Halving the range of a variable resistor???

[MikeMl]

Just thought I'd point out to you guys that this thread is over 4 years old, and the OP has been inactive for over 3 years.

Who cares. I am very interested in this topic. Here is a sim I did of the circuits Einar posted. I hadn't encountered these before.

Note that they transform the effective resistance of R3 as claimed.

This will not of itself help to transform the resistance of a gas-gauge sender because these two circuits require that neither end of R3 is grounded. Note the voltage plots V(b) and V(c) for the two circuits and the Plots of V(b)/I(I1) = to input resistance at Node b. The example Divide circuit cuts R3 in half while the Multiply circuit doubles R3.

 

[EinarA]

Derstrom8: I had a couple of reasons for reviving this dead thread: I didn't think the answers were very good and it was a chance to experiment with posting a schematic. The two circuits I show are linear, affected only by the offset of the op- amp at zero ohms.

 

[EinarA]

A thank you to mike for posting a sim; I bread boarded the divide circuit. You are right I forgot that you would have to rewire the cars circuit a bit to use them.
At first I could not think how to do this and looked in an op- amp book for ideas and was a bit surprised to find nothing. A few failed ideas later and I saw a solution .

 

[alec_t]

Just in case the OP is still alive and interested, here's how he could effectively halve the resistance of a gas gauge sender with one end grounded (as most are), using a single-supply opamp:

 

[EinarA]

Alec: I tried that idea and found it did not work. It doubles the current at zero ohms which you do not want.
( After more thought I feel Alec's circuit should work; it may have some stability issues.)

 

[DerStrom8]

Don't get me wrong guys, You didn't do anything wrong by reviving the thread--I don't believe there are any rules against it. I just wanted to point it out in case you hadn't noticed, so that you wouldn't hope for an answer from the OP.

 

[JimB]

Old threads are revived for various reasons, not many of them are useful.
Often the revivial just adds some fairly trivial comment.

However, in this case EinarA has added very useful information which is being investigated by people who are interested.
I say well done EinarA.

JimB

 

[crutschow]

Darn, snagged again.

 

[MikeMl]

Einar's and Alec's post got me thinking about this (again). What came out of it is not an end-all, do-all, but might add some insight...

The attached sim shows a way of thinking about the problem. Older cars and airplanes use a fuel gauge system like shown in the "original circuit" below. There is a voltage regulator that creates a fixed voltage lower than the car's system voltage (12 to 14.5V). I'm not sure exactly what the voltage is, but I used 8V for the purpose of this discussion. The gauge is just an ammeter, which has an internal resistance R2 Again, I don't know exactly what the resistance of the meter is, but am using 50Ω. For the sender, I am using the OP's 0Ω to 360Ω.

For simulation, I step the resistance of the original sender from near 0 to 360Ω in steps of 1Ω, and I plot the current through the meter I(R2). The independent variable in the simulation is the resistance of R1, and the plots show the resistance of R1 along the X-axis.

Note the current through the gauge (meter) I(R2) is somewhat non-linear, but no matter, it is what it is... The meter face on the gauge is marked according to a calibration such that E, 1/8, 1/4, 3/8, ... F make sense with the geometry of the sender wiper position and the shape of the tank. If we replace the sender with one that has half the resistance of the original at every wiper position, the goal is to have the same current flow through the meter as it would have had with the original sender.

So here (thanks to Alec) is a way of doing that. In "the prototype" circuit, I use a current mirror to duplicate the current flow through the meter, and add that current through the new half-resistance sender. Intuitively, that doubles the current through the half-resistance sender, meaning that V(N) is identical to V(S). The current mirror is an LTSpice voltage-controlled current-source with a transconductance of 1/50 mhos, where the 50 comes from the meter resistance (R3,R4). Note that the plot of I(R4) [the meter in the Prototype] falls perfectly on top of I(R2), showing that they are identical.
For interest sake, I plot the expression V(n)/I(R3) just to show that R3 is indeed (R1)/2. Also note that I plot V(s)-V(n) to prove they are identical.

The "Implementation" circuit shows how to implement the current mirror using an op-amp, a resistor R7 equal to the meter resistance R6, and an NPN used as a current booster. Again, the goal is to force V(M) to be the same as V(S), thereby making I(R6) identical to I(R2) and I(R4). This circuit needs to be checked for dynamic stability before someone tries to build it, but the DC transfer curve does the right thing.

Note that to use this approach, you would have to know the internal resistance of the meter movement, and you assume that the geometry (arm length, total arm angle of travel, resistance linearity) of the new half-resistance sender is similar to the original. You would also have to bring a wire out of the instrument regulator inside the instrument cluster. If the total resistance change of the new sender is not exactly half, with the appropriate scaling, this approach could still work.

 

[EinarA]

Hi Mike. Good write up. Looking at the final circuit, since the op amp holds the upper end of R7 at Vreg it's no different than wiring it directly across the meter. Using a different value of R7 would create different ratios. This looks like the simplest way to correct for a half R sender but the original question was to go the other way, correct a 2R sender. That seems much harder but your ideas might point someone in the right direction.
I wonder what would happen if you increased the reference voltage, wouldn't that make the sender resistance appear to be half.

 

[alec_t]

I'm sure the OP has lost interest, but this interests me . It's a subject which pops up from time to time.
Googling 'gas gauge schematic' throws up several images which suggest the traditional gauge is not simply a conventional ammeter. It's a dual-coil gizmo which may or may not have an additional balance coil and/or series/parallel resistor. Whatever the actual topology, the gauge presents internal resistance (100Ω in my sim) which is in series with the sender and which determines/limits the current when the sender is at zero resistance. I don't see why, therefore, my proposed method "doubles the current at zero ohms" as EinarA says (but perhaps there are some topologies where that could happen?).
My proposed method was intended to add a parallel current path to replicate whatever current the 360Ω sender passes such that, regardless of the gauge internal resistance, whatever voltage is at the sender terminal effectively sees a 180Ω sender. There are stability issues; hence the presence of R3,R4,C1,C2 acting as low-pass filters to suppress oscillation. This filtering may need tweaking in a practical circuit.

 

[throbscottle]

I just re-read the OP. He was replacing the gauge, not the sender. Or am I being extra dyslexic today?

 

[alec_t]

He was replacing the gauge, not the sender.

That's right. The new gauge is expecting to see a 180Ω sender, but he wanted to use the old 360Ω sender.

 

[crutschow]

It is an interesting subject. The suggestions about mirroring the current led to me to just try a simple transistor current mirror. The simulations are shown below compared to a 180Ω sender.

In the first sim the mirror circuit has a slightly higher resistance over the sender range. In the second sim I added a 0.25Ω resistor in series with the emitter of Q3 to bring it closer. In practice the difference of the first circuit from the ideal is likely acceptable.


.

 

[alec_t]

That's a neat solution, crutschow. Considering how inaccurate gas gauges are, either circuit would surely do the job.

Edit: That emitter resistor can even be tweaked to compensate for mis-match of the two transistors.

 

[crutschow]

Note that my simulation is, by default, for identical matched transistors which obviously would not be the case for real transistors. It also doesn't include the effects of any temperature difference due to differences in power dissipation (which could be the more significant cause of current mismatch). To minimize temperature differences and matching problems a dual PNP transistor could (should) be used.

 

[MikeMl]

I also misinterpreted the OP's request. Here is something that I think would solve his problem:

I show four circuits. As shown in the plot, the current through the gauge is the same in each case [I(R2)=I(R5)=I(R8)=I(R9)]. The first circuit would be the case if the gauge is used with the sender it is designed for. The other three circuits are checked against this one.

The next three circuits show a progression toward the final circuit. The goal is create a current-controlled current-source which sinks a current equal to the half-current flowing through the double-resistance sender. That way the current through the gauge is correct even though the sender has double the resistance it should.

To sample the current through the sender, I insert a 0.1Ω shunt resistor in series with the sender to sense the sender current, and convert that to a voltage across R3 using a ZXCT1009 high-side current monitor. The opamp is configured as a voltage-to-current converter, which mirrors (and doubles) the sender current, meaning that the gauge current is double what it would otherwise be.

The plot of V(w)-V(z) shows the error between the voltages at those two nodes. The source of the error is the voltage drop across the sense resistor R10.

 

[crutschow]

Having fun here. Below are three circuits. The first is with the desired 180Ω sender, the second is with a dual-pnp current-mirror and the third is with a rail-rail I/O op amp current-mirror, a simplified variation of Mike's circuit, both with 360Ω senders. The op amp circuit equivalent resistance is basically identical to the desired circuit. The current-mirror has a slightly higher resistance but should be more than adequate for use as a gas gauge.

Edit: The op amp circuit has the advantage of easy adjustment of the sender resistor translation ratio by selecting the ratio of R4 to R5. For example, changing R5 to 2Ω gives an equivalent sender resistor of 240Ω, and letting R5 = 0.5Ω gives an equivalent sender resistance of 120Ω.

 

[EinarA]

Alec: I see now I was too hasty in saying your circuit wouldn't work, it was my version that was done incorrectly, causing me to dismiss that method in error.