Voltage of the magneto?

Discussion in 'Electrical' started by Frankenstein, Jul 31, 2016.

  1. Frankenstein

    Frankenstein Active Member

    Generally what is the voltage and amperage of the magneto while running?

    I'm sure it's not much but still interested in knowing, I'm still on the hunt to make a source of electricity that is easy and powerful enough for lights...
     

  2. jaguar

    jaguar Well-Known Member

    it's 30 volts with no load. amperage is dependent on the load resistance. the correct question is what the wattage is since wattage is volts multiplied by amps.
    using the white wire as a source for lights can shorten the life of the stator coil and change its electrical characteristics which changes spark timing.
    best to look elsewhere. I used rechargable batteries powering LEDs
     
  3. Frankenstein

    Frankenstein Active Member

    Actually I was going to modify a bolt to bolt the bevel gear on to have 2 heads of a bolt welded together, that would hold a mag and spin at the same speed, modify a clutch cover to accept an insulated magneto, then pull the power from that like the cdi pulls from the normal mag... So what kind of wattage can I expect? I can alter my end result with rectifiers and buck boosters to regulate my energy output before the battery.
     
  4. jaguar

    jaguar Well-Known Member

    Estimated wattage? that's a question for an electrical engineer, which I am not.
    Depends mostly on the strength of the magnet and the number of turns of the wire along with its thickness (because thin wire presents resistance)
     
  5. Frankenstein

    Frankenstein Active Member

    Think it's similar to the wattage from that wonderful creations thing on eBay? Do those arms on the normal magneto actually do anything? In a transformer the iron ring that looks similar, minus the second wrap of wire, is the part that transfers the magnetic field from one coil to the second.

    Kinda off topic on that part.

    If wonderful creations can yield about half an amp from the same magnet with using a different number of winds to get 12 volts then wouldn't it be safe to presume that the normal mag would also yield half an amp when the wattage is calculated from the voltage? Same magnet so same energy field?
     
  6. jaguar

    jaguar Well-Known Member

    yes same energy field from the magnet but the design of the coil determines how it transforms that magnetic energy into electrical energy.
     
  7. Frankenstein

    Frankenstein Active Member

    And presuming the coil is being wound by somebody or something who/that can actually wind a good coil, then what would be the wattage? They are both roughly the same size, if anything the wonderful creations are behind a few million coils as far as production goes, the Chinese have some reliable system to make them..

    What is the wattage/amps/volts needed for the cdi to produce a reliable spark, or perhaps what is the wattage of a reliable spark?

    10000 volts is good but with such a low amperage it's not going to do too much, too little amps (like 0.00amps) and the spark won't exist
     
  8. jaguar

    jaguar Well-Known Member

    your last sentence only pertains to the high voltage coil, not the stator coil.

    I just now put a 1K (1000 ohm) load on my tester coil and rev'd it to around 1000 rpm and got 34 volts at 34mA which equates to 1.15 watts.
     
  9. Frankenstein

    Frankenstein Active Member

    Hmm... That's not much, but almost seems flawed?
     
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  10. jaguar

    jaguar Well-Known Member

    I think that is typical for a stator coil. A lighting coil with a heavier gauge wire and less turns produces less voltage but more current.
     
  11. Frankenstein

    Frankenstein Active Member

    But same wattage, it's not the volts I'm worried about, it's the total power I can derive from the system. Are you sure that 1k olm resistor, with the way you hooked up that tester, didn't alter your calculations or something? Those values are scarey close to being 1k ohms.. In fact it's just 29ohms off..

    You're saying normal voltage is about 30-35, if I cut my wire and make my meter part of the circuit I could get the current right?
     
    Last edited: Aug 2, 2016
  12. jaguar

    jaguar Well-Known Member

    V=IxR (voltage equals current in amps times resistance)
    W=VxI (watts equals voltage times current in amps)

    V/I=R
    34v/34mA=1K exactly, not 290 ohms off
     
  13. Frankenstein

    Frankenstein Active Member

    I said 29 not 290 ohms, time to shop for reading glasses jaguar ;)

    Did you make your meter part of the circuit, or cross it over the circuit making it parallel to the mag?
     
  14. jaguar

    jaguar Well-Known Member

    sorry, my eyesight isn't too good. But the answer is still 1K.
    yes parallel to the coil and the resistor. all three are in parallel
     
  15. Frankenstein

    Frankenstein Active Member

    Lol no wonder... Amps must be measured in-line, like you would have to cut the blue connection to the cdi, and place the meter ends between the too, it would appear you measured the wattage being used by the resistor (it probably took the resistors reading simply because it was the shortest path, and your current being passed across the 2 wires wouldn't want to do more work than what it was already doing, electricity is a bitch,) but that's OK, because I think we are getting onto something. I'm busy with work tomorrow at 8am, and will be working for 8 hours till 4:30, so can't go out and start a motor with a meter in series to my cdi, what kind of meter do you use? Like digital r analog?

    If you could tinker around for me it would be much appreciated, it could also open a lot of doors for other people too, as far as getting power for lighting, which is very important at night.
     
  16. Frankenstein

    Frankenstein Active Member

    Just to be clear, the meter would have to be in series, so it would interrupt the cdi if the lead was taken off a wire, the current being used by the cdi will reveal the minimum wattage you'd expect from the magneto.
     
  17. jaguar

    jaguar Well-Known Member

    the resistance that the CDI presents is different for the positive half of the wave than it is for the negative half of the wave.
    So that complicates things. Pure wattage reading of a coil is done the way I did it.
     
  18. Steve Best

    Steve Best Active Member

    Maximum wattage for these motors would be about 1400w which is the 2hp or so that they produce.
    Big enough magnets and lots of coils and you can drag all the hp out of the motor electrically.
    Glue super magnets to the clutch rim and coils to the cover and you have a pretty efficient generator.
    I'd go for 3 phase windings (6 coils) and 6 diodes to rectify it, with a simple voltage capping regulator off a dirtbike or sled.
    You could use 6 permanent magnets glued into an aluminum ring bolted on the clutch.
    I oughta take out a patent.

    Wait a minute, clutch is going 1/10th speed of crank.
    Put it on the crank, either end. If not enough room, go 4 diodes, magnets and coils.
    Single phase. Cheaper, lighter.

    Steve
     
  19. Frankenstein

    Frankenstein Active Member

    Are you being serious about putting it directly on the crank? Doesn't seem like a fantastic idea at those speeds... But that clutch plate idea had a thought going (watch out!) some careful machining of a dead crank, modifying the crankcase and cover, a second bevel gear could be mounted up along the clutch gear, which would spin the new small gear at the same rate as the crankshaft, and would continue spinning after the clutch was disengaged too... Hmmm...
     
  20. Steve Best

    Steve Best Active Member

    Yeah I'm serious. It would take less copper and steel if driven directly off the crank.
    I'm looking at using the motor to charge batteries on a hybrid. The rotor and mag are pretty much maxed as is but there is a lot of room in there. A redesign of the pickup coil could provide spark and volts.

    Steve
     
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