Increase battery pack voltage with a DC step up converter?

[rg12]

Hey Dudes,

I'm using those little step up buck modules to raise a 4.2v battery to power a 12v low amp draw electronic thingies and I was thinking about a theory...
If hub speed is set by the voltage (KV) and I want to increase the speed, why not find a simple circuit module to raise the voltage from let's say 54.6v to 100v (considering my controller supports 100v of course).
Would that work?
I never understood how that voltage raising thing works...will it take more amps in to create that extra speed? (considering my battery C rate is high, say LiPo)
Is there a difference between real voltage and "fake"/raised voltage"?

 

[Alan B]

A device like this is a Voltage Boost Converter.

Power out = Power in * efficiency, so if the output voltage is higher then the input current must be higher to balance.

There is some loss.

It takes a high power converter to handle a motor.

You always pay some efficiency loss for conversion.

For modest increases you will likely be better off with a sinewave controller that can do field weakening, which is sort of like voltage boosting inside the motor, though not exactly.

 

[rg12]

So you say that it's possible but will require a giant converter as big as the whole battery pack or something?

How much of a loss am I getting with boosting volts?

 

[liveforphysics]

So you say that it's possible but will require a giant converter as big as the whole battery pack or something?

How much of a loss am I getting with boosting volts?

Voltage doesn't turn your wheel.

Think in terms of power. When you think in terms of power, every stage of added conversion is merely unnecessary loss.

 

[Alan B]

The size of the converter depends on the technology. It is two things - power switching, which is what your motor controller does, so about that size, plus magnetics to store energy. The size of the magnetics depends on the frequency, but should be smaller than your motor. Bottom line is - not small. And losses, as Luke points out, are additive. Perhaps 10% additional loss. It still might be worthwhile, like afterburners, if you don't use it much.

As I said, better to buy a more sophisticated controller, or get a motor that operates at lower voltage effectively making your battery pack "higher" in voltage compared to the motor.

 

[macribs]

Rather then spending money on a converter what if you put the same amount of cash into your battery pack and just up the voltage by adding more in series?

 

[rg12]

It's more of a theory question than a money saving question as I always wondered about that voltage increasing thing...
Took me about 20 years to understand how my dad's taser gun could produce what seems to me like a hand held LIGHTNING machine from a 9v battery :lol:

Thanks for the info

 

[liveforphysics]

Essentially, it would always be lower cost, much lower weight and greatest system efficiency to simply get a motor wound for higher Kv than to play with any series conversion.

 

[rg12]

Sounds nice to have this as an "afterburner" option like Alan B said, but it includes having a very large converter so it will stay in theory I guess...

btw, is the theory of higher voltage really true? for example producing 1000w with 20A and 50V or 1000w with 50A and 20V, that the 50V setup will be faster?
shouldn't it just be the bottom line which is watts?

 

[Alan B]

With lower voltage the power falls off as the speed increases faster than it does with higher voltage. So they are never equal in power.

 

[rg12]

You mean voltage sag when under load?

 

[Alan B]

Not talking about sag, talking about using different voltages for supply, but also applies for sag.

When you change the voltage the ability to stay at 1000 watts continues to a higher speed, so the systems are not equal even if you never exceed 1000 watts at either voltage.

 

[markz]

I got a quick question in regards to the converter.
Lets say you got a 10V 1A input from a laptop adapter (or battery in terms of the OP) and you install a boost converter to get your desired voltage, like the OP. So now you tune the output of the boost converter to say 20V, what loss would there be for the ampere's at the output, taking into consideration the efficiency of the converter (for arguments sake, lets say its 95%) ?

 

[DVDRW]

Input 10W
loss 5%
output 9,5W @20V = 0.475A

You never gonna find 95% efficient converter at decent price.