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I have to open/close two wire from a closed device. I cannot open due to it's under warranty, and the only part that I can open and modify is an external button that connects the two wires.

Firstly I wrongly thought that I can simulate the button press ion with only optocoupler (4N25), but thanks to this answer I understood that this piece of hardware is not only an opto-isolator but an opto-transistor: it is better to use a SSR instead (a mechanical relay will not work because of the noisy "tick"). I don't have the GND signal but only Vcc and Vo (thanks to Peter R. Bloomfield for this elucidation).

Just to recap, my needs, I have in one side an Arduino Due that runs at 3.3V. In the other side of the circuit I have a signal that, when closed has 0-6V.

I have seen some other questions about how to simulate a SSR like this and this bit as I got this is not tailored to my case: I don't need a zero-cross detection because I have to drive a DC circuit, and I need to drive it from Arduino Due that runs at 3.3V.

What can I do? Are there some cheap SSR that fit my 3.3V driver and DC lead? Is there some simple opto-isolator that I can use without the GND signal (no opto-transistor)? Or something I can add to the 4N25 that I already tried? Is there some other alternatives that I ignore?

Please sorry my entry level in the electronics so correct me in terminology if I have used unsuitable words.

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nkint
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2 Answers2

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I have used a Photovoltaic relay for this in an Arduino driven application, the IR PVT312.

It is designed to work as a substitute for a Relay, and is rated to switch up to 250V at 190mA. It is suitable for AC and DC switching.

As you can see from the diagrams below, it's an opto-isolated MOSFET switch.

enter image description here

akellyirl
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  • this seems to be perfect thanks, I will look into it! – nkint Dec 18 '14 at 12:41
  • Then another question, with the 190mA do you mean the driving circuit or the load? – nkint Dec 22 '14 at 19:17
  • @nkint The load. 190mA in 'A' connection, 210mA in 'B', 320mA in 'C'. – akellyirl Dec 22 '14 at 20:48
  • it seems to me that it is not so much. thank you anyway! – nkint Dec 22 '14 at 20:50
  • last thing: is it possible to drive it directly from a 3.3v arduino right? it is only an led.. I don't need any driving transistor neither a resistance, right? – nkint Dec 22 '14 at 20:52
  • @nkint It can be driven directly from a 3.3V Arduino, no transistor required, a 100 Ohm resistor in series would be advisable. At 6V 320mA is 1.8 Watts ; that's a reasonable amount of power. At 240VAC it can drive 240V x 190mA = 45 Watts. – akellyirl Dec 23 '14 at 11:20
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I'd use a small reed relay. This way you avoid a small voltage drops on solid state relays but it uses more power.

If reed relay is not fast enough for switch it on/off, look for DC voltage solid state relay ICs. Plenty of these to choose from.

Alexxx
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