Low fuel level alarm/shutoff

Thread Starter

SimpleJoe

Joined Mar 22, 2016
35
Ok I will start off saying that it has been a very long time since i did any major theory work with op-amps so excuse my rustiness but here goes.

So I am going to install a secondary fuel tank underneath the tray of my ute and plumb a pump into the OEM tank operated by a switch in the cab for refueling (all above board in my state). however all the pumps i have found can take anywhere from 15minutes to a full hour to empty the proposed secondary tank so i will most likely not remember to switch the pump off and probably burn it out on the first couple of uses.

So my idea is to use a fuel level sender to not allow the pump to run when the tank nears empty. As it will be under the tray of a ute that see a fair bit of water and off roading I have my doubts about the longevity of a home made circuit (but if anyone can suggest a diagram for one with common prts that would also be much appreciated). So i there some form of off the shelf voltage comparator for cheap or am i limited to building my own?


Edit: I forgot to mention that i have seen some fuel level senders with a built in low level alarm but all of them seem rather expensive here is Aus

Thanks,
SimpleJoe
 

crutschow

Joined Mar 14, 2008
24,343
If you have a fuel level sender that you want to use and know it's output when you want the pump to stop, we should likely be able to gin up an appropriate (and reasonably simple) circuit for that.
 

Thread Starter

SimpleJoe

Joined Mar 22, 2016
35
If you have a fuel level sender that you want to use and know it's output when you want the pump to stop, we should likely be able to gin up an appropriate (and reasonably simple) circuit for that.
I haven't nailed down specifics yet but the majority of the fuel senders range from 10Ohm (empty) to 200 Ohm (full) so I guess around the 20 ohm mark?
 
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crutschow

Joined Mar 14, 2008
24,343
I haven't nailed down specifics yet but the majority of the fuel senders range from 10Ohm (empty) to 200 Ohm (full)
Then you could use a simple comparator IC, such as an LM339, to detect the fuel level desired trip point and use that to operate a automotive relay that controls the pump.
It can be built on a small perf board and housed in a small project box if desired.

I can gin up a circuit for that, if that sounds ok to you.
 

Thread Starter

SimpleJoe

Joined Mar 22, 2016
35
Then you could use a simple comparator IC, such as an LM339, to detect the fuel level desired trip point and use that to operate a automotive relay that controls the pump.
It can be built on a small perf board and housed in a small project box if desired.

I can gin up a circuit for that, if that sounds ok to you.
I believe I might even have some 339s on hand from the last project I attempted (battery voltage alarm), hopefully this one turns out better, Thanks a lot
 

crutschow

Joined Mar 14, 2008
24,343
Below is the LTspice simulation of a circuit that should do what you want:
The relay can be a typical 12V automotive type.

The simulation is for a level sensor (Rsensor) that goes from 10 ohms (empty) to 200 ohms (full) [blue trace].
0s is empty and 95s is full on the plot.
The trip point is adjustable by pot U2 from empty to a little over 20% full.

upload_2019-10-14_9-29-47.png
 

Thread Starter

SimpleJoe

Joined Mar 22, 2016
35
After thinking some more about this, the level sensor is going to have to be pretty accurate otherwise a large amount of the fuel will be left behind.

I do remember reading somewhere about old style bikes (and Mitsubishi monteros?) using the self heating properties of a thermistor to tell when it is either under or above the fuel level (from what I read fairly commonplace on petrol tanks, mine is diesel so even safer). The only problem is I can only find the following bung type thermistor:
https://au.rs-online.com/mobile/p/thermistors/8199173/

I'm not too sure about wiring one of these up and if I'm after a NTC or PTC type.

Cheers
 

Thread Starter

SimpleJoe

Joined Mar 22, 2016
35
Ok so i did a little more research and found this thread: https://forum.allaboutcircuits.com/threads/help-with-low-fuel-level-circuit.47707/page-2

from which the main take away was this diagram:
zt6jpe.png

Which I only sort of understand, the VR1 is placed at the top of the tank to get a reference temperature for hot days and VR2 at the low level mark so when the fuel drops below it heats up and matches VR1 making an even voltage divider (which would give anywhere from 6-7ish volts on a car battery). C1 and R1 are just in as a delay due to the OP using this system on a bike and won't be needed in my case as I will add a small surge tank section to the tank so it won't slosh (diagram). This is then input into a Op-amp and turns on a led, which is easy enough to convert to a transistor and relay for my needs.
upload_2019-10-17_11-12-1.png

Where I get lost is on the calculations of the resistors and using that to work out what temperature the thermistors will run at. obviously I want them to run as low as possible so i don't have 2 glow plugs in my fuel tank but still hot enough to be able to tell the difference between the air and the fuel in the Australian desert after driving 700km on the main tank (the aux tank will be near the exhaust as well to make things even better).

This also brings another problem with the dual sensor style, if I brim the tank, both sensors will be under, give the same resistance and not allow the pump to run which means i will probably have to find a slope to part on to trick the sensors, not ideal. I could replace VR1 with a fixed resistor to resolve this but that brings me back to my main problem, not knowing how to read the datasheet to tell me what resistance to replace it with.

Currently I am tossing up between these two transistors,

ideally I would use the first as it is built for fluid levels but i will have to machine a bung for it, but i am unsure about the suitability for option 2 as it is for 3D printer hot ends as best as i can tell.

Option 1: https://au.rs-online.com/web/p/thermistors/8199173/

Option 2: https://www.amazon.com/Modular-Screw-Thermistor-Screw-Thermistors/dp/B00ZAMN3IE


EDIT: late game player: https://au.element14.com/amphenol-advanced-sensors/ge-1711atm/thermistor-ets-brass-10k-screw/dp/2779018?gclid=Cj0KCQjw_5rtBRDxARIsAJfxvYChWg96zfkdf2LlHcFpYjbI5FLdjhhwswppk1GNpP25td6m1tHjifAaAkWhEALw_wcB&mckv=sJbKEcOwV_dc|pcrid|380892800521|pkw||pmt||slid||product|2779018|pgrid|76493923343|ptaid|aud-466289539647:pla-333726873286|&CMP=KNC-GOO-SHOPPING-2779018
 

Thread Starter

SimpleJoe

Joined Mar 22, 2016
35
That thermistor could be a little difficult to cause a reliable delta sense voltage at low temperatures using a 12V supply.
What's the lowest temperature at which you will want to transfer the fuel?
From memory diesel turns into a gel at -10 degrees (Celsius) but being in Northern Queensland I doubt I will ever see any thing below 15 degrees plus as i mentioned in my last reply, this tank is going to be in the vicinity of the exhaust so if anything, I'm worried about the fuel being too hot.
 

crutschow

Joined Mar 14, 2008
24,343
Where I get lost is on the calculations of the resistors and using that to work out what temperature the thermistors will run at.
The temperature rise of the thermistor in air is determined by the power dissipated in the thermistor and its thermal resistance to air.
The data sheet for the Vishay NTCAIMME3 gives the thermal resistace as 2.8mW/K (or °C).

Since the thermistor resistor goes up as the temperature goes down, the least power will be dissipated at the lowest temperature.
At 15°C the thermistor has a resistance of 15711 ohms.

The dissipation for 6V across it is then 6^2/15711 =2.29mW.
This means the temperature rise for the unit out of the liquid is 2.29/2.8 =0.82°C.

The resistance change at that temperature is 3.24%/K so the resistance for the unit in air will drop by .0324 * 0.82 * 15711 = 417Ω or to 15294Ω.
The voltage at the junction of the two is the 12V * 15711/(15711+15294) = 6.807V, giving a 80.7mV change from the 6V when they are at the same temperature, small but detectable by the circuit you posted.

You likely will want to replace R2 or R3 with a trimpot to allow adjusting the trip point to compensate for thermistor tolerances.
(For good resolution, use a pot equal to the original resistor value in series with a resistor equal to 1/2 the original value)
You can use a can of water on the bench to test the circuit operation before you install it.
Worst case is at cold so you may want to test it in the fridge or cold box (allow time for the water to reach the fridge/cold box temperature).

Removing the diode is a good idea, since its forward voltage changes with temperature and will affect the trip point.
 
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Thread Starter

SimpleJoe

Joined Mar 22, 2016
35
Its a shame i can't find the thermal resistance for: https://au.element14.com/amphenol-a...la-333726873286|&CMP=KNC-GOO-SHOPPING-2779018 it'd be much easier to mount but I will persist with my understanding none the less.

The dissipation for 6V across it is then 6^2/15711 =2.29mW.
This means the temperature rise for the unit out of the liquid is 2.29/2.8 =0.82°C.

The resistance change at that temperature is 3.24%/K so the resistance for the unit in air will drop by .0324 * 0.82 * 15711 = 417Ω or to 15294Ω.
The voltage at the junction of the two is the 12V * 15711/(15711+15294) = 6.807V, giving a 80.7mV change from the 6V when they are at the same temperature, small but detectable by the circuit you posted.
So my understanding is to edit the circuit like so:
upload_2019-10-17_16-58-29.png
If i change the top thermistor to a fixed resistor I will avoid my 2 sensors being submerged problem that i mentioned earlier.
Also if i drop it to 2kΩ the thermistor still in the system will see 10V and dissipating 6.36mW
giving a temp change of 6.36/2.8 = 2.27°C
so the calc changes to 0.0324 * 2.27 * 15711 = 1155Ω or to 14556Ω
then the junction voltage is 12V * 14556/(2000+14556) = 10.55V, giving a 550mV change (my soldering and circuit making skills are hit and miss, the bigger the target the better)

And then just update the trim pot as you have stated and all should be sweet? I haven't glossed over anything vital?

Edit: I also did a quick high side check, at 70 degrees it will have to dissipate 57mW and the probe will self heat to +20 degrees so 90, well below diesel ingnition but still not too desirable, this is the absolute worst case I can think of so I may bite the bullet and order myself a couple
 
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crutschow

Joined Mar 14, 2008
24,343
It's probably similar to the other unit.
If i change the top thermistor to a fixed resistor I will avoid my 2 sensors being submerged problem that i mentioned earlier.
You may avoid that problem, but that leaves a much larger one. :rolleyes:
Two thermistors are used for a very good reason.
Two thermistors can cancel the ambient change in resistance/voltage, leaving only the delta change from ambient to submerged.
With only one, you will have to detect the voltage change delta from ambient air to submerged, while the absolute voltage can vary significantly with ambient temperature change, which is a lot larger than the delta change between the two thermistors.

I don't think it can be done with only one thermistor unless you use a microprocessor circuit to ignore the ambient change and detect only the delta voltage change.
 
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