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I was given an older fluke 98 multimeter. I know it's a dinosaur but it features scope displays. The manual shows the proper lead hook-ups and alittle pictured examples, but doesn't explain troubleshooting. I guessing it's for more experienced people in the trade that actually understand what is being shown. Anyway my question is on watching the ignition secondary on the scope. The reading show mosts of the cylinders at 15kV. with some variation except for on cylinder giving a reading of about 20kv. What does this mean? Is a higher voltage display mean more or less resistance? Also why does part of the voltage drop alittle below the zero line. The vehicle runs fine with no problems. I just wanted to familiarize my self with this stuff. Thanks for any inputs.
The higher firing voltage is likely due to a plug that has an excessive gap or is otherwise getting "weak", perhaps due to contaminant buildup or a weak plug wire.
The voltage drops below the zero reference due to back-EMF of the collapsing electro-magnetic field after the plug fires.
If you can get a current (amp) reading you can use Ohm's law to figure out the resistance. Your meter probably only handles 10amp in direct contact, but an induction pickup should show higher.
He's working with 15-20 kilo-volts. I doubt that he needs actual values, besides, where are you going to get a meter that woks on that scale except by going to specialty equipment. The scope gives a good view of what is going on.
Went to tech. school in 74 found old notes on patterns.If youput address in email I would copy and send. But the basics to firing lines are that they should be withen 3 kv. high could be from excessive plug gap,break in plug wire if its small it will jump the gap check with ohm meter,deposit build up in cylinder, higher compression needs more volts. Low reading can be from small plug gap short from carbon on plug or wire. There is more to the rest of pattern scope can allso check diodes in alt
If getting the amperage is going to be trouble, then you can work on the other side of the coil(s). Input voltage and amperage become the output. Multiply I*E=P Then P=E*E/R. I would guess your input is about 25watts. 255,000,000/r=25 R=9803915ohm (9.8m Ohms) or there about.
Please double check this, because I don't get to use this often enough, but that is one idea to get the information your wanting.
He wanted to know if one wire had more or less resistance. This is how you get real values rather than just saying amperage decreases with resistance, therefor voltage must increase.. This is actually more than just plug resistance, it should also account for the wires and dist, everything from the coil to the short. I like to know why and how things work not just that they do, and I passed that information along.
He wanted to know if one wire had more or less resistance. This is how you get real values rather than just saying amperage decreases with resistance, therefor voltage must increase.. This is actually more than just plug resistance, it should also account for the wires and dist, everything from the coil to the short. I like to know why and how things work not just that they do, and I passed that information along.
They are in a totally different world compared to the normal E=IxR. Do a little research on high voltage scope analysis for ignition systems.