Is pin S4 on the Photon 2 5V tolerant?

I have been through the Photon 2 docs and I don't see any warnings about S4. So I am wondering if I can use a 5V signal for something.


Hi @JGist -

As this used to be A4 on previous Gen 3 devices, my immediate guess would be no it is not. To be safe, you can pass the signal via a voltage divider or optocoupler?

What type of signal are you expecting and where does it originate from?


No pins on the Photon 2 are 5V tolerant.

On the P2, VBAT_MEAS is; that pin is connected to Li+ on the Photon 2.

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Yeah fair question. I am looking for a workaround to the CHG variable not working. More details in this thread P2 charge indicator. I have a project that I need to stay awake if the USB is connected. I was going to connect VUSB to S4 to know if the Photon is plugged in. I will probably just use a voltage divider.

Hi @JGist -

Ah I see... a divider should work. Of course there are other options. You can trigger the gate on a MOSFET sending 3V3 from the device to this pin. Or you can even use a LDO stepping the output down to 3V3 (probably what i would have done) . Or a Zener diode (probably my least favourite option to be honest) I guess the diver might be the cheapest way.

Hope this helps.

Regards, Friedl.

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Yeah super helpful.

I am embarrassingly weak when I comes to transistors. I just glaze over when doing a digikey search. An LDO is an interesting thought. At first glance they seem pretty straightforward and not very expensive. I will poke around.

Thanks @friedl_1977 & @rickkas7

Hi @JGist -

No problem at all, glad I could share some ideas!

FET's are very useful, but there are some things you need to know. LDO's are very straight forward... well, the simple ones. There are some that are more involved, but for you purpose a very simple one will suffice.

As long as you provide power to it's VIN pin, it will provide regulated output.

If you need some help selecting a LDO, do not hesitate to ask. Alternatively, a simple voltage divider will work as well, just keep in mind the voltage on the output will fluctuate with the as the input voltage varies.

Regards, Friedl.