# Fuel Gauge converted to Volt Meter

Discussion in 'General Electronics Discussion' started by sabre170, Aug 22, 2014.

1. ### sabre170

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Aug 22, 2014
Hi all.....New to this forum, but not new to electronics. I'm doing some motorcycle work, and as part of that, I'm swapping out my instrument cluster. My new cluster has a fuel gauge on it (digital segments), however, my tank doesn't have a fuel sensor.

I was thinking it would be cool to convert that fuel gauge to act as a battery monitoring volt meter. In other words, "Full tank"=13v, half tank = 12 v, and empty= 11v (or something to that effect.....a different range would be fine as long as it was in that general ball park)

The fuel level sensor works as a variable resistor. Full tank sends (0-10ohms). Half Tank-50ohms, Empty-100ohms. Assuming that the sensor is receiving a constant 12V, the fuel gauge receives 2.4A (used 5ohms) for full tank, .24A for half tank, and .12A for empty.

Thus, I am looking for some sort of sensor that when it gets 13v, it outputs 2.4A; 12V outputs .24A; and 11V oututs .12A. This means it would need resistance of 5.4, 50, and 91 ohms respectively........A variable resistor that is voltage sensitive.....Varistor like, but not quite.

I've racked my brain and I can't think of any single device or simple circuit that would easily do this. Am I chasing an impossible solution here or am I just not thinking of something.

5,164
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Dec 18, 2013
So the fuel sensor is connected to a 5 Ohm resistor and this is what produces the voltage level needed by the cluster?

3. ### sabre170

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Aug 22, 2014
Sorry....the fuel sensor is connected to a variable 100ohm resistor. When full tank, the resistor is near 0. (I used 5ohms for math purposes). When empty tank, the resistor is at 100 ohms.

Half tank = 50ohms

The changing resistance with a "constant" 12v results in varied current being sensed by the the fuel gauge.

I'm wanting to in turn make it so that it is voltage sensitive......if a variable resistor existed but rather than the turning knob that an individual can twist, it is regulated by voltage source.

4. ### sabre170

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Aug 22, 2014
my wording may be poor here....my apologies. I'm ultimately looking for something that will act like a 100ohm potentiometer that in stead of being "varied" by twisting the little knob, it is "varied" by sensing a DC input voltage (this would be my 11-13 volt goal range).

5. ### OLIVE2222

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Oct 2, 2011

5,164
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Dec 18, 2013
What about a battery level monitor made from op amps which switches in a different value resistor using say a MOSFET depending on the voltage level.

7. ### sabre170

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Aug 22, 2014
Sounds possibly a bit complex.....but maybe not. You are definitely on the right track though with what my goal is.

5,164
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Dec 18, 2013
Oh sorry I thought you had knowledge of electronics. To what level would you say you are? I will look at a circuit for you if you are not comfortable with this.

9. ### Gryd3

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Jun 25, 2014
Can't you just put a 550Ω in series?

Vbat --- \/\/\(TankSender) --o-- \/\/\(550Ω) --- GND

Read voltage from the mid-point. Forms a rough voltage divider.
24mA draw at Full tank.
20mA draw at Empty.

13V at Full. (Assuming 0Ω at tank sender)
11.92V at Half. (Assuming 50Ω at tank sender)
11V at Empty. (Assuming 100Ω at tank sender)

Can anyone see a problem with this?

Edit: Would drift with battery voltage...

Last edited: Aug 22, 2014
10. ### sabre170

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Aug 22, 2014
If you had sent me this info 10 years ago, I would have had no problem working this up in a second. Hell....my degree is in Electrical Engineering.......I however, haven't been in the field for 10 years and have grown quite rusty with this. Dusting off the cobwebs as I type. haha!
'

11. ### BobK

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Jan 5, 2010
You don't need to do that. A single op amp can change any voltage range into any other voltage range, linearly. That and a single resistor would do it.

Bob

5,164
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Dec 18, 2013
Thanks Bob. I thought he wanted a few steps to switch automatically in. How can one opamp do this without intervention?
Thanks

5,164
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Dec 18, 2013
Its like riding a bike you soon pick it up again.

14. ### BobK

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Jan 5, 2010

The equation for a differential amp is:

Vout = (Vin+ - Vin-) * g

So if you want to convert the range (V1 to V2) to the range (V3 to V4), put in the two endpoints to the equation above to get two equations: Because the voltage range is being inverted, we are going to feed the input voltage to Vin-. So our two equations are:

V3 = (Vin+ - V1) * g

and

V4 = (Vin+ - V2) * g

First, lets calculate the gain:

Subtract second equation from first to get:

V3 - V4 = (Vin+ - V1) * g - (Vin+ - V2)g
V3 - V4 = g*Vin+ - g*V1 - (g*Vin+ -g*V2)
V3 - V4 = g*Vin+ - g*V1 - g*Vin+ + g*V2
V3 - V4 = g*V1 - g*V2
V3 - V4 = g(V1 - V2)
(V3 - V4) / (V1 - V2) = g

Now substitute g back into either equation to get Vin+, which is the bias voltage we will put on the non-inverting input.

So, let's do an example.

Map a voltage ranging from 1 to 2 onto a voltage ranging from 5 to 3.

i.e.

V1 = 1
V2 = 2
V3 = 5
V4 = 3

So:

g = (V3 - V4) / (V1 - V2) = 2 / -1 = -2

So now we make Vin- the input voltage Vin and we make Vin+ the bias voltage Vb.

From the original equation:

Vout = (Vb - Vin) * g

We use 2 for the gain since we have reversed the two inputs.

Vout = (Vb - Vin) * 2

Substitute in the V1 for Vin and V3 for Vout:

5 = (Vb - 1) * 2
5 = 2 * Vb - 2
7 = 2 * Vb
3.5 = Vb

To check our work, let's also substitute the other end of the range. I.e. make V2 Vin and V4 Vout:

3 = (Vb - 2) * 2
3 = 2 * Vb - 4
7 = 2 * Vb
3.5 = Vb

Eureka!

So we make a differential amplifier with a gain of 2. Put the input voltage to the inverting input, and a bias voltage of 3.5V on the non-inverting input.

Let's see how that works out in LTSPICE. Input is blue, going from 0 to 1V, output is green, going from 5 to 3V.

Now, the biasing for the + input is just a voltage divider that is going to put 3.5V * 2 / 3 on the + input. So we can actually do this with a voltage divider from V+ to ground, as well. We need:

3.5V * 2 / 3 = 2.33V.

So We need a divider with a ratio of 2.33 / 10 = 0.233

So R2 / (R1 + R2) = 0.233

Set R2 to 10K

10K / (R1 + 10K) = 0.233
10K = 0.233 R1 + 2.3K
7.7K = 0.233 R1
7.7K / 0.233 = R1 = 33K

Sure enough:

Bob

Last edited: Aug 23, 2014
Arouse1973 likes this.

5,164
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Dec 18, 2013
Nice work Bob. Quite a bit to digest I'll study this in more detail later.

16. ### sabre170

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Aug 22, 2014
I agree.......mind blown!

5,164
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Dec 18, 2013
Just thinking about this. The new cluster has an input for the fuel gage right? I know you did the tests with a 5 Ohm resistor but why did you think such a low value was needed. I struggle to understand this as in normal operation you would have a considerable current. Which would be wasting power. Is the input resistance to the cluster really that low.

Can you try it with a higher resistor value? Also can you measure the resistance from the input on the cluster to both + and - supply wires. Makes sure it's powered off first. I am going to see if I can find more info on this. Do you have a part number for the new cluster?

Thanks

18. ### sabre170

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Aug 22, 2014
Below is from the people who make my cluster (Acewell model number 2853). NOTE: as mentioned, I do NOT have a sensor/sender in my tank. The lack of sensor in tank is why I'm looking to re-purpose that part of my gauge cluster. The motorcycle is 12V.

The Acewell meters that include a fuel gauge have LED 7 bars. The output of the sender determines how many bars are displayed, as follows:

Fuel Gauge display
Bars displayed..... Sener Resistance (ohms)
7..... 0-10Ω
6..... 11-20Ω
5..... 21-35Ω
4..... 36-45Ω
3..... 46-60Ω
2..... 61-75Ω
1..... 76-90Ω
1 flashing..... 91-100Ω

19. ### BobK

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Jan 5, 2010

I tried to do this right after you asked, and I kept getting the wrong answer. I woke up at 3 AM thinking about it, so I had to get up and work it out. Turns out I was flipping a sign before and once I fixed this, the example worked.

Bob