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2I've bought a smoke alarm with a relay for my 3D printer. If the smoke alarm detects smoke it switches its internal relay. With this I thought I could switch off the printer.

The thing is that after buying it I found out it only goes to 1A at 30V. Then I figured out I have a spare 5V ywrobot relay laying around.

When running the 5V though the relay of the smoke alarm (NC,) can I use it directly to switch the 5V relay for the printer? Do I need something in between or does the relay shield take care of the rest? (So no resistors in between or anything.)

enter image description here

-- EDIT 2 --

I think I've created some noise :) I've made a new diagram that removes the extra devices, just a simplified setup.

After edit one I just winged it and tried to connect 5V to the Pin IN1. This did not work. But connecting Pin IN1 to GND did switch the relay shield. So I guess I have to rephrase the question:

Can I safely connect Pin IN1 directly to GND (Ground)?

I currently got it working like the diagram below. Should I add a component between Pin IN1 and GND? Or would this work just like this.

The rest of the devices are within the volts and amp ratings of both relays mentioned at the start

I've removed the coil from the diagram because it did not represent the entire shield. Now they are labeled by pins.

enter image description here

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2 Answers 2

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Google translate. I assume that the detector power supply is independent.

By using a wireless circuit and remote relay supply , you will not need to connect an additional relay to the 230 V supply network by wires that are always liable to "come loose" ... and your wiring will be simpler. It will be safer and you can also force the shutdown by the remote control if necessary.

No direct or indirect connection of smoke sensor with main, only through remote control.

But verify your battery some time ...

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  • \$\begingroup\$ Why would it be safer? The smoke alarm has a dedicated module to switch external devices. And the relay shield is already switching the printer. I only change the power supply of the shield from a raspberry pi to a usb power supply and a wire to/from the alarm's module. The rest is already existing and working. \$\endgroup\$
    – Tim
    Jul 26, 2021 at 7:12
  • \$\begingroup\$ Google translate. I assume that the detector power supply is independent. By using a wireless circuit, you will not need to connect an additional relay to the 230 V supply network by wires that are always liable to "come loose" ... and your wiring will be simpler. \$\endgroup\$
    – Antonio51
    Jul 26, 2021 at 7:26
  • \$\begingroup\$ I would even more like a powerstrip with a relay build in. But I can't find those in Europe. (with terminals for the input) Your suggestion would definitely work. Although I might use an optocoupler to isolate the devices. Since they are only 1m apart. \$\endgroup\$
    – Tim
    Jul 26, 2021 at 14:37
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Electromechanical relay has a coil and contacts. Coil rated for some work voltage and contacts has max voltage(AC or DC) and max current. If your printer's voltage and current inside those parameters, you can connect it straight to sensor. Additional relay you need in two cases:

  • sensor's output only NO but you need NC
  • printers power supply over the sensor's contact rates.
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  • \$\begingroup\$ Thanks, I think I phrased my question wrong. Altough all components are within the ratings of both relays. The biggest load is the printer with roughly 5apms, where the relay is 10amp rated. I think this would also keep a margin for spikes. \$\endgroup\$
    – Tim
    Jul 26, 2021 at 14:39

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