Coil Voltage
Tech Contributor
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From: I tend to be leery of any guy who doesn't own a chainsaw or a handgun.
If the engine isn't running and the points are closed, you'll get a reading substantially less than battery voltage. Your reading sounds like a good ballpark number. The actual reading is dependent upon the resistance and manufacturing tolerances of the ballast wire, the coil winding resistance, the ignition switch resistance, and the charge state of the battery. Turning the engine over a bit until the points open should give you a reading darn close to battery voltage.
Team Owner
Joined: Jan 2006
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From: Crossville TN
If you have a 'stock' configuration system, it should be in the 7-8 volt range. By design, there should be a ballast resistor in the power feed line from the fuse block to the coil (the other wire comes from the starter solenoid). That 'resistor' is a length of resistor wire in my '71; I think that should be the same for your '69, but earlier vintage Corvettes used a ceramic block resistor mounted to the firewall.
Your coil must be compatible with a 'ballast resistor system' in that it will see lower voltage at the coil [except in "Start" mode when a full 12 volts comes from the starter solenoid]. If you install a standard (12 volt) coil into your system, your spark potential will be substantially reduced due to the difference in the number of windings built into the coil.
Your coil must be compatible with a 'ballast resistor system' in that it will see lower voltage at the coil [except in "Start" mode when a full 12 volts comes from the starter solenoid]. If you install a standard (12 volt) coil into your system, your spark potential will be substantially reduced due to the difference in the number of windings built into the coil.
Melting Slicks
Joined: Mar 2005
Posts: 2,838
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From: Austin TX
You should have 12v. Current was/is reduced on our old cars by using a resistor wire to prevent the points from frying, but voltage is still 12v to the coil.
Last edited by 73, Dark Blue 454; Apr 4, 2010 at 11:58 PM.
Melting Slicks
Joined: Jun 2001
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From: CA
Race Director
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From: California
That is incorrect.
the Ballast resistor with coil makes a voltage divider, so yes, if the points are closed, you would have about 6V and even less... when points open, it will go up to the 12v.
HEI is always 12V
Basic electronics, voltage at the end of a resistor will be the same on the output as on the input, unless the output side is under load...then you will have voltage drop, this drop depends on the load/resistance.
Last edited by pauldana; Apr 5, 2010 at 12:36 AM.
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From: California
If you have a 'stock' configuration system, it should be in the 7-8 volt range. By design, there should be a ballast resistor in the power feed line from the fuse block to the coil (the other wire comes from the starter solenoid). That 'resistor' is a length of resistor wire in my '71; I think that should be the same for your '69, but earlier vintage Corvettes used a ceramic block resistor mounted to the firewall.
Your coil must be compatible with a 'ballast resistor system' in that it will see lower voltage at the coil [except in "Start" mode when a full 12 volts comes from the starter solenoid]. If you install a standard (12 volt) coil into your system, your spark potential will be substantially reduced due to the difference in the number of windings built into the coil.
Your coil must be compatible with a 'ballast resistor system' in that it will see lower voltage at the coil [except in "Start" mode when a full 12 volts comes from the starter solenoid]. If you install a standard (12 volt) coil into your system, your spark potential will be substantially reduced due to the difference in the number of windings built into the coil.
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Tech Contributor
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From: I tend to be leery of any guy who doesn't own a chainsaw or a handgun.
not 100% accurate, but closer.... a point distributor and coil (1) because of the ballast resistor your coil never sees 12v except (2) when you first start, afterwards the voltage is dropped through the Ballest resistor, thus (3) your primary voltage is lower, which will produce a lower secondary voltage...
1) The coil sees 12v between each dwell period, and at the start of every dwell period. Every time.
2) The ignition system always sees less, not more voltage when cranking due to the starter motor pulling large currents out of the battery, causing the battery and system voltage to drop. The ballast resistor shunt wiring is designed to remedy/bandaid this voltage reduction during cranking, by reducing the series resistance in the coil primary circuit. This reduction in resistance allows the coil primary current to remain at near normal levels despite the reduction in ignition system voltage.
3) The secondary voltage is set/determined by the arcover voltage of the spark plug gap. The ballast resistor does not set any voltage parameters. It limits the primary current, which then sets/limits the amount of coil energy which is then transferred to the plug gap.
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From: California
Not 100% accurate, but...oh well, you get the message.
1) The coil sees 12v between each dwell period, and at the start of every dwell period. Every time.
2) The ignition system always sees less, not more voltage when cranking due to the starter motor pulling large currents out of the battery, causing the battery and system voltage to drop. The ballast resistor shunt wiring is designed to remedy/bandaid this voltage reduction during cranking, by reducing the series resistance in the coil primary circuit. This reduction in resistance allows the coil primary current to remain at near normal levels despite the reduction in ignition system voltage.
3) The secondary voltage is set/determined by the arcover voltage of the spark plug gap. The ballast resistor does not set any voltage parameters. It limits the primary current, which then sets/limits the amount of coil energy which is then transferred to the plug gap.
1) The coil sees 12v between each dwell period, and at the start of every dwell period. Every time.
2) The ignition system always sees less, not more voltage when cranking due to the starter motor pulling large currents out of the battery, causing the battery and system voltage to drop. The ballast resistor shunt wiring is designed to remedy/bandaid this voltage reduction during cranking, by reducing the series resistance in the coil primary circuit. This reduction in resistance allows the coil primary current to remain at near normal levels despite the reduction in ignition system voltage.
3) The secondary voltage is set/determined by the arcover voltage of the spark plug gap. The ballast resistor does not set any voltage parameters. It limits the primary current, which then sets/limits the amount of coil energy which is then transferred to the plug gap.
1, when points open, no voltage divide, 12v, when points closed ballest resistor causes a voltage diver circuit, thus the lower than 12v,,, like 6V...]
2. are you 2h1t1ng me??? I am not going to teach a basic electronics class here but BS a resistor can not bring up voltage...
3. the secondary voltage of any transformer is determined my the coil windings of the primary and the secondary windings, and the primary voltage.
Last edited by pauldana; Apr 5, 2010 at 11:01 AM.
Pro
Joined: May 2003
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From: London England
Much as it pains me to say it, I'm with Paul here.
So some pedant doesn't try and flame me I shall be very carefully with my wording.
During cranking the coil receives full battery voltage, once the key is in the run position the voltage goes via the ballast resistor which reduces it, normally by around 3-4 volts below battery voltage. This is because the coil is designed to output it's maximum safe continuous ignition voltage with a lower than battery voltage feed (6-9V normally). The increased output created in the cranking position is theoretically achieved by overloading the coil for a short period, if full battery voltage were fed to the coil continuously it would burn out. In real terms the coil isn't actually overloaded while the engine is cranking as the starter load reduces battery voltage to close to the normal running voltage of the coil.
The coil certainly does not receive full battery voltage everytime the points open and close with the key in the run position.
With regard to the actual secondary output voltage, how much voltage is drawn is governed by the resistance in the secondary circuit, such as compression pressure, plug gap, rotor arm gap, lead resistance etc
This is only true up to a point when the maximum output of the coil is determined by the number of windings and the voltage run through them.
So some pedant doesn't try and flame me I shall be very carefully with my wording.
During cranking the coil receives full battery voltage, once the key is in the run position the voltage goes via the ballast resistor which reduces it, normally by around 3-4 volts below battery voltage. This is because the coil is designed to output it's maximum safe continuous ignition voltage with a lower than battery voltage feed (6-9V normally). The increased output created in the cranking position is theoretically achieved by overloading the coil for a short period, if full battery voltage were fed to the coil continuously it would burn out. In real terms the coil isn't actually overloaded while the engine is cranking as the starter load reduces battery voltage to close to the normal running voltage of the coil.
The coil certainly does not receive full battery voltage everytime the points open and close with the key in the run position.
With regard to the actual secondary output voltage, how much voltage is drawn is governed by the resistance in the secondary circuit, such as compression pressure, plug gap, rotor arm gap, lead resistance etc
This is only true up to a point when the maximum output of the coil is determined by the number of windings and the voltage run through them.
Tech Contributor
Joined: Jun 2004
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From: I tend to be leery of any guy who doesn't own a chainsaw or a handgun.
Much as it pains me to say it, I'm with Paul here.
Mr. Golden: I'm disappointed with you. Several of your previous posts have indicated that you have a decent understanding of vehicle dynamics. I was impressed with that. In this particular post I'm responding to, unfortunately you have several errors, the biggest is agreeing with an error filled post. I fear that some posters are unable or unwilling to listen to facts or physics, so I'll reserve my time and comments to you. Unless I'm wrong, I think you'll listen to reason.
So some pedant doesn't try and flame me I shall be very carefully with my wording.
During cranking the coil receives full battery voltage, Correct. But, what is that voltage? Depending on ambient air, oil, and battery temperatures, the starter motor will drag the battery/system voltage down to 9-11 volts. I think you would agree with me that 9-11 volts is less than 13.6-14.2 volts that the alternator sets for a system voltage when the engine is running. Correct? once the key is in the run position the voltage goes via the ballast resistor which reduces it, normally by around 3-4 volts below battery voltage. No, No, No. A resistor is NOT a voltage regulator. It merely limits current. Depending on where in the dwell cycle you measure, you will get different measures of primary current, and an accompanying variation in the voltage drop across the ballast resistor due to this current. The "charge current" is constantly changing (in an exponential fashion), resulting in continuous changes in the voltage drop across the resistor. It is not regulated by the ballast, as a ballast is not a voltage regulator. It is a current limiter. This is because the coil is designed to output it's maximum safe continuous ignition voltage with a lower than battery voltage feed (6-9V normally). No again. The coil doesn't give a crap what the voltage is across it. Twelve volts is nothing. In normal operation the coil terminals repeatably see several hundreds of volts across the primary terminals, and thousands of volts across the secondary side. Again, twelve volts is nothing. The increased output created in the cranking position is theoretically achieved by overloading the coil for a short period, if full battery voltage were fed to the coil continuously it would burn out. Wrong. See previous explanation. How about a real world example: I converted my '69 distributor to use the GM HEI module about 25 years ago. For packaging reasons I initially used my factory oilcan coil. The engine ran great. The car (and coil) has run flawlessly for two and a half decades, all without a ballast resistor. The coil has not burned out, nor will it. Why? Because the voltage doesn't matter!The HEI module limits the current, but my coil sees 12v all the time. And it's stone reliable. In real terms the coil isn't actually overloaded while the engine is cranking as the starter load reduces battery voltage That's what I've been saying, and other posters are having difficulty with that concept. to close to the normal running voltage of the coil.
The coil certainly does not receive full battery voltage everytime the points open and close with the key in the run position. Despite the fact that I'm starting to wear off the W on my keypad here, that's wrong! What is the voltage at both coil terminals (C+ and C-) between dwell periods (when the points are open)? It is 12 volts (actually 13-14 volts, more than the voltage when cranking, correct?). Now, here is the key question to see if you, or a previous poster, truly understands how a coil works. What is the voltage across the coil (C+ to ground) the moment the points close? I'm too tired to type a bunch of ballast resistor specs or coil specs, so I'll let you choose what size coil and ballast you want to use. I posed this quiz in a post several months ago, and only one person got it right. He actually looked at how an ignition system works, and figured it out. He did not just parrot old wives tales. So I ask you, what is the voltage?
With regard to the actual secondary output voltage, how much voltage is drawn is governed by the resistance in the secondary circuit, such as compression pressure, plug gap, rotor arm gap, lead resistance etc Correct.
It has nothing to do with the ballast resistor, as some have claimed.
This is only true up to a point when the maximum output of the coil is determined by the number of windings and the voltage run through them.
Mr. Golden: I'm disappointed with you. Several of your previous posts have indicated that you have a decent understanding of vehicle dynamics. I was impressed with that. In this particular post I'm responding to, unfortunately you have several errors, the biggest is agreeing with an error filled post. I fear that some posters are unable or unwilling to listen to facts or physics, so I'll reserve my time and comments to you. Unless I'm wrong, I think you'll listen to reason.
So some pedant doesn't try and flame me I shall be very carefully with my wording.
During cranking the coil receives full battery voltage, Correct. But, what is that voltage? Depending on ambient air, oil, and battery temperatures, the starter motor will drag the battery/system voltage down to 9-11 volts. I think you would agree with me that 9-11 volts is less than 13.6-14.2 volts that the alternator sets for a system voltage when the engine is running. Correct? once the key is in the run position the voltage goes via the ballast resistor which reduces it, normally by around 3-4 volts below battery voltage. No, No, No. A resistor is NOT a voltage regulator. It merely limits current. Depending on where in the dwell cycle you measure, you will get different measures of primary current, and an accompanying variation in the voltage drop across the ballast resistor due to this current. The "charge current" is constantly changing (in an exponential fashion), resulting in continuous changes in the voltage drop across the resistor. It is not regulated by the ballast, as a ballast is not a voltage regulator. It is a current limiter. This is because the coil is designed to output it's maximum safe continuous ignition voltage with a lower than battery voltage feed (6-9V normally). No again. The coil doesn't give a crap what the voltage is across it. Twelve volts is nothing. In normal operation the coil terminals repeatably see several hundreds of volts across the primary terminals, and thousands of volts across the secondary side. Again, twelve volts is nothing. The increased output created in the cranking position is theoretically achieved by overloading the coil for a short period, if full battery voltage were fed to the coil continuously it would burn out. Wrong. See previous explanation. How about a real world example: I converted my '69 distributor to use the GM HEI module about 25 years ago. For packaging reasons I initially used my factory oilcan coil. The engine ran great. The car (and coil) has run flawlessly for two and a half decades, all without a ballast resistor. The coil has not burned out, nor will it. Why? Because the voltage doesn't matter!The HEI module limits the current, but my coil sees 12v all the time. And it's stone reliable. In real terms the coil isn't actually overloaded while the engine is cranking as the starter load reduces battery voltage That's what I've been saying, and other posters are having difficulty with that concept. to close to the normal running voltage of the coil.
The coil certainly does not receive full battery voltage everytime the points open and close with the key in the run position. Despite the fact that I'm starting to wear off the W on my keypad here, that's wrong! What is the voltage at both coil terminals (C+ and C-) between dwell periods (when the points are open)? It is 12 volts (actually 13-14 volts, more than the voltage when cranking, correct?). Now, here is the key question to see if you, or a previous poster, truly understands how a coil works. What is the voltage across the coil (C+ to ground) the moment the points close? I'm too tired to type a bunch of ballast resistor specs or coil specs, so I'll let you choose what size coil and ballast you want to use. I posed this quiz in a post several months ago, and only one person got it right. He actually looked at how an ignition system works, and figured it out. He did not just parrot old wives tales. So I ask you, what is the voltage?
With regard to the actual secondary output voltage, how much voltage is drawn is governed by the resistance in the secondary circuit, such as compression pressure, plug gap, rotor arm gap, lead resistance etc Correct.
It has nothing to do with the ballast resistor, as some have claimed.This is only true up to a point when the maximum output of the coil is determined by the number of windings and the voltage run through them.
I'm tired of typing. I'll check this for typos later.
Race Director
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From: California
before I wast my breath and time,,,, do you know ohms law? do you know what a voltage divider is? what is the formula to find the voltage at different points in the divided circuit.... if you do not know this then you do not need to further comment on this problem.
Tech Contributor
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From: I tend to be leery of any guy who doesn't own a chainsaw or a handgun.
Tech Contributor
Joined: Jun 2004
Posts: 20,868
Likes: 961
From: I tend to be leery of any guy who doesn't own a chainsaw or a handgun.
Just for kicks, would you like to enlighten me with your formal training and experience regarding this particular subject? And, would you like to offer up an answer to my quiz? If you're shy, just PM your guess to me.
I'll be waiting for your reply.