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I am using a relay to control a dead simple circuit comprised of a buzzer and a 9V battery. I read the entire datasheet for the relay I purchased, and couldn't understand the schematic. The function of the circuit boils down to this:

send pulse to relay -> have buzzer ring

My circuit works, but it feels very wrong. Behold my beautiful diagram: circuit diagram

The pulse is sent to pins 11/12, and the buzzer rings. but the battery is getting extremely hot (I assume because the buzzer is wired in parallel with the battery...). What is the correct way to wire this? what on earth does pin 2 do?

edit: link to data sheet http://www.mouser.com/ds/2/164/mz-469330.pdf

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    \$\begingroup\$ your buzzer is always connected to the battery. are you shorting it out to turn it off? \$\endgroup\$ – Joren Vaes Aug 25 '17 at 16:50
  • \$\begingroup\$ No data sheet, so this is only a guess, Pins 1, 2 and 7 are probably the pins for a Form C contact (a normally open, and normally closed and a common contact.) From you schematic when you pulse the relay you are shorting the battery. The battery should be in series with the buzzer and the relay. \$\endgroup\$ – Tyler Aug 25 '17 at 16:50
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    \$\begingroup\$ A woeful schematic will lead to woe. \$\endgroup\$ – Andy aka Aug 25 '17 at 16:51
  • \$\begingroup\$ @Andyaka better than none :D \$\endgroup\$ – Arsenal Aug 25 '17 at 16:51
  • \$\begingroup\$ That's not a solid state relay, it's a regular electromechanical relay. \$\endgroup\$ – marcelm Aug 25 '17 at 16:52
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What you build is a Vespa-Circuit. Well I call it that.

What your circuit is doing is shorting the battery out to turn off the buzzer. So while your buzzer is silent a huge current, limited only by internal resistance of the battery and contact resistance of the relay is flowing. That's why your battery is getting awfully hot.

Shorting out the battery would be the right thing to do if your battery was a current source. But it is a voltage source.

So the switch should be in series with the battery and the buzzer.

schematic

simulate this circuit – Schematic created using CircuitLab

That way, when the switch is not closed, the current path of the battery is interrupted and no current will flow. And the battery will stay cool.

As noted in the comments, the MZ-24HS-K-U is not a solid state relay.

As you can see, pin 1 is normally closed and pin 2 is normally opened. You can choose depending on your application which is more suited for you.

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Something like this is probably what you are after, but without a data sheet it is impossible to know for sure.

schematic

simulate this circuit – Schematic created using CircuitLab

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