Auto ElectricsPart 1 



... on 12 to 6 volt reduction  
The subject of running 6 volt devices from a 12 volt battery often comes up.
From the general discussions I've seen, some important information is overlooked. This is very simplified, but I hope this helps some. I drew these diagrams with the intent to explain using a ballast resistor with a 6 volt coil, but the electrical theory applies equally to any voltage reduction application. 

There are 2 fundamental laws of electricity that apply here:
Kirchoff's law states that the sum of all voltages in a closed circuit is always zero. Ohms law defines the relationships between voltage, current, resistance and power 

Starting with a 6 volt system, consider the circuit here. The resistance could be anything (horn, headlights, gauges, coil, etc).
I've chosen to use a 1 ohm value for simplicity. Ohms law states: V = I x R (Voltage = Current x Resistance), or in this case, I = V ÷ R We know the voltage (6 v) and the resistance (1 ohm), so by laws of physics, the current must be 6 amps. You can rearrange the equation any way you want, so knowing any 2 of the 3 variables, you can always solve for the 3rd. 

Now consider the same circuit with a 12 v battery.
With no change to the "load" resistance, I = 12 ÷ 1, so the current has increased to 12 amps. (2x the voltage = 2x the current) The Power laws defines power dissipation: P(watts) = V x I With 6 volts, P = 6 x 6, or 36 watts; with 12 volts, it's 12 x 12, or 144 WATTS!!! 4 x the original! That poor resistor is going to get DAMNED hot, and probably die quickly! This is the major problem running 6 volt devices from 12 volts. While they might seem to operate OK, they'll get far hotter than they're designed for, and might well burn up. 

Now let's keep the 12 volt battery, and reduce the current through the load so it won't burn up.
To get the current back down to 6 amps, you'd need to double the resistance. That's what a "ballast resistor" or "voltage reducer" is for. By adding a 1 ohm ballast resistor, total circuit resistance is now 2 ohms, an since I = V ÷ R, the current is back down to 6 amps. Total power dissipation is still twice what it needs to be though: P = 12 x 6 = 72 watts. Half this power is consumed by the load, so the load device is back to it's normal values and power dissipation. The other half though, is simply wasted as heat in the ballast resistor. 

To understand how the ballast resistor functions, we apply Kirchoff's law. In our circuit, voltage across the ballast (V1) plus the voltage across the load (V2) must add up to 12 volts.
And we know the current (6 amps) and the resistances (1 ohm each), so applying ohms law, V = I x R, we get 6 volts across each resistance, which do indeed add up to 12 volts. What should be clear now, is that the VALUE of the ballast resistor must be EQUAL to the resistance value of the load! Say we changed the ballast resistor to 2 ohms, current would be 4 amps (possibly too little for the load), and voltage across the load would only be 4 volts (again possibly too little). Decreasing the ballast value would have the opposite effect. So there's NO SUCH THING as a universal ballast resistor or voltage reducer. They are tailored to specific applications, makes and models of parts. At the least, you MUST know the resistive value of the load you're trying to reduce the voltage to. 

You may have seen these things advertised as voltage reducers.
Unlike a purely resistive reducer, these are actually a solidstate voltage regulator. You can apply any voltage from 8 to 30 volts to the input, and the output will always be 6 volts. These are as close to a universal reducer as you'll get, but there are drawback and limitations:
To save money, go to Radio Shack and buy the part for @ $1.00, and wire it up yourself. For heavierduty applications, you can buy voltage regulators up to 5 amps (for @ $5.00), but they need a heat sink. Still cheaper and more accurate than a resistor. 


