Halving the range of a variable resistor???

[mike_ie]

Hi all,

I have a problem that I hope you can help me with. The fuel level sender in my car works on the same principle as a variable resistor - when the tank is full, it registers 0Ω, when empty it registers at 360Ω, and varies in between depending on the level of the tank.

However, the replacement fuel gauge that I want to use only reads from 0Ω full to 180Ω empty.

I need to set up a resistor network that effectively halves the value of the 0-360Ω variable resistor at ANY point in its range - within reason of course. If I simply put a 360Ω resistor in parallel, the results are accurate at each end, but the error gets bigger towards the middle of the range, so I need something a little more complex.

Any help would be appreciated.

 

[badscr]

Lets get this right,
The Old Sensor: 0Ω when full and 360Ω when empty.
The New Sensor: 0Ω when full And 180Ω when empty.
so 180 goes into 360, two times, ok.


I think, you could use an op-amp with a gain of 2. then drive an mosfet ..... I need to think some more. hold off on the op-amp idea.

 

[badscr]

What year is the car?

OK, some where in the car there is a resister, that sets-up a voltage divider with the sensor. And the car reads the voltage drop across the sensor. (maybe even current ?)

So it is a matter of sending the right voltage range to the car, without blowing something up.




Anyone else have any ideas?





Note: I am in no way responsible for anything that may result from the information that I give.

 

[MikeMl]

I'd hate to become the guru of fuel senders and indicators, but read this thread, or this other thread. It will take an active circuit to linearize it.

 

[Boncuk]

I'd hate to become the guru of fuel senders and indicators, but read this thread. It will take an active circuit to linearize it.

That would make a perfect fuel indicator. Car manufacturers don't care for linearization.

All the cars (about 20 different models and brands) I drove so far indicated a full tank for the first 100km, decreasing slowly for the next 200km and decreasing rapidly for the remaining 150km.

A super fuel indicator requires two flow sensors - one for the forward fuel transport and one for the reverse fuel and a precise counter circuit. (till now available in formula-1 racing cars - one unnecessary box stop might decide about winning or loosing)

Boncuk

 

[tcmtech]

That has been a common problem I have seen on many Ford vehicles as well. The problem they have is because the sending units are fairly linear but the fuel tank is wedge shaped from about half way down. The bottom half of the tank only holds around 1/4 to 1/3 of the fuel. Many other manufactures and models have that same odd shaped fuel tank issue as well.

But my 1994 Mercury and my 1999 Ford F250 pickup have rectangular tanks and the fuel gauges have always been reasonably accurate from full to empty.

 

[MikeMl]

...
A super fuel indicator requires two flow sensors - one for the forward fuel transport and one for the reverse fuel and a precise counter circuit. ...

I have **broken link removed** in the Cessna. Since mine has a carburetor, it only requires only one fuel flow sensor. In fuel-injected engines, the same instrument uses two fuel sensors, and it subtracts the fuel returning from the injector rail to the fuel tanks from the fuel flowing toward the engine. Interesting, the instrument uses a PIC.

 

[mike_ie]

Lets get this right,
The Old Sensor: 0Ω when full and 360Ω when empty.
The New Sensor: 0Ω when full And 180Ω when empty.
so 180 goes into 360, two times, ok.


I think, you could use an op-amp with a gain of 2. then drive an mosfet ..... I need to think some more. hold off on the op-amp idea.

Not quite.

The sender in the tank goes from 360Ω to 0Ω empty to full.
The new gauge requires a sender that goes from 180Ω to 0Ω empty to full.

Therefore I need to turn the 360Ω to 0Ω sender into a 180Ω to 0Ω sender. Effectively I need to half the resistance of the sender at whatever position it happens to be at. I was hoping for a simple, solid state, robust solution that won't let me down on the road.

Hope this clarifies things.

 

[mneary]

A 360 ohm resistor in parallel with the sensor will give the correct reading at the end points. In the middle the linearity might be unacceptable but you would be the judge of that.

 

[Mike odom]

the fuel guage is probably just an ammeter. Think back to electronics 101 guys (actually high school electronics). An ammeter is placed in series so it sets the full range deflection by putting a resistor in parallel with the meter movement. This shunts (divides) the current so when the input current is at your wanted full range, wants going thru the movement is whatever the full range deflection current of the meter is. So, to double the current going through the loop (half the loop resistance) but get the same deflection current, you need to crack open the meter and replace the shunt resistor inside with half the resistance value you find inside. K.I.S.S.

oh wait, I think I have it backwards. It wants the 180, not 360 ohm, so it wants MORE current than you are feeding it so you have to double(???) the internal resistance.

 

[Mike odom]

I guess the advanced study lesson here would be how to calculate the meter movement full deflection current and then you could compute what the shunt resistor would be for that movement and full scale current (12v/360ohm).

added: with a little further study, you have to know the movement resistance also to pick the shunt resistor value. I don't think it can be calculated from the information provided. Of course, we could find the parallel resistance by measuring. But I don't think that gives us enough equations for our unknowns.

 

[Mike odom]

so knowing a fuel gauge is just an ammeter, an opamp soln presents itself (maybe, it is rather late, afterall).

Has anyone heard of a current halfer????

 

[MikeMl]

Mike IE,

here is what you can do: Power the new gauge as it would be in the car. Use a 14.3V DC power supply (alternator set voltage). Substitute a 1K Pot wired as a rheostat where the sender would normally be connected. Diddle the pot to get Full scale, remove and measure the pot resistance. Do same at 3/4, 1/2, 1/4, and just above Empty. Post the resistance at each point.

 

[EinarA]

I didn't look at Mike ML circuits, but assuming one end of the pot is grounded, an op amp can divide or multiply by almost any ratio. The circuits below show both methods. The amps must have bipolar supplies , an inconvenience in a automotive application , but small inverters a available at moderate cost. The divide circuit will need a +12V to -12V, the multiply can use a +12V to -5V. A PNP buffer at the output will also probably be needed.

 

[crutschow]

A 360Ω resistor in series with the 360Ω rheostat will give a non-linear readout such that a 50% gas gauge reading occurs when there is still 67% fuel remaining. That is a significant non-linearity but I would just live with it rather than try some complex circuit to linearize it (and I don't think there is a simple circuit).

 

[throbscottle]

I'm with Mike, but why crack open the meter? It just wants an extra shunt across it's terminals - assuming it is actually an ammeter.

 

[DerStrom8]

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.

 

[throbscottle]

Doh! How did that happen, then?

 

[tronitech]

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.

He ran out of fuel in Death Valley, drank all his water and died.

 

[DerStrom8]

Doh! How did that happen, then?

Looks like EinarA is a master in the art of necromancy--bringing threads back from the dead