I am not very good at electricity things so any help anyone could give me would be fantastic. Here goes:

  • I have a battery (for power tools); it is 18v and 1500 mAh (imagine DC).

  • I'm trying to get it to power a tool that is designed to take 120V and 6 amp (AC).

Is there a way to calculate how long the battery would last? I read the article on "How to calculate Battery Life" and it was extremely helpful... but I have more questions. Does the AC/DC conversion have any effect on battery life? What affects do volts have on battery life? Is there an equation to calculate this?

  • \$\begingroup\$ A couple of minutes, if you can get 100% conversion efficiency. \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 25 '14 at 4:01
  • \$\begingroup\$ how do you get his number? is there a formula or is it just experience? \$\endgroup\$ – user50051 Jul 25 '14 at 4:32
  • \$\begingroup\$ That battery is waaay too small for your load (120V*6A=720W). \$\endgroup\$ – Kamil Jul 25 '14 at 5:54
  • \$\begingroup\$ DC to AC converter required. But probably not a good way to do things if battery suitable drill is possible. | Time = Energy/Power = (18V x 1.5 Ah) / (120C x 6A) x Efficiency. For (undoable) 100% = 27/720 =0.0375 hours ~= 2 minutes . Less in reality. Half perhaps. And battery current = 720/18 = 40A which it may or may not like. Most wont. | Drill power will be at max load and will run longer at lower load UT don't do it. \$\endgroup\$ – Russell McMahon Jul 25 '14 at 7:23
  • \$\begingroup\$ Simply rewind the motor for 24V and fit 2 car batteries in series. 24V 60AH should give you about 2 hours. If you don't want to rewind the motor, look at a car battery and an inverter to 120V. You'll get a bit less than an hour because the inverter is less than 100% efficient. \$\endgroup\$ – Brian Drummond Jul 26 '14 at 11:55

Look at the power required = 120V * 6A = 720W using P=V*I.

Your battery which is DC has to run through an inverter to get to 120V AC. There will probably be about 10% loss in efficiency, so add 10%

Lets call that power = 720+72 ~ 800W

How much current would the battery need to put out to generate 800W?

800W/18V ~ 44A. (yes that is amps from I=P/V)

Your battery rated at 1.5 A-Hr would last about 1.5/44 = 0.034Hr or about 2 minutes.

This of course would never work. As the internal resistance of your little battery would never allow such a massive discharge, and your whole system would most likely fail to even start turning the motor.

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  • \$\begingroup\$ so basically the internals of a Battery powered drill and drill that plugs into an outlet are more different then i could have possibly imagined. \$\endgroup\$ – user50051 Jul 25 '14 at 4:44
  • \$\begingroup\$ AC motors vs DC motors. Corded tools can offer higher powered motors because they are not concerned with battery life. \$\endgroup\$ – user6972 Jul 25 '14 at 4:47
  • \$\begingroup\$ so there is no logical way to make a corded tool cordless? The motor used is simply to different correct? \$\endgroup\$ – user50051 Jul 25 '14 at 4:49
  • \$\begingroup\$ Cordless tools use a completely different motor in the first place. \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 25 '14 at 5:01


The Amp-Hr rating represents the stored energy at a relatively low current.

As current increases, the total energy the battery can deliver will fall off.

At "high" current, the actual Amp-hr delivered at depletion can fall to 10% of the Amp-hr label. An exponential curve will well represent the actual Amp-hr capacity vs. current.

Much depends on battery internal resistance, in dependent on battery type and size.

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