0
\$\begingroup\$

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?

\$\endgroup\$
  • 1
    \$\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
2
\$\begingroup\$

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

\$\endgroup\$
  • \$\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
  • 2
    \$\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
  • 2
    \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.