Elusive volt

[posplayr]

Interesting. I have never done this so I thought I would try it too. Battery on, headlight on, engine off, I have 11.88 at the battery. Move the the positive probe over to the o/w wire on the coil and I initially have 9.91v. After several seconds, I noticed the voltage climbs. After about a minute it was up to 10.00v. I think I may have some crummy wire and/or connections right? Bike seems to fire ok though. Recommendations?

Charge to battery as it is low on charge. If the voltages at battery will not stay above 12v the likely you need a new one soon

[Sci85]

Yeah it's a brand new AGM BATTERY. I just put it on the charger and will repost results once it's fully charged. Regardless, I think I'm losing a lot voltage through the harness wouldn't you say?

[Sci85]

Ok, after charging, with key on, headlight on, battery is 12.23 and positive side of coil is 10.22. 2 volts drop. I guess that's pretty bad huh? So I need to check the on/off switch and the coil connectors right? Is there anything else in that circuit that I should be checking?

Thanks!

[posplayr]

Ok, after charging, with key on, headlight on, battery is 12.23 and positive side of coil is 10.22. 2 volts drop. I guess that's pretty bad huh? So I need to check the on/off switch and the coil connectors right? Is there anything else in that circuit that I should be checking?

Thanks!

The ignition switch is the primary culprit; if you have a fuse box check that as well. Of course the SSPB will solve all these issues.

[Sci85]

Which I am eagerly awaiting! Eagerly...waiting...

[D'Ecosse]

You will always have a voltage drop at various points of the system as compared to the battery - anywhere it flows current you get a voltage drop. It can be made worse by its path through small gauge or oxidized wires, connectors, switches, relays etc.
The OP was measuring a drop to the coil negative because there is resistance across the coil; however when the connector was disconnected from the igniter there was no current flow - no current flow* means no voltage drop regardless of how high the resistance is
If you want to improve the voltage to the positive of the coil, feed it via relay directly from battery with a better gauge of wire, say maybe 16 ga.

* on a related note, I am advocate of adding relays to headlight circuits to improve the voltage at the lamp and thus improve the output light. I explain to measure the voltage at the lamp socket to determine how much drop they are actually getting compared to the battery; can't tell you how many times I get report back saying their bike is perfect, virtually zero drop. That is because they disconnect the headlight connector to measure across the terminals! Same story - no current flow, no voltage drop! The connector must be connected to the lamp and of course it must be ON! On same subject, it is every bit as likely that the loss is in the negative return (yes, you can also have a voltage drop between the negative terminal of the lamp socket and the battery negative terminal!) - the answer there is, as well as adding relays in the positive legs (one each for both high & low beam) with decent gauge wiring, you should also run new ground return wire back to the battery.

Incidentally if you just get the headlight current out of the KeySwitch circuit (by bypassing with relays), you will automatically improve the voltage to the coil (lower current means lower voltage drop)

[Sci85]

Well, I started work on the coil relay mod last night just as everyone suggests. It's almost done. One question though. You mentioned running a new negative lead. With the coil relay mod, is that still necessary or does the relay ground take care of that problem? Do you leave the harness grounds connected to the coils?

[posplayr]

You will always have a voltage drop at various points of the system as compared to the battery - anywhere it flows current you get a voltage drop. It can be made worse by its path through small gauge or oxidized wires, connectors, switches, relays etc.
The OP was measuring a drop to the coil negative because there is resistance across the coil; however when the connector was disconnected from the igniter there was no current flow - no current flow* means no voltage drop regardless of how high the resistance is
If you want to improve the voltage to the positive of the coil, feed it via relay directly from battery with a better gauge of wire, say maybe 16 ga.

I generally agree with what you say, but there are some specifics here that don't make any sense.

The OP says he is measuring the negative side of both coils and losing 1 volt on either coil(-). He says that the voltage is restored when the ignitor is unplugged. Something doesn't make sense, which is why I requested more measurements.

Generally depending upon where your crank is at either the 1-4 side of the ignitor is shorting the (-) side of the coil or the 2-3 channel of the ignitor is shorting the other coil (-). When shorted the voltage drop across ignitor is only about 1 volt, so the negative side of the coil will also drop to 1 volt above ground. Depending upon which side is shorted the other side will be open and the coil (-) will be at the same voltage as the coil (+).

So don't understand how both (-) sides of the coil can both be dropping only 1 volt (below the battery +12V) and at least one of them is not at only 1 volt above ground. I don't know for sure but perhaps the dwell is such that they can both either be shorted or open at the same time. That is the first quandary.

With the Key on there is typically about 1/2 a volt drop in the battery (from say 12.8V down to 12.3V) due to a nominal 11 amp load from lights and ignition and the normally installed 14 Amp Hr battery). And then depending upon how dirty the ignition switch and other wiring is there can be anything from 1-3 volts of drop from there to the coil (+) side. So figure that you are probably seeing between 10-11V at the coil when the ignitor is not shorting that coil. In an attempt to match the observations, I will assume that the drop in voltage to both coil (+) sides is 1V and so instead of 12.3V the OP is seeing 11.3V at the coil(+) (both sides).

But remember he claims to be measuring the (-) sides of the coil which to only drop 1V means that this ignitor channel is not shoring that particular coil. That is OK and if I assume it is possible for both channels of the ignitor to be off at the same time then we would expect to see the same voltage on both coils and on both sets of posts. Everything on the coil primary is reading 11.3V. This is because the igniter is not shorting either coil and neither coil is pulling any current. Well I guess this is possible if the dwell is such to allow both coils to be off at the same time. However this would seem to be at odds with the next observation. The voltages all return to 12.3 V when the ignitor is unplugged. How is that possible when the ignition is not sinking any current through the coils ???

Reviewing this, 11 amps typically takes a healthy 14 amp hour battery down 1/2 V from 12.8V to 12.3V. Of that 11 amps, about 3.5 amps is the ignition. Based on the measurements above, it would seem neither coil is pulling current (i.e. the extra 3.5amp) but we some how loose a volt with the ignitor unplugged. That would suggest an even larger load then 11 amps from a plugged in ignitor.


The only explanation I can find is that the ignitor is somehow drawing excessive current above and beyond what the coils draw, but that would typically indicate a fault but the bike seems to run.

So I am much more inclined to suspect that there is something wrong with the measurements which gets me back to why I requested more measurements to see if we can find any consistency with the theory of operation of the circuits.

[D'Ecosse]

I should have prefaced my previous with the following but will apply it now - my comments are to general theory and not specific to the GS of which I have limited first-hand knowledge. So apologies if that limitation affects my interpretation.

.... don't understand how both (-) sides of the coil can both be dropping only 1 volt (below the battery +12V) and at least one of them is not at only 1 volt above ground. I don't know for sure but perhaps the dwell is such that they can both either be shorted or open at the same time. That is the first quandary.

I would think that as the igniter has absolutely no clue where the crank is with just ignition on (it must be rotating before it can determine the timing) then it will not prepare to generate a switch to create spark regardless of where it is physically (even if perfectly aligned to where it might normally spark on one of the cylinders).
So I would presume that both coils start out in 'open' state and not in a charging state until it begins to rotate.

The OP says he is measuring the negative side of both coils and losing 1 volt on either coil(-). He says that the voltage is restored when the ignitor is unplugged. .. When shorted the voltage drop across ignitor is only about 1 volt, so the negative side of the coil will also drop to 1 volt above ground. Depending upon which side is shorted the other side will be open and the coil (-) will be at the same voltage as the coil (+). ...

You are saying "open" yet remember the ignitor is an electronic circuit and there is likely still some impedance to ground. There has to be some impedance to ground from the coil positive, otherwise you would measure zero loss at the coil positive relative to battery (unless it was 'on' of course, but then it would be much higher as you suggest).



There will be some loss through the bike's wiring/switch system to the coil positive (most of this is attributed to the lights however) and then a further drop across the coil to the negative and a third drop from the negative, through the igniter to ground; there are effectively three series impedance paths between the battery and ground

Really, the only way to determine what is his real loss to the coil circuit, would have been to measure the voltage at the positive (which is only thing that really matters here anyway) and you could measure at the coil positive terminal when both connected to the coil and unconnected, which would determine whether there really is some small current flow even in the 'open' state, which is what would produce the small (~1V reported) drop compared to the battery. The 1V seems excessive to what I might have expected however and would think some of that 1V at least attributable to the headlight current and the loss to the positive terminal.

The 'spoiler' & what is contradicting in the OP case reported data is why the voltage at the coil negative would actually restore to full battery voltage - there should still be some loss from the current consumed by the headlight circuit (at least to the point where the headlight & coil circuits merge) - so hard to see why there would be FULL battery voltage at the coil negative (and therefor the positive) with the igniter disconnected.

[posplayr]

I would think that as the igniter has absolutely no clue where the crank is with just ignition on (it must be rotating before it can determine the timing) then it will not prepare to generate a switch to create spark regardless of where it is physically (even if perfectly aligned to where it might normally spark on one of the cylinders).
So I would presume that both coils start out in 'open' state and not in a charging state until it begins to rotate.

I'm pretty sure it is possible for an engine to be not turning and have one coil pulling current and not the other. I just don't if there is always just one on or one off.

You are saying "open" yet remember the ignitor is an electronic circuit and there is likely still some impedance to ground. There has to be some impedance to ground from the coil positive, otherwise you would measure zero loss at the coil positive relative to battery (unless it was 'on' of course, but then it would be much higher as you suggest).

There is probably some leakage current but none that anybody here will be able to detect. I doubt there would be any voltage difference across the coil either although that is probably easier to detect. For for most intents and purposes you can assume when the ignitor is "open" it is like an "open" mechanical switch.

To complete the picture for the OP you should add the second coil and probably show both ignitor channels inside on a single enclosed box.

There will be some loss through the bike's wiring/switch system to the coil positive (most of this is attributed to the lights however) and then a further drop across the coil to the negative and a third drop from the negative, through the igniter to ground; there are effectively three series impedance paths between the battery and ground

Really, the only way to determine what is his real loss to the coil circuit, would have been to measure the voltage at the positive (which is only thing that really matters here anyway)

This is correct as you are showing there are at least three variable resistances and it takes more than 2 measurements to determine and explain those things. For the OP to not provide the full voltage (rather just the changes of 1 V) even really makes this a guessing game. If he persists then It will just have to remain a mystery.


P.S. there is a fourth series resistance internal to the battery. It is not exactly a constant as it is a function of state of charge. The less charge the more the battery voltage drops when you pull current. This is called the batteries "internal resistance". As mentioned when the key is switched non the 11 amp load will typically drop a good battery from about 12.8V down about 1/2 a volt.