# Snubber Network Over a Switch

I'm a newbie to electronics even though I have been working on them for a few years in mostly a hobbyist setting. I'm posting here for the first time so please be gentle.

The problem I'm having with my old car is that the headlight switch's contacts are getting eaten away (it's apparently a common problem with my model). I'm not sure if a relay is being used on the circuit or not, I will have to check, but I assume the lights are being driven directly.

I have a 12 V to 15 V power supply (car voltage) driving 2 headlights (65 W * 2). So if I test the Load I should be getting around 9 amp draw (still have to test). If this is true, this will create a big spark that will eat away the contacts.

I read on the Internet that a possible solution is to attach a Snubber Network Over the Switch to reduce arcing.

So my questions are:

1. What values would I use, and how do I calculate it
2. Can I use something like high voltage home appliance electronics components to do it relatively safely?
3. Would polarity be a issue? Eg. the order in which the capacitor and resistor is connected in reverence to the positive and negative terminals.

Here is an image I made with some of the values that I expect to have on hand.

I would like to know how to calculate the values, even if you suggest other options, just to be able to protect simple switches and relays in the future against arching in DC circuits.

I see other uses for a Snubber Network would be in AC circuits and across MOSFETs and inductors. If you want, you can also explain the calculations for these because I read that they have more to do with the frequencies of circuits that these can produce, and this information could be helpful to me and others looking for it.

• If you assume 1 ms switching time and you want say 50 % of the current to pass over your RC link and you allow 100 % overvoltage, an slight oversimplification would require just under 1 mF - a huge capacitor. If you try a circuit simulator online or download for example LTspice, you can play around with some values and see how much effect it has. Commented Jun 21, 2018 at 19:36
• @winny a cap is definitely the wrong approach to handle lamp surge start current Commented Jun 21, 2018 at 21:01
• @Tony Yes. It would take an added series L to limit the inrush combined with overvoltage RC across the switch. Like you say, just get a better switch. Commented Jun 21, 2018 at 21:35
• Series L then stores energy and zaps it on open circuit in order to be big enough to have L/R > bounce time. So also bad. But perhaps slight improvement or slightly worse. Or much worse. Commented Jun 21, 2018 at 22:01

Snubbers are only used to absorb inductive and motor Back EMF now acting as a generator when turned off.

In this case the tungsten lamps have a PTC effect of about 1:10 ratio in Ohms and 10:1 in power when turned off so the surge may start at 90A then reduce to 9A when white hot. For this reason the relay contact current rating must be reduced to handle the turn on surge. Depending on contact quality with silver alloys perhaps with tungsten or titanium added to alloy plating to withstand high arc temps from contact bounce if that exists.

It may need to be at least rated for 25A or more. MTBF is greatly affected by number of cycles and surge current. Omron has app notes and MTBF specs on this application. (Not handy)

Typical use is turning on low beams and thus much less current and going from low to high beam is less of a surge since the filament will already be warm so perhaps 1:5 change in high beam R .

The contacts can handle surges greater than ratings but frequency of switching and over current ratio are factors car manufacturers decide on what relay to use and who is qualified to meet those conditions.

# Conclusion:

Choose a more reliable automotive relay for lamp surges. Much higher current rating and better contact alloy.

Lookup datasheet contact materials.

Omron and TE recommend : AgSnO2 in special cases: AgNi20 or Tungsten pre-contact

• Correct me if i'm wrong.but does a arch not occur then the switch is turning off, when the contacts leave each other they create a spark because the current wants to "stay on".So adding a capacitor over the switch will reduce the current flow over time because the capacitor's plates doesn't actually connect. adding a resistor to this will reduce the current being drawn out of the capacitor. but the values are important because if you choose too big a resistance the spark at the switch would occur anyway.And choosing too big a capacitor may delay the switch of period notably. Commented Jun 21, 2018 at 18:59
• I was thinking of using a higher rated potted ceramic capacitor that is used in some 240v equipment. but it has a very low capacitance (150nF) so i'm not sure if it would work especially with the resistor i chose. Commented Jun 21, 2018 at 19:03
• @DarkPh03n1X for a resistive load, the current doesn't want to do anything. When its on its on, when its off its off. When you have an inductive load, the current wants keep flowing in the same direction it was moving along. So when you open your contact, it still wants to keep flowing, the voltage will increase and rise large enough to overcome the breakdown voltage of air and arc and provide a path for current to flow (albeit for a short time). Commented Jun 21, 2018 at 19:08
• @efox29 thank you for clarifying. what could be the cause of the switch failure then? too high current or maybe age of the contacts on the switch ? Commented Jun 21, 2018 at 19:19
• so is this video's information incorrect? youtube.com/watch?v=Xr5_gUrUZxY Commented Jun 21, 2018 at 19:24