Quote:
Originally Posted by DannoXYZ
If... #2 was direct 12v source, how could he have measured 6.9v on that terminal? Even more puzzling would be how 6.9v became 12.6v on terminal #1 AFTER it went through 330-ohm resistor?
No, it's other way around. #1 terminal on resistor is input. Confirmed by him measuring +12.6v on that line. #2 terminal is output as confirmed by measuring +6.9v. This shows that resistor is working properly and dropping +12.6v down to 6.9v.
We can calculate how much current is flowing in that circuit by V=IR and plugging in voltage-drop:
12.6v-6.9v = 5.7v drop = I/330-ohms
I = 5.7/330 = 0.017 amp = 17 milli-amps
Then we can play what-if and see what output-voltage would be on resistor output terminal #2 if he was to replace that resistor with factory/OEM 100-ohm resistor:
Vdrop = 0.017a * 100ohms
Vdrop = 1.7v
Vout = 12.6v - 1.7v = 10.9v output
So voltage at output terminal would be 10.9v if he swaps in 100-ohm resistor. This is actually higher than 10v maximum test for anti-theft detection and bike may not start. Other people have had to install larger resistors to drop voltage below 10v before bike would start (EFI expert Greg737?).
This is NOT power-source line that powers Ignitor, it's purely for theft/hot-wiring detection. There's no minimum required voltage on that line, just maximum allowed. Which is roughly 10v, anything above it on grey-wire will cause Ignitor to refuse starting bike. Now we just have to figure out what's wrong with wiring that it's delivering +12v on that line to Ignitor instead of proper resistor-output of 6.9v.
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Unless there both are actually 12v pin so that would explain why the resistor end up showing 12 volt on both sides but that too end up being a bigger puzzle and if was the case all he had to do was solider one end of resistor onto pin 1 and other end of the resistor direct to the gray wire itself then maybe it will work.
I know some car their dumb detection is the same like Chevy S10 was one them