GS Charging CONTEST

[Mad GS 750 E]

I think its #5 ,all the currents return to R/R ,and is burned off as heat.Thats why in your GS system health tread you tell us to do the single point ground as close to the R/R as possible,from the harness,frame, and battery.Posplayr your thread on the electrical system really helped me out,you are the man. (I do get extra points for A$$ kissing right) LOL!

[tkent02]

The correct answer is to clean all of your electrical connections and grounds when you get a bike and again every ten tears or so whether it needs it or not.

[posplayr]

The correct answer is to clean all of your electrical connections and grounds when you get a bike and again every ten tears or so whether it needs it or not.

Freudian Slip?

[bakalorz]

if you read the text I have a current clamp alternating between a stator leg and the R/R(+) OUTPUT. I left out the 2 amp charging plot because it was identical to the 10 amp in terms of ripple and I could vary it +/- an amp or so with RPM. I was mesuring the R/R out at "X"


http://www.thegsresources.com/_forum...9&postcount=81

Yes, as you said, measuring at X is what I was assuming at first, due to the text ...

which is why fig 3 in http://www.thegsresources.com/_forum...5&postcount=82
didn't completely make sense.

If you look at the blue line (putatively R/R current out) it goes to Negative 10 amps ...

I.e. that would mean 10 amps going the wrong way into the regulator for part of each clamping cycle ...

Since I don't think that could happen, I figure there could be 3 possibilities:

1) I am mis-reading your scope ... (I think "0" is on the axis, is that right)

2) There really is 10 amps going the wrong way into the R/R (I don't believe that to be possible, but convince me it's true if you believe it)

3) You were actually measuring at Y (i.e. battery charging current) and forgot to mention that you moved the probe. I still believe this to be true, since this is the only way the trace makes sense to me. (I also want it to be true, since this would basically mean I win a R/R)


If not, can you show me where I am wrong ?

[posplayr]

Yes, as you said, measuring at X is what I was assuming at first, due to the text ...

which is why fig 3 in http://www.thegsresources.com/_forum...5&postcount=82
didn't completely make sense.

If you look at the blue line (putatively R/R current out) it goes to Negative 10 amps ...

I.e. that would mean 10 amps going the wrong way into the regulator for part of each clamping cycle ...

Since I don't think that could happen, I figure there could be 3 possibilities:

1) I am mis-reading your scope ... (I think "0" is on the axis, is that right)

2) There really is 10 amps going the wrong way into the R/R (I don't believe that to be possible, but convince me it's true if you believe it)

3) You were actually measuring at Y (i.e. battery charging current) and forgot to mention that you moved the probe. I still believe this to be true, since this is the only way the trace makes sense to me. (I also want it to be true, since this would basically mean I win a R/R)


If not, can you show me where I am wrong ?

I'm resting up for a trip out Hwy 33, so dont have a bunch of time. The current plots (using the current clamp) were alternatively between three places but only the first two were shown.
a.) The stator wire measuring AC current (clamping a yellow wire at the R/R connector) ,
b.) The R/R (+) red output (again clamping the total current from the R/R your drawing "X") and
c.) not shown (because it looked identical to a. except for the differences in average current ) I moved the clamp to the Red wire between battery and fusebox (Your point "Y").

[Roger P.]

All return current flows through the arc created by the regulator frying itself to ground.

[posplayr]

All return current flows through the arc created by the regulator frying itself to ground.

So all the current is consumed in power and light?

[gs road racer]

All return current flows through the arc created by the regulator frying itself to ground.

Ok this entry wins....it happens!!Watch out for that arc.......too funny

[Roger P.]

So all the current is consumed in power and light?

No. A good deal of it is used up in creating the smell as well.

http://www.realclassic.co.uk/suzukigsx550.html

[posplayr]

OK I think we got enough entries so first lets go with what the correct answer is. Bet just before we do this lets have a little explanation of how electricity works. This is in the form of an analogy but it really is closer to electrical flow and helps explain what is going on.

Here is a picture of a tube laying flat on a table (no gravity effects) and it is filled with many small metal balls that allow them to roll very easily inside of the tube. A magnet is passes close to the tube and pushes on one of the balls (at least it only pushes on the ones in proximity to #1). The push on that one ball causes the next to be pushed which in turn pushes the next until the last comes around and pushes from behind. But in reality the push from behind can't be any more than the push on #1 ball because it is the #1 ball that started the push to start with.

The only thing that keeps these balls from whizzing around is a small resistance (shown as some foam in the tube). In order to keep the balls (electrons) moving you have to keep pushing (applying voltage). As soon as you stop pushing, the balls stop because the foam grabs the balls (electrons). The balls have very small mass (like electons) and there is little momentum.

I'm hoping this is a useful visual isation tool to understand how voltage (or pressure or magnetic force) pushes electrons (little balls) around a circuit creating current which is balls per second or electrons per second.

So other than this being a useful visualization tool it should be clear that electricity runs in CIRCUITS or loops. Electrons never leave the wire, they just move from one copper atom to another (silver and copper wire or even steel could be interchanged or mixed).

Electrons DO NOT get pushed down a wire to be expended (into a heap on the ground ) into HEAT.

Electrons DO get pushed down a wire and when going past a resistance cause HEAT.

Current always has to run in a closed CIRCUIT




OK next and we are almost there, in electrical CIRCUIT analysis there is something called "cut sets". A cut set is a way of analyzing circuits by drawing a dividing line (i.e. a cut) across a circuit to divide the circuit into two sets. In the case of our circuit above if we divide the page in two into a top and a bottom, we have an upper and a lower cut sets. This may seem trivial but the total current flow into the "cut set" is zero, and from either top or bottom side the total flows going into that side are equal to the total flows coming out.

This is really the only way to have a circuit.

So it doesn't matter how complex of a circuit we have, if we can draw a line dividing all current flows between a top and a bottom, a left or a right or any other division then the total flows into that set has to equal the total flows out of.


OK by this time, it should be no surprise that this concept of a circuit of electrical current I'm belaboring must apply to the contest.

So as was stipulated , the R/R (+) RED is generating enough current to both supply the GS electrical loads as well as charge the battery. It is the source of all the current (see the little red arrows). If we draw a cut set across the R/R (there are no other sneak paths for current from anything behind the R/R to bring current to the battery or GS ) then all the current coming out or the R/R(+) RED must return back to the R/R (-) in black.

So looking at all the 6 options below, the answer has to be #5 as it is the only one with a balanced current flow . Everything coming out of the R/R(+) red returns back to R/R(-) black.

In fact it should be no suprise (as Steve pointed out) #5 is the only one that will balance out all the flows if we draw three different cut sets across each device in the diagram.





Oh I'll describe this next

P.S. HINT : The Suzuki Manual has the answer under "Charging System"

[posplayr]

I said the answer was in the Manual and so I pulled a page from the 80-83 GS1100E manual. I think it is in most of the 16V manual I have. The electricity works the same weather it is 16V or 8V or whether the R is separated from the /R in the R/R.


Below I have drawn in the complete circuit of how electrons flow in a circuit. The stator actually has 3 legs that are would so that each leg reaches a maximum voltage (push) at different angles of the rotor (motor) and so they are nominally out of phase by 120 degrees (360/3=120). However we can look at the basic flow of current through the full wave rectifier. The diodes simply block current from flowing if they are reverse biased. That means hey will only flow in one direction and only when the voltage is higher on the back side of the arrow.

Without knowing anything about the 3 phase, we can draw a cut set at the output of the R/R and see that anything coming from the R/R has to return and this is shown explicitly by the little arrow coming up from ground in the manual into the R/R(-) lead.

Makes you think a little...... So if all the current has to get back to the R/R (-), how is mine wired? Is it grounded to a rubber mounted side plate? If I'm running all of my grounds back to the negative side of the battery, How does the current get back to the R/R(-)? More on this later.




Here is another picture from the GS1100E manual (Charging System). I have also drawn in the electrical circuit and cut sets when the R/R is regulating. Again I'm simply highlighting what the manual is already describing just to make it more obvious. So when the R/R sense too much voltage at it's output, it crowbars the stator. That means it shorts the stator and just sends the current back letting it get hotter and keeping all of its own juice. So while there is a "short to ground", the short only exists inside of the R/R and never leaves the output terminals of our cut set. Does the R/R get hotter yes, because now we have not only two diodes dropping voltage but also a SCR. But the stator is getting even hotter. And a nice cool oil bath at 220 degrees is very cool and refreshing to keep it from burning up it's insulation.




So what do you think happens when there is a bad connection between the R/R (+) and the rest of the system? Well the R/R(+) tries to push current but the high resistance causes the voltage at the R/R(+) to go to the stator output because it is as if there is no load attached and no voltage drop. So in response the R/R simply shorts the stator to say "sorry we dont need any more from you". In which the stator responds "man is it getting hot around here?" to which the R/R say "Yea I'm feeling it too" . Of course this is only temporary as your battery is dying from lack of charging and you hope you shut the engine off before you fry the stator and the R/R together.

[posplayr]

I'll bite.

#5

all (unused) currents go back to the power source, hence in your drawing it would be the source of rectification. (R/R)

So without further adieu, RustyBronco is the Winner of the Duanage Honda Regulator.

This part of the answer is what i was looking for as the explanation

currents go back to the power source

And he was the second respondent so I think he is the clear winner.

Lets all give him a big applause.



We also have some honorable mentions.......

[kinnet]

Congratulations RustyBronco
Good work!

[posplayr]

I also say #5, for the same reason as BassCliff (although I am sure that reason disqualifies me haha).

as an aside, it's just "current", no "s", sorry I am a terminology whore, blame my father...

cool contest idea!

The reason I use the term currents is in reference to multiple paths not the multiplicity of electrons flowing in a single wire. In the case of the GS there are multiple currents trying to return to the R/R(-)

• Load Current from the harness
• Charging currents from the battery
• Load currents returning from the frame

Once of the objectives of single point grounding designs is to keep these current returns separate until they have to be brought together at true reference ground. This way there is no current sharing in the ground returns which can cause ground loops or ground bounce (e.g.if a big motor has a current flow , and my ground shares the same current path will cause my ground to rise in voltage as the current passes through).

[posplayr]

OK. May be I am tired and should look at the schematic better but this is what I remember from electronics.

The heat generated by the r/r is caused by it converting ac to dc and the amount of current required to run the gs electrics. The more load (ie. headlights, dead battery etc.) on the system will require the r/r to provide more current and the r/r will have to dissapate more heat. If you disconnected every electrical load on the bike except what is required to run the engine, the current load would go down and the r/r would not run as hot.

Please ignore the first paragraph. It is relevant to power supplies just not the type they use on motorcycles.

The return path of a electrical or electronic circuit is ground. Thereticaly there should be no resistance along your ground path and therefore no current flow. This is not possible (yet) in real life. For example there should be no potential resistance between the ground at the r/r and the ground of the battery ( this would make 4 and 5 right ) . If we are tring to measure current flow along a ground then the path of most resistance would have the most current flowing through it.

Or did I screw up understanding the question?

It was a long day at work>

Chris

If you think about this as a circuit rather than just a return, it will answer your own question. $5 is correct.