I need to measure the voltage of a Li-ion battery that is connected through a relay, on the NC pin, to the rest of the circuit (so BAT+ - relay - circuit). The coil of the relay is connected to a usb 5v port used for charging the battery. Basically when charging, the battery is disconnected from the circuit (relay open). The analog pin of the arduino is connected downstream of the relay.

I know that a mechanical relay arc when opening and closing the contacts.

Relay: 5VDC, 3A, SPDT

Max current through contacts at opening: 2.5A

Unknown value for voltage spikes when arcing; i would assume max 50-100v

  1. Do i need to worry about arcing affecting the analog pin on the arduino (both when opening and closing the contacts)? Do those voltage spikes also carry high current?
  2. If so, would a RC in parallel to the relay contacts be enough?
  3. Do i need a voltage divider on the analog pin? What would be the resistors values?
  4. Would a 5v 700mA TVS diode between analog input and ground be better? Sorry for asking dumb questions, i am new to electronics.

Thank you

  • \$\begingroup\$ Usually to protect circuits from voltage induced on a relay we need only a diode in parallel electronicsclub.info/relays.htm When relay is disconnected the EM field collapses and induces several hundred volts on the coil. For that moment the coil is another voltage source in the circuit. The diode in that case acts as "short to ground" so most of the spike goes through the diode and not through the circuit. \$\endgroup\$ Nov 19, 2018 at 14:04
  • \$\begingroup\$ I know about the flyback diode,thank you. But that is used on the relay coil. What i am suggesting is over-voltage protection for arcing on the contact terminals of the relay, which are separate from the coil \$\endgroup\$
    – C. Andrew
    Nov 19, 2018 at 14:10
  • \$\begingroup\$ Sorry I misunderstood the question. I'm not an expert, but I can give you a term that you can Google and read more about. It's "RC Snubber" circuit. (elprocus.com/rc-snubber-circuits) \$\endgroup\$ Nov 19, 2018 at 14:19

2 Answers 2


Do you have inductance in line with the current you wish to break? If not then you should not have any significant higher voltage generated by the opening of the relay. If there was any concern from stray inductance, a TVS should take care of it.

As far as the current being broken. While the relay may have a 3 amp rating it would be desirable to decrease the amount of damage to the surface from the arcing. And the assumption that because you are under the contact rating everything is OK is probably true, for a while.

I think there is good reason to go further to minimise any wear that would occur by not going further. My assumption is that you are using a single cell lithium battery. Voltage from 4.2 volts down to around 3 volts. Any voltage drop you have is a lot more significant at these lower voltages compared to switching 100+ voltages. Most relays are not specified to work at the low voltages you will be using. There often is a minimum voltage and a minimum current level specified.

Because you are switching DC current that means that you will have metal migration of the contacts in one direction. With AC the migration tends to balance out with random switching.

It is desirable for the opening of the relay contacts to happen as fast as possible to break the arc. You have stated that you know about reverse polarity diodes across relay coils to control the back emf pulse when the field collapses. While it is a cheap and easy solution, it is far from ideal. The diode will cause the magnetic field to collapse slowly. (relatively speaking) The switch contacts will also open slowly as a result of the slow decay. This will extend the time of the arcing and that is not desirable.

Automotive applications are mostly 12 volts dc. The switched currents are almost always dc. One of the things they do to decrease the switching time, the arc time, is to use resistors across the coils to absorb the back emf pulse. They balance the resistor's value with the coil current in order to get a faster decay time with minimal flyback voltage.

Take a look at these two links for further explanation. On a relay coil, why use an MOV instead of flyback diode?

Why don't relays incorporate flyback diodes?

  • \$\begingroup\$ Thank you for the note about the flyback time. Regarding arcing and stray inductance, would adding a RC snubber on the contacts be better than a TVS? I think it does both things at one: absorbs inductive spikes and prevents or diminishes arcing \$\endgroup\$
    – C. Andrew
    Nov 20, 2018 at 14:01
  • \$\begingroup\$ The problem is not one of inductive caused voltages. Even if there were no inductive caused voltages you still would have the problem of arcing. When there is enough potential across the circuit, electrons will continue to flow. In your case that isn't a lot of voltage so there wouldn't be much potential to hold an arc, but it still will happen. \$\endgroup\$
    – Rudy
    Nov 20, 2018 at 18:02

As long as the relay has sufficient braking capacity, you don't have to worry about it. And this seems to be the case for you, if the max. current is 2.5A and the relay contacts are rated for 3A

The arcs stay inside the relay. They send out some radiation but it doesn't damage electronics. If the braking capacity is not sufficient the arc will weld the contacts and they won't open anymore. Still it won't directly damage anything else.

It is not possible to avoid the arcs with mechanical relays, if there is current flowing when the relay is opened.

  • \$\begingroup\$ Thank you, but as far as i know a RC parallel to the contacts prevents or at least diminishes arcing. \$\endgroup\$
    – C. Andrew
    Nov 20, 2018 at 14:03
  • \$\begingroup\$ The arc happens because of voltage. When the contacts are very near to each other, but not touching, there is an arc if there is voltage. The arc stays on as long as there is enough current to hold it. When the contacts go further away more current is needed to keep the arc burning. \$\endgroup\$
    – TemeV
    Nov 20, 2018 at 14:54
  • \$\begingroup\$ This deals with arc suppression jaxxeninc.com/pdfs/pcb-relays/tech_info.pdf I am not saying you are wrong but maybe what you said applies to this in the sence that the capacitor absorbs the current (?) \$\endgroup\$
    – C. Andrew
    Nov 20, 2018 at 15:00
  • \$\begingroup\$ Yeah, I googled a bit, and it seems I might be slightly wrong here. RC might indeed be helpful there. \$\endgroup\$
    – TemeV
    Nov 20, 2018 at 15:31

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