Ballast resistor Ohm value

[1952m37lw]

I am trying to figure out the working voltage and ohm reading for an external ballast resistor and coil setup. I have a power wagon distributor running on 24 volt and need a external ballast. Does anyone know what ohm rating the ballast should be on a 24volt system? Thanks

[NAM VET]

I have no idea how to do this, as he whole Electron thing is a mystery to me. But hopefully someone will be along to suggest what is necessary. Good Luck, Hal

[1952m37lw]

I did the math and it takes about 6 ohms of total resistance to get around 21-22 volts. The coil had about 5.2 ohms so I put a 1 ohm resistor and idles about 21.5 volts. Hope this info helps someone else out.

[vit16]

thanks for the info!!.
can you add a pictureans the resistor wattage?
Ronnie

[1952m37lw]

Wattage is above my knowledge all I did was use ohms law then guess and check with resistors I already had.

[PoW]

You need to find out whatever coil you are using needs.

Most coils over 12 Volts use a resistor that is calculated by the actual coil specifications for voltage and current (amps).

Without that data there you can't derive a 'fixed' resistor value.

PoW

[just me]

You need to know the voltage the coil wants to see to determine the resistance.
E/I=R
Then you need to know the current to determine wattage of the resistor.
E*I=P (watts)

So, if it draws 3 amps at 24 volts and you want the voltage at 18 volts you would need a 6 ohm resistor in series.
Now, you can determine the wattage the resistor needs to dissipate. (approximately 75 watts) I would use a 6 ohm 150W resistor in that case.

You need to make some measurements to get the value you need. Then the value of the capacitor in the distributor needs to be determined. This is black arts. Somewhere I have a chart to determine this.

[PoW]

Isn't that what I just said?

PoW

[vit16]

thanks guys

[John Mc]

You need to know the voltage the coil wants to see to determine the resistance.
E/I=R
Then you need to know the current to determine wattage of the resistor.
E*I=P (watts)

So, if it draws 3 amps at 24 volts and you want the voltage at 18 volts you would need a 6 ohm resistor in series.
Now, you can determine the wattage the resistor needs to dissipate. (approximately 75 watts) I would use a 6 ohm 150W resistor in that case.

You need to make some measurements to get the value you need. Then the value of the capacitor in the distributor needs to be determined. This is black arts. Somewhere I have a chart to determine this.

You may be missing something there. If the coil is drawing 3 amps at 24 volts without the extra resistor in there, the coil's primary resistance is 8 ohms: E/I = R gives 24/3 = 8.

The voltage drop across a resistor in series is proportional to that resistor's proportion of the total resistance of the circuit. If you want to drop from 24 volts to 18 volts, that's a drop of 6 volts. The total drop across the entire circuit is 24V. 6/24 = 0.25 (or 1/4 of the total). So if R is the desired resistance for the added resistor and 8 ohms is the primary resistance of the coil, you need to solve R/(8 + R) = 0.25. If you solve for R, you get R= 2.67 Ohms. (Also note that if you were getting a 3 amp draw before you insert the extra resistor, you will no longer be getting 3 amps once the resistor is installed.)

Of course, these are just example numbers. A running engine is probably closer to 28+ volts (at least with a modern alternator. I'm not sure what the original generator was set for). Resting voltage of two good, fully charged 12 volt flooded lead acid batteries in series should be close to 26V (though that really only applies when first starting the truck.) I'm guessing the 3 amp load mentioned was just an example. You could try to measure the actual current draw of the coil, but my guess is that would be problematic, since the draw is intermittent when the engine is running. It might be easier to measure the resistance of the primary of the coil directly with an ohmmeter. Note that the NOS coils have a significantly different primary resistance than the aftermarket coils, so if you actually do need a voltage dropping resistor, the value needed will change if you change from one type of coil to another.

Having said all that, I'm pretty sure my Pertronix ignition is just connected directly to the ignition switch with no resistor (I will double check next time I'm at the truck.) It's running just fine -presumably because it is designed for a 24V electrical system