When you click on links to various merchants on this site and make a purchase, this can result in this site earning a commission. Affiliate programs and affiliations include, but are not limited to, the eBay Partner Network.
I'll start with a little history. I've been working on my 72' 442 X-code just doing a bunch of detail clean up and odds and ends. I haven't started it since Thanksgiving. Since then I have done a lot of stuff but with regard to this problem I pulled the starter and put a new solenoid on it. Did some wire routing and cleanup. Hooked everything up exactly like it was. I pulled a set of crappy old headers off and put on new stock exhaust system, etc., etc. Today I decided to start it up and let it run for a bit and listen to the new exhaust. It cranked over fine but wouldn't start. I cranked it quite a bit. I probably should of given it a little help with some ether from the top but I didn't. I cranked it for bit but no dice. Then something happened I haven't experienced before. When I let off the key and turned it to off the engine and starter continued to crank over without the key in the ignition. It was a holy **** moment. I rushed out to the battery and disconnected it. I did not know what to think. I did add a factory type heat shield over the starter. I was baffled to say the least. Maybe the new solenoid has shorted out somehow? I got under the dash as a precaution and disconnected the three plugs on the steering column for the safety key switch module. I got under the car and pulled off the heat shield thinking maybe it was shorting a wire out on the starter. However, when I got it out there were no heat marks. Still wasn't sure what to think. After doing those things I hooked up the battery again and away went the starter again but with out the solenoid engaged. The starter just spun freely. I disconnected the battery again and that's where it sits. Don't know how I could have gone from an easy starting car to totally screwed up. Any ideas would be appreciated. How is the starter running all by itself?
Thanks for the wiring info, Norm. There was no wire on the post that is supposed to be the yellow wire. There wasn't a wire there before I changed the solenoid. Only the post with the purple wire and of course the main battery wire and a smaller wire with the battery wire on the main post. I changed the solenoid because the old one looked hammered. Figured while I had the starter off I would put a new solenoid on. I don't remember the brand, but I am thinking like you Eric, it got fried. I'll pull the starter out and check the wiring, and probably put the old solenoid back on, at least to test it.
It's been a while since I made a comment on this thread. Been enjoying some sun in the islands. Got a new solenoid put on the starter and am ready to put it back in and try again. However, with regard to the yellow wire that is supposed to hook to the right connection on the solenoid, and then is connected on the other end to the coil. I didn't realize that the GM style HEI wasn't normal in these cars. Whomever built this car put in GM HEI and did not hook up the yellow wire because it doesn't have the old style coil. Does anyone know why they didn't hook it up and does anyone know if it is an issue. The car ran fine and started good before I burned up the solenoid. From the wiring diagram provided by Norm, there is a wire that is connected to the same spot on the old coil that ends up at the ignition switch. From what I can tell the yellow and that resistance wire attached to the old style coil are not connected in this car with the GM HEI setup. Notice the yellow wire tied to the black wire that went on an old style coil, not not used. Is that an issue.
I believe an HEI setup requires 12volts all the time to operate correctly. Therefore, the normal ignition wire with the ballast (which purpose was to lower the voltage to provide longevity for the points) is not used. Whoever installed your HEI must have provided a new ignition wire without the ballast to provide 12 volts all the time. The purpose of the Yellow wire from the starter was to provide 12 volts to the coil during starter cranking to help start the engine. Once the engine is running, the voltage to the coil uses the ballast circuit. Since your HEI setup already is getting 12 volts from the new ignition wire, the Yellow wire from the starter is not necessary.
The yellow wire from the starter solenoid to the ignition coil isn't needed on HEI. Do confirm that it isn't energized including when cranking and if not tape it off and tuck it away.
Just FYI, 12 volts is only present at the starter ports with the ignition switch on your steering column in the start position. Once the car starts and the key is in the run position, no voltage is provided to those ports. So if you had the yellow connected to the coil, it would not create a problem because your coil is already getting 12 volts from your new ballast free ignition wire. But it is not necessary to use the Yellow wire with HEI.
I strongly suspect the solenoid that stayed engaged was caused by low voltage from excessive cranking causing the battery to be low. When there is low voltage, the amps will increase causing the contact plate to weld itself to the contact stud. That is why it stayed on in the cranking position,as the drive was already engaged. After you disconnected the battery, the drive retracted, but the contacts stayed welded together causing the starter to spin when reconnected. The drive didn't engage when the battery was reconnected because there was no power to the S post (key in start position). Be sure to keep the battery charged when doing excessive cranking even if you have to keep a battery charger on it .
When the voltage drops, the current increases in the same proportion.
Yep, voltage and current are inversely proportional to each other.
Ohm's Law (Equation), I=V/R, is the incorrect equation to define the relationship between Amperage & Voltage when either Amperage or Voltage increases or decreases. Ohm’s law states that current in amps (I) is equal to the voltage (V) divided by the resistance(R) in Ohms [Ω]; or I = V / R. Ohm's Law (Equation) will state what the Amperage, Resistance &/or Voltage is of a given circuit, but it doesn't explain the relationship between Amperage &/or Voltage in performing a given amount of Work. Instead, the Power Equation is needed, P (Power) = Volts x Amps. If we express P (Power) = W (Watts) or any other mathematical attribute i.e. HP (Horsepower), Joules, etc., as representing the amount of work performed, it's apparent there is an inverse relationship between Voltage & Amperage (when Resistance remains constant in a DC circuit).
To perform the exact same amount of Work (746 Watts =1 HP), if you increase voltage amperage decreases; conversely if you increase amperage voltage decreases.
1 Watt = 1 Volt x 1 Ampere
Example: 1 HP = 746 Watts
Employing the Power Equation (P = V x A) where P = Watts (Electrical Work)
Solve for V &/or A when P = 746 Watts
When V = 12.6 Volts (R = 100 Ohms=Constant) Amperage = 59.21 Amperes
When V = 6.2 Volts (R = 100 Ohms=Constant) Amperage = 120.32 Amperes
or
746 Watts = 12.6V x 59.21 A;
746 Watts = 6.2V x 120.32 A
Last edited by Vintage Chief; March 13th, 2023 at 04:53 PM.
Keep in mind that when a battery is discharged, its voltage decreases and its current sourcing capability decreases as well because its output impedance increases. It's a dynamic system that is a bit more complex than the formulas noted above.
Keep in mind that when a battery is discharged, its voltage decreases and its current sourcing capability decreases as well because its output impedance increases. It's a dynamic system that is a bit more complex than the formulas noted above.
Indeed. However, impedance = resistance in a DC circuit.
Yep, voltage and current are inversely proportional to each other.
Ohm's Law (Equation), I=V/R, is the incorrect equation to define the relationship between Amperage & Voltage when either Amperage or Voltage increases or decreases. Ohm’s law states that current in amps (I) is equal to the voltage (V) divided by the resistance(R) in Ohms [Ω]; or I = V / R. Ohm's Law (Equation) will state what the Amperage, Resistance &/or Voltage is of a given circuit, but it doesn't explain the relationship between Amperage &/or Voltage in performing a given amount of Work. Instead, the Power Equation is needed, P (Power) = Volts x Amps. If we express P (Power) = W (Watts) or any other mathematical attribute i.e. HP (Horsepower), Joules, etc., as representing the amount of work performed, it's apparent there is an inverse relationship between Voltage & Amperage (when Resistance remains constant in a DC circuit).
To perform the exact same amount of Work (746 Watts =1 HP), if you increase voltage amperage decreases; conversely if you increase amperage voltage decreases.
1 Watt = 1 Volt x 1 Ampere
Example: 1 HP = 746 Watts
Employing the Power Equation (P = V x A) where P = Watts (Electrical Work)
Solve for V &/or A when P = 746 Watts
When V = 12.6 Volts (R = 100 Ohms=Constant) Amperage = 59.21 Amperes
When V = 6.2 Volts (R = 100 Ohms=Constant) Amperage = 120.32 Amperes
or
746 Watts = 12.6V x 59.21 A;
746 Watts = 6.2V x 120.32 A
Well I don't know any more, it's been too long since I used any of this knowledge. I apologize if it seemed I was being smart @ss because I wasn't, just wanted to ensure correct information was out there and instead I did it myself. I declare guys this is making me feel useless, here I am at 67 years old and can't remember the application of basic electronics that I used for thirty years. Sheesh!
I think it was great watching all of this old electronic theory flying around. Sometimes going back to the basics helps get to the root of the problem. It's been 40 years since I studied basic electronics in the first quarter of college. It's nice to revisit it. None the less the solenoid is changed out and ready to go back in as soon as I get off duty. I'll get the battery charged and a little cheater fluid down the chute. Should fire right up. All you guys have been a huge help, as usual. Be in touch.
Just an update to all you guys that responded and helped out. Got the new solenoid with the starter put back in. The solenoid btw was fried. Got everything wired back up and ready to go. Still wouldn't start. No spark. The comment made earlier in the thread about the HEI needing 12V saved me. Bet you can all see where this is headed. After about two hours of screwing around I finally remembered that when I was cleaning up all the wiring and making things nice, there was a wire that accompanied the main cable going from the battery to the starter. It didn't go anywhere so I threw it away. OOPS! It had come undone from the distributor cap battery connection and I didn't know it. If you look at the pics I added to this thread of the distributor cap assy' you'll notice no wires going into the square end end where the batt. and tach connector go. DUH!! Sometimes I wonder about myself. I hooked up battery to the cap and boom, it started right up. Learning is important with all of this. These forums are priceless. On to the next six projects with this vehicle. Thanks again everybody.
To the above electrical discussion, that's a nice textbook discussion, but I want to draw notice to the disclaimer "assuming the same amount of work is done." In a failing automotive electrical system, that is not the case. The mechanical analog to this is calculating ballistics then saying "neglecting air resistance."
I'm back with another problem related to the original problem. I got the car started and running fine. But now the car will not shut off. When the key is in the ignition and turned to the on position everything lights up and the buzzer turns on. The car ignition fires up the car and away it runs. However, when I turn the key back to the off position the engine does not turn off. I have to cut the battery to get it shut down. I've obviously done something to it because it didn't do this before. I'm wondering if the neutral safety & back up light switch could be causing me problems? Before all of this **** happened That switch which is located on the steering column under the dash was loose and the car ignition was intermittent. I could reach under the dash and wiggle the switch and the ignition would engage. When I had the solenoid melt down I wasn't sure what was happening when the starter kept turning over when the key was turned off so I reached under the dash and removed the three connectors from the switch. That didn't help stop the starter and I still had to cut the battery power. Since then with everything replaced and rewired I got under the dash and tightened down the switch. It has two 1/4" screws one on each side. I centered it my best and tightened them down. Then I put the connectors back on. One connector is idiot proof because it's a one way only plug. The other two plugs are such that you could put them on either direction by simply rotating them 180 degrees. Does anyone have any advice here. Obviously it's not good not to be able to shut the car off.
You probably have the HEI battery terminal hooked to constant battery power, it needs to be hooked up to a switched source like the IGN terminal in the fuse block. You need to disconnect the battery until you get this sorted.
In your previous post, you mentioned the car wouldn’t start and you remembered you removed the wire from the battery and you had no power connected to the HEI system. Then it sounded like you wired 12 volts from the battery directly to the HEI and the turned the ignition to start and car started. When I read this, I was concerned that you bypassed the ignition switch all together for power to the HEI system. If this is the case, you would have no way to stop the engine from running without disconnecting the battery. When I replied earlier and said an HEI system needs 12 volts all the time, I meant 12 volts when the key is in the run position not just when the key is in the start position. The factory ignition wire that supplies power to the coil has a ballast resistor in line from the ignition switch on the steering column to the coil and reduces the voltage to the coil to make the points last longer. Since you are using an HEI system which requires 12 volts, you will need to replace the factory ignition wire from the ignition switch on the steering column to the HEI with a new wire that does not have a ballast resistor so that 12 volts will be present. Then when you turn your key to off, the power to the HEI will be switched off and the engine will stop. I would refer to the schematic diagram that was provided in an earlier post to make sure you are replacing the ignition wire and have the proper connections.
Sorry it took me a bit to respond. I'm on duty and trying to take a break for a minute. You got me, it's hooked to straight up battery power. When I figured out that I needed 12V to the HEI BATT connection I should have known it should be an IGN sense line. The only thing that swayed me was the wire I threw away some time back. I'm not certain it was hooked to the HEI but it was hooked to the positive post on the battery. None the less I will get the connection source put in the right place as soon as I get home and can get to it. I will be in touch.
All is well with the ignition. I went straight into the fuse block directly to the IGN sense lug. Starting and stopping the engine is now working nicely. On to the next set of issues. Thanks again, all.
February 19th, 2023, 06:33 PM
