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mattd860

Voltage Drop at Ignition Coil

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mattd860

On my Kohler K301 I noticed yesterday that the voltage at the postive terminal at the coil was approximately 2 volts less than the battery when the tractor is not running. The battery reads 12.1 volts, but with the key switch in the On position, the ignition switch reads 10.0 volts.

I have the carb and other parts off the motor so I can't test the voltage when the engine is running but I imagine there will still be a voltage drop.

Is this kind of a drop normal? Can it prevent my engine from starting and functioning properly?

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rmaynard

When my K241 is not running, the battery voltage reads 12.74. With the key on, the voltage at the coil reads 11. A drop is normal since the coil is a resistor.

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mattd860

yup you're right. I pulled the wire off the coil and tested it and there was no voltage drop at all from the wire. As soon as I attach it back to the coil it shows a voltage drop. Thanks

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Save Old Iron

Matt,

you are on to something here.

Electricity is very simple in its needs. If a battery produces 12 volts and is connected to a circuit, that circuit must consume 12 volts - no voltage can go "unaccounted for"

In your case, your missing 2 volts.

I'll bet if you make this measurement with BLACK meter lead on the battery + cable (picked off the starter solenoid in the pic below) and the RED meter lead to the + terminal of the ignition switch

whignwirevoltdrop.jpg

you will find a voltage drop either across the ignition circuit wiring or across the ignition leg of the ignition switch. Replace any connectors, wiring or ignition switch that may be corroded and causing the increased resistance in the ignition circuit.

The drop could be across the battery + to starter but if that were true, your tractor would probably not crank over.

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mattd860

Alright I will test that later.

I am bit confused because when I disconnect the ignition + wire from the coil and measure the voltage at the wire terminal (that connects to the coil), there is no voltage drop at all. But if I connect the wire to the coil and then measure the same wire terminal, I see a ~2v drop. So I only see a drop in voltage if the wire is connected which just doesn't add up in my brain.

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iicap

I'm sure SOI will add more clarity to your question, but the ignition coil is just an accessory in your electrical system. Turn the key on and current will flow to the coil, but the ignition points have to be closed to complete the circuit. If the engine stopped rotating in a position that the points were open, there would be no connection to ground and there would be no draw on the battery. Pertaining to what I've said, I'm not sure if the condenser might have some small draw even if the points are open, SOI can prob answer that.

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rmaynard

I'm not sure if the condenser might have some small draw even if the points are open

Did a quick check with the condenser connected and disconnected, and there was no difference in the voltage drop. Coil has an internal resistance of 4.1 ohms.

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mattd860

I'm not sure if the condenser might have some small draw even if the points are open

Did a quick check with the condenser connected and disconnected, and there was no difference in the voltage drop. Coil has an internal resistance of 4.1 ohms.

So just to be clear - you also have a voltage drop at the positive terminal of the coil correct? I know you said you did yesterday but just want to be 100% sure. :thumbs:

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iicap

I would say that the voltage drop at + coil term, if it is only in hundredths of a volt maybe caused by how far the current is traveling from the + battery post, to the solenoid stud, thru the wire to and thru the key switch and or ammeter too, and finally the wire that connects to coil+. There is going to be SOME loss to resistence and that is ok. If your difference is 1/2 volt(.50) or more then I'd look for the source of resistance, be it a bad wire connector(corrosion or loose), worn contacts in the ignition switch. You would be doing these searches with the Ohms section of you test meter. Don't be meaning to be long winded, but trying not to leave you guessing what I'm saying.

rmaynard thanks for the clarification on the condenser :thumbs:

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rmaynard

So just to be clear - you also have a voltage drop at the positive terminal of the coil correct? I know you said you did yesterday but just want to be 100% sure.

Yes I do.

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WH nut

When you have the wire unhooked you are just checking voltage, to measure voltage drop the cucuit must be under load. What you are checking is to make sure the circuit can handle the load. You can have a wire with all strands broken and still carry 12 volts if the circuit is not loaded. Voltage drop is a much better test than resistance for the same reason.

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iicap

WH nut, I think I understand what you are saying, I am a student here also.

eg; if all is well, the voltage at the point of load should equal the voltage at the batt. If the voltage at the point of load is less then batt voltage there is a drop. Then you would go looking for the weak link?

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Save Old Iron

I'm not sure if the condenser might have some small draw even if the points are open, SOI can prob answer that.

think of a capacitor as two sheets of tin foil sandwiched over a sheet of wax paper

attach a wire to each sheet of aluminum foil and roll this "sandwich" into a tubular form

caplayers2.png

if you check out the standard electronics symbol for a capacitor, you can immediately see how the symbols truly represents the actual construction of the device

capschematicsymbol.png

2 metallic "plates" separated by a dielectric "wax paper".

capacitors do not actually use wax paper or aluminum foil, but the real components are very close in composition to these common household items.

To answer the question of current flow thru a capacitor, NO DC CURRENT WILL FLOW thru an intact capacitor. The "waxed paper" dielectric prevents electrons from one plate to the next so no DC current passes. So condenser in or condenser out, no difference of current flow or voltage drop will be noticed in troubleshooting the ignition circuit. If you do see a measurable difference, the condenser is probably "leaky" and considered defective.

In the real world, a minuscule but measurable electron flow does take place from damaged dielectric. old dielectrics or damaged dielectrics where arcing has "punched thru" the dielectric and created leakage thru the capacitor.

Further description of the multiple functions of the condenser in the ignition circuit goes way beyond the scope of this post. I believe The Old Wizard has a great post started on how ignition circuits work. I encourage you to check it out.

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Save Old Iron

Matt,

the voltage drop / no voltage drop concept has been answered by WH nut.

no current flow = no voltage drop so each point in the circuit powered by 12 volts will read very close to the battery voltage.

I'll try to post an illustration of why this is so in the next day or so.

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WH nut

Actually no voltage drop would read 0 volts. Look at it this whatever the load is must cunsume all available voltage in order to operate properly. Lets look at a light buld. its hungray and wants 12 volts to eat. If you measure from the switch to the input of the buld it should eat all available volts. if it has a weak stomach and doesnt eat it all what is not eaten shows up in the meter, that would be voltage drop. All you are doing is measring a piece of wire from one end to the next. Now if you measure on the ground side of the wire quess what your meter will show. 12 volts is correct buecause if the wire to ground is good it would be just like measuring fron battery+to battery -.. When measuring voltage drop you load the circuit and measure from switch output to load input and you should read 0, anything other than 0 = voltage drop.

Is that clear as mud

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iicap

A little murky, but 3 or 4 more reads and clarity wll appear. Thanks SOI, WH nut

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WH nut

Go out and play with your meter on a simple lighting circuit then take a jumper and cut most of the strands and yopu will see how the whole voltage drop thing works

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mattd860

Sorry guys - I've been very busy this weekend installing a new wood boiler and chimney in my home and haven't got a round to testing anything yet. I hope to have some free time tomorrow after work - HOPEFULLY. Thanks for all the info you've posted so far - it will give me something to read and comprehend during the workday tomorrow :thumbs:

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