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I am trying to create a circuit to drive a hand drill motor as seen in this drawing with just relays. The current formation to make it go forward, switching 2 and 3 at the same time to make it go backwards, and switching 1 to make it stop. I expect it to be ~18V at ~20A when running. Do I need flyback diodes for this to work correctly and if so how would I go making that calculation and selecting that part? When switching 2 and 3 at the same time is there anything I should be concerned about?

Bbb
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    change the state of one of the relays 2 or 3 ... what happens? – jsotola Dec 02 '20 at 02:41
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    lose relay 1 ..... swap the motor and battery positions ... what do you get now? – jsotola Dec 02 '20 at 02:42
  • If you connect the motors to the same terminals at once, that would create a short circuit which would be bad and might happen if they don't switch at exactly the same time so I should switch 1 to open when switching 2 and 3 from forward to reverse positions which I figured would be a good idea, but forgot to mention. And for the second question, if you switch doesn't that get the same functionality? – Bbb Dec 02 '20 at 03:05
  • No - it gives better functionality because of you swap the motor and battery in your circuit then no matter how you switch relays 2 & 3 you can't short out the battery and it removes the need for relay 1. – brhans Dec 02 '20 at 04:02

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