| EdB | -- 02-19-2018 @ 7:57 AM |
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Installed Super's recommended starting enhancement; ran 10 guage wire, with a diode inserted, from the starter side of the solenoid to the positive wire on the coil. Fastest start ever, but when I moved the ignition switch to Stop, the engine kept running. Stopped it by pulling the lead from the coil to the distributor. Where did I go wrong this time? Thank you, EdB
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| cliftford | -- 02-19-2018 @ 12:01 PM |
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Unhook the circuit you added at the coil, put a volt meter on it and see if you are getting current through it to ground, with the engine running. If so, it sounds like a defective solenoid. There should be no current at the solenoid starter post [the one that goes to the starter] except when the engine is cranking. Also check the ignition switch and make sure it is shutting off every time. This message was edited by cliftford on 2-19-18 @ 12:39 PM
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| Drbrown | -- 02-19-2018 @ 6:50 PM |
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Depending how your system's wired (or mis-wired) your generator or alternator could be feeding your coil directly (by passing the ignition switch). This problem is more common with alternators because they put-out more at lower rpms as long as the engine is rotating. Because of this possibility I installed a 4-terminal Master/Kill Switch .... two for battery and two for the Alternator. Key is to check your wiring and as said, your solenoid and ignition switch.
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| cliftford | -- 02-19-2018 @ 8:06 PM |
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After re-reading your post something just hit me. Assuming you still have a 6v positive ground system, it sounds like you may have the coil wired backwards. It wouldn't cause the problem you describe, but the coil would put out a weaker spark.
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| TomO | -- 02-20-2018 @ 7:22 AM |
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It would help if you gave more details about the car. Year? 6 or 12V? What direction is the arrow on the diode pointing? Tom
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| supereal | -- 02-20-2018 @ 2:01 PM |
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Here is my original instructionsM This message was edited by supereal on 2-20-18 @ 2:06 PM
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| supereal | -- 02-20-2018 @ 2:08 PM |
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Apparently my scanner is playing games. I'll try again.
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| MG | -- 02-20-2018 @ 2:15 PM |
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supereal - Thanks for posting that again....
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| supereal | -- 02-20-2018 @ 2:23 PM |
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Finally found the original page. Follow the instructions carefully and it should do as mentioned. If you connect the circuit to the battery side of the solenoid, instead of the starter side, it will override the ignition switch, the engine will not stop as mentioned in the post. A word of caution: Pushing the starter button without the switch will run the engine as long as the button is pushed. This is important if you are "bumping" the starter button to turn the engine while you are under the hood. Disconnect the circuit at one end or the other if you are working on the engine. I devised this "quick start" more years ago than I want to remember. The purpose is to bypass the resistor to compensate for the voltage drop due to the starter and provide full voltage to the coil. then automatically return the system to stock when the engine starts. For cars in the 50's and 60's this type of circuit was built into the system It really helps starting hot or cold.
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| EdB | -- 02-20-2018 @ 5:21 PM |
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The car is a 41. much of the wiring is original. 6 volt positive ground. Has a pancake distributor and round coil; both the coil and the condenser are new from NAPA. The diode is installed so that the end with the band is closest to the coil. Used 10 gauge stranded wire. Diode was soldered in place. Connections were covered with shrink wrap. After the wire was connected to the starter side of the solenoid, the coil end touched some metal and a small spark resulted, so I assume that the diode is correctly placed. Thanks for considering this. Good driving weather is approaching.
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| cliftford | -- 02-20-2018 @ 6:11 PM |
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You stated that the coil end of the wire touched ground and you got a spark. If the engine was not cranking when this happened, there's your problem. Current is being fed to this circuit some other way. Check the previous posts.
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| 40cpe | -- 02-20-2018 @ 6:44 PM |
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It sounds like he is connected to the battery side of the solenoid.
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| supereal | -- 02-21-2018 @ 1:58 PM |
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I included a test in the page to help diagnose to see if the diode was "facing" the right direction. The positive end must be facing the input of the coil. While I use the "top hat" type of diode, follow the third paragraph of the instructions to be sure. The diagram assumes that the battery is grounded at the positive pole in the original configuration, or the diode will be backward. Setting this up isn't "rocket science". That is why I put the diagram in the page.
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| ibshane | -- 02-22-2018 @ 11:23 AM |
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From the description of how this Quick Start Circuit works..., I am assuming that it only works with the OEM coil with external ballast resistor...., but will not do anything if you are using a remote 6v coil with internal resistor/resistance?
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| sarahcecelia | -- 02-22-2018 @ 12:57 PM |
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Who or what is 'Super?"I would like to look into this fast start system.Where can I get info about it?? Regards, Steve Lee
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| MG | -- 02-22-2018 @ 1:42 PM |
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Go to page 1 of this thread and scroll down to the 2nd post by supereal and the jpeg he posted.... This message was edited by MG on 2-22-18 @ 2:24 PM
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| Bill E Bob | -- 02-22-2018 @ 3:27 PM |
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My '47 has the original coil with what looks like an external resistor as part of the unit. Will this affect how it would be wired for the "Quick Start" system?
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| cliftford | -- 02-22-2018 @ 3:49 PM |
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You are correct. this circuit won't do any good using a coil with internal resistance. But have you solved the original problem yet?
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| ibshane | -- 02-22-2018 @ 4:41 PM |
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Sounds like the OP had the circuit to the wrong side of the coil?
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| ibshane | -- 02-22-2018 @ 4:44 PM |
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Is this circuit intended to enhance spark at ignition.., or increase the cranking speed of the starter?
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| cliftford | -- 02-22-2018 @ 7:21 PM |
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It makes a stronger spark during cranking only. It has nothing to do with engine cranking speed.
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| supereal | -- 02-24-2018 @ 12:40 PM |
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If you are using the common round coil in place of the genuine Ford coil, it is likely that coil has a builtin resistor. That requires bypassing the Ford resistor. It is important o remember that the Ford resistor also functions as a regulator. As the system voltage rises due to increased generator output, the nichrome winding of the resistor heats up, increasing resistance. As Ohm's Law says, voltage is the product of current and resistance, the points and coil are protected from damage. As for who I am, just a very old guy who loves our old Fords, and spent too many years working on them.
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