Revision of homemade electric diagram

[pdqford]

I think I am learning something here.
But here is where my head is at.

When you have the new wiring harness installed and running,
place your amp clamp at A and see what direction the current is flowing.

If the current is flowing towards the fuse on its way to the battery,
then the battery is being charged.

The line at B feeds the ignition switch, relay, and ?
When the battery is being charged by the R/R,
where will line B get its current to feed the ignition switch?


amp clamp.jpg

[Diirk]

I'll do my best to explain, but to make it easier to understand I'll make it brief and as such less technically true. So if any engineer comes across this bit of information, don't shoot me for not mentioning the real flow of electrons, importance of magnetism or including circuits exclusively being powered by the battery or what have you.
Also please keep in mind that I am really merely a novice that couldn't have written this answer down a couple of months ago because that's how new I am to all of this.


You are correct that a battery can not simultaneously charge and supply. However, it doesn't need to. The battery is only used for starting the bike and supplying power at idle. This is because for the rest of the time, the stator (through the r/r ofcourse) will supply power to the bike.

1. When starting the bike, obviously an external source of power is needed because the stator isn't able to provide anything "standing still".
   So we draw current from the battery to power up the starter motor.
   .
2. Once the bike is running, it idles and the stator supplies less than the 12V the battery has.
   As such, the bike's loads (lights, heated grips etc) are running off the battery.
   .
3. Now we ride the bike: the increased rpm of the engine spins the flywheel around the stator faster, thus making enough power to supply the needs of the bike's loads.
   • Because the output of the stator is now greater than that of the battery, the potential difference (difference in voltage) between them makes the battery a load. It is now being charged.
   • When fully charged, the draw of current will be lower as the battery doesn't "ask" for more and the r/r will lower the voltage to whatever is needed.
     .
4. At the next traffic light, the potential difference will be in favor of the battery (it supplies more volts than the stator at idle/low rpm) so it will take over the function of supplying power to the bike's loads again.
   .
5. Rinse and repeat steps 3 and 4

[pdqford]

Diirk, I really do appreciate the time and effort you have taken to help me understand your planned wiring diagram in non-technical terms.
It is obvious that you as a novice have a much better understanding of this electrical stuff than I do.
(And much better verbal and computer skills than me.
BUT I will select some of your explanation as a final attempt to explain where my head is at.

You are correct that a battery can not simultaneously charge and[*]Now we ride the bike:[*]Because the output of the stator is now greater than that of the battery, the potential difference (difference in voltage) between them makes the battery a load. It is now being charged.]

That is exactly my point. When the stator voltage is greater than the battery's voltage, the wire going up to the ignition switch is not fused.

Enough.

[salty_monk]

It's fused on the individual circuit fuses shown in orange on the other side of that switch. I don't think any current would flow until the switch is powered on so there would be no smoke in that red wire to leak out in that instance..... It would do nothing until you switched it on & at that point the fuse would blow.

I may be wrong though.....

[Diirk]

That is exactly my point. When the stator voltage is greater than the battery's voltage, the wire going up to the ignition switch is not fused.

Enough.

https://ibb.co/4MgKX5D

In the link above you can see I highlighted the circuit starting from the r/r. The red wire going to the ignition switch does seem to be unfused.
However, when that wire leaves the switch again as an orange wire, we can see #1 goes to the NO relay and #2 goes to the RNC fuse.
You always want to look at the circuit as a whole and not a single wire going from one place to another (given the wire gauge is correct, of course).

1. The wire marked #1 going to the relay is "signal" and therefore will carry a very low current.
   This part of the circuit is indeed unfused but because of the very low current I'm not too worried about it.
   .
2. The wire marked #2 going to the fuse is, well, fused.
   If anything should happen that fuse breaks and the red wire will also be protected because if one part of the circuit is interrupted, the whole circuit is -> as salty_monk correctly stated.
   .
3. Now if you look at the right you'll notice the KOSO instrument highlighted in green that is completely unfused. This is a potential hazard.

This happens to be the last (minute) thing I added to the diagram and even though this is again a low current (0.15A I think?), the instrument itself is exposed to the elements which increases the risk of a short happening. Hadn't really seen the risk it could pose until I started looking at it so you have my thanks for that. I could pop a little 0.5A inline fuse in there just to be on the safe side.


PS: thank you for complementing my "verbal skills". It's a nice compliment for a non-native speaker

[Who Dat?]

The FH012 should be a series type (although there's some debate about it) so the stator shouldn't turn crispy... again


Rich82GS750TZ and Who Dat? maybe you can weigh in now the diagram is clear?

I'll have to take a look at the diagram a bit closer after I get home from work, but felt it necessary to comment on another statement.

1. Now we ride the bike: the increased rpm of the engine spins the flywheel around the stator faster, thus making enough power to supply the needs of the bike's loads.
   • Because the output of the stator is now greater than that of the battery, the potential difference (difference in voltage) between them makes the battery a load. It is now being charged.
   • When fully charged, the draw of current will be lower as the battery doesn't "ask" for more and the r/r will lower the voltage to whatever is needed.

The r/r does not lower voltage. It accepts whatever is coming in from the stator, rectifies it to a pulsed DC current, then limits (regulates) it to whatever its setpoint is, usually around 14 volts. As the engine slows down from riding speed, the input voltage is not going to be enough to need regulating, so the voltage does drop, but not because the r/r lowers it.

As you are (hopefully) aware, a shunt-style r/r does its regulating by shunting any excess back through the stator (essentially to ground), meaning that the stator is ALWAYS producing as much as it can. Some of it will get used by lights, ignition, accessories, etc., but the rest is simply wasted. A series-type r/r does its regulation by interrupting the circuit. It does it very quickly, without an oscilloscope it might apear to be steady, but it will actually be a series of pulses, which will average out to whatever the voltage is supposed to be. This means that overall, there is less current going through the stator, allowing it to run cooler.

And, I am not so sure that the FH012 is a series r/r, I belive it is still a shunt-style. Might be a newer MOSFET style, but that is just the type of transistor used in the shunting circuits.