# Converting fuel level sender reading

I have a fuel level sender installed in my custom built car which supposedly should read ~50 Ohms when empty, and 2 Ohms when full. I've measured it 30 Ohms empty, but this could be because the arm/float is resting on the bottom at half-engagement.

I want to read this sender with a 0-5V input and from what I've gathered so far I need some kind of voltage divider for this to work. I have access to 5V and 12V.

My EE skills are nowhere what they used to be, can anyone help me out with this?

• Is either end of the sensor grounded or connected to +12 V? May 8, 2020 at 17:48
• The only thing connected to the sensor now is two wires into it. I can connect it however it needs to be connected for this to work. May 8, 2020 at 17:51

simulate this circuit – Schematic created using CircuitLab

Figure 1. A constant current source of 166 mA will give an output of 0.33 V at 2 Ω and 5 V at 30 Ω.

A constant current source prevents variations in fuel gauge readings at varying engine revolutions. A typical constant current source is shown in Figure 1. How it works:

• R3 draws base current from Q2 turning it on.
• As the current through Q2 rises the voltage drop across R2 increases and turn on Q1. This happens at about 0.75 V.
• Q1 then turns on pulling the base of Q2 towards +12 V and preventing Q2 turning on any further. The circuit settles down at $$\ I = \frac {0.7} {R_2} = \frac {0.75} {4.5} = 166 \ \text {mA} \$$.

There are a few things to watch out for:

• Q2 will have up to 13 V across it when the engine is revving high. $$\ P_{Q2} = V_{Q2}I_{Q2} = 13 \times 0.166 = 2.1 \ \text W \$$ so it will require heatsinking.
• Beware that as drawn the voltage to your 0 - 5 V ADC will get 14 V if the sensor is ever disconnected. You would want to add some protection there to prevent frying the ADC. Maybe a 10 kΩ resistor in series and a 5.1 V Zener diode to ground.
• I have no idea what current the fuel sender can handle. The power dissipated in it will be VI = 5 × 0.166 = 0.83 W.

Here is a simple circuit that does what you request. D1, C1, C2, U1 form a 5V regulated supply, capable of withstanding +/-35V transients on the input (not really professional automotive grade but better than nothing). It takes the 10-14V electrical automotive power and converts it to a regulated 5.0V. Maximum power is about 400mW in U1.

R2 and Rv (the sensor) form a voltage divider and OA1 amplifies the 0-1V to 0-5V. Keep the wire length short from OA1 or add a 1K-10K series resistor.

Output voltage is approximately (Rv/30$$\\Omega\$$)*5.0V, more exactly it's nominally

25.1*(Rv/(120+Rv)) so at mid-scale (15 ohms) you get 2.78V rather than the ideal 2.50V, plenty good enough for a gas gauge but you can correct it digitally if you like.

Note that if the high side of Rv gets shorted to +12 it will back feed through R2 and destroy OA1. It might damage Rv as well if the other side is grounded.

simulate this circuit – Schematic created using CircuitLab

ADC cannot read resistance which is varying with fuel level. Hence you need to convert it to voltage. The voltage divider is with a known resistor and the fuel rod in series does this trick.

Please refer to the attached pics. The fuel rod voltage drop varies with the fuel level, hence you will be able to sense the fuel level.

Note: My fuel rod varied from 33 - 240 ohms. Hence the circuit is for that.

• Thank you! How do i best calculate R? May 8, 2020 at 17:48
• Technically, R could be of any value as long as you know it's value. However, Higher the R-value, more voltage across it, so the range of voltage you read from the fuel sensor will be limited. Lower the R-value, less voltage across it, so the range of voltage you read from the fuel sensor will be wide. Pick a value which makes sure that you get a voltage of 5V when the tank is empty, 0V when the tank is full. I'm guessing this is what you want. May 8, 2020 at 17:53