Originally Posted by Scopeman
Hi mugaliens, this meter reads empty 0 ohms and full at 128k ohms using 12v the sensor in my tank is around 17 ohms full and about 80 ohms empty with 12v.
just trying to keep this as simple as possible! your idea sounds good, do you have a diagram I can look at?
Thanks, ☺
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To summarize:
Tank
- Full: 17 ohms
- Empty: 80 ohms
B-17 Gauge:
- Full: 128,000 ohms
- Empty: 0 ohms
This required a bit of research.
Background:
Automobile fuel gauge sensors are floats connected to potentiometers. As the fuel level in the tank drops, the float drops, and the potentiometer increases in resistance.
This checks with your Jeep fuel tank readings, but does not check with your B-17 gauge readings.
More Background:
Fuel sensors in large airplances use low-voltage capacitors, arranged vertically in the tank, and connected in parallel. In fact, there are 30 such capacitors in the Airbus A320. The fuel acts as a dielectric, so the circuit capacitance is proportional to the height of the fuel.
Since fuel gauges are important, it's important to design them to be "fail-safe." If the gauge or sensor is broken, the gauge will read zero.
In the case of your car, if the circuit is broken (infinate resistance) the gauge will read empty.
Regardless, I think it's a mismatch between your B-17 gauge, which works on capacitance (I think), and your Jeep's fuel tank sensor, which works on resistance. If this is the case, a simple resistor bridge will not work.
The reason I said "I think" is because I couldn't find any information to determine how the B-17's fuel tank sensor works. Fuel is a slight conductor, so for all I know, it could simply have two vertical wires. The deeper the fuel, the greater the conductances (less resistance). However, that doesn't match your gauge, which is reading backwords.
Therefore, I suspect it uses capacitance, not resistance. If so, it's probably not a straight hookup, as there's no way for a simple DC signal emitted by the B-17 gauge to measure capacitance!
This leaves us with one of two possibilities:
1. The B-17 gauge includes its own circuitry to measure the capacitance of the tank sensor. However, I find this very unlikely, given what I know of avionics.
2. I suspect there was a second device, either one at each tank or a central unit in the cockpit or cabin which translated the fuel sensor's capacitance into the 128k/0k resistance range the gauge is able to use.
Before we proceed, we have to remember op-amps didn't exist back then! So the solution would involve a tube-based LRC circuit (the first practical application of the transistor didn't occur until 1947).
But a transistor would work just fine.
So, the problem boils down to the following LRC components:
1. DC power supply: 12V
2. Transistor
3. Variable resistor: 17 ohms (full) and 80 ohms (empty)
4. Gauge: 128k ohms (full) and 0 ohms (empty
Task: Design a simple circuit using these four components such that when the variable resistor is at:
A. 17 ohms the Gauge sees 128k ohms
B. 80 ohms the Gauge sees 0 ohms
It's ok to add extra resistors and capacitors, but no op-amps! They hadn't been invented yet!
I'll play around with this, but I'm sure there will soon come along either an electrical engineer or a circuit tinkerer who'll whip out a schematic for you.
ETA: Just thought of something: The B-17 tank could have used a simple float potentiometer, but with the readings reversed, such that it was an open circut (or at 128k, a nearly open circuit) when it was full, and a closed circuit when empty.
