How can I make 20-200 ohm variable resistor look like a 20-240 ohm resistor?

Question

I have a fuel sender in my kit car that goes from 20R to earth (full) to 200R (empty). Unfortunately the gauge (constant current source, measuring resistance to earth?) expects empty to be 240 ohms, so displays '1/4 full' when empty. What is the simplest analogue way to make 20-200R look like 20-240R?

(40R in series gives a full tank showing '3/4 full' on the dial: Still not ideal!).

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So, a great suggestion from Transistor (below). I think I'll change R4 in his/her Figure 1 to a constant current source since car volts can vary quite a lot!

I believe the value for R3 (above) will be fairly independent of the transistor specs?

Update: Yes, this works nicely, gauge now goes from full to zero. I put R3 on a little trimmer - 720R for 1.6mA. Many thanks transistor!

Answer 1

I have a fuel sender in my kit car that goes from 20R to earth (full) to 200R (empty). Unfortunately the gauge (constant current source, measuring resistance to earth?) expects empty to be 240 ohms, so displays '1/4 full' when empty.

Status   Spec.   Actual   Reading  Voltage
------------------------------------------
Full       20 Ω    20 Ω   100%      160 mV
Empty     240 Ω   200 Ω    25%     1600 mV
                  Required voltage 1920 mV

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. (a) What you've got. (b) The voltage reading can be boosted to the correct level by feeding an extra 1.6 mA through the sensor. (c) A simple current source based on 12 V supply and 1920 mV for empty.

What is the simplest analogue way to make 20-200R look like 20-240R?

R4 looks simple to me.

I suspect that the 8 mA source is constant current so that the gauge doesn't fluctuate with revving of the engine (other than that gradual downward trend as you burn up the earth's carbon fuel reserves). To avoid R4 introducing variation you might want to feed it from a stable voltage source - as high as you reasonably can - and recalculate for your new voltage.

At full the extra 1.6 mA will increase the voltage through the 20 Ω by 32 mV. On a span of (1920 - 160 =) 1760 mV this represents an error of 36 / 1760 = 2% so with a full tank it will read about 98%. This should be acceptable.

Answer 2

I think the easiest way would be just getting a new variable resistor. You could use a transistor to switch in an extra resistance but potentiometers are pretty cheap, so swapping out would be the easiest option.

Answer 3

Per your comment "It's a digital gauge that runs about 8mA into the resistor in all positions", you don't actually need to make it "look like a resistor" from the gauge's point of view - you can just provide the gauge with a low-impedance voltage source, where that voltage is a linear function of the sensor resistance. A fairly straightforward op amp circuit should do the trick. You'll want to output a voltage that's 20*.008=0.16 V at one end of the scale and 240*.008=1.920 V at the other end of the scale.

Answer 4

Here's a resistance multiplier with a 240/200 ratio. No need for reference voltages, the ratio is set with two resistors. The negative resistor end can float up to several volts above ground. The multiplier works across a wide range of currents, from a fraction of mA - where it gets quite slow - to about as much as M1 and R1 can dissipate.

^{simulate this circuit – Schematic created using CircuitLab}

The parts can be jellybean: LM324, 1N4148, BSS138. 2N7000 might work too. C10 can be potentially increased for better stability.

The circuit is fairly accurate. Accuracy is mainly determined by resister tolerances. Op-amp's input offset adds a +/-3 ohm resistance offset worst case.

The transfer curve, from R1's resistance to the resistance seen at the output, is as shown below, for -10mV, 0, and +10mV op amp offset (worst case limits).

Answer 5

There's no easy way to do it, anyway. You could add a 20ohm fixed in series with a 20-220 ohm sensor, to give 40-240. Not quite full, but empty would be empty.

If the sender truly is 20-200, something is wrong for the gauge to expect 240. Bend the float arm up?

Are you sure true fuel "empty" is the full travel of the sensor? The simple solution is to add fixed resistance to achieve true empty, and just recognize that 90% represents full.

It's doubtful the gauge will respond in a perfectly linear fashion anyway, unless your gas tank is a perfect square or vertical cylinder.