I replaced the ballast resistor on my 67 L79. I ran my motor for 10 minutes. The ballast resistor mounting bracket was very hot. Is this normal? The coil was not hot. Thanks for any help.
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Re: Hot Ballast Resistor
yep ! -
Re: Hot Ballast Resistor
yep !Comment
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Thanks Clem *NM*
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Thanks Clem *NM*
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Re: Hot Ballast Resistor
How hot can the ballast resistor run? Mine is so hot that it will boil off a drip of water in a second. That's with the engine off and the ignition switch on for a couple minutes.Comment
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Re: Hot Ballast Resistor
How hot can the ballast resistor run? Mine is so hot that it will boil off a drip of water in a second. That's with the engine off and the ignition switch on for a couple minutes.Comment
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Re: Hot Ballast Resistor
i know from experence that they will raise a blister on your finger. with the key on and the engine not running but the points closed i would think it would get very hot because you are drawing current all the time.Comment
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Re: Hot Ballast Resistor
i know from experence that they will raise a blister on your finger. with the key on and the engine not running but the points closed i would think it would get very hot because you are drawing current all the time.Comment
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Re: Hot Ballast Resistor
Total current consumed in the ignition can be as high as 100 watts, just imagine a 100 watt light bulb attached to your firewall! John McGrawComment
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Re: Hot Ballast Resistor
Total current consumed in the ignition can be as high as 100 watts, just imagine a 100 watt light bulb attached to your firewall! John McGrawComment
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Clem's right on!
This is the worst case scenario for the ballast, engine off, ignition on, and points closed. The time ensemble for current draw is MAXimum and I2R power dissipation is peaked out. When the engine is running the current waveform that flows is VERY complex, but it's essentially a ROUGH sawtooth shape with zero crossings when points open and energy is transfered from coil primary to secondary. This causes the average power dissipated by the ballast resistor to drop + you have car-in-motion convection cooling effects. So, you're challenging the ballast to dissipate heat under these conditions! It'll take it but it will get HOT!!!!Comment
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Clem's right on!
This is the worst case scenario for the ballast, engine off, ignition on, and points closed. The time ensemble for current draw is MAXimum and I2R power dissipation is peaked out. When the engine is running the current waveform that flows is VERY complex, but it's essentially a ROUGH sawtooth shape with zero crossings when points open and energy is transfered from coil primary to secondary. This causes the average power dissipated by the ballast resistor to drop + you have car-in-motion convection cooling effects. So, you're challenging the ballast to dissipate heat under these conditions! It'll take it but it will get HOT!!!!Comment
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Hum.....
Have you ever seen a wire wound 100W resistor of 0.3-2.0 ohms? They're HUGE -- approaching one foot in length with POUNDS of ceramic to absorb the dissipated heat. The key to sucessful DC circuit analysis (AC operation is VERY complicated), is to take several other factors (some not so obvious) into consideration:
(1) Winding resistance in coil primary that's in series with ballast, it dissipates heat too hence the coil transformer is immersed in an oil bath inside its metal case).
(2) Ignition points -- they have an ohmic surface contact resistance that's not insignificant. Put a VOM across a set of points, let 'em close, then read the low voltage resistance someday. You'll be surprised at the magnitude of the reading as well as variance from actuation to actuation vs. fresh/new and old/worn condition.
(3) The magnetic storage capacity of the coil's primary. When points are initially closed (or ignition is turned ON with points closed), the lion's share of power is transfered into the magnetic flux stored in the coil's primary section (V = L di/dt). Of course, once the coil's transformer core is saturated (di/dt = 0), the transient solution to the circuit's differential equation 'dies out' and we're left with the steady state solution as the only active participant. That's when the ballast starts to really dissipate heat!Comment
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Hum.....
Have you ever seen a wire wound 100W resistor of 0.3-2.0 ohms? They're HUGE -- approaching one foot in length with POUNDS of ceramic to absorb the dissipated heat. The key to sucessful DC circuit analysis (AC operation is VERY complicated), is to take several other factors (some not so obvious) into consideration:
(1) Winding resistance in coil primary that's in series with ballast, it dissipates heat too hence the coil transformer is immersed in an oil bath inside its metal case).
(2) Ignition points -- they have an ohmic surface contact resistance that's not insignificant. Put a VOM across a set of points, let 'em close, then read the low voltage resistance someday. You'll be surprised at the magnitude of the reading as well as variance from actuation to actuation vs. fresh/new and old/worn condition.
(3) The magnetic storage capacity of the coil's primary. When points are initially closed (or ignition is turned ON with points closed), the lion's share of power is transfered into the magnetic flux stored in the coil's primary section (V = L di/dt). Of course, once the coil's transformer core is saturated (di/dt = 0), the transient solution to the circuit's differential equation 'dies out' and we're left with the steady state solution as the only active participant. That's when the ballast starts to really dissipate heat!Comment
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