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Power supply for hobby motor

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unseen wombat

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Hey, I have a hobby motor for a automatic fishing reel project and I'd like to power it with the smallest number/size of batteries possible. This is the motor: https://www.amazon.com/gp/product/B072N84V8S/ref=ppx_yo_dt_b_asin_title_o00_s00?ie=UTF8&psc=1

It needs 12V and 100 mA according to the specifications.

I tried to use a joule thief with a AA battery, but as I'm sure you guys could have told me, it didn't work. (I still don't really know why though, current?) A 12V A23 battery works, but those are very low capacity, only 55 mAh, so if there were some way to boost one or two AA's or coin cells up to the required voltage and current, I'd rather do that.

Would two coin cells and a voltage doubler circuit work?

Thanks.
 

Just remember if you boost voltage, you're going to cut current. A decent AA battery should be able to produce about 1A at 1.5V, but if you want 12V, you'll only get about 100mA, depending on efficiency, etc. WE can't tell you why your joule thief didn't work. Did you actually MEASURE current? Did you measure the current out of the battery? Also, although the RUNNING current of the motor may be 100mA, you need a LOT more to get it started; that might be your problem. A capacitor across you power supply output might help.

You need to look at your circuit efficiency, the capacity of the battery(s) you want to use.
 

Thanks, but unfortunately I don't even know what I don't know. I hoped someone could recommend a circuit for me that would accomplish the goal of powering the motor with as few/small batteries as possible. Do you know what would be the most likely thing to work that I could try first?
 

Hi,

Ah is just charge, but you need to calculate with energy.
Energy = Wh = Ah x V.

Example.
A 12V, 55mAH may deliver the same energy as a 2V, 330mAh.

To get more precise results you need to calculate the conversion efficiency in and the internal battery resistor loss.

Klaus
 

One possibility is a voltage multiplier, made from an array of diodes and capacitors. To get 12V your supply can be three 1.2V cells in series (since rechargeables are a popular way to go).

Dickson V multip 7 diodes 4 transis 3_6V supply 120 ohm load gets 13VDC.png

4 transistors make a full H-bridge applying AC to the voltage multiplier. This arrangement is sometimes called a half-wave parallel multiplier, and sometimes a Dickson multiplier.

When the motor is disconnected, output reaches a maximum of about 17V. Minimal current is drawn from the batteries.

When the motor is turned on, it rapidly draws down the voltage to 13V. Current draw becomes an average 1A.

The clock can come from a 555 IC pulse generator circuit.
 

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