1970john

My old starter has two smaller terminals R and one is S. The new starter has only one smaller spaded terminal and the Battery. Does anybody know how I can wire the old wiring to the new starter?

S489

i connected the wire from the wiring harness to the S terminal, i think. it's the one that engages the solenoid. the other wire i wrapped in the harness. have had no starting problems. if memory serves, the R terminal sends 12V to the coil instead of the normal lower value during cranking.

69427

The starter/coil connection shunts the ballast resistance. It does not change any ignition voltages.

Theiskell

Check out this thread there is some good info here.
http://forums.corvetteforum.com/c3-t...-wiring-q.html

Also do a search on “starter wiring” Lots more information in the search feature. The bottom line is, if you are going to use your old points distributor and coil with a mini starter you’ll need to do one of two things. Eithere put a diode in line with the wire feeding power back to the coil or put a relay in line. If you send 12v back to your ignition it’s not going to last long. The diode or a relay can solve that problem.

S489

thanks for correction

1970john

Thanks for the info

S489

i guess this is what i was trying to say:
"Ballast Resistor: This is an electrical resistor that is switched in and out of the supply voltage to the ignition coil. The ballast resistor lowers voltage after the engine is started to reduce wear on ignition components. It also makes the engine much easier to start by effectively doubling the voltage provided to the ignition coil when the engine is being cranked. Not all car manufacturers used a ballast resistor in their ignition systems. . ."
http://autorepair.about.com/cs/gener.../aa052502a.htm

midyearvette

..couldnt have said it better....this is all controlled by the ign switch and solenoid during cranking..when the key is released to run, the br takes over and provided 5-7 volts to the coil...while cranking, the post off the solenoid "hot wires" the coil......

WESCH

Hi

I also got a high torque mini starter from SUMMIT andf faced the same problem with my 68 BB. There was a instruction sheet with the starter mentioning the possible requirement of the diode.

Unfortunately, the new starter smoked from the solenoid very shortly after installation and the solenoid burned out.

Summit replaced it instantly without any arguments or questions.
Great people to deal with. They even reimbursed return shipment from Europe which came to 80 USD for a starter that costs 125 USD.

The new starter is in Europe in customs now. Have to wait till after the holidays to get it and will try the 2nd install.

I had slight problems with my first one, installed all 4 shims , but this was still not enough to clear the gears. Anybody else having this problem ? Hope, the replacement one fits better.

Rgds. Günther

69427

Sorry, but Mr. Autorepair is incorrect. "He's" merely parroting an old wive's tale that's been circulating for years.
My dad explained the starter wiring reason years ago when I just started to drive. His explanation made sense to me. I then learned the physics of it in (electrical) engineering school a few years later.
The ballast resistor limits the current passing through the coil and points. It does not magically change any ignition voltages. For all intents and purposes, the coil and points don't give a crap what the system voltage or battery voltage is (6, 12, or 24 volts). It's the current going through the system that burns things up (similar to burning out a fuse).
At slow speeds (idle, or even cranking) the points and coil can handle a steady current of about four (4) amps. At a nominal 12 volt system voltage, we find that the coil windings have a resistance of about 1.5 to 2.0 ohms. If we place a 1-1.5 ohm ballast resistance in the system, we get a coil charging current (dwell current) of 12/ (1.5 + 1.5) = 4 amps. Voila. We now have enough coil current to allow a sufficient amount of energy to be stored in the coil to generate a strong spark at ignition time. Life is good.
Now, what happens on a cold midwest or northern January morning? As the starter is trying to turn over an engine full of extremely thick oil, the starter will pull a lot of current. Lots of current out of a battery will cause its voltage to drop. Compounding this, a cold battery is a crappy source of energy. These two factors (lots of current out of a cold battery) will cause the battery and system voltage to drop significantly. What happens when the battery voltage drops to 6 or 7 volts? Lets look at the spark energy. Six volts divided by (1.5 + 1.5) ohms gives 2 amps. Chances are the plug isn't going to fire when we've cut the current in half (the energy drops by a factor of 4 when we do that.) As the vehicle manufacturers don't want a bunch of unhappy customers every winter, they have to do something to cure/bandaid the battery issue. As the manufacturers can't know which vehicles are going to be started in warm climates, and which are going to be started in cold climates, they just put the same compensation/bandaid circuit in every car. If during cranking we add some wiring that will shunt (short out) the ballast resistor during cranking, what will happen? At the same low battery voltage (6 volts), and only the coil winding resistance (1.5 ohms), we get 6/1.5 = 4 amps. We now have the same spark energy at 6 volts as we had previously had at 12 volts. Nothing magical happened to change the coil/system voltage back up to 12 volts while it was cranking.
The down side of this simple bandaid is if you have a large battery on a warm day, trying to clear out a flooded engine. If the battery only dropped to 10 volts during all this cranking, we have a coil/points current of 10/1.5= 6.7 amps. Probably burn up the points, and possibly overheat the coil. Notice that this damage can occur even though the battery is lower than 12 volts. It is the current that burns up stuff (I squared R equals wattage).
That's it. Mr. Autorepair should do a bit more proofreading on his publications.

1970john

I have a HEI distributor so it looks like I don't need to connect the R terminal up at all since I don't have a coil.

69427

If someone has converted your engine to an HEI unit, you are correct that you don't need the extra starter connection. I assume you have made certain that the ballast resistance has been bypassed with this distributor. The extra energy available with an HEI is only available if it isn't choked off with the ballast resistance.

(ps: you still have an ignition coil.)

WESCH

Hi

69427, what you write is correct, but nothing else as was written B4.
The only constant is the resistance of the coil, so the Voltage drop by installing a resistor in series with it will regulate the Amps passing though.

Of course the spark strength is measured in Watts, which is depending on Voltage and Amps.

Why are you so strict trying to say that the Voltage drop does not do anything to the ignition system ?

But as I wrote, your version is OK.

Can anybody explain why a HEI does not need this resistor ? Also in a HEI, there is a coil with a given fixed resistance and during cranking, the Voltage drop with weaken it's spark strenght, or ?

Rgds. Günther

69427

I had the good fortune on my first job out of college to be mentored by the guy who designed the HEI electronics. He was, and continues to be, a very bright guy.
The HEI module has circuitry in it to measure how much current is actually passing through the coil (and module). Once sufficient current has built up (it takes time), the module will slightly shut down the module power transistor (that replaces the points) so that the current cannot build up any higher and possibly damage the coil. When the correct advance point occurs, the transistor shuts off completely (analogous to the points opening), and the coil fires. All system current limitation is done electronically (and accurately) despite the battery voltage at that particular moment. With the elimination of the ballast resistance, the coil will charge faster (good for higher RPM use compared to points), and the coil can charge to full current during lower battery conditions. Battery voltage can affect how quickly a coil will charge, so the HEI will automatically adjust its internal dwell time. At low RPM or high battery voltage, the module might only need 20 degrees dwell to charge the coil. At high RPM or low battery voltage the internal circuitry might bump the dwell angle up to 40 degrees to maximize the available time for the coil to charge up to the desired current level.

Best Regards,
Mike

noonie

One of the best things in life was when I got a new 75 Olds with hei on those subzero temps at 6 in the morning. Fired up everytime like it was summertime. Never went back to points. Hei is still my favorite.

flyingman21

So what wires do I cut and where do I bypass? I have a new engine harness with HEI and mini HI comp starter on a 69 SBC. Thanks.

WESCH

Hi

69427, thanks for the explanation on the HEI. Makes sense that the electronic circuit regulates the HEI .

And I stand slightly corrected on the ignition energy power being Joules , or ?

Rgds. Günther

69427

You are correct. Joules is the unit of measure of energy in ignition systems.