2 questions: Boosting current and step down converters


Revisiting an old project as I'm going to have a few days off work.
First off, I'm a bit of a hardware noob, so apologies on this if it sounds stupid or is completely wrong.

I wanted to try and build a robotic arm that uses 2 stepper motors and 2 servos. The arduino will run off a battery, but the servos/motors will need some more power and should also not be connected up to the same battery (apart from where you share the common ground). So the plan is to run the motors off the AC mains so I don't have to connect 15 batteries or an expensive RC battery to run all of this.

So, I was thinking about getting a wall wart that supplies 5V and 3A, but I realised the 3A may not be enough for all the motors as they can (depending on the load) drain up to 1A each.

Is there a way to boost the current when I have a 5V DC 3A wall wart? Will connecting capacitors in some way boost current or do they only store voltage?

Altenatively, I could get a different Wall wart that provides 12V and 5A, but then I need to step down the 12V power without generating lots of heat or needing large heatsinks. So can I connect two of these DC-DC converters in parallel (so I would then have 2 rails supplying 5V at 2.5A which then each power 2 motors)?

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I'd like to avoid work bench power supplies if possible so I can have it as a permanent completed project.

Thanks in advance to anyone who can help or takes the time to read this!

"Is there a way to boost the current when I have a 5V DC 3A wall wart?"
No. Find a higher capacity output supply.

You can use a switching DC-DC converter to bring 12V down to 5V. Use two if you want to have separate 5V supplies.
Or get a couple of 5V, 4A supplies. Are not expensive. I've used this one before. Pololu.com carries it also I believe.

Is there a way to boost the current when I have a 5V DC 3A wall wart?

Not without breaking the laws of energy conservation.
Forget the idea of using capacitors they do not store enough energy for what you want.

If the current is increased the voltage must be reduced so that the product ( voltage times current ) is the same, well slightly less actually to account for losses.
A switching regulator sometimes called a DC to DC converter is what you need to do this.

Responses much appreciated. I knew I was probably thinking things over wrongly.

My search results never turned up 5V 4A plugs, but now they are. Must have been having a blind moment!

One of those should be enough (only if all motors and servos are under heavy load would each one draw 1A+ simaltaneously, so this should cover things I hope as most of the time they should only draw somewhere around 500mA-750mA each).

If it's not enough, I'll go down the DC-DC converter route with a 12V 6A power supply (though I'll probably need a large heatsink on the DC-DC converter to take it from 12V to 5A I assume?).

Thanks again!

though I’ll probably need a large heatsink on the DC-DC converter to take it from 12V to 5A I assume?)

No it depends on what components you have in the circuit. However you can’t design that one yourself, it is rather a tricky thing to get going.

If your motors take 5V then it is better if you feed it with 12V and use a switching driver to limit the current. In that way the current gets into the coils quicker and you can step the motor faster than you otherwise would do.