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I have salvaged a DC motor from an old (10-15 years) food processor and would like to use it for a project. However, in the process of taking it apart the controller board was broken so I can't measure the output to find out what voltage it needs. Are there any ways I can determine what voltage was used to drive it in the first place?

These are the things I know:

  • It is definitely a brushed DC motor.
  • The circuitry of the food processor is - Input from the mains (240V AC) goes through what I believe is an isolating transformer board (240V AC -> 240V AC, measured with a multimeter) which then goes immediately into a rectifier on the main board, followed by some components whose purpose I don't know, then to the motor.
  • The resistance of the motor (across the motor terminals) is about 15 Ohms.
  • The food processor is rated for 500W.
  • When driven at 5V from a PSU, it turns quite slowly and can easily be stopped. Same for 12V. 24V drives it faster but it still doesn't have enough torque, and I can't get any higher voltages from the PSU.
  • The motor is 7cm in diameter and weighs about 1kg.
  • The motor has no identifying markings.

From the second point, it would seem that because the transformer doesn't step down the voltage and 240V AC gets rectified into 240V DC, the max voltage of the motor is 240V DC but this seems much too high. The other circuitry probably used PWN to reduce the effective voltage but still... Does this assumption seem correct?

The project I want to use it in is a tabletop disc sander. For this purpose I guess the motor doesn't need to run at full speed but maybe ~60V? I'm interested in building my own DC motor controller circuit to just supply a constant voltage, nothing too fancy. Does this seem achievable? Could anyone point me in the right direction? What do I need in the circuit other than a transformer, rectifier & flyback diode?

I have some experience with digital electronics but this is my first time using high voltage.

Thanks

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  • \$\begingroup\$ Surprising that it is DC... Those are ususally AC motors. In any case, unless you can find a part number and a specification, there isn't much else I know to do. Don't forget, a motor drawing that kind of power is not just using a high voltage, but a lot of current as well. \$\endgroup\$ – Kurt E. Clothier Nov 20 '15 at 15:06
  • \$\begingroup\$ It would surprise me if it really is a DC motor. I would expect it to be a brushed universal motor. I see those a lot in kitchen devices here in Germany (230VAC.) The rectifier (if it is just a single diode) is probably for speed control. \$\endgroup\$ – JRE Nov 20 '15 at 15:09
  • \$\begingroup\$ Have you got a picture of the busted controller? \$\endgroup\$ – JRE Nov 20 '15 at 15:10
  • \$\begingroup\$ Can be a DC servo motor with permanent magnet. Look if it has stator winding or just rotor winding, does it have a feedback like tacho or encoder? \$\endgroup\$ – Marko Buršič Nov 20 '15 at 15:51
  • \$\begingroup\$ @MarkoBuršič: In a food processor? Really? google.de/… \$\endgroup\$ – JRE Nov 20 '15 at 16:03
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The voltage across a DC motor is made up of two components. A component from the resistance of the coils and a component from the inductive behaviour of the moving coils. When a motor is stalled it behaves like a resistive load.

Most motors if stalled while supplied by their rated voltage will draw considerably MORE than their rated power (and will overheat if a protection circuit doesn't kick in).

For a resistive load:

$$P=\frac{V^2}{R}$$ $$V^2 = PR$$ $$V=\sqrt{PR}$$

To get 500W into a 15 ohm resistive load would need about 80V. So that means that this motor is likely designed for a supply voltage of considerably more than 80V.

I would guess that this motor is designed to run at full power when fed with rectified mains and then there is some control circuitry to reduce power on lower power settings.

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  • \$\begingroup\$ What voltage you would need (using this logic) to supply a motor made of supercondcting wire? \$\endgroup\$ – Marko Buršič Nov 20 '15 at 20:38
  • \$\begingroup\$ Considerablly more than 0V ;) \$\endgroup\$ – Peter Green Nov 20 '15 at 20:53
  • \$\begingroup\$ Yes, almost the same, lower for the voltage drop only : current * 15ohm. If nominal voltage is 230Vrms rectified then In=500/230=2.17A, voltage drop is 32.55 therefore a same motor with superconducting winding would require 198V. The resitance is looses only, back EMF plays the role not the power dissipated on winding resitance. \$\endgroup\$ – Marko Buršič Nov 20 '15 at 21:46
  • \$\begingroup\$ A motor with superconductors would just rotate at a speed where the back e.m.f. was equal to the applied voltage. If the supply was strong enough (could supply sufficient current), you couldn't slow down the motor by loading it -- the current would just increase. \$\endgroup\$ – jp314 Nov 21 '15 at 5:03
  • \$\begingroup\$ @jp314 And what you want to say? What is different from normal motor, except that it doesn't have copper losses? \$\endgroup\$ – Marko Buršič Dec 3 '15 at 13:03

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