R/R Testing - Voltage through R/R

[Steve]

In this scenario, if you turn off the headlight and the charging system is still operating at full capacity, it is possible to overcharge your battery and boil off the water.

Are you sure about that? I thought that is where the regulator part of the R/R kicks in and shunts the electrons to ground.

That's what I was thinking, too.

I don't think the voltage would go any higher, it would just get there sooner.

Because there is less load on the charging system, it will start charging basically at idle and will reach the regulation phase very quickly.
When that happens you will simply be shunting MOST of your output instead of A LOT of it.

.

 

[posplayr]

OHMYGOD!!
That brings back bad memories. Hopefully it works better (and faster) than it did in the 80s. A Monte-Carlo simulation on an 8088 (even with a math co-processor) could take hours.

That computer would have been from about 1984, I had a clone PC (from PC shopper) with an 16 Mhz 8088 with 20 Mb HD back then as well .

At Delco Electronics (where i worked) layout was still being done by the drafting department, so I'm not sure P-spice was really widely avaliable in 1984-85. Computers are stupid fast these days compared to back then, throughput almost seems to be an afterthought these days.

 

[posplayr]

Pos, what is the value of those resistors?

no idea, Suzuki doesn't provide that info.

 

[081dbx64]

LOL...back in 1985 I paid $5,000 for a 10 MEGAbite hard disk. Yikes

 

[raistian77]

The resistors and the zener are there to fire the SCR when the voltage on the Red (+) output is too high. The SCR shorts the winding so that the current stays in the stator and doesn't go to the battery.

The SCR is like a latched diode. It is turned on by a "gate" (controlled by the zener and resistors) and then is shuts itself off when you try to drive current backward through it (when the AC reverses).

My buddy explained it to me, but can't say I followed it ; it is a little confusing so I won't even try. The complication though seems to be that the control circuit is not measuring the output with respect to the ground but rather the one phase.

I'm assuming it works as drawn. If someone has a P-Spice they could try and simulate it.

I can see it now that you explained it. Simple design, I don't care for the shorting the leads together to shunt the current. Would be much better to open the phases up.

 

[koolaid_kid]

Computer Shopper, geez. I heard they finally went out of business. Back then it was a huge book, and pretty much the only way to get any decent prices.
I built my first PC from parts purchased at a local computer store. It was an 8 MHz 8088 back when everyone else was running 5.7 MHz. Started with a whopping 256 Kb of ram and a single floppy. I was going to school at UTD and the computer labs were impossible to get into, so I built one so I could do homework at home. It just cascaded from there.
I won a sales contest at work and purchased a hard drive and controller card. 30 Mb back when everyone else had a 20. Woohoo!
O.K., hijack over.

OP, if anything, I would expect the extra power being shunted to ground to create more heat in the R/R. IMHO, the stock one on my bike did not have enough cooling capacity; the Honda R/R I replaced it with appears to have about twice as much (bigger unit, so fin area greatly increased).

 

[081dbx64]

Computer shopper...god I loved that book.................a pervibal computer skin mag.

 

[koolaid_kid]

I can see it now that you explained it. Simple design, I don't care for the shorting the leads together to shunt the current. Would be much better to open the phases up.

I agree, it must be that way for cost reasons. However, my GPz (and probably my former 81 KZ750) has the same design and never had any problems. I should post a pic of my GPz stator, it looks brand new. I suspect Suzuki just has a faulty design in the R/R area, but I have no data to support my suspicions.

 

[posplayr]

LOL...back in 1985 I paid $5,000 for a 10 MEGAbite hard disk. Yikes

they were cheaper in Computer shopper. Mine (8088 clone with 20 Mb ST-220 IIRC) was about $1500 with a monochrome Hercules video card.I also got one of the first IBM PC's at work and it was probably about $4000 (same time frame)

 

[posplayr]

Computer Shopper, geez. I heard they finally went out of business. Back then it was a huge book, and pretty much the only way to get any decent prices.
I built my first PC from parts purchased at a local computer store. It was an 8 MHz 8088 back when everyone else was running 5.7 MHz. Started with a whopping 256 Kb of ram and a single floppy. I was going to school at UTD and the computer labs were impossible to get into, so I built one so I could do homework at home. It just cascaded from there.
I won a sales contest at work and purchased a hard drive and controller card. 30 Mb back when everyone else had a 20. Woohoo!
O.K., hijack over.

OP, if anything, I would expect the extra power being shunted to ground to create more heat in the R/R. IMHO, the stock one on my bike did not have enough cooling capacity; the Honda R/R I replaced it with appears to have about twice as much (bigger unit, so fin area greatly increased).

Last time I looked , Computer shopper was still around? Must be 10-15 years since I looked

your probably right on the clock speeds., it has been a while.

 

[raistian77]

Are you sure about that? I thought that is where the regulator part of the R/R kicks in and shunts the electrons to ground.

I don't belive it actually shorts to ground, more like shorts the phases across each other. The stator still produces power it just cycles around untill unshunted and connected back to the load.

 

[posplayr]

I don't belive it actually shorts to ground, more like shorts the phases across each other. The stator still produces power it just cycles around untill unshunted and connected back to the load.

yep.................

 

[raistian77]

getting there lol. Explains burnt stators, resistance in the circut (corrosion loose connectors, etc) must make the stator act like it is under a load at all times even when shunted. If the resistance is bad enough the load could easily pass 15 amps and never trip a fuse do to the load being in a unfused section. Heck bad enough the stator could be under a 30 or 40 amp load and smoke itself trying to 'keep up'

 

[posplayr]

getting there lol. Explains burnt stators, resistance in the circut (corrosion loose connectors, etc) must make the stator act like it is under a load at all times even when shunted. If the resistance is bad enough the load could easily pass 15 amps and never trip a fuse do to the load being in a unfused section. Heck bad enough the stator could be under a 30 or 40 amp load and smoke itself trying to 'keep up'

I measured the stator currents for SHUNT and SERIES R/R's here. The summary charts are scaled to total current even though I measured stator leg current.

SHUNT_vs_SERIES_RR_Compare_Tutorial

The issue with resistance at the output of the R/R is that it causes the regulator to over control and shunt the stator more than it should. That is because if there is excessive resistance (e.g. 0.5 ohms) there will be a large voltage drop at the output of the R/R.

For example If the R/R tries to push 10 amps and the 0.5 ohms is on the positive leg, then we are looking at a 5 volt drop. Since the battery is already at 12V+ then output of the R/R is at 17V + which will keep the R/R regulator circuit shunting causing the R/R and the stator to get hot.

The battery is not charging, and there is barely enough current going to the GS, because the R/R is constantly shunting current in the stator.

 

[081dbx64]

Question

I'm measuring the current drain on my battery in series between the positive battery terminal. My meter is set to 200m

My reading is 10.2.. Is this .102 amps or .0102 amps?

Any ideas

 

[posplayr]

Question

I'm measuring the current drain on my battery in series between the positive battery terminal. My meter is set to 200m

My reading is 10.2.. Is this .102 amps or .0102 amps?

Any ideas

Probably 10 mA but what meter is it?

 

[Steve]

Question

I'm measuring the current drain on my battery in series between the positive battery terminal. My meter is set to 200m

My reading is 10.2.. Is this .102 amps or .0102 amps?

Any ideas

In the 200m scale, your limit is 200 miliamps.

Your display shows 10.2, so that is 10.2 miliamps, or 0.0102 amps.

.

 

[081dbx64]

Ok...so divide by 1000...got it.