This is the first time I ask on the Electrical Engineering site of Stack Exchange, please feel free to dismember my question.

Please take into account: I am thinking about the hypothetical ideal case (100% efficiency, no energy loss) and the voltages and currents stated here are hypothetical values. If there are implications on the specific voltages (like types of batteries, or anything else) I would really appreciate if you let me know in your answer.

Let's say I have the following requirements regarding power for an electrical application that I am developing:

  • 2.5V DC
  • 1000A (all the time, not peak).
  • Portable. Duration of the battery: 1 hour

In an ideal world I would buy a 2.5V 1000Ah battery and it would be fine. However, imagine that I found a 5V 500Ah battery (and can provide 500A continuously). My assumption is, if I downregulate the voltage to 1V, it will be able to provide higher amperage at the outside of the regulating circuit (while not surpassing the previous power output which was 5V*500A = 2500 W).

My doubts are:

  1. How would you regulate the voltage? I am pretty sure a voltage divider won't do the trick, and I guess a voltage regulator would work, but with such high amperage I am not sure about the limitations, I have little experience with power supply circuits.
  2. With that solution, if the voltage is halved, would the amperage at the output of the regulating circuit be doubled? (that means, I could use the 5V 500Ah voltage source). Again, on "ideal" components. I understand there would be energy loss.

Thanks in advance!

  • \$\begingroup\$ Ampere hour represents how much continuous current a battery can deliver in an hour before it dies( or need recharging). Therefore, if you use more current, the usable time of the battry will be less. \$\endgroup\$ – Long Pham May 25 '18 at 17:10
  • \$\begingroup\$ Why would you regulate to 1 volt when you need 2.5 volts? \$\endgroup\$ – Andy aka May 25 '18 at 18:28
  • \$\begingroup\$ You cannot define the party player results unless you know who else is supplying the load, how much and how fast \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 25 '18 at 18:48
  • \$\begingroup\$ Buck convertor. But for 1000A, that's big bucks. \$\endgroup\$ – Brian Drummond May 26 '18 at 12:12

Portable welder with a small hand truck of batteries?

The item you're looking for is a "buck converter". That performs low-loss conversion of a voltage to a lower voltage at higher current. 2.5kW is a bit of a design challenge though.

Note that these can be made efficient for a large stepdown, so you could have 12V or 48V battery stacks and reduce the width of wiring required.


The short answer is YES, it does work like that. However all regulators are not created equal.

Some regulators, like linear regulators basically convert all the extra voltage into heat. This will not work. You will only get 500Ah out at 2.5v, and the regulator will dissipate 1250W!

A passive regulator like a voltage divider will be even worse.

What you need is a switching regulator. This type of regulator will make use of the extra voltage instead of just turning it into heat. If it was 100% efficient, it would work exactly as you describe, supplying 1000Ah at 2.5 from your 5v 500Ah pack. In reality it will not be 100% efficient.

Now, returning to reality.. you may have difficulty finding a switching regulator that can put out that kind of current. I think you may have better luck converting your battery voltage into AC, and then running that through a big transformer (like a microwave oven transformer). You can rectify it after that, or if you're just using it for welding you may be able to leave it as AC.

Edit: Also, side note. You're probably aware of this, but the battery voltage will not stay at the nominal voltage (5v) through the whole 500Ah, and then suddenly go down to zero. In reality it will decline slightly as the battery discharges before "falling off a cliff" near the end. The regulator will compensate for this, so it shouldn't cause a problem, but just something to keep in mind when you're looking at dropout voltages and whatnot.


There's no such thing as a free lunch, and you can't get more energy to your load than the amount of energy you take from the battery. Energy (in joules) is equal to power times time $$P\times t$$ or in your case you could use $$V \times I \times t$$ For your application you want to deliver $$2.5\times 1000 \times 60 min/hr \times 60 s/min = 9 \mathrm{MJ}$$

An ideal battery of 5V and 500Ah could provide $$5 \times 500 \times 60 \times 60 = 9 \mathrm{MJ}$$ So if you have an ideal battery and a switching voltage regulator that is 100% efficient then you can meet your requirements.

In the real world, you need to know the specified discharge rate (usually given as a fraction or multiple of C) that was used to come up with the mAh rating of the battery. If you draw a larger discharge current then the capacity of the battery will be decreased.

Likewise, you need to determine the achievable efficiency for the switching regulator. If you really need 1000A rather than 1000mA then you won't get that off-the-shelf from the usual hobby sources, so I really don't know what you should expect.


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.