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I have made a PMDC motor. The resistance of the coils is 0.25 ohms. But the commutation system is very resistant. It adds more 0.85 ohms to the armature resistance. So, the total resistance becomes 1.1 ohms through the commutation system. The motor needs 14.80 volts and 5.75 amperes to work properly, using the same commutation system. If the commutation system is replaced by electronic sensors, would the over resistance added by the commutation be eliminated? If so, what amount of lesser voltage would be needed? my calculation (according to ohm law) concludes that this motor needs only 3.363 volts if the over resistance caused by the commutation system is removed. As: Total resistance/ coils resistance= 1.10/0.25=4.4; Present voltage/4.4=14.8/4.4= 3.363 volts. Is this calculation wrong?

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  • \$\begingroup\$ How will you connect to the rotor/armature if not using a commutator and how much resistance will this have? PMDC motors usually have a spinning magnet to avoid such problems. \$\endgroup\$ – Andy aka Apr 19 '20 at 12:25
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Assuming you can overcome (with zero resistance using slip rings or some other magical method) the 0.85 ohm resistance in the commutation, this reduces the effective motor resistance from: -

$$\dfrac{V}{I} = \dfrac{14.80}{5.75} = 2.574\Omega$$

To....

2.574 ohms - 0.85 ohms = 1.724 ohms

But, because you will still need to have the same current flow of 5.75 amps, the terminal voltage required to deliver this is: 9.913 volts

If your proposed "new" method to eradicate commutation results in less torque required to turn the motor at operating speed then, the current (5.75 amps) will also reduce a little bit. However, assuming that the real load torque is much bigger than the "friction" torque this won't reduce the 5.75 amps by very much. You might get away with 5 amps for instance and the terminal voltage might become 8.62 volts.

Your calculation failed to recognize the full resistance of the armature/rotor as being 2.574 ohms. This is probably because you forgot that the rotor coil produces a back-emf that appears to increase the stall resistance of the rotor/armature. The stall resistance of the armature/rotor is 0.25 ohms but, when running in the conditions you list it will be 1.724 ohms.

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  • \$\begingroup\$ I am grateful to you for your detailed answer. It will help me a lot. \$\endgroup\$ – Umer Farooq Apr 19 '20 at 14:24
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You haven't allowed for the back EMF in the voltage calculation. There are 3 sources of voltage drop opposing the supplied voltage :

  • I (5.75A) * R (0.25 ohms) in the winding, = 1.43V.
  • I (5.75A) * R (0.85 ohms) in the commutator/brush, = 4.88V.
  • Back EMF which is speed dependent, and currently 14.8 - 4.88 - 1.43 = 8.49V

Eliminating the commutator losses (and maintaining the same speed and torque) reduces the voltage requirement by 4.88V, giving 9.92V.

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  • \$\begingroup\$ Thank you very much for your answer. Your answer leads me to the right calculation. \$\endgroup\$ – Umer Farooq Apr 19 '20 at 14:01

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