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iwasnevy 12-30-2010 06:13 AM

Help with low fuel level circuit
Hi all
Sorry if this has been asked before, I searched and all I could find were discussions dealing with tying into a typical float-on-an-arm type factory sending unit.

I'm not afraid of electronics, I can read a schematic, I understand the basics, and can design and build simple circuits, but I could use some help here cause this is a bit over my head and I don't know exactly what I'm looking for.

I need to make a low fuel light circuit (automotive type application, in case it's not obvious!) which drives an LED. This application DOES NOT use a typical fuel level float on an arm supplying a resistance which varies with the fuel level, but instead uses a thermistor (?) which is cooled by the fuel surrounding it. When the fuel level drops below the thermistor, it heats up and it's resistance goes down (or up?)

(Before anyone jumps about using something that heats up in a fuel tank, it's perfectly safe - simple chemistry here, it can't explode or incinerate...)

Hondas have historically used this method to illuminate a standard filament bulb in the dash. One problem with the old honda circuits that drove a bulb - it was a direct driven current/resistance circuit, so as the thermistor heats, the bulb comes on slowly, dim at first and then brightening. If you take a fast corner, the bulb goes out as fuel sloshes over and cools off the sensor. This will be used on a motorcycle, where fuel will be sloshing around even more. in 2010, we can do better than that...

Here's what I have in mind - as best as I can explain - a comparator circuit, where when the voltage drop across the thermistor reaches a threshold, it latches a circuit into an ON state to illuminate an LED, and stays latched until voltage is removed (ignition cycled). The reason for this is to avoid the light continuously turning on and off when the fuel level is right at the level of the thermistor.

Ideally, the voltage drop is "buffered" by a capacitance circuit of some sort so it's not latching on falsely when fuel uncovers the sensor for a short period of time (such as when taking a corner or going through a long curve on the road) - so the voltage drop would need to stay low (high?) consistently over the course of maybe 2+ minutes before the circuit would latch on.

I understand principles and generally what the basic electronics components do and how to use them, but I don't have the direct experience or knowledge to know what specifically to use or how to to put it together, so saying "use a voltage divider" doesn't do much for me, (while I can imagine that it's probably two resistors in series, I really wouldn't know it if it bit me!) I need hand-holding, more specific help or pointers to existing schematics...

beenthere's suggestion in this thread:
was the closest I could find, but while I followed what he was saying, I don't know how to use his suggestion to do what I'm looking for.

Any help, direction or resources would be greatly appreciated, and I'd even pay someone to design this *relatively* simple circuit for me.


marshallf3 12-30-2010 05:33 PM

I just rely on the reserve switchover on my motorcycle petcock. When I have to hit it I know I've only got about 40 or so miles to go so I just fill up as soon as I can. This sometimes presents a difficulty as I have to seek out a station that sells GOOD 100% gasoline, the E10 ethanol stuff doesn't run as well nor is it very carb friendly on the 32 year old bike I have. It will work in a pinch but I tend to keep it out of there as much as I can - same as with my car. Luckily we've got that option down here, may cost 5 - 10 cents more a gallon but after years of calculating mpg I come out ahead by using 100% gasoline and my car's OBD-II computer will actually sense if I'm using Premium gas and advance the timing a few degrees thus I benefit from that as well, again offsetting the additional cost.

You present a problem that's very difficult to deal with. I've been trying to come up with a gas gauge for my bike but I'll be darned if I'm going to make a hole in the tank to run wires through. Given time I'm going to try to mount some sort of ultrasonic device on the exterior of the tank hidden "under the frame hump" so to speak. The idea is that through hitting it with some pulses and looking at the return results I may be able to calculate how much of the total volume of the tank is air and how much is liquid by the change in the resonant frequency.

Some people have made a Tee connection to the supply line and run it up in an inconspicuous area with a tiny orifice as a termination for the tube then made marks on it. Of course you've got to stop and look but it will give you a general idea of where you're at. Another method some have looked into has been a very precise pressure sensor but I've never heard if it was ever even tried nor put into practice. In theory it should work but again, without a large damping circuit, you'd need to stop before you'd get an accurate reading.

The thermistor idea has extreme drawbacks on a bike as you've noted, and just a small splash onto the device will cool it back down.

Motorcycle tanks present a problem as they're formed quite differently from that of a car but I really think that, once I get some experience in programming microcontrollers, my ultrasonic method will work out pretty well. It's been used on a much larger scale to do the same thing so I don't see why it couldn't work on a cycle tank as well.

iwasnevy 12-30-2010 07:06 PM

Thanks for the reply. The whole point of doing this is to eliminate the reserve and not have to deal with running out of fuel and switching the petcock. Trying to bring this into the 20th century here...

As far as implementation, that's easy - there are provisions there already in the factory petcock with very little modification. The "reserve" pickup (really, just a short tube) will become the sole fuel outlet. The thermistor will go inside an aluminum tube which replaces the main "pickup", at the same height. (These pickup tubes are soft plastic, so I don't want to put a warm thermistor in them.) The tube will have a few smallish holes drilled in it's length so it essentially acts as a baffle. Hopefully this will eliminate most of the physical splash-over effect, and the capacitance circuit should buffer the rest so that once the light is on, it means it.

marshallf3 12-31-2010 02:12 AM

By all means give it a try if you can sneak in through the petcock. After that all you'd need is a simple comparator circuit. You could even use a tiny LM35 or LM335 in the TO-92 case and achieve the same results without the heating but then you'd be dependent on the outside air and gas still in the tank temperature differentials.

iwasnevy 12-31-2010 03:09 AM

Comparator circuit - got it. So the thermistor would be one of two resistors in a voltage divider, which would feed into a comparator circuit. The comparator circuit would then trigger a latching circuit, right? So far so good :)

I know the thermistor takes a bit of time to heat up, from my experience, about 15 - 30 seconds or so depending how much current I'm drawing through it. In my mind, that's too short of a time - a long curve in the road would be enough to allow the thermistor to heat up and trigger the circuit.

So how do I get a buffer in there so that it's not triggering the comparator every time the thermistor gets uncovered? Enough so that the thermistor has to be uncovered for, say a couple of minutes.

I need to study a voltage divider circuit I guess. What I would need is for a capacitor to trigger the comparator once it's discharged, because it'll take no time to charge, but some time to discharge. So I need a voltage divider to charge the capacitor when the thermistor resistance is high (cold) and discharge the capacitor when it's low (warm).

Am I on the right track? Help?

I'll draw up a schematic of how I *think* it should be, once I get a chance to mull this over a bit.

KMoffett 12-31-2010 07:16 AM

What's the possible temperature range of your fuel? With a thermistor and a resistor as a voltage divider feeding the comparator, the actual temperature of the fuel might be an issue in setting a reference voltage level. If you can use two heated thermistors in series as the divider, one always in the fuel and one at the low-alarm fuel level, the circuit will self-compensate for running in a Death Valley summer or in an International Falls, Minnesota winter. ;)


marshallf3 12-31-2010 07:18 AM

You are on the right track but remember you'll need to isolate the divider network from the buffer cap with a diode, then you can program the discharge time of the cap by putting a little resistance in parallel with it.

KMoffett's suggestion is also a valid one as you're dealing with a complicated problem here.

iwasnevy 12-31-2010 03:58 PM


Originally Posted by KMoffett (Post 314822)
What's the possible temperature range of your fuel?

I suppose theoretically, it could be anywhere from < -10șc to 50șc, although I'll guarantee I won't be riding if it's that cold! The thermistor is 'reading' an upper temp threshold, so if I set the voltage trigger to be anything above, say 65șc, I think I would be ok. I have a feeling it's not going to be an issue - I remember the honda ones, when they were uncovered, they got very hot to the touch, enough so that there was the hissing sound of evaporation when it was submersed in the fuel. I could always just experiment with some warm water to find that threshold. For that matter, I could just run to the junkyard and yank a thermistor out of a honda and measure it's range and just use that. Then I know I have the "right" one and can work from there.


Originally Posted by marshallf3 (Post 314823)
you'll need to isolate the divider network from the buffer cap with a diode...

You're saying so that the cap can't feed voltage back into the divider network, when the voltage in the divider drops below the voltage in the cap, right?


I think I'm getting this. hopefully I'll have some time to sit down this weekend and wrap my head around it and draw up a schematic. Then redraw it once you all tell me all the ways it won't work :)

I can't believe I couldn't find a discussion like this on google. Seems that this would be something someone's done before.

Thank so far for the help - I love a good challenge!

And happy new year!

iwasnevy 01-01-2011 01:48 AM

Ok, here's my first attempt, tell me if I'm thinking of this capacitor thing wrong.

(Just to get an idea of how this should work, I used the specs for the thermistor from beenthere's suggestion in the other thread I referenced earlier.)

Here's how I figure it - When the thermistor R2 is cold, it's resistance is high (>250 Ω) and the voltage at Vout 1 is high (>485 mV), charging the capacitor C1.

When the thermistor R2 is hot, it's resistance is low (<100 Ω) and the voltage at Vout 1 is low (< 175 mV), and the voltage divider's output is below the potential in the capacitor C1 and C1 begins to discharge through Vout 2.

Once C1's voltage at Vout 2 drops below the threshold (175 mV), it triggers the comparator input at Vout 2 (which I don't have in the schematic yet).

So what am I missing? Anything? A resistor to ground between C1 and the comparator so it has somewhere to discharge? am I placing the capacitor in the circuit correctly?

marshallf3 01-01-2011 03:19 AM

First of all the diode is backwards in your schematic, the cap should go to ground and there should be a high resistance bleeder resistor across it. It was also mentioned that it might be of help if you had a dual thermistor setup so you could better sense the temp difference between the air and the liquid.

Start with this but it's going to take some time to get all the values correct, especially since we don't know what your output will be driving.

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