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Hot Dog Piggy Tails
Guest
So your not squaring the wave (turning it to dc)?
Flapdoodle
Member
If that question was addressed to me, yes it is rectified. The circuit consist of only a resistor and a diode. The band on the diode goes toward the white wire.
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duivendyk
Guest
I think 6V SLA's might be a suitable target battery,rather bulky but fairly robust and can be protected with a zener based protection,the key is to come up with a circuit that does not waste precious input power,I'm thinking of a zener + overcharge indicator LED prompting the operator to turn off the charge switch or alternatively turn on the lights.That's pretty straightforward and does not waste any charge current when it matters.
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duivendyk
Guest
The reason you would like an RC parallel circuit for testing is that in my experience some of the inexpensive digital readout multimeters are not all that good at measuring the short high-peak current pulses to give an average value you can rely on.The safer way to proceed would be to start out with a 100 Ohm load resistor with at least a 1000 micro farad cap across it.A 2200, 3300 or 4700 mF 25V electrolytic cap would be even better, plus side at ground!!) and to measure the output voltage at running speed, say 5k rpm and check if the engine runs OK.Then you put another 100 Ohm in parallel,and repeat (50 Ohm load) then three (33 Ohm), four or five (20 Ohm).By the way 6V across 20 Ohms is 300 mA. Then the current can be arrived at by V/R=I . VxI gives us the output power in Watts. If you dont have a cap, set the multimeter to measure the current going into the resister combination & hope for the best ,or if you have a 6Volt SLA lying around,measure the current going into the battery from the rectifier diode.(band on it towards the white wire). Put at least 20 Ohm resistor in series with it to limit the current.Don't use a completely discharged or dead battery,they can have high internal resistance.That will lead to erroneous results.Good luck ,give something along these line a try.Hopefully we can get useful design info for arriving at an adequate battery charge circuit.
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Flapdoodle
Member
Well said duivendyk.
I still wonder if there are models made with different output voltages for the white wire. Mine measures 33 VAC, so a 50 volt cap would be needed.
I still wonder if there are models made with different output voltages for the white wire. Mine measures 33 VAC, so a 50 volt cap would be needed.
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duivendyk
Guest
Is that open circuit?.I'm pretty sure that there is a lot of EMI around these engines,picked up by meter leads,stick a 0.1 mF ceramic cap.right at the meter input.Putting a couple of kohms in one of the leads to act as a lowpass filter will keep the electrical crud from screwing things up. The old fashioned analog types are less suceptible than the digital sampling types. What you are measuring could well be due to short duration voltage spikes.This is a nasty RFI environment.Using a 50 V cap does not hurt anything of course, if you've got one lying around.
Flapdoodle
Member
Yes, that is open circuit with a low impedance meter. Best way would be a scope, but mine died a horrible death years ago.
A diode and cap will charge to the peak voltage if there is no load. However, peak does not tell us how much of that is usable.
I had another thought... *if* the rated amperage for the white wire is actually .5 amps, at 12 volts we may be able to squeeze out 6 watts.
A diode and cap will charge to the peak voltage if there is no load. However, peak does not tell us how much of that is usable.
I had another thought... *if* the rated amperage for the white wire is actually .5 amps, at 12 volts we may be able to squeeze out 6 watts.
loquin
Well-Known Member
Yes. But, you're wasting half the available power. A bridge rectifier will route both the positive and negative halves of the AC cycle to the DC buss. In addition, a bridge rectifier will reduce the DC ripple, since it's charging the capacitor twice as often. Shottky diodes are also recommended, as they only drop 0.3 volts, as opposed to 0.6 volts for standard silicon diodes.
Finally, you do not want to put too large a cap across the dc output in this instance... Until the cap is charged up, it presents a very low impedance to the AC source. In effect, initially, it's a short circuit. The larger the capacitor, the harder it will be to get your motor started.
Flapdoodle
Member
Yes, it wastes half the power cycle. That was deliberate to prevent robbing from the positive going half wave going to the CDI. Because both windings are on the same core, load on one robs from the other if you use either an incandescent light (wired directly across the white and black wires), or a full wave rectifier (bridge) on the white wire. For those more technically minded, it is not actually stealing voltage, but rather magnetic flux which in turn drops voltage in the other winding. If they were wound on separate cores, this would not be the case.
The amount of time to charge the cap is very small. When I made my test the bike started faster than it ever has. That is most likely coincidental. But as I mentioned before, there IS no capacitor in the finished circuit because the battery smooths the supply. The batteries do not care if it is 1/2 wave for charging. I have been using either half wave or pulse width modulation to charge SLAs for decades without the slightest problem. If I decide to design a regulator for my bike it will be PWM.
As it stands now, I expect I may have to use two diodes in series to drop the voltage another .6 volts once I have evaluated the circuit sufficiently, so there was need to go to schottky (the correct spelling) diodes.
The amount of time to charge the cap is very small. When I made my test the bike started faster than it ever has. That is most likely coincidental. But as I mentioned before, there IS no capacitor in the finished circuit because the battery smooths the supply. The batteries do not care if it is 1/2 wave for charging. I have been using either half wave or pulse width modulation to charge SLAs for decades without the slightest problem. If I decide to design a regulator for my bike it will be PWM.
As it stands now, I expect I may have to use two diodes in series to drop the voltage another .6 volts once I have evaluated the circuit sufficiently, so there was need to go to schottky (the correct spelling) diodes.
heathyoung
Member
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Sounds good to me 
I wonder if you increased the power of the magnets you could increase the power output of the magneto - its probably running close to saturation though anyway...
Those neodium magnets out of old HDD's would be interesting to use...
You also raise an interesting point - if the CDI only uses the positive cycle (and a >6W load will cause it to droop/stall) Maybe grab <6W by current limiting on the positive cycle, and >6W on the negative - two different value current limiting resistors would do the trick
What is the maximum output you can draw on the negative cycle? Is it still only 6W?
Another thought - is it really a CDI ignition system? - it must be triggered by the negative going pulse (as most use a separate coil to 'trigger') - or is it just a coil?
I wonder if you increased the power of the magnets you could increase the power output of the magneto - its probably running close to saturation though anyway...
Those neodium magnets out of old HDD's would be interesting to use...
You also raise an interesting point - if the CDI only uses the positive cycle (and a >6W load will cause it to droop/stall) Maybe grab <6W by current limiting on the positive cycle, and >6W on the negative - two different value current limiting resistors would do the trick
What is the maximum output you can draw on the negative cycle? Is it still only 6W?
Another thought - is it really a CDI ignition system? - it must be triggered by the negative going pulse (as most use a separate coil to 'trigger') - or is it just a coil?
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