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I am trying to fix a brake on a mitre saw. I am yet unsure of its operating principle. The switch (shown taken apart) has one terminal on the left side with grey wire going to the motor housing, and two terminals on the right side (black/live wire on the upper terminal, and white/neutral wire on the lower). In the “Off” position the sliding carriage is pushed to the left side, connecting gray and white wire. When saw is turned on, the carriage is thrown to the right side, connecting live and neutral, and starting the motor.

saw switch internals

After the lever is released, the switch returns to the “Off” position. At that point the motor supposed to be forced to slow down. I do not know for sure, but I suppose it reverses direction of the magnetic field to resist rotation, until it stops, and because the live wire is not connected, the motor does not spin in the opposite direction, but only stops.

Currently the motor is not being forced to stop, but slowly coasts to stop. I have tested the switch with an Ohm-meter: infinity between black and white terminals in an “Off” position, short in “On” position, reverse situation between grey and white terminals. Upon physical inspection, switch appears to be in perfect order. Therefore, the common purported causes of malfunctioning brake, worn brushes (they look good, and so is the commutator) and broken switch, do not seem to apply here.

How do I troubleshoot the motor? What is the schematic to help understand the operation? What is this switch called (I could not find proper terminology)?

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    \$\begingroup\$ I think you need to draw the full wiring diagram, since the on position can't be connecting live and neutral together as this would short the power supply and blow a fuse. \$\endgroup\$
    – HandyHowie
    Feb 12, 2020 at 9:31
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    \$\begingroup\$ I am trying to fix a brake on a mitre saw. — This rules out an induction motor pretty certainly. It's an universal motor, and this brake switch most likely cuts the mains and shorts the rotor winding instead. If it fails to to the latter, the contact doing that is burnt out. \$\endgroup\$
    – Janka
    Feb 12, 2020 at 9:32
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    \$\begingroup\$ All the internet posts say the same thing: your motor brushes need to be replaced . These motors brake by shorting out the motor leads. If the brushes are bad, it can't short the motor. \$\endgroup\$
    – Natsu Kage
    Feb 12, 2020 at 9:36
  • \$\begingroup\$ I would expect that if the brushes were bad, then the motor as a whole wouldn't run? \$\endgroup\$
    – SomeoneSomewhereSupportsMonica
    Feb 12, 2020 at 9:44
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    \$\begingroup\$ @SomeoneSomewhereSupportsMonica No, because the connected HV voltage breaks the air gap insulation, producing a conductive arc. At lower voltage the worn brushes don't conduct. \$\endgroup\$
    – Marko Buršič
    Feb 12, 2020 at 10:09

2 Answers 2

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As noted by others you should try changing the brushes. I had a washing machine with a universal motor stop working. (No notice, just went from working normally to stopped mid-wash.) The brush and commutator surfaces showed no signs of damage. The only visual indication was that the brushes were much shorter than a new pair. Fitting new brushes restored normal operation. So compare your brushes to a new pair and replace them if noticeably shorter.

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The lever switches the contacts as a 2P2T but the stationary contacts are dark and not visible but likely oxidized from arc heat. The typical universal brake motor has high current from BEMF.

Sometimes repairable surface if sufficient material exists. It may be an alloy of copper. Better to replace worn parts.

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  • \$\begingroup\$ I think it's a 1P2T switch, because it has 3 terminals, not 6. Contacts on it were fine, there was some soot in "on" position terminals, it only needed light filing. I would like to learn, and be able to test and troubleshoot to make certain the parts need replacing, before replacing them. \$\endgroup\$
    – theUg
    Feb 15, 2020 at 2:34
  • \$\begingroup\$ Basic Ohm's Law can be used for Switches , apply a rated or limited current and measure voltage drop. \$\endgroup\$
    – Tony Stewart EE75
    Feb 15, 2020 at 5:05

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