DISTRIBITEUR COIL PROTECTION MWO Nr1

[SV2IPW]

Here, is my simple solution, how to protect the coil.
You can add a 1Ω/50Watt wire resistor in series with distribiteur....so the current will be limited by the resistor but the voltage will be the same.
more in the photos:
http://to4x4.gr/phpBB3/viewtopic.php?f=26&t=1676

[svramselaar]

hello


the coil has a resistance about 6 ohm secondary
if the mutt is running the voltage a about 27 volts

the voltage at the coil go back 1/7 of 27 volt is 3.8 volt lower left 23.2 volt over the coil
the coil is 6 parts and the resistance 1 part of 27 volt 3.8 volt a part

the law of Ohm = volt = amp x resistance

in holland I =Amps U = Volts R = Ohm
U=I x R R = I : U I= R : U

also the amps go back from 4.5 till 3.8 amps

the ignition works as wen your generator is not working

the resistance will give 3.8 V x 3.8 Amp = 11.4 Watt on heat

greetings george

[SV2IPW]

Υοu mean that the resistor wattage is too hi???

[svramselaar]

hello

no only tad the voltage at the coil go down with 3.8 volt and the max amps go down with 0.7 amps

the resistor can have the heat is is 50 watt and it gets 11.4 watt so it not melt down

the spark will be a bit lower as normal

i think a good coil go bad if let the ignition on and the engine not running
if the points are closed at tad time there will be a current over the coil 4.5 amps and it will burn out
the pointless electronic ignition i don t know if it set itself powerless (a new car do this ) otherwise the same as above


george