Cellphone Battery Booster Built At The Checkout Counter

When you’re away from home and your cellphone runs out of juice it can be a real downer. Sure, you could find a store and buy a wall charger, but wouldn’t it be more fun to build your own battery booster without using tools? [Spiritplumber] did just that, popping into a Radio Shack for the parts, then making his how-to video (embedded after the break) while standing at the checkout counter. You can see he hust set his camera on top of the battery display case and got to work.

He’s using four D cell batteries to provide 6 volts of power. Assuming your phone charges at 5 volts this is going to be just a bit too high, even though there’s some tolerance with most phones. To overcome that obstacle he added a diode to the circuit, taking advantage of the 0.7 volt drop that it brings to the mix. Grab a plug adapter for your model and then just hand twist the connections. [Spiritplumber] admits it would be better to solder these, but in a bind you can get away with it. We looked up some prices for this method and we figure this would cost around $18 (batteries included) depending on the price of the plug adapter for your phone.

Of course if you’re just looking for a way to charge your phone without paying consumer prices there are ways of accomplishing that as well.


[Thanks Mkb]

40 thoughts on “Cellphone Battery Booster Built At The Checkout Counter

  1. Great hack, simple but the use of the diode makes it.

    Maybe I’m missing something; but is there a reason why he says to use Alkaline rather than rechargables? What’s the problem with using rechargables for this?

    1. Very strange. USB chargers are 5V… alkalines are 1.5V, rechargables are from 1.2 to 1.3V

      4*1.5 = 6V
      4*1.2 = 4.8 to 5.2V

      But! the diode is essential. I tried charging my G1 using 4 AAs without the diode and i think that it made my phone discharge even faster than without it :-)

      BTW I understand why owners of G1 are doing desperate batterypacks like this one :-D

    2. The problem with rechargeable batteries is the capacity, not the voltage (although the terms voltage and power seem to have been mismatched in the article anyway).

      A standard battery has more capacity – measured in mAH. This is an indication of how many hours the battery can supply a certain current for. Alkaline have more capacity than zinc carbon batteries and both types generally have more capacity that a rechargeable battery.

      You can find more details on Wikipedia.

      And by the way, no need for the diode. Your phone would work fine from 6V. Just stick a volt meter on your standard mains charger to get some idea. Certain brands go up as high as 10V.

      1. yes, certain brands do. but if the phone charges from usb, it needs 5v. the problem is that phones who charge from 5v usually have a linear regulator and that would be further stressed if powered from higher voltage.

      2. “You can find more details on Wikipedia.”

        And that is why the information is bunk.

        Alkalines do not have more capacity than modern rechargeables. The wikipedia article itself admits that: “An AA-sized alkaline battery might have an effective capacity of 3000 mAh at low drain, but at a load of 1 ampere, which is common for digital cameras, the capacity could be as little as 700 mAh”

        That is because of the high internal series resistance of the alkaline cell, which is by design to stop it from emptying itself on the shelf. The efficiency of the battery at anything above tens of milliamps is so poor that it won’t put out very useful amounts of current. Charging a cellphone takes on the order of half an amp, at which the rechargeable cell delivers at least twice the energy.

  2. The reason rechargables won’t work is voltage. Alkaline or even old fashioned carbon-zinc batteries are 1.5V each. Times 4 makes 6V. Rechargables are not 1.5V, but 1.2V for NiCad and 1.25V for NiMH. 4 times 1.2 is only 4.8V, not enough voltage! 4 times 1.25 is 5V, which is right on the edge of usability but will quickly drop under load to below 5V.

    1. Ah, if that’s the only reason; could this be done with 5 rechargables; taking it up to 6 volts? Or is it likely to kill the batteries?

      Would a voltage regulator would be a better bet to use? I have a few 5v ones lying around…

      1. That is the only reason yes :)

        While RS has 7805s that you can use, they tend to have a higher dropout voltage than 1v, so 4 batteries wouldn’t work to give you 5v. And you’d need to buy two battery holders because RS doesn’t carry holders that take more than 4 cells.

        The challenge was “build it quickly and now”.

      1. As he states in the wiki – 4*15=6V. Diode of this type makes 0.7 drop, we are left with 5.3V, which is in tolerance (you get another small drop after switching the load on). Rechargables would be too low even without the diode.

    1. i think you will find your theroy wrong and the phones recharging circuitry will take about any voltage bellow 5V with only a loss of effectiveness.but would be noticeable for what your doing.but it would have a great gain by being rechargeable cells

    1. I don’t think that diode is there to act as regulator. IMHO it protects the alkalines to be charged from phone when they run low…

      BTW stupid “Report comment” link. I always click on it when i want to reply. Can’t it be moved next to the date/time of comment? Also aligning the “Reply” button on the right (to current place of “report” button) would be great

      1. No. It’s there to drop the voltage by .7 volts like stated in the article.

        In order for the phone to charge the batteries, it would need a higher voltage than the batteries, in which case it wouldn’t be able to draw any current from the batteries.

        However, if you use a very small diode, it does actually work as a sort-of regulator because the forward drop depends on the current running through it. It may start at .5 volts for small currents, and end up at 1 volt at higher currents before it burns out.

    2. Because silicon diodes have a forward voltage drop of 0.7V. Germanium diodes have a forward voltage drop of 0.3V. Diodes can be used as a series voltage regulator if you need a drop in a multiple of 0.7 or 0.3.

  3. Fantastic hack. He put forth a real-world scenario and how you could throw a modest amount of cash and a bit of wire twisting to solve the problem.

    Then he made a short vid to show everyone how to do it.

    I could see this being useful after flying into a city, as the TSA isn’t going to let you carry anything like this in carry-on (just ask Star Simpson)

    Although if I found this scenario happening more than once I’d probably start packing a 4 cell Maglight in my carry-on bag so I’d not have to keep re-buying the same shit in every city.

    Perhaps one could get a few alligator tipped test leads through the TSA “gate rape” so you would get more reliable connection than wire twisting? If not, you could also re-buy the test leads from any Radio Shack in the nation.

    1. “Fantastic Hack”…

      Umm, NO. IMHO this is STUPID. Why?

      OK, well first of all, there is no enumeration through the data or ID pins. What does this mean? You will charge at a MAXIMUM of 500mA – But more likely, without this enumeration , 100mA. It is quite apparent that NOONE here (including author) is familiar with the USB-IF Battery Charging specification (nor Apple specs for that matter…).

      So, what, you charge for 100/500mA until the battey voltage drops to the low threshold (likely around 4.75V) under load and then you… Throw the batteries out?

      The proper solution: Hardware enumeration/handshaking (via ID/data pins) + boost/buck converter, or better yet, 4×1.2V rechargeable batteries with a boost converter. Smaller, simple, charges at higher current, and much better utilization of battery capacity.

      Oh, FYI, what I just mentioned has been done a thousand times over and can be found all over the internet.

      So tell me… WHY did this site publish someone who connected 4 batteries in series and called it a “Hack”??? Have you guys never seen batteries before?

  4. I am sure I’m not the only one to do this, but one time when our power was out for a long time and I needed to charge my phone, I duct-taped some batteries together, pulled the replaceable battery off the back of the phone, and used alligator clips to connect the batteries to the spring-terminals underneath where the battery fits. It worked great, and the total cost to do this at the register would be noticeably less — that adapter plug probably wasn’t cheap!

  5. I’ve used 4x AA Ni-Mh to charge my phone a few times. It seems the best way.
    I would also make it universal, work with rechargeable and alkaline. Just using a simple PMOS regulator. It doesn’t have to be really precise, just to allow a low drop.

  6. I saw a Masai do this in Tanzania some years ago. He cut a stick length-wise, whittled out some of the inner wood, put the batteries in, connected the leads of the charging plug to the batteries, and wrapped it up with a strip of rubber from an bicycle tire’s inner tube. I was impressed so I tried it too. It worked the first time. Then when I tried to demonstrate it to my neighbors I got the polarity wrong and blew a fuse in my phone. It cost about $3 to get the phone repaired.

  7. The *only* reason this works, as the author says, is this: “Cheapie wall-wart phone chargers output 5V nominal, but the actual voltage varies wildly, going as far high as six volts and usually stabilizing between 5 and 5.5 volts while under load. Cell phone manufacturers know this and design some tolerance into the phones.”

    Alkalines are 1.7-1.8V when fresh. That gives 1.8*4=7.2V. Assuming a 0.7V drop across the diode, that’s only 7.2-0.7=6.5V, a dangerous level for a phone designed for 5V.

    Furthermore, using a diode, you get a 0.7V drop only when you reach a certain amount of current. That is, the forward voltage drop of a diode is current-dependent. If your phone’s current draw drops temporarily, there will be voltage spikes.

    This is NOT a suitable circuit for sensitive 5V applications! My guess is his phone has a built-in regulator that’s dropping the “5V” input down to another voltage, so it is able to tolerate voltages slightly higher than 5V. For anything requiring a regulated 5 volts, do NOT use this method and expect it to work!

    Sincerely, an electrical engineer.

  8. Took me less than an hour to build my own version.


    There are only two major differences between mine and his. I’m charging a much older phone that does not use a USB connector and instead charges at 5.7 volts via a barrel jack. Also, I’m using a regulator so that I can use a variety of different batteries, including 9V, a six pack of AAs, Cs, or Ds, a car, truck, or motorcycle 12V, or a pair of 6V lantern batteries in series. Plus, it, too, could be assembled at a RadioShak counter. You would need a little more money and a Leatherman tool for wire cutting and stripping, but you could do it almost as easily.

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