Electronic ignition install help needed

[VFilms]

Looking for a little help installing a Rick Ehrenberg electronic ignition with the new distributor. According to the instructions I need to run the blue wire from the Hirev7500 module to the ‘run’ side of the ballast resistor. I know the ‘start’ side of the resistor feeds 12 volts to the coil while cranking after which the ‘run’ circuit in the ignition switch takes over and reduces the voltage through the resistor. Here’s my question. As there is no power to the ‘run’ side of the resistor while cranking what feeds the module while cranking. This is my first electronic ignition install and I don’t want to damage the module. Thanks

 

[VFilms]

You're okay, just follow the wiring instructions. The ECU doesn't get or need full constant 12v when cranking. And as is often preached (for good reason!) make sure your box has a good ground.

@beanhead Thanks. This is how I've wired it, as per the instructions. I'm just curious, however, what powers the ECU upon cranking? If the ECU is connected to the 'run' circuit of the ignition switch and on the resistor (blue wire) there is no power going to the ECU until it comes off the 'start' circuit and is in the 'run' side of the resistor. You say the ECU doesn't need 12v while cranking. Just trying to understand.

 

[Ronnman]

Films,
The power is supplied from the start side through the resistor and on to the ECU. The resistor can allow power and current to flow in either direction.
Ron

 

[beanhead]

A little clarification to my original post...During cranking, no voltage is sent from the ignition switch on the 'run' wire but there will be a voltage present at that point due to the 'start' circuit backfeeding through the ballast resistor. What I meant by "it doesn't get 12v" is that it doesn't get the full 12v until the key is in 'run'.

 

[VFilms]

A little clarification to my original post...During cranking, no voltage is sent from the ignition switch on the 'run' wire but there will be a voltage present at that point due to the 'start' circuit backfeeding through the ballast resistor. What I meant by "it doesn't get 12v" is that it doesn't get the full 12v until the key is in 'run'.

Thanks
Makes sense now.