freely use the charger with my smartphone without causing damage(Would you use it on your phone?)

EDIT:I understand that the two-switches-activated-simultaneously problem and the differences between the batteries caught your eye,but I would like an answer more focused on the current and voltage supplied to the smartphone,as well as the way the regulator outputs power(is it ok if it's continous?should it be pulsed so it won't cause any damage?)

  • 5
    \$\begingroup\$ Bad idea to have batteries of different capacity / charge in series, which happens once some are somewhat discharged and others aren't. Also it is easy to short the top two batteries, which may cause violent self destruction of the cells. I wouldn't use it on my phone. \$\endgroup\$ – jippie Mar 6 '16 at 20:28
  • 6
    \$\begingroup\$ This goal should really be accomplished with a switching regulator. In practice many emergency charges use a switching boost converter run from a smaller number of cells. Questions which ask people to enumerate points are not really a fit for stack exchange sites. \$\endgroup\$ – Chris Stratton Mar 6 '16 at 20:43
  • 1
    \$\begingroup\$ It would also be a good idea to guard against both switches being "ON" at the same time... \$\endgroup\$ – Brian Drummond Mar 6 '16 at 21:59
  • 3
    \$\begingroup\$ This won't damage your phone, but it is a rather silly design. As others have mentioned, get a boost converter instead of that ancient 7805. Most designs these days only require a few external passives. If you're looking for recommendations, search/start a new question. \$\endgroup\$ – Jay Carlson Mar 7 '16 at 8:06
  • 4
    \$\begingroup\$ If your going to use 6~8 batteries for this, you may as well just buy a car usb charger and wire the batteries to it. A linear regulator will be so much more inefficient. At 500mA, you are drawing 6 watts at 12V, only 2.5 watts goes to the phone. That's 40% efficiency, 60% wasted power. With a car charger, typically a MC34063 or better, you'd get closer to 80~90% efficiency, aka battery life \$\endgroup\$ – Passerby Mar 7 '16 at 8:11

Can I freely use the charger with my smartphone without causing damage(Would you use it on your phone?)

Yes and yes.

The 7805 does all the work for you by setting the output voltage at 5v. You don't have to worry about the current output because the phone will take care of that by limiting its current draw such that output remains at 5v. Worst case (ie the phone doesn't control current draw), the 7805 will protect itself with its short-circuit protection.

Now just because it's safe to use and I would be fine using it, doesn't mean you should use this design. As other have pointed out, it's going to be very inefficient and you're going to have a bunch of batteries at differing states of charge. You want to try to keep all batteries at a similar charge so that you have an output you can count on. If you have some dead cells mixed with charged ones you might not realize it at all. Then when one starts leaking acid and corroding, your backup battery pack is no good anymore.

What you really want here is a switching boost regulator. Not only will that allow you to use all your batteries consistently and maintain equal charge, but it will reduce the number of batteries (smaller pack) needed for the same purpose. It will also have the benefit of being able to function on a lower total battery charge, giving you a longer charge from the same batteries.

Depending on your budget (shoe-string or super shoe-string), switching regulators might be too pricey for what you want. A neat little DIY circuit you should check into is the Joule Thief. It's not very efficient, but is extremely simple and very affordable to build. Better still, it can be used in conjunction with your 7805 to give you a stable and phone-safe output. There are literally hundreds of sources for circuits like this on Google if it's a route you're considering.


Yes you can connect it. But take a look at the temperature of the regulator and eventually mount a heatsink to it.

The Power dissipation of the regulator can be calculated by the current flowing to the phone x the voltage drop on the regulator. If you take an eye on the input voltage using the switch, I still would assume 1,5 - 2 Volts for it. Assuming a loading current of 1,3A we get heat dissipation of 2,6 Watts on the regulator. That would require a small heatsink or it really gets hot.

Besides the wasted power on the hot regulator I don't think anything bad could happen to your phone. And of course don't close both switches or you short 2 batteries, but that's obvious.

You really should use a DC-DC converter because they have a much higher efficiency factor and you don't waste a lot of your precious battery energy in pure heat. Search for DC-DC step down converters (non-isolated). They aren't expensive and by using a step down converter you have one big advantage:

The input current drawn from the batteries is lower than the output current, nearly by the same factor you lower the voltage. Means: If you convert from 10V to 5V and your phone draws 1Amp, your batteries only have to deliver 0.5A. It is a little more in real, because of efficiency factor, but the point is: For the first time it makes sense to add more batteries in series and connect them all because the current they have to deliver gets lower and they will last longer.

PS: I also would think about using Lipo Batteries (like the one's used in RC models) because they have a much higher capacity, but that's a different story.


Use a 7805 drop in compatible DC to DC converter, its rated for 2A. you can ditch the switch on your input to change the regulator voltage (All you have to do is keep the input 0.7V above the 5V output.) The DC to DC converter converts power with +90% efficiency. The regulator is going to drop 4V (on the lowest switch setting so it will be ~40% efficent, and stepping up the voltage to 12V will not buy you anything.

To prevent damage to your phone, make sure the voltage does not exceed 5V by design.

Batteries should be matched, the reason behind this is if they are not matched, they start dissipating power. Ideally they should be matched with age and voltage and manufacturer to prevent power dissipation. If they dissipate too much power they explode or leak.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.