I am planning on using the following circuit for my motorized value.

enter image description here

But I'm bit skeptical about how this might continue to work. Although I'll be using two identical relays, my fear is that at one point in time down the line, one of the relays may start switching earlier than the other (wear and tear may be) and it'll short circuit the battery.

And that'll obviously will lead to shut down the relay and again it will turn back on and so on creating an endless loop.

Firstly, is that possible? Or am I thinking too much? And if my struggle is real, how could I possibly solve this?


1 Answer 1



You have changed the schematic. Yes, in your schematic, at any point, out of the box or "aged" you can potentially be shorting the battery at every single switch over. It may be unlikely if the relays are exactly the same, but it is a very real risk, at any point in time.

This is why the schematic you posted earlier, with the relays connected the other way around, with the motor at the centretaps, is the way it is. Because the designer of that circuit had thought ahead about the implications of production margins. Or it was by accident, but the result is very much the same: Their design is safe, because only the motor can ever be shorted, while your new schematic isn't safe, because the supply can be shorted.

To make sure people coming by later understand what we're all on about, I shall redraw your originally posted schematic, followed by my original answer:


simulate this circuit – Schematic created using CircuitLab

If you carefully trace all the possible connections with these relays you will see that your battery cannot be shorted.

Your motor can be shorted, but that's the main point, that's how you turn it off (and also put the brake on). If both relays are up, as drawn, the motor is shorted to the battery +, while the battery - is disconnected. If both are down, the motor is shorted to battery -, while battery + is floating. If they are in different positions the motor rotates left or right.

  • \$\begingroup\$ Sorry the circuit I've pulled up from the net seem to be flawed. Let me get back with the correct circuit. \$\endgroup\$ Sep 16, 2015 at 7:11
  • \$\begingroup\$ I have edited the question to reflect the circuit I have designed for this. \$\endgroup\$ Sep 16, 2015 at 8:57
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    \$\begingroup\$ @ΕГИІИО Well, yes, the circuit that you have made has that risk, and not only when they become old, but right out of the box. So the one you pulled up from the net, in fact isn't flawed, but made that way very intentionally \$\endgroup\$
    – Asmyldof
    Sep 16, 2015 at 13:38
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    \$\begingroup\$ @ΕГИІИО For future reference, don't completely rewrite the question in such situation, because that may invalidate answers. Instead, keep the original question along with its schematic, and add new material to it. Something long the lines of: "update: I realized that the original circuit was flawed. [why flawed] [improved version] " \$\endgroup\$ Sep 16, 2015 at 20:36
  • \$\begingroup\$ @NickAlexeev Thanks. Sure I'll take more caution when editing the question next time. Cheers. \$\endgroup\$ Sep 18, 2015 at 7:00

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