Free Amps...
- Thread starter 70_West
- Start date
70_West
Member
Ha! Chapter 7.2, pg 7.4 shows Amperes 35-50 @ 3000 RPM in the Service Manual as well.
R_W_B
Senior Member
Well I learn by exposing my ignorance. So having said that I want to say this. Hoop I can't really see how the amp reading would escape the stator input. Even if you separated the Maxi. In other words the system amps are going to be reflective of system voltage (including charging current potential) whether the stator - regulator is outside or in parallel with the system. The increase in volts (when the regulator opens) is going to affect amps regardless. Please correct my errs where I missed the concept.
Hoople
Account Removed
The test was not to determine how much amperage the charging system can provide. It was to determine how much current the bike needs to operate. If the charging system was connected BEFORE the Maxi, everything after the Maxi would be what the bike required to operate and the test would have worked.
(entry point #5, 2nd paragraph)
R_W_B
Senior Member
Yea I understand what the test was for, I'm saying I don't see how the charging input can be eliminated from the test. If the charging system is wired 'TO THE POWER SOURCE' which is the battery, (it seems to me) it has to cause a change in either the voltage or the amps of the system that's also hooked to the same power source.
Hoople
Account Removed
It doesn't matter if the power source is a battery,, a battery with a charger attached,,,or an A/C power transformer rectified to DC that you plug in the wall. The current used by the load will not change.
For Example:
Lets say we already know the answer to the question and the bike uses 20 amps with the headlights on & running.
We now unplug the stator plug from the regulator so there is no longer a working charging system. All we have is the battery. We now start the bike. The bike is running only off the battery. The battery by itself is supplying the 20 amps to run the bike.
We now plug in the stator plug into the regulator. We now have a battery and a working charging system. At idle the stator can only generate 10 amps but the bike still needs 20 amps to run so the battery is supplying 10 amps and the charging system is supplying 10 amps. Nothing has changed, the bike still consumes 20 amps.
Now we increase the RPM to 3000 RPM. The stator is now creating more current then we need. The stator supplies ALL of the required 20 amps to run the bike, along with enough extra current to charge the battery. We may even have to ask the regulator to cut back on some power in order not to overcharge the battery,,,But the bike is still drawing 20 amps to run.
In all 3 cases the load current (20 amps) remains the same.
R_W_B
Senior Member
Well I guess I'm gonna have to re-read some of my old electronic books cause I remember vaguely depending on whether the 2nd power source is wired in series or parallel with the first power source the volts stays the same on one and the amps stay the same on the other. But I also remember that the 'other' one (volts or amps) of each scenario does not stay the same. Something changes no matter which way you wire. Now granted it appears from what you say it's wired for the amps to remain the same.
Currently I having trouble visualizing this since on most standard parallel circuits (like a bike is and also a home wiring) if you increase the voltage source (lets say a dramatic one like a line voltage spike) the amps is gonna rise in the total system since I=E/R and R being the same when E rises then I rises. Can you give any more clues what I'm missing in the scenario ?
R_W_B
Senior Member
Ok (answering myself here and saving Hoop fooling with me) I have found some clues in one of my old books. Haven't found the forumlas (or laws) yet to back it up in calculations, but have found the theories that confirm what Hoop was trying to tell me.
For 2 voltage sources wired in parallel the volts stay the same but amp potential (not load amps) doubles.
For 2 voltage sources wired in series the volts would double but the amp potential stays the same.
A bike or automotive circuit is parallel so it is the former scenario.
2 voltage sources (say two jumped 12volt batteries) in parallel will still only deliver 12 volts but have the capacity for twice their rated amps when together.
Now if you change the voltage potential for either you end up with a somewhat precarious condition that cannot last for long or something is gonna get hot (if I reading my text correctly) since the larger potential wired in parallel with the smaller potential is going feed the difference in potential at the speed of a 'short'. My puzzle is the stator pumping the battery voltage up from 12 to 14 volts, hence the need for the voltage regulator, which I know from experience the old car units would close and open many times a minute in actual operation. The newer (like my Dyna) digital diode rectifier regulators I am not so familar with the characteristics or ramifications of them. Anyhow don't won't to get to far off topic but just trying to apply the laws I was fuzzy on these test scenarios.
Hoople if you see any areas you can enlightment my vision on this (thereoms or laws or forumlas to experiment with) feel free to throw them in.
In closing and still this part very much on topic, wouldn't (as dangerdan previously said) an amp probe around the either the negative or postive battery cable read system amperage ?
Lets try this explanation.
In a parrellel circuit the total amps drawn equals the sum of all the loads. It=I1+I2
The total voltage drop across each parrellel load equals the voltage source. ET=E1=E2
Regulator connected at 1000 RPM
Total Idle Current = Regulator current + Battery current
20 = 10 + 10
Regulator removed at 1000 RPM
Total Idle Current = Regulator current + Battery current
20 = 0 + 20
Regulator connected at 3000 RPM
Total Running Current = Regulator current + Battery current
20 = 20 + 0
Regulator removed at 3000 RPM
Total Running Current = Regulator current + Battery current
20 = 0 + 20
As Hoople says "In all 3 cases the load current (20 amps) remains the same".
In a parrellel circuit the total amps drawn equals the sum of all the loads. It=I1+I2
The total voltage drop across each parrellel load equals the voltage source. ET=E1=E2
Regulator connected at 1000 RPM
Total Idle Current = Regulator current + Battery current
20 = 10 + 10
Regulator removed at 1000 RPM
Total Idle Current = Regulator current + Battery current
20 = 0 + 20
Regulator connected at 3000 RPM
Total Running Current = Regulator current + Battery current
20 = 20 + 0
Regulator removed at 3000 RPM
Total Running Current = Regulator current + Battery current
20 = 0 + 20
As Hoople says "In all 3 cases the load current (20 amps) remains the same".
R_W_B
Senior Member
Thanks for the reply. Well yes that's a neat explanation of ohm's law but it doesn't really explain why we can't obtain actual system current (amps) by probing the battery cable.
When I ask (same as you said in an earlier post) if reading the amps by AmpProbing the battery cable during operation gives system amps ?
Further I am trying to decipher really how system amps can be classified as 'not' inclusive of the system charging current supply. I.e. the battery has a nominal voltage and it has a maximum output voltage which it is never going to reach without the charging system applied current (and 14.x volts since you can't have current flow without sufficient potential).
With normal resistant loads (lights) the current will drop with the supplied voltage drops. With induction loads (the starter) the amps will actually go up with a voltage drop due to the amature turn speed never pickup up speed from lack of voltage.
Further (I'm still trying to verify this by relearning Kirchoff's laws) I don't really see why the amp reading at the battery cable is not what we want anyhow since it would appear to me if you have 14.x volts (coming down the line from the regulator) any excess amps over the load requirement is being consumed (charged) back into the battery plates (it having the weaker potential).
But regardless to me (in my ignorant arrogance) battery amps + charging amps ARE the system amps (I.e. amps at the battery cable). Without the charging amps included the battery output is going to drop and not register a true systems amps.
So I guess I've said enough to either be right or so wrong Hoople will give up on trying to teach me anymore. In any case I've got my electronics book and reading on it when I have time. Fortunately at the current time I really have no need to know any of this so it is not a pressing issue, just curiosity.