> Hi, Ok I tired your method and it worked. I ended up needing around 24 ohms. 
> The only concern I have is that the resistor runs a bit warm and it's a 20 
> watt. (Actually two 50 ohm, 10 watt resistors in parallel) I measured the 
> voltage drop across them and got 9.6 volts. (taken with the sending unit 
> grounded to simulate an empty tank.)

Does that mean they're dragging the battery down to 9.6 volts, or is there 
something else in the circuit that could account for the remaining 2.4 volts or 
so?  That is a scary amount of power dissipation - automotive stuff is famously 
low-impedance and power hungry (a blower moter can often eat 30 amps), but that 
seems a bit extreme for a dash gauge.  Maybe measure your sender to see if it's 
really 0-30 ohms or perhaps it doesn't go all the way down to zero?

> I should have measured the resistance of the gauge...

I'm guessing it's about twice your resistance, which works out to 48 ohms or so.

>    I'm concerned about the damaging the sending unit.. should I be?

It could be the lower-resistance sending unit you have is intended for use with 
a higher-resistange gauge.  If so, it'll be carrying more current than it's 
designed for.  Unfortunately, without knowing the gauge resistance it's 
supposed to work with, it's hard to tell.  If you want to be really safe, 
you'll have to go for a more complicated solution that measures the sender 
resistance and then drives the gauge directly.  Alternatively, you could power 
the gauge with a lower voltage, which would also lower the current and allow it 
to work without the parallel resistors.  This would have an unknown effect on 
the linearity, and there's still an outside chance the sender would want even 
lower current.

- John

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