Topic: Ignition question

[Lost and Confused]

When I bought my "74 750k, it came with a aftermarket ignition. I have read in our forum that most aftermarket ignitions are rated lower in amp power, and can melt their wires. Can any of our members recommend a good alternative? Our do I try to find a stock one? I do have new harness, rectifier and regulator from Oregon parts, Pamco ignition.



Thanks, Bob

[bjbuchanan]

If it is the cheaper emgo switch, which is the cheap replacement one then yes slowly over time the contacts will diminish. It is the same idea of what low ohm coils do to the killswitch

You can get plenty of time out of it, it isn't gonna die immediately but an oem switch would be better($$). You would be better off keeping what you have and installing a relay. Hondaman sells the kits for like 18$ which is much cheaper than the keyswitch and keeps everything in place. The kit comes with small jumper wires so now cutting crimping splicing

[pamcopete]

The stock points coil has a primary resistance of 4.5 Ohms so when the coil is on, the current would be 14/4.5 = 3.1 Amps, but the coil is only on for slightly more than half the time with a 190 degree dwell angle, so the average current is 190/360 = .53 X 3.1 = 1.64 Amps. The average current is what would affect the contacts of the kill switch, not the peak current.

The 17-6903 "Ultimate" coil has a resistance of 2.5 Ohms, so its peak current is 12.8/2.5 = 5.12 Amps but the PAMCO rotor has a 120 degree dwell angle so the coil is only on for 120/360 = .33 of the time so the average current is .33 X 5.12 = 1.69 Amps

However, in both cases, there are two coils powered by the same kill switch, so the average current that the kill switch sees is 1.64 X 2 = 3.28 Amps for points and 1.69 X 2 = 3.38 Amps for the PAMCO.  That's only .1 Amps more than points.

I use 14 Volts for the points because the kill switch typically drops about .5 Volt. The transistor in the PAMCO, as in all electronic ignitions, drops about 1.2 Volts so I use 14.5 - 1.2 - .5 = 12.8 Volts for the PAMCO.

[bjbuchanan]

The stock points coil has a primary resistance of 4.5 Ohms so when the coil is on, the current would be 14/4.5 = 3.1 Amps, but the coil is only on for slightly more than half the time with a 190 degree dwell angle, so the average current is 190/360 = .53 X 3.1 = 1.64 Amps. The average current is what would affect the contacts of the kill switch, not the peak current.

The 17-6903 "Ultimate" coil has a resistance of 2.5 Ohms, so its peak current is 12.8/2.5 = 5.12 Amps but the PAMCO rotor has a 120 degree dwell angle so the coil is only on for 120/360 = .33 of the time so the average current is .33 X 5.12 = 1.69 Amps

However, in both cases, there are two coils powered by the same kill switch, so the average current that the kill switch sees is 1.64 X 2 = 3.28 Amps for points and 1.69 X 2 = 3.38 Amps for the PAMCO.  That's only .1 Amps more than points.

I use 14 Volts for the points because the kill switch typically drops about .5 Volt. The transistor in the PAMCO, as in all electronic ignitions, drops about 1.2 Volts so I use 14.5 - 1.2 - .5 = 12.8 Volts for the PAMCO.

Your ignition is a great design and the input is nice but this is an ignition switch thing pamco. I was just trying to give something he could maybe relate to

[pamcopete]

If it is the cheaper emgo switch, which is the cheap replacement one then yes slowly over time the contacts will diminish. It is the same idea of what low ohm coils do to the killswitch

I was referring to this part of your post. The 17-6903 "Ultimate" coil is 2.5 Ohms vs the 4.5 - 5.0 Ohm stock coil, but the average current through the kill switch with the lower Ohm 17-6903 coil is only .1 Amp more when using the PAMCO due to the drop across the transistor and the 120 degree dwell angle.

[bjbuchanan]

I get what you mean I was just trying to give something to relate to that has probably been seen commonly on these bikes