Help building a Relay Switch

I need a 5 VDC .3A NO relay switch, that will be used to close the circuit feeding a 12 VDC 5A device.
As soon as the trigger current stops, the switch should open again.
I ordered part 2449-J1021AS55VDC.45-ND and I assume it will do the job, but I’m not too much of a techie so was hoping it would be easier to put in the circuit. It looks like this needs to be mounted on a PCB with terminals added to attach the trigger device and the load. I’m at a loss on how to do that.
Are there any all-in-one relay switches that would work i.e. with terminal lugs. And preferably I’d like a solid state device so it’s easier(?) to maintain.
I’d really appreciate anyone’s help with this. Thank you in advance!!!


Welcome to the Technical Forum. Yes. I answered the e-mail and provided SSR Relays on the link that were PCB mounted. Due the low voltage and current they were the only options I had. I did find some din rail options on this link: . I am not finding anything else. Sorry.

Yes, that referenced option looks like it would work!
Is it normally open and only closed while a current is being applied from the trigger device, and will open again when the trigger current falls to 0.
Thanks again!

Just a note: The SSR’s linked above are only for AC loads and not DC loads. If you need to turn on a DC load, you’ll need something else.

I’m not completely clear on your circuit description. Are you planning to use a solid-state relay to turn on the coil of another relay? If so, at what voltage and how much current does the coil require?

It’s a DC load that will be used. Here’s what I’m trying to do:
I have an alarm system at our 2nd home in Honduras, but the sirens supplied are not loud enough to be heard from the guard station. So I bought a 12 VDC 30 watt outdoor siren.
The alarm control panel will produce a 5 VDC .3 A current when it goes off, that I want to trigger a relay switch that will cause the circuit to close between a 12V battery and the siren. The siren would continue until the alarm is turned off at which time it would return to NO state.
I’m not sure a solid sate or mechanical device would be best. And although I have rudimentary soldering skills would prefer an all-in-one switch. But if nothing like that is conveniently available I will try and build my own module

OK, much clearer now. You can source 5V at up to 0.3A and you need to use that to enable the 12V 30W siren.

Final question:

Is your siren located near your 5V 0.3A source, or located some distance away?

If it’s nearby, then you just need a single relay which can be triggered by 5V at a smallish current to turn on the roughly 2.5A 12V siren load.

If it’s far from that 5V source, then you may want two relays – the first, a small relay capable of driving the coil of a remotely located second relay which in turn, powers your siren. This is because a long cable running higher current is less desirable.

Assuming the first case (siren relatively close to the 5V source), then I would suggest something like the DC60S7 as a solution.

(3A version shown)

I agree that solid state is a reliability advantage, especially in that humid climate. This 7A version allows you to operate without a heatsink, as it will handle at least 3A at 80°C without any heatsink. It will take less than 5mA from your 5V source to trigger, so that’s no problem.

The battery that runs the siren is only about 12 feet away from the alarm panel. So running a wire from alarm panel to relay then connect that to battery/siren circuit would work? I like the idea of that all enclosed part you referenced with terminal lugs. Much easier for me to deal with :slight_smile:

And to confirm, it’s NO, so that when the current from alarm panel goes off, it will open the circuit again and siren will stop?

Correct. It’s normally open, so it only closes the circuit when the 5V source is active. When no 5V on the relay input, the siren will be off.

That should be no problem. If you have the relay near the siren, you’ll only need a small 2-conductor wire to supply the 5mA 5V signal to and from the input of the relay.

Thank you so much for your help. I ordered the unit and per the datasheet, it looks pretty straigtforward to connect

You’re welcome! Let us know if you have any further questions.

You’ve been so very helpful, but I have one more hopefully small request: I can’t figure out how I should wire the load to the relay. Ive attached a photo of my parts.
Is it possible to show me how to complete the circuit, so the siren comes on only while receiving a single from the alarm panel?

Thank you!!!

No problem. See drawing edits below.

Note that you do not connect contact 2 of the relay to the negative contact of the siren. The connections are shown in blue with little embedded arrows showing the current flow direction from battery to #1 relay contact, out #2 relay contact, into positive contact of siren, out negative contact of siren, and then back to negative battery terminal.

I added an optional diode to the diagram connected between the positive an negative terminals of the siren. This would only be necessary if the siren happens to be an inductive load, which is not too likely but possible, and depends on the internal construction of the siren. The purpose of the diode is to prevent voltage spikes, which would be created when an inductive load is turned off, from damaging the solid state relay’s internal components.

The diode is a good insurance policy if you are not sure about your siren. Something like an SB5200 would be suitable. You would wire it with the stripe on the body of the diode connected to the positive terminal of your siren.