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I've got a motor that's driven by a 1.5V battery. The shaft turns slowly and that's the way I'd like after replacing with a USB cable (5V) source.

In the current circuit there's no resisters (just a diode) - so using ohms law - not sure what the current draw is. It's V/0 right? Infinity?

If I can find the current draw - I plan to use 5/I to add an appropriate resistor and I should be good to go - right?

Please help the nube boob.

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  • \$\begingroup\$ no, not right: A diode has a specific current vs voltage curve. Its "effective" resistance is not 0! \$\endgroup\$ – Marcus Müller Nov 21 '19 at 12:24
  • \$\begingroup\$ Also if you add a resistor to reduce the run-time 5V to 1.5V your motor might not start. \$\endgroup\$ – Oldfart Nov 21 '19 at 12:38
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Just adding a resistor is not the right way to go. The current draw of a simple DC motor changes depending on the load. In an unloaded condition the current will be at a minimum. In a stalled condition it will reach a maximun. To run a 1.5V motor from a 5V source you should step down the voltage with a regulator IC or a string of 4 or 5 silicon diodes, (the diodes will need to be rated for at least the maximum motor current, or best guess in this case). Using a USB from a PC might be risky if you do not know the motor's current draw, best to test out the idea using a USB wall adapter first.

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  • \$\begingroup\$ Thanks. It's a micro motor - so I doubt the current draw is more than 0.1A - but what do I know. I've read that most Si diodes the V drop is around 0.7 - so to drop from 5V to below 1V - I'll need 6 of them. \$\endgroup\$ – dashman Nov 21 '19 at 13:26
  • \$\begingroup\$ @dashman - You were initially asking for 1.5V. That's a 3.5V drop, (3.5/.7=5). I listed 4 to 5 because at a higher current levels the diodes can have an even higher voltage drop. \$\endgroup\$ – Nedd Nov 21 '19 at 14:57
  • \$\begingroup\$ Having 5-6 diodes in serial doesn't seem right...I think I'll use a resistor value about 5V/0.1A= 50 ohms - that should do the job right? \$\endgroup\$ – dashman Nov 21 '19 at 15:19
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    \$\begingroup\$ @dashman: If the motor runs on 1.5 volts, and you have a 5 volt supply, you only want to drop 3.5 volts in the resistor (or diode string), not 5 volts, so you would want a 35 Ohm resistor. However, as others have said, using a resistor to drop the voltage to a motor probably won't work, as the motor current will vary with load, so the voltage dropped in the resistor will also vary with load. \$\endgroup\$ – Peter Bennett Nov 21 '19 at 17:20
  • \$\begingroup\$ @PeterBennett thank you for the correction. So use a diode? \$\endgroup\$ – dashman Nov 21 '19 at 18:26

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