>>> I need to make a fuel tank sender and it's gauge compatible. The gauge is >>> expecting to see a sender (basically a variable resistor) that has a value >>> 0 to 90 ohms. The current sender in the tank has a value of 0 to 30 ohms. >>> The original gauge, changed long ago is NLA and while an appropriate sender >>> is available, changing it is going to be an issue.
>> The easiest approach is to make the existing gauge look like the gauge the >> sender expects. To do this, parallel it with a resistor. In effect, you >> want a resistor that's half the resistance of the existing gauge, so it in >> parallel with the existing gauge, will be 1/3 the resistance. Then the >> (approximately 3x) current from the sender will be divided unevenly, with >> about 1/3 the current (equivalent to a sender with 3x the resistance) going >> through the gauge. Making part of this resistance adjustable will give you >> the adjustment you're after. Note that this is the approach I would use if >> it were mine. Note that there might be some odd effects on linearity, but >> fuel gauges aren't famous for linearity anyway (I had a VW where full to >> half barely moved the pointer, but it had lots of resolution near empty, >> where it mattered more). > In your first solution, I think you have your circuit backwards. If I read > the issue correctly, the new sender has a resistance of 0 - 30 ohms and what > the gauge expects is 0 - 90 ohms. I will be honest though, I don't have a > clue of how a gas gauge and sender work. I could be wrong, but I don't think I am in this case. The way I look at it is that the signal is a current, driven through the meter and the sender from the vehicle's battery. The meter is basically an ammeter - the amount of current flowing through it determines the position of the pointer (many gauges do this by running the current through a resistance element that acts as a heater wrapped around a bimetallic strip attached to the pointer). So what the sender is doing is adjusting the amount of current flowing through itself and the meter, and the meter is measuring this current. The current is the vehicle voltage divided by the sum of the meter resistance and the (varying) sender resistance. The maximum current flows when the sender is at minimum resistance, and the minimum current flows when the sender is at maximum resistance. The old gauge (for the 90Ω sender is expecting a larger range of current, as its minimum is less), so you need to find a way to have less current flow through it somehow - the 30Ω sender has less maximum resistance, and therefore a smaller range of currents (its minimum current is larger). You just need to translate this larger minimum current into the smaller minimum current your meter expects, or you'll only get part of the range. My approach was to divert the "extra" current around the meter, so the meter would see the larger range of currents that it expects. It's not a cut and dried solution, however, for three reasons: 1) the meter resistances could be different, causing a different range of currents to flow. 2) there are no rules on whether maximum current means "full" or "empty", so your meter could read backwards. 3) since the current/resistance relation is reciprocal, it's non-linear, and some senders and some meters are made so the change in resistance or reaction to current is also nonlinear (many senders use a variable pitch resistance winding to accomplish this). As you can see, there are rather a lot of unknowns (in particular, the resistance of the meter for the 90Ω sender and the resistance of the meter for the 30Ω sender). - John -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send an email to [email protected]. To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/BA88128F-7103-468A-AD39-E2947DFB361D%40mac.com. For more options, visit https://groups.google.com/groups/opt_out.
