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I wanted to use an old electric driller that it's battery has worn down and i couldn't find a new battery. on the battery it says 7.2V. So i took a AC-> 7.5V DC 1A Transformer and used it in place of the battery. it worked, but it didn't had a lot of power, and I'm trying to figure out way.

here it is

can it be that the driller need more then 1A? it doesn't make sense cause in this current it probably drained the battery in a few consecutive minutes.

so WTF?

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2 Answers 2

up vote 3 down vote accepted

It's very possible it's more than 1A. As a very simplistic example, if the batteries it were using had 2000mAH in them, then it means they could supply 4A for 0.5 hours, and I'd say a half-hour of continuous use from a drill like this is being pretty generous.

It won't hurt to get a bigger supply: as far as current goes, the motor will just pull what it needs. If you don't have one you can use/scavange for free though, I think you'll probably find it's cheaper just to buy a 120V drill: as you increase in current, the power supply will get a lot bigger and more expensive (note: there is a reason wired drills use 120V AC motors, and don't switch to DC).

Save the motor/gears, maybe you can use it for some other project, but from a cost perspective, it's just not economical to fix something like this.

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Good point, I did a search for 230VAC to 7.2VDC at greater than 1A, and everything got expensive fast. –  Doresoom Mar 28 '11 at 19:38
    
o.k. thanks. i'll keep that in mind. –  Asaf Chertkoff Mar 28 '11 at 20:23
    
@Doresoom: i still would like to know what is the formula for that calculation you did there... –  Asaf Chertkoff Mar 28 '11 at 20:24
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@Asaf: I edited my answer to include the formula. It uses power (Watts) as the key to bridge the gap between mechanical and electrical. Just make sure you keep your units consistent! (IMHO, a metric standard forced upon the world would be awesome. Silly imperial units...) –  Doresoom Mar 28 '11 at 20:30
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I checked this list of capacities for Dewalt batteries, and most of them were above 1 amp-hour. Assuming they're not capable of running continuously for an entire hour, then they're discharging at a rate greater than 1 amp.

EDIT: I did some calculating off of rough numbers I could find concerning 7.2V drill operation, and here are the results:

A max 44 in-lb torque operating at 200 rpm for a 7.2VDC drill requires 2.3A.

The formula is:

Torque*RPM=Power
Voltage*Current=Power
Torque*RPM=Voltage*Current
Current=Torque*RPM/Voltage

if you've got torque in in*lbs and rotational speed in RPM, multiply your answer by 0.00188308048 to get the correct units of Amperes. (1 inch pound is 0.112984829 Joules, and 1 RPM is equal to 1/60 Hz)

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o.k. so i need to look for a 7.5V with more 2-5 AMPS, right? –  Asaf Chertkoff Mar 28 '11 at 18:37
    
@Asaf: Yeah, probably in that range. See above edit for a ballpark figure. NOTE: I couldn't find any torque vs. rpm curves for drills, so I'm not sure if I got the max torque speed correct. Generally, the max torque is the stall speed, but you can't calculate anything using 0 for the shaft speed, since power goes to zero. –  Doresoom Mar 28 '11 at 18:42
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BTW, nice MacGyvering. –  Doresoom Mar 28 '11 at 18:45
    
first of all, it would be lovely if you could share the formula of the calculation you've made. second, is it matters that the drill isn't as good as makita's drills? it is a very cheap one. and third, on the drill i found this prhase "300/540rpm". is that means that the lowest rpm it can preduce is 300 and the highest is 540? and also - is this information helps us in understanding the range of currents? –  Asaf Chertkoff Mar 28 '11 at 18:52
    
@Doresoom: thanks, i live for this stuff! –  Asaf Chertkoff Mar 28 '11 at 18:53
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