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#1
06-23-2010, 02:24 PM
 tgarrity Junior Member Join Date: Jan 2009 Posts: 10
Replace Transformer for Cordless Drill Battery Charger

Hello there. I have a cheapo 18v cordless drill from Menard's and the battery charger supply went bad. I determined that the primary winding of the transformer is open. The specs on the outside of the transformer shows 24vDC output. My question is, if I can find a replacement transformer, does it need to be higher than 24vAC output in order to produce the required voltage to recharge the 18v battery? (I'm assuming U.S. AC line voltage input.)

I remember the .707 RMS calculation from my electronics school daze, but
I looked at the lessons here in AAC and it shows DIVIDING by .707 for a full wave bridge rectifier (my case) produces a HIGHER voltage than the AC secondary transformer output. So I'm confused. I would think the average DC output would be LOWER due to loss in the sine wave output, and I would think you would MULTIPLY by .707.

In any case, what size transformer do I need? Any help is appreciated.

Thanks,

Terry
#2
06-23-2010, 03:34 PM
 SgtWookie Expert Member Join Date: Jul 2007 Location: In the vast midwest of the USA; CST Posts: 22,038

Yep, it's RMS/.707107 - or RMS * 1.414
You'd need to subtract the Vf of the rectifier diodes as well; roughly 0.7v at their rated current for standard silicon diodes. If you are using a full wave bridge (4 diodes) then it's roughly 1.4v. If you're using a CT secondary and 2 diodes, then it's just the 0.7v loss.

It would help a lot if you posted clear photos of your existing circuit board, top and bottom.

It's hard to take good photos of boards. Best lighting is outside on an overcast day; if it's clear weather drape a white sheet over yourself and your subject to diffuse the light. Don't use a direct flash; the picture will be very harsh and contrasty.
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#3
06-23-2010, 03:38 PM
 Bernard Senior Member Join Date: Aug 2008 Location: Tucson AZ USA Posts: 3,192

I would try for a 24V secondary @ 1/2A to 1A & not worry about the math as we do not know how much filtering or regulation is used.
 The Following User Says Thank You to Bernard For This Useful Post: tgarrity (06-23-2010)
#4
06-23-2010, 04:00 PM
 tgarrity Junior Member Join Date: Jan 2009 Posts: 10

Wait a minute. You want me to go OUTSIDE with a white sheet over my head???? Are you trying to get me killed??

Quote:
 Originally Posted by SgtWookie Yep, it's RMS/.707107 - or RMS * 1.414 You'd need to subtract the Vf of the rectifier diodes as well; roughly 0.7v at their rated current for standard silicon diodes. If you are using a full wave bridge (4 diodes) then it's roughly 1.4v. If you're using a CT secondary and 2 diodes, then it's just the 0.7v loss. It would help a lot if you posted clear photos of your existing circuit board, top and bottom. It's hard to take good photos of boards. Best lighting is outside on an overcast day; if it's clear weather drape a white sheet over yourself and your subject to diffuse the light. Don't use a direct flash; the picture will be very harsh and contrasty.
#5
06-23-2010, 04:03 PM
 tgarrity Junior Member Join Date: Jan 2009 Posts: 10

Thanks Bernard. There is NO filtering that I can see. Just a small circuit board with 4 diodes attached to the secondary. That's it. Just didn't know if I needed a higher secondary output to compensate for the effective RMS voltage. Thanks!

Quote:
 Originally Posted by Bernard I would try for a 24V secondary @ 1/2A to 1A & not worry about the math as we do not know how much filtering or regulation is used.
#6
06-23-2010, 04:22 PM
 SgtWookie Expert Member Join Date: Jul 2007 Location: In the vast midwest of the USA; CST Posts: 22,038

Ahhh, better get the current rating from the original transformer, and match it up.

What's the current or VA or Watt rating of the original transformer? It's a wall-wart, right?

The likely problem with your transformer is that the thermal fuse opened up due to overheating. The overheating might be caused by too much of a load on the output (batteries shorting internally) or the transformer windings are starting to short out, or one or more of the diodes in the output bridge are shorted.

Better check everything out before you slap a new transformer in there and burn it up as well.
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#7
06-23-2010, 04:39 PM
 tgarrity Junior Member Join Date: Jan 2009 Posts: 10

Thanks SgtWookie. I don't have it with me, but I believe it was somewhere in the neighborhood of .5 A. It's a Coleman drill from Menard's (home improvement store), but el cheapo. I dissected the top layer of teh transformer and there appears to be a 2K resistor that is open. A tech I talked to here at work said it was probably to act as a fuse. I checked the diodes, they are OK. I can believe the batteries might be toast. They are NiCd. Or maybe it was CHEAP. There's a surplus store here that sells just about every kind of transformer you'd ever need, and they are cheap, so worth a shot. Thanks guys!
#8
06-23-2010, 05:05 PM
 SgtWookie Expert Member Join Date: Jul 2007 Location: In the vast midwest of the USA; CST Posts: 22,038

Try replacing the 2k fuse with one of the same or next higher wattage.

If you simply use another transformer without the current limiting, you'll fry your batteries.
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#9
06-23-2010, 07:30 PM
 tgarrity Junior Member Join Date: Jan 2009 Posts: 10

I can't get inside of it enough to replace the resistor. It looks like they encased it after soldering the fuse inside. I can't see of a way to get at it without destroying it.

Last edited by tgarrity; 06-23-2010 at 07:40 PM.
#10
06-23-2010, 08:32 PM
 SgtWookie Expert Member Join Date: Jul 2007 Location: In the vast midwest of the USA; CST Posts: 22,038

Yeah, they do that.

Well, if it was in fact a 2k resistor, that's a bit odd - but it would serve to limit current on the primary side, rather than having to regulate the secondary side.

What kind of batteries does the drill use? NiCD? NiMH? LiPO? Any clues?
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 Tags battery, charger, cordless, drill, replace, transformer

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