So I'm looking into buying a portable fan, most I've looked at run on D batteries (8) aka 12V. Now I don't feel like buying 8 D batteries every time I want to run this thing and I can't find rechargeable Ds (im sure they would be expensive). I was thinking however, I do have number of 18v drill batteries, is it feasible to fun the fan on off of those?

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    \$\begingroup\$ Which fan, which batteries exactly? That's important so we can give you a relevant answer. \$\endgroup\$ – AndrejaKo Aug 22 '12 at 13:59
  • \$\begingroup\$ Someone told me that D-type and the other similar batteries are some of the lowest cost for available power batteries that you can get. Even some packaged batteries, maybe your drill, are just repackaged D or similar. \$\endgroup\$ – kenny Aug 22 '12 at 14:03
  • \$\begingroup\$ The battery is a Dewalt 18v drill battery, the fan is an O2-Cool 10" fan that takes 8D's ... is that enough info? \$\endgroup\$ – user379468 Aug 22 '12 at 14:17
  • \$\begingroup\$ You may need to add a battery protection circuit so you don't risk over discharging the battery. The battery may have one already, or it may be built in to the tool. \$\endgroup\$ – John La Rooy Aug 18 '16 at 2:49

Yes. You just need to lower the voltage to 12V. A simple PWM circuit is all that you need. Something like this: enter image description here

From http://www.dprg.org/tutorials/2005-11a/index.html

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  • \$\begingroup\$ excuse my noob question, but I may actually also have a crappy 12V drill, in theory would that work? do you have to also match the amperage? or is that just something with wall current \$\endgroup\$ – user379468 Aug 22 '12 at 14:21
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    \$\begingroup\$ @user379468 With a 12V battery, You'll have no problems (just be sure to match the polarity between motor and fan). Current is not an issue: if it works with 12V from D cells, it will works with 12V from a NiCd/NiMh/Li cordless drill battery pack. \$\endgroup\$ – Axeman Aug 22 '12 at 14:26
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    \$\begingroup\$ Might be worth mentioning that the peak voltage will still be 18V (in case someone tries to use this circuit for something other than a simple load) and also the duty cycle will need to be set to (12/18) * 100 = 66%. Maybe add a fixed resistor to make it able to vary from 0% to 66% (rather than up to 100%) \$\endgroup\$ – Oli Glaser Aug 22 '12 at 16:23

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