16 vac coil relay

[Reloadron]

The relay you linked to can switch 15 amps, however, I don't see a reference to AC or DC current? The reason I suggested an automotive relay is because these relays are designed to switch DC current. That is important, very important! Thus I would use an automotive relay. The fact the contacts are rated for 30 amps is fine. Most common automotive relays, generic in flavor are 30 to 40 amps rated. That is great for your application. They can also be had for about $5.00 give or take from any auto parts supply store.

You could likely use AWG 20 for the relay coil. That coil will not draw much current at all.

This link should be to a Tyco pretty generic automotive relay. Look at the VF4 series. The data sheet(s) on these relays shows the coil voltage at 12 volts (with plenty of room) and the coil resistance of about 90 Ohms. That tells us at 12 volts the relay coil will draw about 133 mA. Not much at all. So lets say at the cap where we get our DC we measure about 22 volts (E peak at the cap less a few volts) So I want 22 - 12 = 8 / 133 = so I want about a 60 Ohm maybe 2 Watt resistor.

The reason for all the "approximates" and "abouts" is because this is not a very exacting science. The 12 volt coil actually will run fine at 14.5 volts in an automotive system and actually pull in as low as about 8 volts.

Just make sure you place a diode like a 1N4004 across the relay coil with the diode band (cathode) on the positive side. Place the resistor on the positive side. Keep all commons at common. Use the relay to switch the 12 volts from the battery or your other 12 volt source for the outside lights.

Hope that helps
Ron

 

[abukazoobian]

Ron-

Thanks for all of the help.

Question,

what determines what watt resistor you would use?

and why does 22-12=8? or does the missing 2 volt get accounted for by the 2 watt on the resistor?

Finally, if we want the lights to last longer we would replace the batteries with ones that have a better amp hour rating?

 

[Reloadron]

Well the theory works like this. If the transformer had exactly 18 volts and went into the full wave bridge we would have some voltage drop across the bridge. As the diodes in pairs conducted on alternate half cycles of the AC input they would drop about 1.2 volts. Thus we would have about 18 - 1.2 = 16.8 volts. Now the cap acting as a filter to smooth the DC unloaded would charge to E peak or the peak value of the RMS value we have and that would be about 1.414 times RMS or 16.8 * 1.414 = 23.755 Volts DC. That assumes unloaded which it won't be. However, we can work with that. I used 22 volts earlier.

Back to the relay coil. In the example we had a coil that would draw 133 mA at 12 volts. The coil resistance was 90 ohms. Now if we were to directly apply 22 volts to a 90 ohm coil it would draw about 244 mA and the coil would smoke. So we need to limit the current and get it down to about 133 mA. Thus we take our 22 volts and subtract the 12 volts (coil) and that leaves us in this case 10 volts. Now we have 10 volts / 133 mA = 75 ohms. Short of buying a precision resistor we need to work with something off the shelf and common. That leaves us a 62 or 68 ohm resistor. Now to the power. One power formula is the current squared times the resistance. So we have (.133 * .133) = .018 * 68 = 1.2 watts therefore a 2 watt resistor. A 62 ohm 2 watt would also work. Remember automotive relays are pretty forgiving. They typically see between 13.6 and 14.5 volts or more. Thus I say it is not a very exacting science. The voltage drop across the resistor would be .133 * R or in the case of a 68 ohm resistor .133 * 67 = 8.9 volts. I would just find a 67 ohm 2 watt resistor and run with it. Again, this is pretty forgiving.

I would buy a common automotive relay like I linked to. Measure the coil resistance. If it is around 90 to 100 ohms I would just use a 67 ohm 2 watt resistor.

Ron