GS Charging CONTEST

[kinnet]

Okay - Here is my answer to the question, at the risk of sounding like a complete and total idiot...

#5 All return current to R/R

The return current would not go directly to the battery, as it already being fed by going through the R/R to convert the AC to DC that the battery needs. The R/R also regulates the amount of voltage to the battery, so if the voltage is above say... 14.4 the flow will be shut off (hopefully) so the battery doesn't cook.

[Steve]

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.


Now, to try to clear up some mis-information:

... 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.

Although there is some heat generated in the rectifier, there is a lot more generated in the regulator portion.

... 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. ...

Exactly backwards here.
The regulator does its job by shunting the current to ground until it senses that the voltage is too low, then switches it back to the load until it senses that the voltage is too high. The switching itself generates a bit of heat, but the shunting straight to ground creates a LOT of heat. If you could load the system to the point where the regulator is sending all the current to the load and there is not quite enough voltage to trip the shunt, it will be running the coolest. The problem is that the load would have to vary with engine speed, as the output is also controlled by engine speed.

The problem is compounded if you have a stator with smaller wire that starts producing charging voltage at a lower RPM. It will continue to produce even more voltage as the RPMs rise, all this excess needs to be shunted to ground (causing heat). A better solution would be slightly larger wire in the stator and an FET regulator, which has fewer losses in the rectification and regulation processes, meaning that it charges sooner and runs cooler.

.

[chef1366]

What do I win?

[Steve]

What do I win?

You, sir, probably don't win anything, but the prize was stated as a Duaneage R/R.

RULE #1.) The prize is shown below. delivered to the winner via USPS.

[Mekanix]

# 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 gauge 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


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.



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

[mottyl]

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.


Now, to try to clear up some mis-information:

Although there is some heat generated in the rectifier, there is a lot more generated in the regulator portion.



Exactly backwards here.
The regulator does its job by shunting the current to ground until it senses that the voltage is too low, then switches it back to the load until it senses that the voltage is too high. The switching itself generates a bit of heat, but the shunting straight to ground creates a LOT of heat. If you could load the system to the point where the regulator is sending all the current to the load and there is not quite enough voltage to trip the shunt, it will be running the coolest. The problem is that the load would have to vary with engine speed, as the output is also controlled by engine speed.

The problem is compounded if you have a stator with smaller wire that starts producing charging voltage at a lower RPM. It will continue to produce even more voltage as the RPMs rise, all this excess needs to be shunted to ground (causing heat). A better solution would be slightly larger wire in the stator and an FET regulator, which has fewer losses in the rectification and regulation processes, meaning that it charges sooner and runs cooler.

.

Exactly Backwards!

So if you read it in reverse does it make sense?

This morning I looked at the charging schematic and see what you mean. At little or no load it is shunting to ground.

Last night my head was comparing it to a power supply. If it is sitting there at idle there is little heat. Once you apply a load and draw current through it then it heats up.

Happy Friday everyone.

Chris

[2BRacing]

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.

[posplayr]

OK looks like we got a bunch of responses, and the they are starting to slow down.

Get your answers in, we might end this a little early to keep up interest.

[bakalorz]

OK I have decided to have an Educational GS Charging Contest, however it is not completely educational because I'm offering to deliver this Brand Spanking New "Duanage HONDA R/R" to the winner. Duane hand picks these babies, connectorizes for easy install and tests them. If it is busted (or not) he even takes them back.

The Rules:

RULE #1.) The prize is shown below. delivered to the winner via USPS.


RULE #2) First person to correctly answer the following question by posting the CORRECT answer and the CORRECT explanation to this thread.

Determining if an answer is CORRECT should be unambiguous; it is #1,#2,#3,#4,#5, or #6.

On the other hand, you need to explain in at least simple term why the answer is correct. I'll be the judge of what is a suitable explanation.

RULE #3)
The question is:

Rule #4) Chef1366 already knows the answer and is disqualified.

Rule #5) Contest ends 6:00 PM Sunday Nite April 18th 2010




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

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
When the Regulator part of the R/R is regulating, it does so by briefly shunting all the power of the Stator internally.
(this occurs in the connections between the stator and R/R and those currents are not shown in any of drawings 1-7 above)
During that time, the stator and R/R are effectively not in the external circuit (battery and GS) at all, and all the loads on the bike draw their power directly from the battery.
This corresponds to drawing 7.

--

Now I suspect you might argue that my description is either not one of the provided posibilities, or is getting too detailed and complicated, but I would counter with four thoughts.

1st, You used and capitalized the word "CORRECT" several times above when describing your desired answer.
Usually when choosing between several alternatives which are all more or less correct, the simplified case may be used to explain things to beginners, but the more detailed answer is the one considered more "correct"
For example: are the orbits of planets circles or ellipses? is it more correct to assume frictionless motion or motion with friction included?
Using these guidelines my answer is clearly more "correct" than the provided potential answers.

2nd, while all this may seem like a small nitpick compared to just averaging over time and saying the answer is only #5, it actually is pretty important depending on what you are analyzing.
During the shunting part of the cycle, ALL the power requirements of the bike are met by the battery, and the pulse of current the battery supplies is an order of magnitude larger (and in the opposite direction) than the average charging current.
Depending on what you are analyzing, that pulse will have significant implications.

3rd, you said you wanted this to be an educational contest ... I would say that learning additional details of how the R/R functions is educational. Even if someone would choose to use the simpler model in a given situation, they would now do so as an educated choice and understanding the possible implications of that choice, rather than just not realizing that there is additional complexity

And finally, while the rules of this contest steer one to specify a simple answer, REALITY is not so simple. In this case the actual physical operation of the circuit says the answer really is not one of the preformed choices provided us. If you actually want the "correct" answer ... It is more complex than the choices which were provided ... And if you actually want to analyse how a circuit behaves, you will need to use the REAL analysis, not the simplified one ... If you want to calculate the voltage dropped in some of the wiring, or the heat rise, you WILL have to use drawing seven as well in your answer ... Or your answer will be wrong ... And for predicting the actual behavior of the actual wiring on the bike, my answer will give more accurate numbers than the other options provided. Which as far as I am concerned makes it "CORRECT"

---

In conclusion, it's fine to just use drawing 5 to provide a simplified explanation when desired; but if a detailed/correct analysis is required, then both drawing 5 and drawing 7 must be included.

I look forward to your reply.

[gggGary]

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.

[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
When the Regulator part of the R/R is regulating, it does so by briefly shunting all the power of the Stator internally.
(this occurs in the connections between the stator and R/R and those currents are not shown in any of drawings 1-7 above)
During that time, the stator and R/R are effectively not in the external circuit (battery and GS) at all, and all the loads on the bike draw their power directly from the battery.
This corresponds to drawing 7.

--

Now I suspect you might argue that my description is either not one of the provided posibilities, or is getting too detailed and complicated, but I would counter with four thoughts.

1st, You used and capitalized the word "CORRECT" several times above when describing your desired answer.
Usually when choosing between several alternatives which are all more or less correct, the simplified case may be used to explain things to beginners, but the more detailed answer is the one considered more "correct"
For example: are the orbits of planets circles or ellipses? is it more correct to assume frictionless motion or motion with friction included?
Using these guidelines my answer is clearly more "correct" than the provided potential answers.

2nd, while all this may seem like a small nitpick compared to just averaging over time and saying the answer is only #5, it actually is pretty important depending on what you are analyzing.
During the shunting part of the cycle, ALL the power requirements of the bike are met by the battery, and the pulse of current the battery supplies is an order of magnitude larger (and in the opposite direction) than the average charging current.
Depending on what you are analyzing, that pulse will have significant implications.

3rd, you said you wanted this to be an educational contest ... I would say that learning additional details of how the R/R functions is educational. Even if someone would choose to use the simpler model in a given situation, they would now do so as an educated choice and understanding the possible implications of that choice, rather than just not realizing that there is additional complexity

And finally, while the rules of this contest steer one to specify a simple answer, REALITY is not so simple. In this case the actual physical operation of the circuit says the answer really is not one of the preformed choices provided us. If you actually want the "correct" answer ... It is more complex than the choices which were provided ... And if you actually want to analyse how a circuit behaves, you will need to use the REAL analysis, not the simplified one ... If you want to calculate the voltage dropped in some of the wiring, or the heat rise, you WILL have to use drawing seven as well in your answer ... Or your answer will be wrong ... And for predicting the actual behavior of the actual wiring on the bike, my answer will give more accurate numbers than the other options provided. Which as far as I am concerned makes it "CORRECT"

---

In conclusion, it's fine to just use drawing 5 to provide a simplified explanation when desired; but if a detailed/correct analysis is required, then both drawing 5 and drawing 7 must be included.

I look forward to your reply.

I cant answer anything specific while the contest is going on but you might want to refer to this link.

http://www.thegsresources.com/_forum/showpost.php?p=967435&postcount=82

[tkennedy]

I'm going to say #4 because I know nothing about charging systems other than mine is no good and comes from my living room wall. I feel like I have read somewhere that the current goes back to the battery, thus creating a charge for itself.

[bakalorz]

I cant answer anything specific while the contest is going on but you might want to refer to this link.

http://www.thegsresources.com/_forum...5&postcount=82

Ok, I looked at that, and I have questions ...

The first two figures look pretty much as I expected.

The third does too, except for one thing:
According to the text, the blue line is the current coming out of the regulator.

So that would be measured at point X (or alternatively, W) in this figure.





But if I am reading your scope right, the blue line in fig 3 is spiking down to -10 amps during shunting.
As I understand the regulator, it shouldn't do that.

So I wondered, what if the current probe were at point Y (or Z) ...
THEN the trace would make sense, because the -10 amps is the battery feeding the bikes loads ... It would also make the quote: "The charging current is 2 amps in this graph as measured by my averaging scope"
make sense.

And further, It would be VERY good news for me ... since that strongly supports the description I gave of the ground currents in the post above (i.e. having a negative pulse thats much larger than the average charging current) ... which would mean I win the contest ...


So ... is all that right ???

[posplayr]

Ok, I looked at that, and I have questions ...

The first two figures look pretty much as I expected.

The third does too, except for one thing:
According to the text, the blue line is the current coming out of the regulator.

So that would be measured at point X (or alternatively, W) in this figure.





But if I am reading your scope right, the blue line in fig 3 is spiking down to -10 amps during shunting.
As I understand the regulator, it shouldn't do that.

So I wondered, what if the current probe were at point Y (or Z) ...
THEN the trace would make sense, because the -10 amps is the battery feeding the bikes loads ... It would also make the quote: "The charging current is 2 amps in this graph as measured by my averaging scope"
make sense.

And further, It would be VERY good news for me ... since that strongly supports the description I gave of the ground currents in the post above (i.e. having a negative pulse thats much larger than the average charging current) ... which would mean I win the contest ...


So ... is all that right ???

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/showpost.php?p=967419&postcount=81

[trippivot]

nope. none of the above.

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.