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Originally Posted by mt_goat
I found it interesting that there is already a resistor soldered between the same contacts on the front side of the tach. So now, if I did this tach fix right, there are two resistors in parallel. Any EEs out there know how that works? I'm thinking that the resistor with the lowest resistance will take more of the current flow.  BTW I've heard that this same fix works on the 4 cyl to 3.4 swap too. |
You know I'm an ME like you, but I do remember this from my Physics and Circiuts classes.
You have the same voltage drop across both resistors (since they go from/to the same pins) so yes, less resistance = more current, but you also have to remember that by putting in two resistors, you're making two paths for the current to flow through (less total resistance). So with the two resistors in || you have less resistance, less voltage drop, higher volts output.
Forgive my ignorance on the tach fix, what is the problem that you're trying to get rid of?
I know that some of the guages on our aircraft simulators are linear (-10V to 10V one one input = full range of needle) and some use a sine/cosine needle driver (the sine and cosine of the physical angle we want the needle to point to are sent to the guage, where they are manipulated w/ firmware).
For the linear guage, a different voltage (higher) would result in a different reading (higher).
For the sine/cosine guage, when we drive the needle with less voltage it goes to roughly the same place but slower (dampened). If you're trying to hold the needle horizontal, a low voltage signal may be too weak to support the weight of the needle (which is pretty small, btw!)