Resistance multiplier?

[jadnhm]

Hi all,

I'm not sure if I'm using the correct terminology here, but I think what I need is a resistance multiplier.

I have a situation with my vehicle that I think I can remedy with a (relatively) simple circuit, but I am not knowledgeable enough to work out how to do it.

My car actually has two fuel level sending units, one in each side of a saddle style tank.

Each sending unit is essentially a potentiometer with a max value of about 200 ohms and a min value of very close to 0 ohms. More ohms == less fuel in the tank, and 0 ohms means the tank is full.

They are wired in series.

sending_units.PNG (6.5 KiB) Viewed 1513 times

There is a short in one of the units and it has no resistance, so the computer thinks this side of the tank is completely full. The other side works fine, and that is the side that has the electric fuel pump so that's the side that really matters anyway.

As a consequence, the fuel gauge reads about half empty when it is actually empty!! This is ok for me for now, and I can watch the mileage on the odometer to know when it needs fuel too, but I would really like to fix this up, esp if it's not too tough to do. The sending units are a lot of work to replace and very expensive, and junk yards don't seem to keep the one that I need, keeping only the 'electric pump' side.

What I'd really like to do is have a circuit of some kind that effectively doubles the resistance of the one sending unit that does work so that completely full is still 0 ohms, but completely empty would be 400 ohms rather than 200 ohms. This way the fuel gauge would read completely full or completely empty based on only the one sending unit. There is an ejector pump that pumps fuel from the other side of the saddle tank, so the levels should be relatively close anyway, and also, as I said before, the electric pump is the one that needs the fuel anyway so in the end I really only care about what's on the side with the working sending unit.

Anyway, I'm sure this is possible, but I'm not sure how to do it, so I put it to the board.

Any ideas?

[pebe]

It's difficult to suggest anything without knowing a bit more about your setup. You say the potentiometer reads between and 200ohms - but that would be using it as a rheostat. It your car uses a micro then it will probably measure voltage using a D/A converter.

One way it might do that would be if the pot has a fixed voltage across it and the moving wiper picks off a voltage that the micro measures.

Another way would be if current is passed through the pot (between one end of the pot and its wiper) and a load resistor.. The varying resistance of the pot would then vary the current which would produce a varying voltage across the load resistor. The micro would then measure that.

Is it possible to explore a bit more to see if either of these methods is used?

[jadnhm]

Thank you very much for the reply pebe. I will definitely explore some more and see what I come up with. Any advice on how to approach the problem is appreciated of course.

I'm pretty sure I remember measuring 8V on the supply side of the circuit. Since the 'base' voltage level for the car is of course 12V (or thereabouts depending on the charge level of the battery) I'm going to venture a guess that the supply voltage of 8V is important.

Also to be clear I don't /think/ this is going through the engine's fuel/ignition computer, but rather being processed by the dash module, which I would guess is something more along the lines of an analog meter ( Ammeter? Voltmeter? ) than any kind of digital circuit.

I hope that extra info helps. I'll report back when I get a chance to look in to it a bit more.

Thanks again.

[I_Daniel]

If there is an arm with a float attached to the resistor then it is usually a wirewound variable resistor. If it measures zero then there is a break or the wiper is no longer making contact. Of course the float could also be stuck. There are only two level indicators I know of and that is the variable resistor with a float the other one works with fibre optics.

[jadnhm]

Thanks for the info I_Daniel.

Yes I have seen this fuel assembly out of the car and it is a float on a wire arm which moves a wiper on a fixed resistor to vary the resistance. I believe this is called a rheostat, as pebe correctly pointed out.

It measures 0 ohms and I'm pretty sure there is a short inside the assembly in the tank. I'm fine with that - I'm trying to make my gauge read normal numbers from only 1 sending unit. As I said in the first post, the two units are wired in series, so if I could make a circuit that would 'double' the resistance of the first sending unit it would work properly.

Here are some additional details.

I think the circuit looks something like this:

sending_units_2.PNG (8.79 KiB) Viewed 1480 times

The gauge is actually measuring the voltage drop over the load on the tank sending units (rheostats).

When the resistance of the sending units is 0 ohms, the tank is full and the voltage seen by the gauge will be very close to 0V (ie: the voltage drop over the tank sending unit rheostat is approx 0V).

When the resistance of the sending units is 200 + 200 ohms the tank is empty and the voltage seen by the gauge will be very close to the supply voltage (which I think in this case is approx 8V) - I haven't had time to pull everything apart and measure again.

What I'm aiming for is the below:

tank 2 has been removed from the circuit

tank 2 has been removed from the circuit

sending_units_3.PNG (8.92 KiB) Viewed 1480 times

As you would expect, the gauge is reading approx half of what it should - ie: the gauge shows approx 2x (twice/double) what it should. Empty is around 1/2 tank on the gauge.

I need a circuit that will give a voltage drop of approx the supply voltage (of 8V).

Does that make sense?

[pebe]

Hi jadnhm,

I think I have got the solution. (I have recently changed over from XP to Win7 and haven't yet mastered the new version of the 'Paint' program that I draw my circuits with. So I hope a description will be enough).

Originally, the gauge was supplied with current via a pot comprising a 100ohm resistor and two 200ohm resistors. According to Thevenin, that circuit can be simplified to an equivalent circuit of a single resistor (100ohms and 400ohms in parallel) fed from the open circuit voltage (without the load of the gauge) at the junction of the two resistors.

With 100ohms and 400ohms across 8v, that open circuit voltage is 6.4V, and the two resistors in parallel equal 80ohms. So if we can get those same two values when using a single 200ohm resistor, the gauge will read as normal.

To get the 6.4V we need to reduce the 100ohms to 50ohms. But 50ohms in parallel with 200ohms will then be 40ohms instead of 80ohms, so we need to add an additional 40ohms.

So, finally, you need to change the 100ohm resistor for a 50ohm one (or parallel another 100ohm across the existing 100ohms), and add an extra 40ohm resistor in series with the gauge.

Hopefully, that will work OK.

EDIT: 19/7 @13.25
This modification will result in some increase in current through all components.
If you can measure the resistance of the gauge, then there may be an alternative way of doing it.

[I_Daniel]

If you have taken the sender out of the tank then connect an Ohm meter to the slide and the one end where the wire to the receiver/read-out is connected. There should be a resistance reading. Slowly move the float up and check if the slide moves up or down and what resistance you read as it moves.
Usually when the float goes up the slide moves to the bottom and thus the resistance becomes higher unless of course it is the end where the output is connected then it becomes less. Zero resistance could indicate a short which I cannot see for the usual fault is that either the slider is not making contact or one of the wires has a break in it.

If the two tanks are actually interconnected at the bottom with a pipe etc. then Pebe's suggestion should work because the level in both tanks would automatically equalize.

By the way the difference between a rheostat and a potentiometer is in the use of a variable resistor. A rheostat has only the wiper and one end of the variable resistor connected while the potentiometer has both ends and the wiper connected.
The rheostat therefore varies the resistance while the potentiometer acts as a voltage divider changing the voltage between almost zero and maximum - as in an audio volume control. Pebe is using the correct terminology while I am also using it correctly when I say a variable resistor