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I am trying to convert my 9.6V NiCad drill to run off a 12V Li-on battery. When I connect the Li-On directly, the motor only just slightly spins and stops. The only way I can get it to work is using a Step-down Buck Converter. Initially I thought it needs 9.6V to start working, but nope works fine at 12.3V too, just not when connected directly to the battery. What am I missing? How do I get it working connected directly to the battery? Does the DC motor need a starting capacitor?

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closed as off-topic by Leon Heller, RoyC, Voltage Spike, JYelton, Dmitry Grigoryev Jun 3 at 10:46

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  • \$\begingroup\$ Seems very strange. Can you draw a diagram (block is probably fine). Devil is probably in the details here. \$\endgroup\$ – winny May 13 at 7:53
  • \$\begingroup\$ It is literally connecting either the motor directly to the LiOn Battery, or putting a buck converter in the middle. Nothing fancy or complex. \$\endgroup\$ – user1270949 May 13 at 7:58
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    \$\begingroup\$ You mention "9.6V NiCad" and "12 V Li-Ion" as if that makes it immediately clear what type of battery that is and for what it is designed. There can be huge differences though. The motor probably needs a high current, the NiCad battery pack is designed to be able to deliver that. Only high current Li-Ion cells can do the same (deliver a high current). As "high current" cells are generally more expensive if your battery pack is "generic" then chances are it is not for high current. Also it might have a battery protection circuit which limits that current to protect the batteries. \$\endgroup\$ – Bimpelrekkie May 13 at 8:02
  • \$\begingroup\$ Sorry, but I am very new at this. The battery protection circuit definitely makes sense as the motor pulls in around 8-9 Amps at startup (via the buck converter), but only 1-2 A otherwise. Not sure what it attempts to pull via the Li-Ion battery directly. My cheapo multimeter doesn't register much of a current pull in that time. \$\endgroup\$ – user1270949 May 13 at 8:11
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Do you have the C rating and the capacity of the Li-Po battery this will give you how much the current the battery is able to deliver continuously.

The buck regulator will cause the current draw to be less on the high side so if it is the over current protection that is stopping you this might be the reason.See this page for introduction

You might overcome this by limiting the inrush current by a NPC thermistor or connecting a load in series and then disconnect it when the coils in the motor has been charged.

Also remember that the motor is rated for a given current at a given voltage. If you connect it to the Li-Po at a higher voltage the current will be greater and the potential to burn out the motor is present it definitely will run faster and hotter.

The best solution would probably be to run with a buck-converter. The protective circuit would hopefully cut out the power when your battery reaches 11.1 V but some circuitry to detect and cut off the battery should be implemented as well as to not damage the battery pack.

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  • \$\begingroup\$ I think this may be it. I am using a 12V 4.4A Liitokala battery rated at max discharge of 1C. Thanks for the link, just realized that this concept of "C" rating exists. That would explain this behavior as the battery seems to have its own protection circuit & I can measure the motor pulling upto 8-10A at startup. At the moment, I would highly like to avoid the buck converter as it is quite large and will not fit in the battery case. In any case, I assume it is the bunch of capacitors, really doing the heavy lifting here. \$\endgroup\$ – user1270949 May 13 at 9:07
  • \$\begingroup\$ droking.com/… This is the buck converter I am using \$\endgroup\$ – user1270949 May 13 at 9:07
  • \$\begingroup\$ It is really a combination of the coil, capacitors and switching transistor that does the heavy lifting. learnabout-electronics.org/PSU/psu31.php \$\endgroup\$ – Morten May 13 at 13:37
  • \$\begingroup\$ Thank you for the PSU link. Will go read up on that. Is there any way to reduce the size of the buck converter? Can I use a smaller one rated for 3A output? Or do you think it has something to do with the construction of this particular converter that is helping me? \$\endgroup\$ – user1270949 May 13 at 19:58
  • \$\begingroup\$ A lower rated one might help, but make sure that it can handle the spike current at startup. Yes there is a way: the limiting factors are the input/output range for the current and voltage, this will result in the tolerances needed for your components so that they will not overheat and let out the magic smoke. Take a look at this for a starter opencircuitinstitute.org/content/…. Find some application notes, online calculators and search for components on digikey, farnell or octoparts to get you started. \$\endgroup\$ – Morten May 14 at 10:24

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