Arduino Due Switching 5v?


I am using an SSR to control a heater. The SSR is rated between 3v and 20v but it will not turn on at 3v. It does turn on at 5v.

I have the arduino due set up to control pull low when the heater comes on. Will it damage the arduino due if I connect the otherside of the SSR to +5v instead of +3v?

(im pretty sure the answer is yes, but i was hoping for an easy solution instead of wiring up a mosfet).

I believe it is risky, it [u]may[/u] be OK. However is it not a recommended condition.

There are two considerations:

1) What is the current required by your SSR input if it were connected to 5V? Does it exceed the recommended for the due. If yes you can stop here and consider the Mosfet

2) If the above is "OK" then we have to look at the input circuit of the SSR. It likely is a diode and a resistor. If the input draws little or no current when connected to 5V - 3.3v = 1.7 Volts then you might get away with it. I'm guessing on what "little" current is but I would guess less than 200 to 500 µA.

I realize that the answers to the above questions require some testing, I don't know a way around it.

Perhaps if you has a cheap china ProMini running a 3.3 volts you could perform a test with it. If you damage the ProMini you only loose a few $$


JohnRob: I believe it is risky, it [u]may[/u] be OK. However is it not a recommended condition.

Yup, I reckon that's the only consideration.

For the sake of one transistor/mosfet just do the job properly.

Yours, TonyWilk

The Due pins cannot provide much current at all. Different pins are rated differently as well, so you need to check the datasheet for the Due's processor and your SSR. You'll likely need a transistor/FET to boost the current.

SSR's are opto isolated, the input is an infrared LED, and typically these are rated 15mA or so, beyond any of the Due's pins if I remember correctly.