Hoping some one can enlighten me on what is acceptable max voltage drop on ECU powers and grounds. I have read 50mv but I have checked some bikes that run fine with more than that. Attached is a waveform for a BMW F650 single fuel injected. It has an intermittent stalling problem usually on idle. It had a stored CKP code but I scoped that and managed to get it to stall and the only thing that was dropping out was the Ignition and Fuel injector. I was wondering what you guys thought of the voltage drop on the ECU Power feed? Is that normal to see it drop like then when the ignition coil fires? Thanks in advance.

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Hi James, good to hear from you. Short answer to your question is 350 mv to 500 mv. That is the range I have seen taught as the max. That is for the entire circuit not parts of said circuit.

Long answer would take a few captures and explanation but I prefer 250 mv to 300 mv to be more my standard on circuits that have heavy loads such as motors that are constant. Momentary on circuits can tolerate slightly higher drops. Such as injectors, ignition coils and such. I still would prefer 350 mv or less.

Your setup of your leads... It seems to me you connected each lead to one side of the circuit. In other words the power side you went with your lead from Batter positive to the positive side of the coil you were measuring. Am I correct in that?

James do you still have the bike there to get a few captures?

Thanks for having a look Spence. As far as connections go yes, positive of the ECU to battery positive etc. I do still have the bike as I have not been able to get it to act up again so far.

James H wrote on Jul 30^th, 2019 at 5:52am:

Thanks for having a look Spence. As far as connections go yes, positive of the ECU to battery positive etc. I do still have the bike as I have not been able to get it to act up again so far.

My pleasure to have a look James!

If you wish to play(I mean work) here are a couple suggestions I would like to see. Hook a lead on both sides of the component we are gonna look at with the leads grounded as normal and the other end into the power feed and another lead on the control side. Have an amp clamp so we can what it is pulling and the 4th lead on the battery.

Give me one on your ignition and another capture on your fuel injector. I normally don't use all these channels all the time for this type of test but for conversations sake it might do us well to see them all.

Remember to use antenuators on the control side do to the voltage spike.

Hope all is well in your world.

Spence

.   [smiley=popcorn.gif]

I took one today if anyone want to play... Gotta speak up though and if I post it do some homework.

Anyone game?

Yes please Spence.

Nikk wrote on Aug 6^th, 2019 at 5:12pm:

Yes please Spence.

Fair enough. Here ya go with a capture similar to what I asked James to get. Lets say we wish to figure out how much voltage drop we have on the power feed wire to the ignition coil.

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The idea I have is to look at the circuit while it is under load to see what the voltage drop. Compare this to the voltage when the circuit is not under a load. I also put the battery voltage in the capture to use as a comparison as well.

So how would you consider a way to arrive at the voltage drop?

3Ms
Measure the battery voltage as a reference voltage.

Measure the voltage on the supply you are testing at the loaded end whilst the circuit is under load.

Subtract the loaded voltage from the battery voltage we are using as a reference. The difference is your voltage drop.

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Nikk wrote on Aug 7^th, 2019 at 2:08am:

3Ms Measure the battery voltage as a reference voltage. Measure the voltage on the supply you are testing at the loaded end whilst the circuit is under load. Subtract the loaded voltage from the battery voltage we are using as a reference. The difference is your voltage drop.

Keep measuring... What is the drop underload?

Could we possibly use math channels?

Hi Spence,
Lets try this, Maths channel C-A. 12.75v voltage drop at the end of the capture.


Sorry slow replying, our 14 year old deaf Border Collie decided to take herself out for a walk.

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Nikk wrote on Aug 8^th, 2019 at 1:32am:

Hi Spence, Lets try this, Maths channel C-A. 12.75v voltage drop at the end of the capture.

Excellent! But... Follow that math channel over to where the load of the ignition coil is and tell me what you get for a reading.


Nikk wrote on Aug 8^th, 2019 at 1:32am:

Sorry slow replying, our 14 year old deaf Border Collie decided to take herself out for a walk.

Hope all is well with the pouch! Everyone reunited?

Anyone else have any thoughts?  [smiley=popcorn.gif]

This what you looking for?

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Nikk wrote on Aug 8^th, 2019 at 12:26pm:

This what you looking for?

Exactly!

Surprising the coil still managed to draw 4amps.
Thanks for sharing Spence.


Pooch made it home, has now lost her free roaming priviliges. Funny she has never wandered before, family reckons doggy dementia.

Nikk wrote on Aug 8^th, 2019 at 2:51pm:

Surprising the coil still managed to draw 4amps.

So while under load what did you see as a voltage drop?

Would you normally consider this good?


Nikk wrote on Aug 8^th, 2019 at 2:51pm:

Thanks for sharing Spence.

My pleasure! But we ain't done yet. Seems the others are letting you do all the work...


Nikk wrote on Aug 8^th, 2019 at 2:51pm:

Pooch made it home, has now lost her free roaming priviliges. Funny she has never wandered before, family reckons doggy dementia.

Good to hear! (no pun intended)

I see a 7.1v voltage drop under load, no this is not good.



I would have expected just under battery voltage with some slight voltage drop whilst the coil was charging.

Nikk wrote on Aug 9^th, 2019 at 4:40am:

I see a 7.1v voltage drop under load,

Very good! That was pretty easy eh?


Nikk wrote on Aug 9^th, 2019 at 4:40am:

no this is not good.

In most cases you would be correct. Keyword "most cases"... And if there's interest we can go further with this as well.


Nikk wrote on Aug 9^th, 2019 at 4:40am:

I would have expected just under battery voltage with some slight voltage drop whilst the coil was charging.

Rule of thumb is 500 mv for entire circuit max.

Very good!   [smiley=thumbsup.gif]

It looks like the coil amperage still got high enough to hit current limiting even though it was only pulling 4amps.
How much amperage did the coil pull after repair.


Always interested Spence.

Sooooooo can we figure out the resistance that is in that circuit?

We can using ohms law V/I but I am not sure I am correct using the unloaded circuit voltage 13.17v divided by current 4.04amps =3.26ohms.

As current starts to flow the resistance is going to change.

Nikk wrote on Aug 11^th, 2019 at 6:46pm:

We can using ohms law V/I but I am not sure I am correct using the unloaded circuit voltage 13.17v divided by current 4.04amps =3.26ohms. As current starts to flow the resistance is going to change.

You want the voltage at where the load is at 4.04 amps not the OCV(open circuit voltage). This is the whole reason for checking voltage drop.

Give it another shot.   8-)

Spence wrote on Aug 12^th, 2019 at 3:44am:

Nikk wrote on Aug 11th, 2019 at 6:46pm: We can using ohms law V/I but I am not sure I am correct using the unloaded circuit voltage 13.17v divided by current 4.04amps =3.26ohms. As current starts to flow the resistance is going to change. You want the voltage at where the load is at 4.04 amps not the OCV(open circuit voltage). This is the whole reason for checking voltage drop. Give it another shot.   8-)

Sorry, Wat are the points in the circuit that are open circuit as opposed to 4.04 amps?

Logic wrote on Aug 12^th, 2019 at 3:56am:

Spence wrote on Aug 12th, 2019 at 3:44am: Nikk wrote on Aug 11th, 2019 at 6:46pm: We can using ohms law V/I but I am not sure I am correct using the unloaded circuit voltage 13.17v divided by current 4.04amps =3.26ohms. As current starts to flow the resistance is going to change. You want the voltage at where the load is at 4.04 amps not the OCV(open circuit voltage). This is the whole reason for checking voltage drop. Give it another shot.   8-) Sorry, Wat are the points in the circuit that are open circuit as opposed to 4.04 amps?

When no amperage is flowing.

We then have just under 1.5ohms while current is flowing.

Spence, very sorry for the disappearing act and no timely reply, had a stressful family emergency that took me away from things for a while. I will read through and try to implement some of the great info you guys have provided, very much appreciated.

James H wrote on Aug 31^st, 2019 at 8:43am:

Spence, very sorry for the disappearing act and no timely reply, had a stressful family emergency that took me away from things for a while.

No worries James! I hope all is well. Sorry to hear you were going through a rough spell.


James H wrote on Aug 31^st, 2019 at 8:43am:

I will read through and try to implement some of the great info you guys have provided, very much appreciated.

Look forward to hearing from you. Let me know if there is something I can help with if needed.

looks like 7.4v

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Sorry so late fellas. It looks like the voltage level when loaded is 5.9V. The amperage is 4.07A. If we divide 5.9 by 4.07 we get 1.45 ohms while under load.
  Is this the actual resistance? Or is this like a dynamic measurement only?

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