After my previous question I had concluded to use the following circuitry to operate this solid state relay:

enter image description here

But now I found out that, from the micro-controller(Arduino UNO) digital output pin to the relay there will be around 3 meters of wire.

Now Im worried because of the length of the wire, noise can induce voltage at the gate of the FET(?).

So I thought better to isolate the grounds of the micro and the relay input.

Im not familiar with optocuplers but could the following be a better solution for such scenario?:

enter image description here

The SSR " Input Impedance" is given as "current regulator" 16mA for 5V. So would this optocoupler need an extra resistor? I have the 4N26 at the moment.


It would be easier to protect the gate rather than adding an opto-coupler.

To protect the gate, you can add a Zener diode with a voltage lower than the Mosfet Vgs max and a 1k resistor in serie.

You can also add a small cap to dampen the noise.


simulate this circuit – Schematic created using CircuitLab

Another simple option is to replace the FET with a NPN Transistor, since the transistor is current driven, noise from the cable will have little effect.


simulate this circuit

Don't forget the flyback diode on the relay.

  • \$\begingroup\$ Gate threshold voltage for 2N7000 is 3V. Your zener is 5.1V. Is that fine? \$\endgroup\$ – panic attack Oct 8 '18 at 13:24
  • \$\begingroup\$ @panicattack The zener diode should be greater than the designed operating voltage of the circuit. \$\endgroup\$ – Hearth Oct 8 '18 at 13:25
  • \$\begingroup\$ You should look at the Vgs Max, it is usually on the absolute maximum rating section of the device datasheet. It is +/-20V for 2N7000 \$\endgroup\$ – Damien Oct 8 '18 at 13:26
  • \$\begingroup\$ The relay is SSR doesnt need a flyback diode \$\endgroup\$ – panic attack Oct 8 '18 at 13:26
  • 1
    \$\begingroup\$ No my aim was if the noise couples it would false trigger at even 4V. zener will not prevent that. \$\endgroup\$ – panic attack Oct 8 '18 at 13:26

Going from your first schematic to your second schematic doesn't really help. If anything it might be worse. It's like deciding whether beer is better than smoking.

What you want is either twisted wires, a shielded cable (coax) or differential signaling.

By twisting the wires, the electromagnetic fields that may be interfering with the loop is being twisted. Imagine just one twist, forming the number 8. Then as electromagnetic interference is interfering with both holes in the 8, the current cancels out. By having more twists, you are making more of the cable like an ooooo (imagine infinity symbol), meaning that you are less sensitive to where the noise is. This is probably good enough for you. If your design is going to be on a vehicle, or something else with a lot of noise, then you will probably want to use coax or differential signaling.

The coax is simply covering your signal with ground. The idea is that any noise that would interfere with your signal wire, is instead shorted to ground, and your signal is unaffected. This is slightly more expensive compared to twisting the wires (free). But this is also great.

If you are for some reason in an environment that is so ridiculously noisy, or working with very high frequencies where everything can be considered as noise. Or you just want to be ~100% certain that you want your signal to work as intended. Then you can have two wires in differential mode, one with your signal, and another that is negative of that value. So if you want to send 5 volt, then you would send 5 and -5 on to wires. And if you want to send -5 volt, then you send -5 and 5 volt. The idea here is that any noise that effects your signal, will also affect the other signal right next to it. So any noise will be cancelled out because you will be, on the receiving end, measure from one signal to the other. You will probably need two chips for this, one on the sending side and one on the receiving side, for your particular setup.

I'm not 100% sure if coax is better than differential signaling, or vice versa. But you can use both together which might be something you'd do for a medical product where something just is not allowed to fail. If something fails, someone dies. Just twisting the wires is probably good enough for you, though I don't know if your product is about life and death, or how noisy your environment is.

  • \$\begingroup\$ For long wires the opto solution is recommended. Since it isolates the grounds of the transmitter and the receiver. Differential signalling is not possible for me for this case. \$\endgroup\$ – panic attack Oct 8 '18 at 13:54
  • \$\begingroup\$ @panicattack I think that is for reasons which are not relevant to you. Here's why: 1) aren't you sharing the ground? Meaning that both ends are close to each other, voltage wise? Meaning that using an opto solution is pointless? 2) The channel (loop/wire) is still there, which is still susceptible to noise. 3) an opto solution is good when the LED side is floating, or on an entirely other voltage potential than your BJT (in the opto-IC), neither of which is the case in your schematic, or problem (I think, I still don't know how you are powering the two different sides (arduino & ssr)). \$\endgroup\$ – Harry Svensson Oct 8 '18 at 14:01
  • \$\begingroup\$ The power supply for the uC and the SSR Relay are different power supplies they dont share grounds. For the opto case I mean \$\endgroup\$ – panic attack Oct 8 '18 at 14:11
  • \$\begingroup\$ My problem is here since I dont know the input impedance of the SSR I cannot be sure whether I need a resistor for the collector. docs-emea.rs-online.com/webdocs/140b/0900766b8140b509.pdf It says for the inout impedance "Typical Input Current @ 5VDC! is 16mA. I dont know what does that mean. \$\endgroup\$ – panic attack Oct 8 '18 at 14:19
  • \$\begingroup\$ @panicattack That is a messy datasheet, yeah, I'm on the same confused page as you. - It looks like you don't need a NMOS / opto. - Don't you have the SSR infront of you, allowing you to test it out? \$\endgroup\$ – Harry Svensson Oct 8 '18 at 14:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.