1
\$\begingroup\$

I'm going to try and build my own battery pack for a portable speaker. It uses 20 volts and 3 ampere as input. I realize I will burn through quite a few batteries if trying to power it with just AA or even D type batteries. That's not a problem, the problem is that I'm a bit rusty when it comes to my electrical knowledge.

What I would like to know is if I need some kind of regulator for something like this, or if it's enough just to keep below the 20 volts and the device will automatically draw the right amount of amperes? This would mean that I could connect thirteen 1.5 volt batteries in serial to get roughly 19.5 volts out. Or do I also need to parallel connect those 13 to get the right amperes?

So, for you TL;DR people out there:

How would I connect AA or D batteries (serial/parallel) to power my 20V/3A device? Do I need some regulator type device for it, or is it safe if kept below those values? If I need some regulator/other for my desired setup, what can you recommend that's suitable for my situation?

\$\endgroup\$
  • 3
    \$\begingroup\$ I wouldn't try to use standard cells for this; you'd need at least 48. Look into using a large Li-ion pack instead. \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 8 '13 at 21:08
5
\$\begingroup\$

Batteries connected in series provide the sum of their voltage but at the current of the lowest rated battery.

Batteries in parallel (of the same voltage) provide the same voltage as one battery, but at the current capacity of all battery current capacities summed.

Your amplifier is rated at 20V/3A which means it draws 3A most likely at maximum volume. I would expect it draws much less at lower volumes. However, I will provide calculations based on the highest current draw.

"D" cells (batteries) store more energy (18000+ mAh) than "AA" cells (~2750 mAh), and therefore will last longer than "AA" cells for a given load.

If you connected 14 "D" cells in series, that would provide 21 volts (a 5% voltage increase should be tolerated by your amplifier, and as the batteries drain, they will be significantly lower than this voltage).

To address the current draw, you will need to connect additional sets of 14 "D" cells in parallel. Each set connected will increase the current capacity. With only one set, you have a theoretical maximum of 18 amp-hours, which means you could operate a 3A device for 6 hours. However, if you look at the datasheet for "D" cells, the mAh capacity is significantly reduced at high current draw. For example, at just 0.5 A, the mAh capacity is given as ~1500 mAh. This means that your 3A amplifier would only run for a half hour! It gets worse - because you are drawing much more than 0.5A, the life of the batteries is much less. In fact I would not expect the unit to work properly with only one set of batteries in series.

Ideally you would connect enough sets in parallel so that you are only drawing 25mA from each set, which gives the best-case mAh capacity. But this would require that you connect an impractical 120 sets; a total of 1680 "D" cells!

With 6 groups of 14 cells, each set would provide 500mA, and come much closer to providing the 1500 mAh capacity shown on the datasheet. I would expect a run time of about a half hour then.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Wow, thats very helpful - if maybe a bit depressing for my goals. I might try to build one anyway - to see for myself how far i could go with different volume and bass settings (as you said, it's likely 3amp only for max). But regarding some kind of regulator, you don't think i would need anything protecting the device as long as I don't go much over 20volts in total? Is there someething like it I could ensure the device doesn't draw above certain voltages&amperes, if so i could find the lowest possible to make it power up? \$\endgroup\$ – Christoffer Bubach Jul 8 '13 at 21:42
  • \$\begingroup\$ also, how could i with minimum energy loss convert a 12v car battery to 20v - and for a 80Ah type battery, about how long could it last? \$\endgroup\$ – Christoffer Bubach Jul 8 '13 at 21:53
  • 1
    \$\begingroup\$ The amplifier should never pull more than 3A, unless there is a fault of some kind. A voltage regulator is used in situations where you need to convert say 6V from four 1.5V cells to 5V for integrated circuits or similar. The extra voltage is converted to heat. Since you are combining batteries to reach 20V, you shouldn't need one. However, if you wanted to use two 12V batteries in series, giving 24V total, you might want to use an LM317 or something to regulate the voltage to 20V. Be aware of the heatsink requirements. \$\endgroup\$ – JYelton Jul 8 '13 at 23:02
  • 1
    \$\begingroup\$ To calculate how long something will operate (in hours), divide its current usage (in amps) into the amp-hour capacity of the power source. A 3A device will operate for 80 Ah / 3A = 26.7 hours. This is provided the power source is not de-rated with a full 3A draw. \$\endgroup\$ – JYelton Jul 8 '13 at 23:04

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.