That's a good test. I'd follow Jim's instructions on rectifier and voltage regulator. I did it myself and it works great.
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I agree.
tmkenney
XS650 Member
Just to follow up on this, I got a replacement regulator/rectifier from Mike's after fighting about it with them for a bit. Was cheaper than building custom units like was mentioned. Bike still charges at 12.4ish volts when revved at idle, and the positive stator brush only gets 11 volts at idle. Im stumped. Is 12.4 and slightly higher at speed just fine to run? I have an LED headlight and tail light, no blinkers or anything. Thanks lads.
Any wiring Schematic we can see ?
12.4 is to low
Is it correct it charges when power to brush . Please describe where power is taken from and
Difference against with regulator.
Is the connector pins all in place not pushed back barb not holding it in place ?
Fuse for charging ?
12.4 is to low
Is it correct it charges when power to brush . Please describe where power is taken from and
Difference against with regulator.
Is the connector pins all in place not pushed back barb not holding it in place ?
Fuse for charging ?
tmkenney
XS650 Member
This is the basics, juat a simple setup for wiring. But power from battery at 12 volts goes into regulator unit, and coming put of the unit on the green wire is 10.8 volts, 11 when running. Same at the brush. So there's no power loss in the wiring, the regulator just isnt putting out proper power. I did also test the battery out of the bike and its good to go. There's no fuses for charging as im still building everything
Attachments
Is it like that below
I would check the ground wire Black from the Regulator
ensure it is solid
Starting with the connector pins
Myself would pull an extra wire all the way to battery minus If possible
I would check the ground wire Black from the Regulator
ensure it is solid
Starting with the connector pins
Myself would pull an extra wire all the way to battery minus If possible
tmkenney
XS650 Member
Yes mine looks like this, but brown only goes to switched power into the regulator, not back to rotors. Ground on one brush with black wire, and green on the other. The ground for the regulator does also go right to the battery, splitting off from the brush ground. Just weird how its only sending 11 volts, not full battery power
Not sure here ..and I don't remember That is another problem with these Rec Reg combos
It is a black box , not known what is inside
Splitting off from Brush ground ( If that is what is happening )
splitting off from the brush ground
is perhaps not enough should some kind of Comparison happen inside the combo upstream from the brush ground -- needing ground inside there at some junction inside as the left most black -- it is going in there. Into the black box
It may well be behind electronics components in a circuit affecting electric values.
It is a black box , not known what is inside
Splitting off from Brush ground ( If that is what is happening )
splitting off from the brush ground
is perhaps not enough should some kind of Comparison happen inside the combo upstream from the brush ground -- needing ground inside there at some junction inside as the left most black -- it is going in there. Into the black box
It may well be behind electronics components in a circuit affecting electric values.
Maybe others will confirm or deny.
If you are only making 12.4 through the regulator, the battery will never b fully charged, the rotor will never get a rest. Burn out with obvious cost related replacement of parts. Saving some money may b costing more in the long run.
This problem causing rotors to fail is partly why the XS had a reputation of a weak charging system
Yes that can be a problem A bit uncertain info and not done the read up
But if I understand correct the regulator Ground is connected at the Brush on the ground side
, splitting off from the brush ground.
So on the upside brush there are ca 12 V and the rotor have ca 5 ohms resistance.
Ohms law U = R x I
I current = U/R = 12 V /5 = 2.4 Ampere
Passing through at the brushes
Kirchoff law
Kirchhoff's Current Law (KCL) states that current entering a junction equals current leaving
So stock at the downside there is one wire only no Junction That 2.4 Amps which for electronics is high
As stock only have one way out.
But with the splitting in there is now a second wire that 2,4 Ampere can escape into the Regulator.
And 2.4 A on a Transistor ( if that has happened ) there can have fried the regulator instantly.
If it is or was 2.4 depends on the circuits in both wires after the junction How that 2.4 Distributes depends on that
Now this is without me thinking it trough or check possible wiring . It would help if you could draw up something
we can have a look at.
I would disconnect that regulator ground before power up And pray Most likely the Regulator is fried.
As the saying goes " Dont ask me how I know Having connected a regulator wrong a couple of times "
No warning no poff no smoke just dead and game over
But if I understand correct the regulator Ground is connected at the Brush on the ground side
, splitting off from the brush ground.
So on the upside brush there are ca 12 V and the rotor have ca 5 ohms resistance.
Ohms law U = R x I
I current = U/R = 12 V /5 = 2.4 Ampere
Passing through at the brushes
Kirchoff law
Kirchhoff's Current Law (KCL) states that current entering a junction equals current leaving
So stock at the downside there is one wire only no Junction That 2.4 Amps which for electronics is high
As stock only have one way out.
But with the splitting in there is now a second wire that 2,4 Ampere can escape into the Regulator.
And 2.4 A on a Transistor ( if that has happened ) there can have fried the regulator instantly.
If it is or was 2.4 depends on the circuits in both wires after the junction How that 2.4 Distributes depends on that
Now this is without me thinking it trough or check possible wiring . It would help if you could draw up something
we can have a look at.
I would disconnect that regulator ground before power up And pray Most likely the Regulator is fried.
As the saying goes " Dont ask me how I know Having connected a regulator wrong a couple of times "
No warning no poff no smoke just dead and game over
tmkenney
XS650 Member
Here is a very basic drawing of my wiring. Mike's reg/rect unit with 3 white wires, black wire and green wire in 1 connector going to stator. The black ground wire goes to stator brush, but is spliced close to the reg/rect to also connect to the battery ground. The bike charges around 12.3/12.4 when running and does not die. It does seem silly to me that adding another ground path would cause any issues at all, to me it would seem like ground is ground.
Attachments
Not sure here tentatively feel free to question
But the electric machine the Alternator has one input the power going to the rotor the regulated input
that affects the output .The white ones
Green and black to the brushes
The rotor gets ca 12 V and 5 ohm 2.4 Amps coming out on the black .. ( I Believe )
That 2.4 A is distributed at the junction
from the lead coming from 9 o clock there is 2.4 a that current has to go somewhere
it can chose the wire at 3 o clock or the wire at 12 o clock
There are most likely some resistance somewhere in the 12 o clock since 2.4 A to ground is kind a big number
Say worst case poor ground the 12 o clock is open Then 2.4 Amps tries to smash into the 3 o clock wire smashing into the regulator
That can perhaps destroy it.
On paper or as the schematic the ground is ground but there will still be some resistance in 12 and 3 o clock wires
Without splice aka no 3 o clock wire then the 2.4 A Cannot smash into the regulator
the black from regulator is grounded somewhere else fex all the way to the battery minus
and
But the electric machine the Alternator has one input the power going to the rotor the regulated input
that affects the output .The white ones
Green and black to the brushes
The rotor gets ca 12 V and 5 ohm 2.4 Amps coming out on the black .. ( I Believe )
That 2.4 A is distributed at the junction
from the lead coming from 9 o clock there is 2.4 a that current has to go somewhere
it can chose the wire at 3 o clock or the wire at 12 o clock
There are most likely some resistance somewhere in the 12 o clock since 2.4 A to ground is kind a big number
Say worst case poor ground the 12 o clock is open Then 2.4 Amps tries to smash into the 3 o clock wire smashing into the regulator
That can perhaps destroy it.
On paper or as the schematic the ground is ground but there will still be some resistance in 12 and 3 o clock wires
Without splice aka no 3 o clock wire then the 2.4 A Cannot smash into the regulator
the black from regulator is grounded somewhere else fex all the way to the battery minus
and
