White wire succses story!!!!

[duivendyk]

ET, are you using the neg. charging arrangement ?.Yes you are getting 2.2W at high rpm.I am guessing the WW source inductance is beginning to affect the output some at high rpm (a bit messy to explain).You could try putting a cap in series with the WW (neg. side towards WW),try 470 mF and work your way down &record output at diff.speeds.Will also give me better idea what the source impedance is ,which is useful to know somethig about.
Impression,It would be best to try out a negative output charger first.As I have explained it seems likely that such a circuit has less effect on the ignition system.Cathode of diode (band) towards WW then your 27ohm resistor& switch and to neg.side of battery, pos.side the ground. If you have Am meter put that in circuit also.If you feel cocky, reduce or get rid of the resistor. Low voltage light bulbs (6-12V) can have low resistance,measure with ohm meter,if available.Good luck.

 

[duivendyk]

ET,I just realized that I goofed, this extra cap deal won't work unless another diode is added to the circuit.Without it you'd end up just charging the cap and that would be the end of it,no current to the batt.Another diode is needed after the cap to discharge it. At the junction where it connects to the series diode going to the battery, another diode needs to be connected to ground, with the cathode (band) on the ground end, and with the anode lead connected to that junction.So three leads come together there, the pos. cap lead and both diodes which should have opposite polarities.Don't short out the series cap,you would kill the engine.This arrangement should have in theory at least more output than the simple circuit,but it's also more likely to mess up the ignition.The size caps could affect its performance, so it's worth while finding out what works best.

 

[etacovda]

ok, will try when i officially get up (is 4am here, cat decided to go crazy and wake me up!) - i see, so white wire inductance increases resistance with more rpm, yes? i vaguely remember inductance from highschool physics, harder you push in one direction, inductance pushes back or some such...

 

[duivendyk]

Red alert,read my latest post !!

 

[duivendyk]

I goofed again,sorry about that, latest poop ,the series cap to the WW needs the PLUS side connected to the WW and at the neg. side goes a diode to ground, with the band (cathode ) to ground,the other diode is hooked up as before (band towards cap& other diode) other end to battery.This is called a peak to peak rectifier circuit.

 

[duivendyk]

I ran some numbers based on your test data and concluded that the inductance effect had to be minor.The ptp circuit gets the most output for any configuration.Could make ignition unhappy.Be careful messing with batteries,sulphuric acid is nasty stuff.Someone else has a 6V battery,don't know if he has suitable Ammeter.

 

[etacovda]

I ran some numbers based on your test data and concluded that the inductance effect had to be minor.The ptp circuit gets the most output for any configuration.Could make ignition unhappy.Be careful messing with batteries,sulphuric acid is nasty stuff.Someone else has a 6V battery,don't know if he has suitable Ammeter.

battery is a power tool battery, a series of small ni-cd cells i think. Easy to pull to bits + rework as needed. Assume this isnt an issue with regard to the circuit power output?

PTP? is that positive to positive?

 

[duivendyk]

No not really as long as the battery is capable of supplying output,ptp stands for peak to peak (electronics engrs jargon), it means the voltage difference from pos. to neg. peak.The circuit I described works as follows: during the positive going part of the pulse the cap is charged to the peak value of the pulsevia the diode to ground,so the neg.lead of the cap is lets sayat -6V with respect to the WW,OK now when the pulse starts going negative say to -17 V the voltage at the series diode input goes to -23V (-6 V+-17), when it goes below -12.6 V ,the diode starts to conduct and dumps current into the battery.Then the whole thing starts all over again,the capacitor which has lost charge (and voltage) is charged up again during the pos cycle.So the output is the peak to peak value of the ac input waveform (Our ac pulse).Start out with 470mF,and than try 220 and see if there is any difference in charge current.Would show up first at idle.You could also try 470 and 330 mF in parallel.

 

[duivendyk]

Continuation,it would be nice if you could kluge up a 6V battery,at least you have a usable meter.My thought on 6V vs 12V batteries.Batteries are rated according to Ah capacity, not as they should be on watt hr capacity.which is the true indication of the energy stored, 1 VA=1 Watt.So a 6V 2Ah battery stores the same energy as a 12V 1h battery.So if you supply 2 Amp at 6V, it's equivalent to supplying 1Amp at 12V power wise.So if we can charge a 6V battery with more than twice the current than for a 12V one we're ahead of the game.I suspect that this will turn out to be the case.But It's more likely that the ignition will act up,esp. if you use the p.t.p. rectifier.But nothing ventured, nothing gained!

 

[etacovda]

indeed, sounds good. I will build the ptp circuit tonight and test. I have two of these 12v drill batteries so i will alter one to be 6v so i can test side by side.