Discussion in 'General Electronics Chat' started by Ingeniir, Jun 29, 2011.

1. ### Ingeniir Thread Starter New Member

Jun 29, 2011
19
0
Hey all,

Is the following a valid path of logic:
I need to switch on a 25 watt lightbulb using mains power. I have an automotive relay whose SPST contacts are rated at 30A at 12VDC. The voltage running through this circuit would be 120V, which is 10 times the voltage rating of the relay, but the current would be less than 1A, which is well below the current rating of the relay. I know that components usually fail because they are passed too much power, which overheats the part and destroys it. Since this relay is rated for 30A at 12VDC, using P=IV, it can handle 360 watts of power, therefore using a 120VAC lightbulb is safe.

My guess is that I made some incorrect statements and conclusions, can someone clear this up for me?

2. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
No, you are not figuring correctly.

The relay contacts must be rated for at least the voltage across the circuit that they will be making/breaking. The contacts in your automotive relay are only rated for the automotive voltage range. If you attempt to operate it from mains voltage, the relay may not be able to break the arc, and a fire may result.

Always use electrical components that exceed the requirements of your circuit, particularly when working with mains power.

Also note that incandescent lamps draw far more than their rated power when electricity is first applied; when the bulbs' filament is cold, it is seen as almost a dead short. As the filament (rapidly) heats up to emit light, the resistance increases a great deal. However, the initial surge of current may exceed 10x the rating of the bulb. This high surge of current as the contacts of the relay are closing will tend to weld the contacts together.

So, 25w x 10 = 250 Watts; 250W / 120VAC = ~2.083 Amperes. You should use a relay with contacts rated for 125v or more, at >= 2A.

Ingeniir likes this.
3. ### Ingeniir Thread Starter New Member

Jun 29, 2011
19
0
Thanks a lot, that was a very informative answer!

4. ### WellGrounded Member

Jun 19, 2011
32
2
The relay that is attempting to be used may be OK.

Upon initial closing of the contacts there is virtually no resistance at the contact point. This is a heavy duty relay if it is rated for 30 Amps, therefore there will be a high current going through the coil to keep high pressure on the contacts. There will be less than 0.1 Ohms contact resistance. The latching is almost instantaneous to make a secure contact. The power going through the bulb is AC so that the initial surge is not a constant current load but 120 cycle pulse making the initial heating up of the filament a series of + and - sine wave pulses rapidly decreasing in current strength.

When the current crosses the first 0.0 Amp flow crossing of its cycle there is no heat threat to contacts. The only threat would be if the contacts closed a fraction(milliseconds) before the current started to rise on its' positive or negative cycle, and even then the heating would only last one cycle = 1/120th of a second = 0.008 seconds. If we use your example of surge current then the temporary contact heat would be 2.0 Amps X 0.1 Ohms X 0.1 Ohms = 0.020 Watts, which is negligible.

When the contacts break is when you have the greatest risk of damage. In order to sustain an arc the current draw is very important because it is the current that generates the heat to keep the air in the gap ionized. Since this an AC resistive load at about 0.008 seconds from interruption of the contact there will be no AC voltage across the contacts, so there will also be no current. In the next 0.008 second part of the cycle the 120 Volts won't jump the separating contacts, otherwise the bulb would come on when the relay was not powered. If the contacts take a bit longer than 0.008 seconds to open fully and the AC goes through several cycles the same situation arises where the the 120 Volts starting up has to jump the widening gap to get continuity which it can't do after a few 0.008 cycles.

There is more of a danger of arcing with a DC source where there is constant voltage source, supplying a constant current source to allow arcing to continue should the relay contacts fail to separate far enough.

When the light bulb is on the maximum current at peak voltage going through the contacts would be about ~0.3 amperes that would last for a duration of about 1/2(the peak) of 1/120 of the cycle time and is not really a lot of current to sustain an arc.

Please note that since this is a car relay it is not expected to have AC voltage going through it and thus there is no reason to give it an AC rating, but that doesn't mean the relay is not durable enough for the contacts to withstand AC use. It is only that we don't know the upper 120 VAC Ampere undocumented limit is, but I suspect it would far higher than 25 watts since it is DC rated for 30 Amps.

I would say try the relay through a 100 ON/OFF cycles where the the light bulb filament has gone off and look at the contacts with a magnifying glass and see if there is anything more than the very tiny black mark that you sometimes see on the exact contact points of the contacts themselves.

Danny

5. ### ifixit Distinguished Member

Nov 20, 2008
639
108
Hi Ingeniir,

IMO the electrical specifications are not relevant.

If the relay is not UL approved for use with 120VAC then I would recommend not using it for that application. You would be liable for unforseen circumstances that result in; property damage, personnel injury, or death.

Experimenting on the lab bench is one thing, but using unapproved products in an application that is expected to work reliably and safely is unwise.

Play safe,
Ifixit