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Another voltage drop question

Virg464

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Tracking down cause for a no spark condition. So, since I'm not a auto electrical guy, I'm learning as I go. I've checked for voltage at the ECU, the Ballast resistor, and the coil + side. I get 12 (plus) vdc at the ECU connector, the Ballast resistor, and almost 12vdc at the coil until I check at the coil with the ECU connector plugged in. Then, voltage at the coil + side drops to 5.5-6 vdc. The DVOM neg lead is connected to the battery neg post.

Question is: Is this normal for the voltage at the coil to drop from 12v to 6v when the ECU is plugged in?

Thanks.
 
Normal. The coil circuit is open till the ecu plugs in - the ballast resistor does not drop any voltage until the circuit is grounded.
 
Electric choke? If so unplug it and see if you get spark back. Just one of the wild things I've encountered ..
 
Tracking down cause for a no spark condition. So, since I'm not a auto electrical guy, I'm learning as I go. I've checked for voltage at the ECU, the Ballast resistor, and the coil + side. I get 12 (plus) vdc at the ECU connector, the Ballast resistor, and almost 12vdc at the coil until I check at the coil with the ECU connector plugged in. Then, voltage at the coil + side drops to 5.5-6 vdc. The DVOM neg lead is connected to the battery neg post.

Question is: Is this normal for the voltage at the coil to drop from 12v to 6v when the ECU is plugged in?

Thanks.
The transistor on the Mopar ECU, with everything connected is turned ON, allowing current to flow through the ballast resistor and coil primary winding to ground (thru the ECU). The ballast resistor limits the amount of current flowing. When the engine cranks and starts the pick up coil assembly in the distributor, creates a pulse that turns OFF the transistor on the ECU, creating the spark at the coil's secondary windings. The ECU controls the ON/OFF time of the coil's primary winding. YES....there will be a voltage drop because the transistor is on, current is flowing thru the ballast resistor and the coil........
BOB RENTON
 
I think 5.5-6v is pretty low.
Do you have a low impedance coil?
 
I think 5.5-6v is pretty low.
Do you have a low impedance coil?
Coils have resistance of the windings not impedance. Impedance is a term used with the term inductive reactance or the rate of change or the time it takes to charge/discharge the windings after the current supply is turned on/off......thought you might like to know.......
BOB RENTON
 
Coils have resistance of the windings not impedance. Impedance is a term used with the term inductive reactance or the rate of change or the time it takes to charge/discharge the windings after the current supply is turned on/off......thought you might like to know.......
BOB RENTON
Sure they have impedance. 2 x pi x f x l. The f is the frequency which is based on dv/dt which affects di/dt. Once charged and at DC level the impedance portion is zero. Capacitance also plays a role in the charging rate because it affects dv/dt.

Decatur Illinois? Original home of the Bears! LOL! I spend a lot of time there on Garfield Ave at Decatur Industrial Electric.
 
Sure they have impedance. 2 x pi x f x l. The f is the frequency which is based on dv/dt which affects di/dt. Once charged and at DC level the impedance portion is zero. Capacitance also plays a role in the charging rate because it affects dv/dt.

Decatur Illinois? Original home of the Bears! LOL! I spend a lot of time there on Garfield Ave at Decatur Industrial Electric.
I appreciate the input...thanx....dv/dt and di/dt calcs are a little beyond the understanding of many......derritives with respect to voltage, current and time....differential equations...too many years ago. A given frequency is involved but due to the constant changing one must pick a designed frequency to accurately define the time....maybe worst case condition...yes....capacitance also effects the charging/discharging time....as it opposes inductive reactance.....it least it was when I learned it......
BOB RENTON
 
Sure they have impedance. 2 x pi x f x l. The f is the frequency which is based on dv/dt which affects di/dt. Once charged and at DC level the impedance portion is zero. Capacitance also plays a role in the charging rate because it affects dv/dt.

Decatur Illinois? Original home of the Bears! LOL! I spend a lot of time there on Garfield Ave at Decatur Industrial Electric.
I did a lot of contract engineering for ADM Glycols division in Decatur and used DIE to rebuild/repair one of the 1000 hp hydrogen gas compressor motors.....4160 v 3 phase Class 1 Division 1 Explosion Proof design because of the manufacturing of H2 gas for the process.....the H2 compressors were 4 stage double acting design to get to ~ 1500 psi. Someone did not add lube oil the the front journal bearing on one compressor resulting in the bearing melted, rotor struck stator windings, tore up shaft journal requiring a partial shaft and seal replacement....and visited DIE many times during the rebuild. I redesigned the lube oil supply for temp, pressure and measured flow as a run permissive with alarm b4 shut down on the network computer system. ......cheers...
BOB RENTON
 
My interest was in finding out if the voltage divider between the ballast resistor and the coil was way off .
But it digressed into semantics and boastful qualifications which did nothing to help solve @Virg464 with his ignition problem.
 
My interest was in finding out if the voltage divider between the ballast resistor and the coil was way off .
But it digressed into semantics and boastful qualifications which did nothing to help solve @Virg464 with his ignition problem.
There is no voltage divider network.....ballast resistor to coil to the ECU to ground. "was way off" is a normal condition..... "way off" was determined by ???? If you think this is too much, what should it be??
BOB RENTON
 
Assuming 12 Vdc at the battery if you have 0.5 ohm ballast resistor and a 1.5 ohm coil.
If it were a pure resistance circuit you would measure 9 Vdc at the coil.
12 X (1.5/2) = 9 Vdc

Now if he has a coil with .6 ohms

Then:
12 X (.6/1.1) = 6.45 Vdc

So I totally disagree.

Again I'm trying to help the guy.
 
Electric choke? If so unplug it and see if you get spark back. Just one of the wild things I've encountered ..
No choke.

Embarrassed to say, I found out what was causing the no spark issue. I somehow (without realizing it) while doing other things, moved the distributor about 15 degrees retarded, so the rotor was well past the post in the cap when the coil was firing! :BangHead:
 
Assuming 12 Vdc at the battery if you have 0.5 ohm ballast resistor and a 1.5 ohm coil.
If it were a pure resistance circuit you would measure 9 Vdc at the coil.
12 X (1.5/2) = 9 Vdc

Now if he has a coil with .6 ohms

Then:
12 X (.6/1.1) = 6.45 Vdc

So I totally disagree.

Again I'm trying to help the guy.
What is the (1.5/2) expression represent? Likewise, .(6/1.1) represent? Just curious.....
BOB RENTON
 
Coils have resistance of the windings not impedance. Impedance is a term used with the term inductive reactance or the rate of change or the time it takes to charge/discharge the windings after the current supply is turned on/off......thought you might like to know.......
BOB RENTON

I did a lot of contract engineering for ADM Glycols division in Decatur and used DIE to rebuild/repair one of the 1000 hp hydrogen gas compressor motors.....4160 v 3 phase Class 1 Division 1 Explosion Proof design because of the manufacturing of H2 gas for the process.....the H2 compressors were 4 stage double acting design to get to ~ 1500 psi. Someone did not add lube oil the the front journal bearing on one compressor resulting in the bearing melted, rotor struck stator windings, tore up shaft journal requiring a partial shaft and seal replacement....and visited DIE many times during the rebuild. I redesigned the lube oil supply for temp, pressure and measured flow as a run permissive with alarm b4 shut down on the network computer system. ......cheers...
BOB RENTON
Cool! DIE is good people......... I spun them around on their heads! LOL! I'm kinda known to that in the electrical motor industry.........
 
What is the (1.5/2) expression represent? Likewise, .(6/1.1) represent? Just curious.....
BOB RENTON
I shouldn't have to explain this.
The terms were pretty clearly defined.

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