can I run a 12v air tyre inflator from a 18v drill battery?

[neiloneil]

What components do I need to step the dc voltage of my drill battery (18v x 4ah) down to 12v to run a lower voltage appliance? What would a circuit diagram look like for this please?

 

[EddieM]

What components do I need to step the dc voltage of my drill battery (18v x 4ah) down to 12v to run a lower voltage appliance? What would a circuit diagram look like for this please?

I would have thought that would flatten the battery pretty damn quick.

 

[echoes]

Probably not worth the hassle. A cursory search for tyre pumps shows power ratings of 120W. That's 10A at 12V, so a 'simple' resistive or linear circuit will need to dissipate (lose) at least 60W.
A switching converter will be better, but maybe think again about the whole idea.

 

[neiloneil]

the inflator is 120 watts , so if a reasonably efficient method be found, my battery should run for half an hour or more.
Its more the theory that interests me than the practical. I have 18v batteries always at my disposal, which could (in a pinch) be used to power 12v radios, flashlights, etc. I think a resister could be used, but dont know whether it should be in series or parallel. you mention A switching converter, Ill search it on the web. Its just out of interest really. Thanks

 

[neiloneil]

Ah, I guess the extra power needs to be lost as heat, rather than recycled

 

[Tigercubrider]

What brand?
https://www.makitauk.com/product/dmp180z

 

[echoes]

If it's of more out of interest, then a switch mode buck converter would be reasonably efficient.
If you're asking, as you are, about whether a resistor is a suitable way to drop voltage at significant currents and if it should be in series or parallel, then you're probably being over ambitious. A 12V battery is the simple solution!

 

[neiloneil]

Its how I
learn things I guess, by asking dumb questions. Logically, i suppose the resister should be in series, and it would need to shed a lot of heat too. I had large ballast resisters on a honda motorcycle's ignition coils which got rather hot I seem to remember. I never really understood their purpose at the time, but getting hot was their intended purpose.

 

[big-all]

if all the energy can be used it will run for 36 mins 4x18x60=4320wm divx120=36 mins
but i suspect you will loose perhaps 20-30% as a rough guess allowing for some conversion and transmission losses but assuming virtually non lost as heat
if you start shedding 1/3 as heat [18down to 12v] on top off say 25% losses would be 50-60 losses or perhaps 15 to 20 mins run time
now off course fully a guess with wild assumptions but more in the real world than in the 90-100% conversion manageable off the sales blurb you often read

 

[flameport]

This is the type of item required: https://www.ebay.co.uk/itm/272811870386

Possible to build one from components, but not worthwhile given the cost to buy the components alone will be more than the complete module.

Any other option such as using a resistor will waste vast amounts of energy.

 

[ericmark]

Since a 7 Ah VRLA battery is around same price as the volt dropper what is the point? So if really want to use 18 volt, then a 7012 voltage regulator and a red LED on the common to negative will give you 13.4 volt which can charge the VRLA battery which will run the tyre pump.

 

[Tigercubrider]

A mate who makes hi end electronics for specialist industrial gizmos that he sells sometimes buys Chinese ready made stuff just to get components or chips that he can't buy at a reasonable price as a small buyer