Farmall 404

I'm still not sure where the power to the coil is coming from. Since you said no power on the black wire until the switch is on, I suggest that you disconnect the white wire from the resistor to the coil. If you then have no power to the coil with this wire disconnected and the switch on, it says that the coil is only going to get 6v (or whatever the resistor is rated for) when connected with the white wire from the resistor. The other white wire connected at the ring terminal at the coil + is just sending power to somewhere else but at a reduced voltage. Looks like the power flow 12v is switch to resistor via the black wire, then out from the resistor at reduced voltage to coil and whatever else the white wire connects to (again at a reduced voltage). If you want to leave the resistor in place and have reduced power to the coil, you probably should just go with a 6v coil. That appears to be the way it was wired below SN 114623 except the resistor would have been in line between the 12v coil and the distributor (coil output - to resistor input then resistor output at 6v or less to the distributor). Without a wiring diagram it's hard to figure it all out by trial and error. Good Luck. Stan
 
staninlowerAL said:
I'm still not sure where the power to the coil is coming from. Since you said no power on the black wire until the switch is on, I suggest that you disconnect the white wire from the resistor to the coil. If you then have no power to the coil with this wire disconnected and the switch on, it says that the coil is only going to get 6v (or whatever the resistor is rated for) when connected with the white wire from the resistor. The other white wire connected at the ring terminal at the coil + is just sending power to somewhere else but at a reduced voltage. Looks like the power flow 12v is switch to resistor via the black wire, then out from the resistor at reduced voltage to coil and whatever else the white wire connects to (again at a reduced voltage). If you want to leave the resistor in place and have reduced power to the coil, you probably should just go with a 6v coil. That appears to be the way it was wired below SN 114623 except the resistor would have been in line between the 12v coil and the distributor (coil output - to resistor input then resistor output at 6v or less to the distributor). Without a wiring diagram it's hard to figure it all out by trial and error. Good Luck. Stan
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The power is coming from the ignition switch when turned on, (it's just a on off switch) it has a (push button for starting) it is the black wire that comes out of the harness that feeds power to the resistor gos through the resistor and turns into a white wire that goes to the positive side of the coil the other white wire that comes from the coil i believe goes to the R terminal on the starter selanoid.
 
My suggestion is to remove the 12v black wire from the resistor and connect it to the new 12v coil + terminal along with the wire that goes to the starter solenoid (I think it's white) so both are connected together via the ring terminal post at the coil input +. Connect the 12v coil - terminal to the distributor via the existing red wire. Then with the switch on, 12v power is supplied to the coil + terminal and to the white wire that goes to the solenoid. With a new 12v internal resistor coil you no longer need an external resistor. Based on what I see from the MESSICKS info that would be the way it was wired starting with SN 114623 and afterwards. JMHO Stan
 
Morris farm said:
. . . the other white wire that comes from the coil i believe goes to the R terminal on the starter selanoid.
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In that case, the white wire connected to the solenoid is probably there to feed a full 12 volts from the solenoid to the coil during starting. This would bypass the resistor to provide a full 12 volts to the coil and increasing coil putout while starting.
 
Jim Becker said:
In that case, the white wire connected to the solenoid is probably there to feed a full 12 volts from the solenoid to the coil during starting. This would bypass the resistor to provide a full 12 volts to the coil and increasing coil putout while starting.
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Jim, that's very interesting, and something I had thought about years ago and even tried it with a 8N wired with a toggle switch to provide a direct line to the coil from the battery (I don't remember any of the details). I'm curious if the second feed should go to the coil input or output and does the applied voltage in some way produce a greater inductive current output. Thanks for any comments and explanation. Stan
EDIT: In this thread I believe both the coils described are 12v with one having an external resistor and the other had internal resistor from the details obtained from MESSICKS website. I have not yet found a wiring diagram.
 
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