GS Charging CONTEST

[posplayr]

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:

Assuming your GS is charging correctly while you are running down the road, then your charging system is producing current to both:
a.) Charge your battery
and
b.) Provide power to the GS loads (light, coils, igniter, bells, whistles, etc).

The power originates in the stator as AC voltage and current, but the R/R transforms this power into DC power at nominally +14V.

The power and current comes from the R/R out of the R/R(+) RED leg and splits part going to the battery and part going to the GS load.

The Question is : Of the 6 option shown below which one is CORRECT? Note the Red Currents are shown correctly, so which return current figure is CORRECT?

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"

 

[kdo58]

#4 all the current comes through rr, and #4 has single ground point which works the best.

 

[rustybronco]

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)

 

[BassCliff]

Hi,

I would say #5 also. The zener in the r/r has to know when to shunt excess energy to ground when the battery is fully charged.


Thank you for your indulgence,

BassCliff

 

[wkmpt]

Is this a trick question? The diagram I have for my 750 shows all current flowing to ground, not towards the r/r or the battery (-) terminal.

Granted, I observe on my bike that there are several wires that connect to the battery (-), so I am going to say that return paths for the lighting, load, and r/r flow to the battery (-) terminal, which is grounded to the frame.

I choose #4.

 

[donimo]

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!

 

[SqDancerLynn1]

#5 also, the reg dissipates the excess current as heat

 

[Guest]

Where's "all the above"?

I haven't been disqualified since war ball.

 

[kinnet]

I need more time! I'm in class... Stupid systems analysis and design. I'd much rather be researching this current flow matter.
Maybe I should change my major.

 

[Sandy]

#5 for me too. Current's burned of as heat at the R/R, that's why it needs fins.

 

[posplayr]

systems analysis and design. .

That sounds like an EE class

 

[kinnet]

It's all IT nonsense. But you could use the SDLC for EE I'm sure.
I will have an answer for this contest by the end of the night. How correct it will be is undetermined. Ha.

 

[Theo]

So I guess we are talking about current flowing from positive to negative rather than electron flow (negative to positive). I don't really know much about electrical theory, and am at work, so can't look at the manual, but I would guess #3. All of the current flows from the positive lead, powers all of the circuits and then makes it's final journey to the R/R and battery. Since both the battery and R/R are connected, I would think that means the current is returned to both from the GS. Even though the R/R is controlling how much voltage the whole system sees, I would think that happens with the alternator connections and not the connections that deliver +14 V. That's my 2 cents.

 

[Guest]

I thought it returned to the tires.
Sorry Jim, I gave it up.

 

[mottyl]

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

 

[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.

.

 

[Guest]

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