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#1
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Ballast Resistor
Ignition switch has a 'run' and a 'start'. Bad ballast resistor would run only when in the start position (bypassing BR), well known. Question: Is the 'run' hot when cranking? Is that the wire to run to the aftermarket ignition that doesnt use the resistor (MSD 'on' wire; small red). I am thinking it is because if you were to use the BR and the wiring that is associated, the car would never start (not a full 12V to the small red wire on MSD?) How much of a voltage drop does the BR have when cold, or is it a current limiter (no voltage drop, just amp drop?) thanks
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#2
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Run only applies voltage in the run position. Start only applies voltage in the start position. That is why you can start a car with a bad ballast, but won't stay running after you release the key to run.
Either retain the ballast (the MSD box doesn't care) or bring both leads together and then to the MSD. |
#3
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Hey, thank you ehostler. First time I've heard anyone come right out and say the resistor isn't needed with MSD-type ignition boxes. Even the instructions are vague on this point. I've run my Mallory this way for a couple of years with no problems.
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#4
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If you actually follow the instructions on the MSD it does state that Ballast resistor is not used in the ignition primary circuit.
BUT************* It also states that the old coil + wire can still be used as the on trigger wire for the msd WITH the ballast resistor left in. The msd is powered directly from the battery(if wired correctly), and the old coil red wire only turns it on. Just like a trigger wire on a headlight relay. There will be absolutely NO DIFFERENCE in performance when using msd with the ballast resistor left in the car, so long as you have wired the MSD directly to the battery, as it says to do in the instructions. As Ed said.....the MSD box doesn't care whether it's there or not I prefer to leave it in, so when the msd lets me down (note I said when, not if) then it's a quick switch back to the stock stuff. |
#5
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Cool. That is what I was thought. Ill just bring them both together and route them to the Hyfire 5 'on' terminal.
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#6
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And, incidentally, it's impossible for a resistor to provide a "amp drop" without a corresponding voltage drop. When current flows through a resistor, there's ALWAYS a drop in voltage, along with the reduction of current.
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#7
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OK, ill go with that. I used to deal with little "current limiters" on some phone stuff. Are those just like little shunts that pop when there is a large draw? They "healed" when you pulled them and reinserted them. Maybe Im confusing amp and current.
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#8
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When running MSD, I just solder a piece of brass across the IGN1 and IGN2 terminals on the ignition switch.
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#9
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Naw, you're not confusing them, because they're the same. Current is the same as amperage.
I believe what you're referring to is similar to a circuit breaker, which opens if excessive current flows through it. It limits the current by opening the circuit and reducing the current flow to zero. It's like a fuse, 'cept it can be reset. When you get a blue fuse, it needs to be replaced. Voltage (E) equals amps (I) times resistance (R) (E = I X R) put another way, Amps equals voltage divided by resistance (I = E/R) put yet another way, Resistance equals voltage divided by amps (R = E/I) If you change voltage (R), you change the current (amps). If you change resistance (R), you change the current (I). If you change the current, it's because you changed either the voltage or the resistance of the circuit. So, using numbers that make the math easy: Say you have a 12 volt system (I know it's more like 14 when it's running, but stay with me here..) Into this 12 volt system, you connect a 2 ohm ballast resistor in series with a 4 ohm coil. The resistor and coil are in series. In a series circuit, the current through the entire circuit is the same at all points. Total resistance of the circuit is 2 ohms for the resistor, and 4 ohms for the coil, for a total of 6 ohms. From the formula above, we can calculate the current through the circuit: I = E/R, or in our case, I = 12 Volts divided by 6 Ohms. So I equals 2 amps. We have 2 amps flowing through the resistor and the coil. Now, we can figure the voltage drop across each item. Let's do the 4-ohm coil first. E = I times R. So, voltage = 2 amps times 4 ohms, or 8 volts. This means of the 12 volts total on the system, 8 of them are working through the coil. The resistor would be 2 amps times 2 ohms, or 4 volts. Wow! 4 volts plus 8 volts equals 12 Volts. So, we know the math is correct. Now, some people are surprised when their ballast resistors get hot. Another formula is used to figure how much power is being dissipated (or consumed) by a device. The formula is P = R times I squared. Our resistor is dropping 4 volts at 2 amps. So, 4 volts times 2 amps (squared), or 4 times 4, equals 16 watts of power being turned into nothing but heat by the resistor. That 16 watts is important. We have to do something with it. Energy can't be increased or decreased, but it can be converted into another form of energy. Some devices turn this power into light, some turn it into torque (like an electric motor) In our case, we can't do any work with the resistor, so the 16 watts of power is converted into another form of energy: HEAT. 16 watts may not seem like much, but it really IS a lot of power for a resistor. If you take the back off a radio, TV or other electronic device, you'll see tiny little resistors on the circuit boards that are rated at 1/8 or 1/4 watt, generally. So a resistor that is rated at 16 watts or higher is really quite a heavy-duty item. Now, if you can find a coil that truly is rated to operate at 12 volts (ON A CONTINUOUS BASIS), you'd be pushing the full 12 volts through the coil, and no power would be wasted in the form of heat by the resistor. Less heat equals less wasted energy. In reality, though, many "12 volt" coils actually have a resistor built into their case, so you're not gaining anything in that case, except a coil that operates hotter (because that power being consumed by the resistor is being turned into heat). This gets me to a pet peeve of mine: People who check voltage on an open circuit. It'll fool you more often than not. Homework assignment: Problem number 1 You have a DC voltmeter with a 119,000 ohm internal resistance. You have a 1000 ohm resistor with one side connected to the positive post of your 12 volt battery. You connect the positive lead of your voltmeter to the other side of the resistor, then the negative lead of your voltmeter to the battery negative post. What will your meter read? Round your answer to the nearest one-half-volt. There is only one correct answer. Show your work. Problem number 2 You have a DC voltmeter with a 119,000 ohm internal resistance. You have a 1000 ohm resistor with one side connected to the positive post of your 12 volt battery. The other side of the resistor is connected to one wire of a 12 volt light bulb that has 6 ohms of resistance. The other wire of the bulb goes to the battery negative post. Will the light come on? What is the voltage across the light bulb? |
#10
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Working on it..... Trying to remember........ Did you go to Pickens?
torch |
#11
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Nope, but I used to live about 5 miles from Pickens Tech. My brother took a Refrigeration course there, though. Mine was USAF tech school at beautiful Keesler AFB in Biloxi, Mississippi, followed by On-The-Job (aka Trial-By-Fire) training with USAF F-4 Phantom maintenance and then cargo, commuter and major airline jobs. Now, I teach the dadburned stuff, so some of it must have "took" during my 30 years of fixin' these crazy flyin' machines.
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#12
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#13
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I went to Pickens in '89-'90, in the diesel program. Ken Condor was my teacher. Good teacher, made you figure it out with what you were taught, instead of handing you the answer.
torch |
#14
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Question 1.... I will have .0012 volts. I am not concerned with the internal resistance of the voltmeter, since it is designed to measure potential in the circuit.
Question 2.... The voltage across the lightbulb is .0012 volts, since the DVOM measures potential. I don't know if the lightbulb will light; how dark is the room? I don't know if I'm right, since Fat Tire can screw you up and I'm struggling with my memory, but let me know if I'm on the right track. torch |
#15
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Quote:
The problem all arises from the statement made that there is no ballast resistor used with MSD.(comes from the msd wiring instructions) The statement is true, BUT misleading. OEM...Ballast res goes in the ignition primary circuit. Ignition primary circuit is the circuit that supplies power to the small + lead on the coil. On a regular ignition system, power goes through the switch, then the ballast, then to the coil ( all when the Key is in the run position ). With MSD.... it is wired from the msd box straight to the pos coil lead. The old pos coil lead isn't used to power the coil anymore. The coil get's it's power from the msd box. Therefore the Ballast resistor is already gone from the ignition primary. You would have to add one for there to be one. The old coil pos wire can then be used as the switch "on" wire for the msd. The MSD get's it's power from the battery direct( with no ballast resistor in there either). The switch on wire does not need to be 12 volts, and it says so in the wiring instructions, so the old coil + wire can be used. The only reason you would have to remove or bypass the ballast resistor in a car with MSD, would be if you were using the old coil pos wire as the MAIN heavy power lead for the MSD. Not wiring it, to the battery direct as specified in the wiring instructions. That's why the MSD instructions seem vague on the topic, and don't specify to remove the old ballast. If you follow the wiring instructions, you have ALREADY removed the ballast from the igniton primary system without even touching it. |
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#17
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.0012 volts is wrong. Try again. |
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