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Suppose I have two brushed DC motors, each rated 3000 RPM and 50 watts. One is rated for 18 volts and the other for 36 volts. How will their sizes differ?

I assume that because of higher voltage the 36 volts one will require two times smaller current in the windings and so four times less cross-section for the winding wires and so the windings will take notably less space.

What other size differences in these motors and their parts (windings and cores) sizes should I expect?

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    \$\begingroup\$ No way to tell, because it completely depends on the torque/current/magnetic properties. Either can be much smaller or bigger because of the magnetic path. The 'watts' are a fairly useless metric with motors. \$\endgroup\$
    – user36129
    Commented Aug 21, 2013 at 9:00
  • \$\begingroup\$ @user26129: Okay, let's assume they are designed for the same maximum torque and use identical core materials. \$\endgroup\$
    – sharptooth
    Commented Aug 21, 2013 at 9:08
  • \$\begingroup\$ @sharptooth Still not so simple: As the coil parameters change, so will the weight of the coils. If the coils are on the rotor and not the stator (you mention brushed DC motors, after all), this changes the inertia of the rotor. I can think of a few other such factors which add exasperating little twists to the problem, e.g. brush shape and size changes to cope with a higher voltage at circuit break. Interesting question, though, and I'm keen to see if any definitive answers appear. \$\endgroup\$ Commented Aug 21, 2013 at 11:53
  • \$\begingroup\$ The simple answer is the number of turns and wire thickness changes to compensate for the higher voltage assuming you want to keep the current the same ... however as user26129 and Anindo Ghosh state a whole lot of other stuff changes... \$\endgroup\$
    – Spoon
    Commented Aug 21, 2013 at 12:15
  • \$\begingroup\$ Find a parts catalog and some examples! \$\endgroup\$
    – pjc50
    Commented Aug 21, 2013 at 12:56

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If you have a motor at voltage V1 and you want to have the same performance at a new voltage V2, then you have to multiply the number of turns by V2/V1 and you have to multiply your magnet wire CMA by V1/V2. So, to go from 18 V to 36 volts, you would double the turns in your motor and half the CMA (circular mil area) of your magnet wire (that is, increase the wire gauge by 3).

That is a rule of thumb. When you go to do the actual ratings tests for the 2 motors, you will see some differences. Some of it is due to more current going through lower voltage motors than higher voltage motors. Some of that is due to the fact that larger wires don't bend as easily and don't lay down as nicely as smaller wires. So motor manufacturers may tweak the number of turns or wire size to get the performance at the different voltages as close as possible.

Doubling the number of turns and halving the wire gauge (or vice versa) ends up giving you about the same slot fill if you use the same lamination. Smaller wire will pack better than larger wire so your effective slot fill will be smaller but overall its usually close enough that a motor manufacturer will use the same lamination.

Here's a little more explanation of why you change the number of turns and CMA of the magnet wire. You are trying to keep the flux density in the magnetic circuit of your motor the same. If you double the voltage, then based on the resistance of your coil, your current will be cut in half. If current is cut in half, then you need twice as many turns to get the same flux density. However, if you double the turns, then your resistance increases, so you need to decrease the CMA of your magnet wire by half to get the resistance (and thus current) back to where it was.

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  • \$\begingroup\$ Assuming CMA is cross-section, why do I make it two times less instead of four times less? \$\endgroup\$
    – sharptooth
    Commented Aug 22, 2013 at 9:13
  • \$\begingroup\$ Yes, CMA is circular mil area which is just cross sectional area of the magnet wire. You are trying to keep the flux density in the magnetic circuit of your motor the same. If you double the voltage, then based on the resistance of your coil, your current will be cut in half. If current is cut in half, then you need twice as many turns to get the same flux density. However, if you double the turns, then your resistance increases, so you need to decrease the CMA of your magnet wire to get the resistance (and thus current) back to where it was. \$\endgroup\$
    – Eric
    Commented Aug 22, 2013 at 12:17
  • \$\begingroup\$ I see now. This is very important and lot at all obvious. Could you please include these design reasoning into the answer? \$\endgroup\$
    – sharptooth
    Commented Aug 22, 2013 at 12:20
  • \$\begingroup\$ I've edited the answer. \$\endgroup\$
    – Eric
    Commented Aug 22, 2013 at 12:29

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