White Wire Voltage - 20+ V ??

[Civalunrest]

I use one of those cheap bell lites from walmart it has three bulbs in it 2 LEDs and a 3 watt in the middle i took out the battery compartment and ran the lights direct to the white wire with a single toggle to cut the lights off. While standing still with lightes on the bike idels ruff but once i get going that light system is brighter than a flood light I Highly recomend these lights

 

[MurraySabrePlus]

I added two components to the circuit I offered earlier.
View attachment 12501

The second diode allows voltage from the magneto to power the lights while the battery is still trickle charging. The capacitor and to some extent, the battery minimize LED flicker at idle. The lights or whatever will still work with the engine off.

Adding a motor to a bicycle creates a dilemma; do I go for spandex, or a black leather jacket?

Sorry but, I don't know how much you know about electronics, but that gif diagram of some circuit won't work. It doesn't make sense at all.
As drawn, that thing would destroy the battery and blow up the capacitor!

 

[MurraySabrePlus]

Sorry but, I don't know how much you know about electronics, but that gif diagram of some circuit won't work. It doesn't make sense at all.
As drawn, that thing would destroy the battery and blow up the capacitor!

EDIT:
now I see you're using some positive grounding.... hmm.
Still, it's overly complicated and un-needed with those diodes and cap.

 

[Flapdoodle]

EDIT:
now I see you're using some positive grounding.... hmm.
Still, it's overly complicated and un-needed with those diodes and cap.

Remember that it is AC coming from the magneto, so the absolute minimum would be one diode to make it DC. Batteries don't like AC. The resistor prevents over charging the battery. The first diode and resistor trickle charge the battery. Eliminate the cap if you like and the second diode if you don't use the light often.

I have been using the circuit since late last summer without any problem.

 

[MurraySabrePlus]

Remember that it is AC coming from the magneto, so the absolute minimum would be one diode to make it DC. Batteries don't like AC. The resistor prevents over charging the battery. The first diode and resistor trickle charge the battery. Eliminate the cap if you like and the second diode if you don't use the light often.

I have been using the circuit since late last summer without any problem.

Thanks, but you're preaching to the choir Flappy, I'm an electronics technician with decades of experience and run a tv repair shop.
But see my simple solution....
http://www.motoredbikes.com/attachment.php?attachmentid=16129&d=1237354275

 

[etacovda]

negative ground ends up with 18v peak, whilst positive ends up with 6v - dont you waste 1/2 of the power by putting the diode 'with' the white wire rather than against it?

 

[MurraySabrePlus]

etacovda, I find your post interesting.
I assume you're referring to flap's use of positive grounding. That stumps me too.

I also wonder how any voltage difference can be electronically possible from a coil of wire with an induced magnetic field, whether it's wired one way or the other for rectification.

It's been documented on here somewhere that the coil of wire involved is about 2 ohms, wound on a C-shaped magnetic stator, along with an additional winding of several hundred ohms, apparently used to fire the sparkplug along with the CDI module.

Now from my electronic background and extensive experience in this area, I don't understand where a substantial difference can occour in output voltage from the 2 ohm coil.

The magnet revolves, creates a voltage in said coil of Alternating current (AC).
This AC voltage has a plus/minus value resulting from the north-south alternating poles from the rotating magnet.
It can only be a symetrical (equal) waveform of the frequency that those magnets revolves, a.k.a. positive 12v/negative 12v (as an example).

Just as a power transformer on an AC line transfers it's power from the primary though the laminations to the secondary.... it's equal pulses, pos/neg.

So I've been scratching my head here when I've read some of these posts about how the voltage can be a different value depending on HOW the AC is rectified from a single coil of wire.
Unless of course, there's something to that magneto that no one knows about.

 

[etacovda]

well, in the small amount of reading ive done here it has been suggested that the white wire is tapped part way down the coil, with the blue wire using the positive to charge the CDI, whilst the negative is untouched - i guess that the 12v difference is headed to the CDI to create spark, whilst the negative remains untouched. My _personal_ testing shows that peak voltage is -18.8v to positive 6.8v on the white/black wire circuit. Using a peak to peak circuit i managed 4.8w at 13.9v.

I was referring to your diode and non-negative grounding circuit that you suggested impression use.

 

[Flapdoodle]

Thanks, but you're preaching to the choir Flappy, I'm an electronics technician with decades of experience and run a tv repair shop.
But see my simple solution....
http://www.motoredbikes.com/attachment.php?attachmentid=16129&d=1237354275

I am using the negative half of the full wave to prevent robbing power from the positive half that is used by the CDI.

BTW, I am a retired engineering tech. I designed and built automated manufacturing equipment.

 

[MurraySabrePlus]

well, in the small amount of reading ive done here it has been suggested that the white wire is tapped part way down the coil, with the blue wire using the positive to charge the CDI, whilst the negative is untouched - i guess that the 12v difference is headed to the CDI to create spark, whilst the negative remains untouched. My _personal_ testing shows that peak voltage is -18.8v to positive 6.8v on the white/black wire circuit. Using a peak to peak circuit i managed 4.8w at 13.9v.

I was referring to your diode and non-negative grounding circuit that you suggested impression use.

Ah, so in effect, there's loading of the overall coil during the "half-cycle" due to the CDI being "fed", thus lowering the WW's output as well.

That can make sense yes.

With that in mind, my circuit, by changing the polarity of the diode and battery should work fine then. I will change it to reflect this phenomenon.
(ugh, another editing) LOL.

---> http://www.motoredbikes.com/attachment.php?attachmentid=16135&d=1237394647