GS Charging CONTEST

[posplayr]

OK, many others have guessed #5, so I have to offer the correct reason to win, right?

It's #5 because it's the only one that shows arrows going BOTH ways at EVERY device.


.

Exactly correct Steve, all circuits have to observe the rules not just the R/R.

[posplayr]

# 5

Reasons, R/R is the highest source of potential, higher than the battery therefore current is highest there.

Batt is nominally 12.8v RR is 14.4v therefore batt is load and GS is load.

So the highest potential Must overcome load in order to raise potential in GS and Batt which is at Batt potential of 12.8v initially, to get it to 14.4v .

And since the R/R only has one ground coming out of it and one positive which are both smaller guage wires than the the many positives and ground wires coming to it actually make,

which means this connection where the many grounds from the GS and BATT meet the single wire from the R/R there is a large difference in over all available current carrying capacity by the wires from the R/R to the initial connections.

For instance there are two positive wires that go to the RR that come from the BATT and the GS. So that is a 2 to 1 ratio of current carrying capacity, and its the same for the grounds. Two grounds to one R/R ground. The R/R's should have wires that are a guage bigger than all the incoming load wires combined.


My 2 cents

great contest


:P

Yes R/R(+) has the higher potential than the battery or the load that is what is causing current to flow from R/R(+) to battery/load.

The stator actually is even higher at times which is how it forward biases the diodes to get current to flow through to the R/R(+) out.

Thanks for the Kudos, I'm hoping this makes it clearer why R/R connections are so important.

I am also surprised by the small size of the R/R (-) Since the wire is small coming out of the R/R it needs to be as short as practical.

# 5



Extra info,,

What we have is a 3 phase Regulated and rectified charging system and on the stock regulator there are six diodes and a zener circuit which acts in a voltage breakdown of 14v. Two diodes are required to convert the incoming sine wave into a pulsed DC positive state. One for the rising part of the wave and one for the falling part. Six diodes are required to rectify 3 phases in order to give an output that resembles steady DC voltage with a little bit of ripple, at a frequency that increases with RPM. However the battery acts as a capacitor and reduces the ripple making it a more stable Dc voltage.

:P

Yep that is right on. I think the zener's break down voltage is actually lower but it is compared to a voltage divided 14V output for ratio regulation.

# 5

Now a FET R/R ( Field effect Transistor) is better because it doesn't shunt, It use pwm (pulse width modulation) which turn the fet's on and off faster if there is a demand for power and slow if its just a low demand. There for no shunting heat and only rectifying heat and no burnt up stator

sorry got on a rant there :P

This is really another topic, I think the current implementations dont open the stator winding connections, but rather stay cooler because they don't use diodes in the lower legs of the rectified, they implement synchronous rectification by switching FET's on and off as a diode would. The FET's have much lower on resistance so it drops less voltage and create less heat.

[posplayr]

Correct answer is #5.

Reason: When GS is running, the stator's output is more than that of the battery, and so the stator (and not the battery) produces the greater amount of power that feeds the GS electrical system. So the current has to return to the stator (and not the battery), and it does this through the R/R (-) via part of the internal diode bridge.

Excellent,

[posplayr]

Hi Posplayr,

I'll submit that none of the available answers is actually correct, and my answer is "none of the above"

The correct answer is that the situation alternates rapidly and continuously between your #5, and my new drawing #7.






Drawing #5 would apply during those parts of the cycle when the regulator is not shunting.
Others have given the rationale for drawing #5, and I don't think I need to repeat any of that.

Regarding the inclusion of drawing #7

Bakalorz,
I think that you are changing the rules a little including an AC v.s. strictly DC analysis, but that is OK. The original current flows for each option are show with the R/R as the only source of current. This is know to be the case when running down the road at say 3-5K RPM. If "Charging System" is operating properly the voltage at the terminals of the battery will be at 13.5V plus and so the battery can't source current. In the link I provided is a 1 Ghz bandwith scope plot using a current clamp to measure the current out of the R/R(+) lead. The average is about 10 amps. Even though there is some AC on top there is no indication of negative flow out of the R/R(+). In addition although I did not provide the plot, I described current going into the battery as being very similar to the plot shown, it simply varied with RPM. The most I could get out of it was 2 amps average.
It is possible of course for there to be a small instantaneous current coming from the battery during a dip in the ripple.

But here is the issue, if I take your two cases, and assume that there is some duty cycling back and forth between these two #5 and #7, your #7 ignores the continuous Amp average output of the R/R. In the plot below is my GS1100ED charging at 3000 RPM. There is no letup or spiking in the current flow on to cause a reversal. and so your #7 really violates the known behavior.

Here is the same thing (idling) but looking at the total current coming from the R/R. You get about 10Amps, the RR is not regulating and so you only see the harmonics associated with a 3 phase full wave rectifier. The label says 2 ADC but that is wrong. The scale is 5 amps per major division.



The charging current moving from the R/R to the battery is basically the same but can be adjusted up and down (positive and negative) by the idle speed; it is near zero and I could only get it up to 2 amps at 3K RPM.

[posplayr]

I'm with wonder Steve the flows must be balanced the hot and ground flows must be in opposite directions. Only #5 shows balanced flows at all three points. Whether the - current flows through a wire or the frame is irrelevant. The amount of current flowing to (or from) each of the three areas can vary extensively. but the - flow will match the + flow The battery can be either load or source. the accessories will always be a load and the alternator will always be a source. The R/R has diodes that prevent the alternator from being a load or at low or no engine speed it would be.

This is all true. The only time the R/R is not a source is when the engine is not turning.

[posplayr]

The paths have to be shown separately for me to get on board with the
question.

1. 3 wires from the stator to the R/R. then 1 main wire to the Battery - first we are creating E.M.F and changing it to D.C. and getting it to the storage unit-battery-.

then

from the battery to the fuse block via key switch on to the GS positive loads (including the other switching wire to the R/R)

Negative Grounds are ALL universal and connected.

hot (+) are not universal and all connected between the R/R and main loads.

the battery is a buffer the volt spikes so spikes do not pop bulbs and other weird electrical problems.

if the battery is not in between the R/R and all the running loads wired properly we will have crazy electrical problems like the forums are full of.

I think the question as posed is accurate and representative of the electrical theory as well as the GS charging implementation. #5 is the correct answer as all current operates in a circuit.

All current coming from R/R(+) has to be returned on R/R(-).

The currents to the R/R(-) complete the crcuit for the currents in the stator.

[posplayr]

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!

It may be A$$ kissing but this is really the best answer to why I started the Contest.

Mad GS 750 You get a whole bunch of EXTRA POINTS (little arrows)

-> -> -> -> ->
-> -> -> -> ->
-> -> -> -> ->

http://www.thegsresources.com/_forum/showthread.php?t=152769

By picking one of the R/R mounting bolts as the single point ground point, the R/R(-) wire can be very short, the mounting plate is automatically grounded and all of the return current wires can be tied securely together at a single point. Here are the connections

• #1: Ground R/R(-) to case and side plate
• #2: The shortest wire to a frame bolt
• #3: The shortest wire to the Battery (-)
• #4: Connects to the (B/W) harness negative ring lugs (typically the one attached at the solenoid mounting bolt). You can leave the other harness ground ring lug (B/W) where it normally is picking up the battery box ground.

Here is a matching picture on a GS750EX, it will work well for most all E's at least. The layout may be a little different on the G's and L's but the ideas will be the same.




Here is a GS1100ED done similarly; Here the harness ground lead would not streach all the way so I just bolted it to the other side of the R/R. R/R should probably be turned the other side around but all wires would fit the same way.
The reason for all these connections is basically the fact that all currents from the charging system (leaving the R/R (+) red wire have to return to the R/R(-) black wire. Some of these currents come back from the harness B/W, some from the engine , some from frame return loads. By connecting all of these as close as possible to the R/R (-) you have created a "single point ground" at this point. In this way the current that runs from the Battery (-) to the R/R(-) is only carrying the battery charging current and nothing else. That is the real reason for having a single point ground. There is no current return sharing of the various loads.

[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.

And the 5 R/R connections especially.

[posplayr]

After reading this thread, it is probably easier to understand why these two ground leads (one to the side plate where the solenoid mounts which leads to the R/R mounting) and the 2nd which goes to the battery box.

Both ground wires are crimped into the harness but the harness did not burn. This shows there was alot of current going from one ground to another ground. It was enough to over heat the ground wires and smoke off all of the plastic insulation.

I opened up the wrap around the harness which is why it is unbundled. None of those wires got hot other than being next to the smoking grounds.

This is how I found my GS750 EX right after I bought it. I read the stator pages and cleaned all my connections and everythiing was fine.

Before doing all this work I had already ordered new parts. Eventually I replaced the stator, but the same OEM R/R is still working with the improved grounds and heat sinking.

[lucabond]

#4 current flows from a higher potential + to a lower potential -.

[kinnet]

I have a question about SPG.
I have read that it is an invitation to corrosion... Is this true? And if so, how can it be prevented or kept to a minimum?

[Mekanix]

Dialectric greace would slow corosion down.

Using all the same type of lugs helps. As in all aluminum or all coper etc.




Its amazing what you learn when people are challenged

This contest was a great Idea

[bakalorz]

Ok, we have a fair amount to go over, and I need to use stuff from several posts, and I'm having issues with the forum getting them all to quote, so anything I write will be red, and anything of yours I'm quoting will be black, also the forum has a length limit so this will be two parts ...

Hang on, its a doosie, and ... Here we go ...


Before we start, just to clarify for any lurkers; the way the regulator regulates is that when the voltage starts to get too high, it briefly shorts the stator internally, and then lets it charge again. It does these cycles many hundreds of times per second. The period during which it is shorting is called shunting.

The first thing I want to establish is that when the regulator is shunting, there is Absolutely NO current ... Zip, Zero ... coming or going from the regulator to the battery or the rest of the bikes systems ...

One of the figures you provided shows this nicely:



Note what you wrote in your fourth sentence:
IT IS IMPORTANT TO NOTE NONE OF THE SHORTED CURRENT EVER PASSES THROUGH OUR CUT SET
BOLDED AND CAPITALIZED by you.

So there is NO current coming out of the regulator.

The three upper bridge diodes prevent any backflow of current the wrong way into the R/R. (with the exception of an insignificant amount that can still flow to the trigger circuitry)

So there is no current crossing the purple cut line you convientiently provided

So do you see that during shunting, there is absolutely no current being provided by the R/R to either the battery or the rest of the loads on the motorcycle ?

So as far as the battery and loads are concerned ... during shunting it's as if everything to the left side of the figure is erased and doesn't exist.

So now lets look at whats on the right side of the cut line, what's over there ???

A battery and some loads ... do those loads still require power ... do the lights, coils and ignitor shut down for 40-50 percent of the time when the system is regulating (you mentioned a duty cycle of about 40-50 percent in one of your other posts) or do they still keep functioning.

Those loads still require power, and convieniently they are still connected to a battery ... so when the regulator shunts and stops providing current, the battery takes over and supplies the loads during the period of shunting ... when the shunting period is over, the R/R starts to provide current to both supply the loads and replace the charge that just came out of the battery (plus a little extra to keep it charged) again.

Which is exactly what I described in my first post, using your figure 5 and my figure 7.

What causes the battery to alternately provide current and be charged as this cycle goes along? basically the voltage at its terminals: When there is no current from the regulator, the loads pull the voltage below the battery's resting voltage at the time (this resting voltage is significantly raised compared to the batteries normal resting voltage because the battery was just being charged a millisecond ago) When the regulator provides current, it raises the system voltage enough to cause the battery to accept charge.

So as the regulator alternately provides current and shunts, the battery voltage alternately bounces up and down by a fraction of a volt, causing the battery to alternately recieve charging current (from the R/R, which is also running the loads at that point)(your figure 5 that everyone likes) and then supply current (to the loads) (My figure 7 in my first post in the thread)

All this is a more detailed explanation of exactly what I wrote before.

Now lets look at your post ...

Bakalorz,
I think that you are changing the rules a little including an AC v.s. strictly DC analysis, but that is OK. Its not only OK, its REQUIRED for the analysis to be very useful at all !!! the DC charging current is 2 amps or so ... the superimposed AC current is 10 amps or about 5 times as much, so its much MORE important to many aspects of the total behavior of the system.
The original current flows for each option are show with the R/R as the only source of current. Well, thats the assumption you used; however, I'm disputing the accuracy of that assumption This is know to be the case when running down the road at say 3-5K RPM. (note that RPM)

If "Charging System" is operating properly the voltage at the terminals of the battery will be at 13.5V plus (average, it bounces up and down around the average as the R/R charges and shunts) and so the battery can't source current.

No, it's pulled down enough that the battery CAN source current ... also, don't forget that the battery was only a millisecond ago being charged.

The resting voltage of a battery is dependant on its charge history. for the brief period the battery is capable of providing current at a much higher voltage than normally.

(this temporary raising of the resting voltage is the reason that you have to wait at least an hour to measure the state of charge with a voltmeter ... and its an exponential thing, so the effect is MUCH more pronounced in the extremely short time periods being looked at here)

Continued next post, forum length limits

[bakalorz]

Continued from previous

In the link I provided is a 1 Ghz bandwith scope plot using a current clamp to measure the current out of the R/R(+) lead. The average is about 10 amps. Even though there is some AC on top there is no indication of negative flow out of the R/R(+). You are correct about the figure from the link that you copied to this post (I moved it to here instead of the end of the post) unfortunately that figure does not apply: it is at IDLE, and the R/R is not regulating, only rectifying. You say so yourself.

P.S. aside from the caption, one way you can tell its at idle is the width of the yellow traces; compare them to the ones in the next figure, they are much wider (slower) (time base is the same for both figures

Your caption for the figure:
Here is the same thing (idling) but looking at the total current coming from the R/R. You get about 10Amps, the RR is not regulating and so you only see the harmonics associated with a 3 phase full wave rectifier. The label says 2 ADC but that is wrong. The scale is 5 amps per major division.





Back to your current post:

In addition although I did not provide the plot, I described current going into the battery as being very similar to the plot shown, it simply varied with RPM. The most I could get out of it was 2 amps average.

Actually you did provide the plot in your linked post, I have copied a marked up version here, the only changes made are the markup in red.

And it shows a TOTALLY different story.
It absolutely vindicates what I am saying, which is what I told you when I replied to your earlier message.



The zero level is shown by the little blue caret (my label "A")
The amps/division is 5 (current is the blue line for any lurkers following)

The points I want to direct your attention to are the ones I have labeled "B"
These show a current of MINUS 10 amps leaving the battery (i.e. going the WRONG WAY to charge it ... yet curiously just the right value to run the loads on the bike ... what a coincidence)

This is consistent with my description, but absolutely not possible according to your description (yet here it is in your scopes picture)

And finally, lets look at area C. This is when the regulator is not shunting.
There is up to 10 amps going into the battery, and if you compare the width of this area to that of area B, you will see that area C is wider.
So there is more current going in than coming out on average (which your averaging scope calculates for you at 2 amps) which is the AVERAGE current charging the battery ... but as your scope clearly shows, the actual currents going into the battery varies from 10 amps to MINUS 10 amps during the course of a cycle as the R/R shunts to regulate.

(And one brief aside, area C invalidates any posibility that you are getting minus readings at area B because you have the probe turned the wrong way around)

Before we leave this figure I want to briefly cover our prior discussion of where you measured this trace (refrencing a figure I called fig 8 at the time)
(I'm not going to include it here, unless you want to continue to argue the point).
I believe it to be at a point Y, and you believed it to be at a point X in that figure (I believe this to be an honest mistake, not trying to accuse you of anything here).
I believe that for several reasons, the most important of which is that you wrote the following in the caption:
The charging current is 2 amps in this graph as measured by my averaging scope
The only way your scopes averaging function could get the 2 amp average is if it was in position Y. If it were at position X it would have averaged the charging current plus the bikes load current, and come up with a much larger number.

However, even if I were wrong about the position of the probe, IT STILL WOULD NOT HELP YOUR CASE. You would still have to explain a 10 amp current spike going the wrong way ... and the only possible source would still be the battery

Oh, and this is mainly a note for the lurkers, there are 3 or 4 other neat and educational things we can see in this figure. They aren't really relevant to the point I'm trying to make, but if anyone wants to learn more stuff, I'll make another (shorter I promise) post to explain some more stuff.
Just post that you want more info ...


Ok, back to the current post

It is possible of course for there to be a small instantaneous current coming from the battery during a dip in the ripple.
But here is the issue, if I take your two cases, and assume that there is some duty cycling back and forth between these two #5 and #7, your #7 ignores the continuous Amp average output of the R/R. In the plot below is my GS1100ED charging at 3000 RPM. There is no letup or spiking in the current flow on to cause a reversal. and so your #7 really violates the known behavior.

As I showed above, you used the wrong plot.
When you look at the correct plot there is huge obvious unmistakeable spiking and reversal in the current flow.

My #7 does NOT violate known behavior, it demonstrates it as shown in your own Scope traces.

Boy this got long, but ... In Conclusion:

I have demonstrated that my description is correct, according to the theory as shown in the Suzuki manual and your scope traces.

And since I was the only one to describe the actual operation of the R/R system accurately, correctly, and to such a level of detail, showing important features not even hinted at by anyone else;
I believe you owe me a shiny new (used) regulator.
(Sorry Rusty Bronco)


If you disagree with any point, I'll be more than happy to discuss any further points of contention.


Take it to your EE buddies at work, see what they say.


Further, I actively invite comments and especially questions from any of the other bystanders in the forum:
This was supposed to be an educational contest, and we learn by asking.


By the way, this contest has been one of the most interesting series of posts in a while, thanks for running it.


I am greatly looking forward to your reply.

[posplayr]

Ok, we have a fair amount to go over, and I need to use stuff from several posts, and I'm having issues with the forum getting them all to quote, so anything I write will be red, and anything of yours I'm quoting will be black, also the forum has a length limit so this will be two parts ...

Hang on, its a doosie, and ... Here we go ...


Before we start, just to clarify for any lurkers; the way the regulator regulates is that when the voltage starts to get too high, it briefly shorts the stator internally, and then lets it charge again. It does these cycles many hundreds of times per second. The period during which it is shorting is called shunting.

The first thing I want to establish is that when the regulator is shunting, there is Absolutely NO current ... Zip, Zero ... coming or going from the regulator to the battery or the rest of the bikes systems ...

One of the figures you provided shows this nicely:



Note what you wrote in your fourth sentence:
IT IS IMPORTANT TO NOTE NONE OF THE SHORTED CURRENT EVER PASSES THROUGH OUR CUT SET
BOLDED AND CAPITALIZED by you.

So there is NO current coming out of the regulator.

I never said there was no current coming from the R/R or crossing the cut set. I posted that there is no shunting current crossing the cut set. I also posted a plot of output current from the R/R(+) RED output that shows there is 10 amp output maintained at 3000 RPM. I'll stop here.