jag1886

I'm putting an electronic ignition on my 65 and the instructions say to get rid of the resistor wire but when I check the voltage I get 12.25 volts on this wire. Does the resistance wear down with age in these wires?

jag1886

I'm installing an electronic ignition on my 65 and the instructions say to get rid of the resistor wire when I check the voltage at the coil I have 12.25 volts. Do these wires loose there resistance as they get old??

henryk8398

No. If anything the resistance of any wire should increase with age. Are you sure you are measuring the right wire? With the key on run or cranking? Measure both. Otherwise, unless you know the complete history of the car, it's possible the wiring has been altered. Does your car burn out points pretty quickly? If so, that indicates the resistance wire has been removed at some point.

MDchanic

The resistance wire will read closer to 8v. 12¼v is normal battery voltage.

If the wire in question turns on and off with the ignition, it sounds like you're good to go, but be careful it's not patched in from some other circuit that's already fully loaded - you don't want to blow a fuse at 70mph and have the motor die on you.

- Eric

jag1886

It's the correct wire it has the factory braiding that is on all resistance wires and the yellow wire is the bypass from the starter and it's dead (because I'm not cranking the starter). This car has not had the wiring altered it's still all in the factory wire wrap.
I'm going to use this wire I just found it odd that there was no resistance in it.

Bluevista

How do you type that small 1/4?
Is that there one of them those keyboards with them there weird furrin accent symbols on it?
We don't take kindly to them there keyboards around these here parts.

Weird that the resistance wire would read 12+ volts?
I wouldn't trust it either, gremlins could pop up later, I'd run a dedicated wire from the ignition that's hot in the cranking and run position just for peace of mind.
Hey!..I found the plus sign on my keyboard, never noticed it before.
Still can't find no small quarters.

BlackGold

When you measured the voltage, the points were probably open -- which means no current was flowing. Crank the engine a touch so that the points close and you should see the voltage drop some at the coil's positive terminal and almost to ground at its negative terminal.

BlackGold

Your points are open and no current is flowing, hence no voltage drop. crank the engine a tad and try again.

MDchanic

If you've got over 9v at that wire, then there is something wrong.
Maybe the wiring was altered, maybe you've got a short somewhere (like inside the harness, not far from the distributor), but resistance wire DOES NOT lose its resistance.

I'd recommend investigating further first, but if the potential sudden loss of motive power doesn't bother you, then that's your call.

Well, see, out here on the coast, we're just more open-minded .

The trick, though, to typing ¼, or ½, or even ⅔ depends on your computer.

On my Mac, I click on "Special Characters" under "Edit" and a neat window pops up with all of those, and also ∛, ∫, ⇒, ☢, ☠, and a few hundred others.
Things I use all the time are just key combos, like ° (shift-option-8) and — (shift-option-"dash") and Ω (option-z).
I can do the, fo-rine accents, too, generally by doing option-[desired accent key] then the letter, such as option-u then u for ü (German timing marks read "Früh" and "Spät"), or option-e then e for é (I've grown rather blasé about these characters, really).

When I'm at work, on that rotten PC, I open Word (it's that hideous newer version with all the menus changed), select some sort of a "Special Characters" tab on the top right (it has a large Ω on it), type the one I want, then "copy" and "paste" it into the browser window.

I'd better quit now - don't wanta rile you up with newfangled dagnabbed information — you might done run me off.

和睦

- Eric

MDchanic

It should read 8-9v regardless.

- Eric

jag1886

The wire was disconnected from the coil and tested with a volt meter.

Oldsmaniac

Is the meter accurate??

jag1886

Battery tests 12.51 volts with the same tester.

MDchanic

Having two identical threads asking the same question is really annoying.

- Eric

jag1886

I'm terribly sorry but I needed to find an answer yesterday not in a week.

Aceshigh

If anything they INCREASE resistance as they get older because of corrosion built up over decades at the open connections.

Aceshigh

Agreed. LOL , I see there's 2 threads on this now.

MDchanic

I've just put cross-links on them, to keep everyone from going insane.

I think it's too late for the mods to combine them (but maybe not).

- Eric

BILL DEMMER

NO, it won't. as stated correctly above, if there is no load on that wire, reading the voltage on it will show full battery voltage.


bill

MDchanic

Okay.

Just got the voltmeter, got the key, went outside, took the cover off the Chevelle, connected the battery, took off the air cleaner, disconnected the points and tach wires, and checked.

12.33v at the battery.

11.94v at the coil.

You're right, I'm wrong. Damn.



- Eric

Lady72nRob71

That was why it was high. A resistor will only resist if there is current flowing through it.

X2. Hook the coil back up and turn the crank to where the points close and the voltage will drop to 7 or 8V.

BTW, I combined the two similar threads as routine housekeeping.

jag1886

I believe you are right I checked voltage with the car running and the voltage slowly drops down to about 9 volts after a couple of minutes. Problem solved.

jag1886

Sorry about posting twice, this is the only sight I belong to that people get wound up about that. Once again Sorry and I WON'T do it again.

MDchanic

Okay, I already put my foot in my mouth once here, but "after a couple of minutes?"
If there's a load on the circuit, the voltage should be low immediately.

Am I wrong on this?

- Eric

jag1886

It did, I started the car and put the volt meter on and continued to hold it there and it just slow click away until it got down to 9 volts. It's beyond me, in all the years I've worked on cars I had never tested one of these.
How knows? But that's what it did.

MDchanic

Did you measure something that was 12v immediately beforehand?
Could it be that you've got a digital meter with a buffer that prevents sudden changes, and you read a 12v source first, then put it on the 9v, and it took a while for the meter reading to change because it's designed to change slowly?
Seems very odd...

- Eric

MN71W30

Huh? This is a strange thread. Maybe the heat would add a little more resistance. But with it running it should go up a little because the charging system would increase the voltage 10% or so.

Lady72nRob71

The charging system could have accounted for at least 1V of voltage drop, as the battery charged up and alternator voltage dropped some.

As for differences between engine on and engine off...
A DC voltmeter is designed to read straight DC voltages.
Jag was trying to measure a pulsed voltage through a resistor, inductor, and capacitor - an RLC circuit. The resistor wire is the resistor (R)
When switched on and off, the coil is a big inductor (L). The points are opening and closing, switching it on and off. Magnetic field in the coil charges and collapses with each switch. Factor in the condenser (capacitor) (C) and you have a pulsed RLC circuit (though C is very tiny). I deal with these every day at work in power supply design. Complex and requires lots o' math (ick...)

The DC voltmeter will only measure the average DC level, here about 9V, esp a digital meter. An analog meter might show some waviness at idle. Reve the engine and voltage may change.
Measure with an oscilloscope and it will be somewhat of a blunt sawtooth waveform. Remember those autoscopes on the big tuneup machines in the 80's? I bet those waveforms were off the coils. You could see the dwell angle and stuff, too. It has been 20 years since I saw one last.

If you shut off the engine and rotate crank to close the points, then the coil behaves like long piece of wire (like a 2 Ohm wirewound resistor). You have straight DC flowing through it then and the DC volt meter will give an accurate reading, about 7V with a stock coil. (The resistor wire is 1.35 Ohm).
Do NOT leave leave current flowing like this (engine off) for more than 20 seconds to prevent coil, points, and wiring damage! Parts were not designed for straight DC!

MDchanic

All true, Rob, but even so, the voltage should be reasonably constant, and maybe fluctuate a bit in the 8-10v range, not start at 12v and slowly go down to 9.

Also, since this is a pulsed DC LRC circuit, regardless of what the precise parameters are, shouldn't the factors of inductance, resistance, capacitance, and frequency just all serve to create a constantly varying impedance that, while not constant is never NEGATIVE? In other words, there's no AC here, and no reverse EMF or reactive currents, as far as I know, so at the very least, the thing should consistently look like a big resistor, and NEVER show its full input voltage (12v) while the engine is running.

Did that make sense?

'Cause none of the rest of this does, and I'm confused.

- Eric

MN71W30

If Jag is planning on running HEI he should consider running a new wire from the panel anyway. I heard of guys running #10 ga wire that seems like over kill to me. I'd run a new #12ga to the HEI from the panel and forget the resistor wire.

BlackGold

Rob's right. I've seen even very high-quality digital multimeters give false readings when attempting to make a DC measurement of a waveform with AC on it -- especially if that AC contains really high frequencies (like the switching power supplies Rob works on). I would expect better results with an old-school mecahnical Simpson meter.

Before anyone tries to tell me that the negative coil terminal doesn't have high frequencies on it, I'll assure you that a mechanical switch (which the points are) opening and closing on an RLC circuit does produce extremely high frequencies. I'll leave the proof to professor Fourier.

Now, I can't guarantee that that's why Jag is seening the voltage drop so slowly while the car's running. I'm just saying that one explanation might be a meter that's fooled.

Eric, you could've saved yourself all that trouble if you just brushed up on Ohm's Law. But I'm glad you proved it to yourself (and the rest of us).

MDchanic

The annoying thing is that I'm quite familiar with Ohm's Law.
I just thought I could remember measuring that voltage in the past, and didn't even think of the theory.

I agree that there are all sorts of low amplitude harmonic frequencies within that system that could confuse a digital meter's computer.

I wonder though whether the meter's computer is loading sequential readings into a buffer, then time-averaging them to damp the "swing" of the output, so that if he measured 12v first, then put the lead right on the ~9v, it would slowly tick down to the new voltage, rather than changing immediately.
My digital meter has a display that skitters around all over the place, which makes it a bit hard to read, and I could imagine some companies trying to make the display more user friendly by smoothing its changes and averaging its instantaneous readings.

- Eric

BILL DEMMER

HEI likes a low impedance supply, 12-gauge is usually the perfect size.
HEI varies dwell with rpm, hence the circuit load varies vs. rpm.


bill

Oldsguy

Did that once, fried the wire! The sad thing is I didn't even think about it when I did it and should have known better.

Lady72nRob71

When the points are closed, everything is somewhat predictable.
Within way less than a second when the points close:
Current begins to flow through resistor wire. Voltage point that Jag was measuring (called point X hereonout) starts at 13V, coil begins to charge. As charge builds up in coil, the voltage ramps down until the coil's DC resistance is met (about 2 Ohm) and now point X now measures about 7V. This is called inductor saturation.
Remember class - a coil (inductor) appears like an open to DC when it is discharged, but a theoretical short when it is fully charged! (Opposite of a capacitor).

Now the points open and fun things happen..
Voltage stored in the coil collapses (goes down VERY fast) and its current flow causes very high voltage positive AND negative spikes at point X for a tiny fraction of a second (the nature of inductors). The condenser capacitor's job is absorb and reduce most of these spikes, a form of EMI (electromagnetic interference.)
After these HV spikes disipate, Point X returns quickly to 13V, as the circuit is open.

Points close again and it repeats. Picture a sawtooth waveform with spikes at the transition points...
As the disty turns, the points are pulsing many times a second. The pulse width is a bit unknown to me (dwell), but regardless, it is a form of alternating current in the (mostly) positive range.

Right. It boils down to using the wrong meter to measure a pulses RLC circuit. The meter itself has a 1 to 10M Ohm internal impedance to also add to the mix...
It is possible to better integrate the meter into the circuit by adding a 100k resistor and 1uf cap in series between point X and ground and place the meter negative to ground and its positive to the point between the above R and C.

An o-scope with a 10x probe is needed to check it correctly (a 1x probe can cause the inductive spikes to damage the scope) The autoscopes probably displayed these well.

It does with the high voltage pos & neg spikes. Here is when Mr. Condenser comes in to ground away much of that high frequency noise.

A meter will often read the highest voltage first. A high quality meter will sample faster and compute the average faster, but still there is a wait time.
The pulses and voltage spikes confuse it as it is not made to measure them. The internal processor finally reads the average. Ignoring the spikes, point X pulses from 13V to 7V, a 6V difference. The median voltage is 10V. The pulse width and spikes accounted for the 1V error.

A long winded (and interestingly detailed) answer to:
Turn key on, do not start engine, attach DC voltmeter from point X (coil +) to ground. Short other side of coil to ground (to simulate points closing). Read voltage. ~7V means resistor wire is good, 12 means it's bypassed, 0 means it is open.
Do not keep coil energized for more than 20 seconds to prevent wire and coil damage.

Lady72nRob71

Given that the circuit will draw 3.58A with engine off, and the resistor wire is 1.35 Ohm, you are dissipating over 17 watts of power (in heat form) in that 2 foot piece of wire. That WILL smoke it until it finally opens up.

Lets keep that "engine off, ignition on" time as short as possible, never over 20 seconds at a time. Give a cool-off time of a couple minutes between times.