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I am replacing a small DC motor that powers a Santa mobile using single AA battery - the only thing I could find was on Amazon, by "ajax" - listed as 1.5-3.0 V. The motor spins the propeller a little too fast, and it seems like reducing the voltage to 1.0 V or so would give the correct RPM. If so, what is the best (easy) way to reduce it?

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    \$\begingroup\$ The obvious starting point would be a small resistor. Another possibility would be a Zener diode, but it's probably overkill under the circumstances. \$\endgroup\$ – Jerry Coffin Dec 5 '16 at 16:50
  • \$\begingroup\$ I found 2 Ajax motors listed as 1.5-3V on Amazon - which one is it, the round one or the rectangular one? What diameter is the propeller? \$\endgroup\$ – Bruce Abbott Dec 5 '16 at 17:11
  • \$\begingroup\$ The round motor (5 pack) prop diameter is 5 3/4" \$\endgroup\$ – Dave Brick Dec 5 '16 at 17:17
  • \$\begingroup\$ @12lapointep ... yes it is. \$\endgroup\$ – Passerby Dec 5 '16 at 17:21
  • \$\begingroup\$ 5.75" propeller is too big for your motor. It will draw excessive current even at the lower voltage (=shorter battery life). I suggest buying the 4.5V version amazon.com/Ajax-Scientific-Round-Mini-Motor/dp/B00EPQKSY8/… \$\endgroup\$ – Bruce Abbott Dec 5 '16 at 17:33
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I'd start by putting a diode in series with the battery and see what you get. A ordinary silicon rectifier diode, like any of the 1N400x series will be fine. These will drop 600-700 mV, which might be enough to slow the motor down to the speed you want. If it's still too fast, try two diodes. It is unlikely that will be too fast.

The advantage of a diode versus a resistor is that it will drop the voltage a more predictable amount, regardless of the current being drawn by the motor. A resistor in series will work, in theory, but that requires knowing the current and voltage the motor draws at the desired operating point. The resistor will also get more in the way when the motor is trying to start up. It takes more current to start than to continue running once up to speed. However, a resistor will drop more voltage during startup, right when the motor needs it more.

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  • \$\begingroup\$ What about the peak current rating for those diodes? What is the order of the ohmic resistance in the motor windings (or corresponding current)? \$\endgroup\$ – Peter Mortensen Dec 5 '16 at 19:49
  • \$\begingroup\$ @Peter: A 1 A diode really should do it. They can do significantly more current for short periods of time, and the source is just a single AA battery. This device wouldn't last long if it was pulling more than a 1 A diode can handle for sustained periods of time. \$\endgroup\$ – Olin Lathrop Dec 5 '16 at 20:43
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Use a diode or diodes, 1N4004 for 0.7V drop, 1N5819 for about half that.

Using a resistor will not lower the no-load speed, it will lower the available torque (so the speed might drop when the motor is loaded, or it might stall).

A voltage divider will lower the speed but it will also unnecessarily lower the available torque, and it wastes a great deal of battery power.

A diode (or diodes) will almost maintain maximum torque at the lower speed without getting into the complexity of a feedback or IR compensated controller.

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Voltage divider schematic I don't know what the load is on your device, but if you put a resistor in series with it, it will bring down the voltage some amount. You could test it with trial and error until you get a desirable rpm.

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  • \$\begingroup\$ The current drawn by the load can vary, it won't be constant. So the voltage will vary at the Vout junction too. \$\endgroup\$ – Bence Kaulics Dec 5 '16 at 17:16
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    \$\begingroup\$ @Bence Kaulics propeller loading increases with rpm, while motor speed reduces as loading (and current draw) increases - which results in a stable operating point. \$\endgroup\$ – Bruce Abbott Dec 5 '16 at 17:25
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    \$\begingroup\$ R2 causes wasted current, which may not be such a good idea since this device is battery-operated. \$\endgroup\$ – Olin Lathrop Dec 5 '16 at 18:08
  • \$\begingroup\$ Thank you everyone for your suggestions, I went with the 1n4004 diode - works perfectly, $8.50 in parts saved BIG money over ebay purchase.ebay.com/itm/… \$\endgroup\$ – Dave Brick Dec 6 '16 at 0:53

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