HELP me build a gas gauge????

browningbuck · Dec 4, 2007

hey guys i have a 12V system and need help to figure out a solution to this problem...okay i dont need help...i need some one to hold my hand through this

**broken link removed**

superbrew · Dec 4, 2007

**broken link removed**

k7elp60 · Dec 4, 2007

I think the LM3914 just may work for you. I would suggest that you start with a three terminal regulator like a LM78L05. Put a resistor from the +5V output to the hot gas gauge lead. This is the junction to feed to the LM3914 input. The reason for the regulator is that the DC voltage on a automobile varies depending upon charge of the battery and the engine running. The junction of the two resistors will ensure a stable voltage for the fuel gauge sensing circuit.

3v0 · Dec 4, 2007

For auto use you can do better then a 7805 regulator.
Try looking at LM2931AZ50R if you can live with a 100mA output. They are a bit spendy at 35 cents each

https://www.st.com/stonline/books/pdf/docs/6740.pdf

Ideal for automotive application the
LM2931 is protected from reverse battery
installations or 2 battery jumps. During the
transient, such as a a load dump (60 V) when the
input voltage can exceed the specified maximum
operating input voltage (26 V), the regulator will
automatically shut down to protect both internal
circuit and the load.

browningbuck · Dec 4, 2007

okay guys yall are GREAT thanks a ton...so am i thinking this through on the first suggestion( is hould have checked here before i got to looking at it and playing on photo shop

)

opps just figured i need six....can i leave LED #10 UN-Xed out

oh **** got to looking at the second link and i have NO clue where i would put that????

can i find that LM3914 at a radio shack???
**broken link removed**

browningbuck · Dec 4, 2007

well i was able to pick up all this stuff over at a local electronic stor...here to hoping it works(and if i can figure out how the hell i wire this up???)

superbrew · Dec 4, 2007

I have only used the LM3914 once, and it was from a kit, but you are going to want to put a resistor in series with the sensor, connecting one of the resistor to +V and the other end to one end of the sensor. The other end of the sensor goes to ground, creating a voltage divider. Then the signal in of LM3914 is connected to the junction of the resistor and sensor. What voltages to use depends on the resistor values chosen. Ideally you could build this on a breadboard and use a potentiometer for your series resistor to "calibrate" the system. Also, you can just use any of the outputs for your leds, just leave the others open. Hope this helps.

browningbuck · Dec 4, 2007

yeah that makes a litthe more sense(thanks for the help) if anyone else wants to chime in i would be MORE than happy to take some advice on these

Leftyretro · Dec 4, 2007

browningbuck said:
yeah that makes a litthe more sense(thanks for the help) if anyone else wants to chime in i would be MORE than happy to take some advice on these

You may have a direction problem in that your sensor goes down in resistance as the level goes up so the voltage across the divider will be lower when full then open. You may have to add a op amp to reverse and scale the measurement voltage.

Lefty

superbrew · Dec 4, 2007

The LEDs could be turned around and connected to ground using pullup resistors if that is the case.

browningbuck · Dec 4, 2007

ahhhh okay so ita all hooked up and works...kinda....i cant figure out what i need to do to get the clibration correct....it will light up and as soon as i connect ANY resistance to the singal pin it turns all off, but i know it works becasue i can slightly toucht the signal lead and use my body as a resistor and can get what ever i want to light up.

SO how do i calibrate it??? i tried switching resistors out with other resistors but NOTHING...WHAT DO I DO?????? i need it to work from 12 to 132 ohms on the signal

superbrew · Dec 4, 2007

How exactly do you have it connected?

browningbuck · Dec 5, 2007

ill take a picture tonight but its all on a nice little bread board

i also got to thinking this morning that maybe i am using too much resistance as my ref and that is causing it to react so easly...just a thought i have no calcs or even an understanding on it so if you know something i would love what EVER help i can get.

SPDCHK · Dec 5, 2007

With reference to the attached schematic.

I used the voltage divider rule and calculated a volt drop resistor of about 8.12 Ohm in series with the gas measurement device. (RV1 in the circuit replaces the gas sensor). I got a voltage of between 0.7V and 5V for a measurement range of 12-132ohm.

This is only for example. I don't know if the fuel you are measuring is flammable or not? (Diesel, Petrol ???). Calculate the resistor wattage to ensure that they don't heat up too much.

If you require a more precise 0..5V input signal, you will need to use an OPAMP.

The LM3914 will light a led for every 0.5V on the signal pin. Assuming a voltage level of less than 1 volt on an empty tank, the IC will start switching on the LED's as the resistance value of the measurement sensor decreases as the level increases.

The diodes in line with the LED's are all 1N4148 diodes. I did not include current resistors inline with the LED. You will need to calculate the required resistors yourself.

browningbuck · Dec 5, 2007

HUH???? okay first i am a total newbe, so is 1k21 the same as a 1.21k ohm resistor???

second why are you putting a 8.12 ohm resistor on the gas potentiometer???

I had mine setup VERY similar to this minus a few little things (like resistance values are just a little off) so i will work on this a little when i get home

WOW why is this so difficult for me and so easy for you guys???okay dont answer that one.

Leftyretro · Dec 5, 2007

SPDCHK said:
With reference to the attached schematic.

I used the voltage divider rule and calculated a volt drop resistor of about 8.12 Ohm in series with the gas measurement device. (RV1 in the circuit replaces the gas sensor). I got a voltage of between 0.7V and 5V for a measurement range of 12-132ohm.

This is only for example. I don't know if the fuel you are measuring is flammable or not? (Diesel, Petrol ???). Calculate the resistor wattage to ensure that they don't heat up too much.

If you require a more precise 0..5V input signal, you will need to use an OPAMP.

The LM3914 will light a led for every 0.5V on the signal pin. Assuming a voltage level of less than 1 volt on an empty tank, the IC will start switching on the LED's as the resistance value of the measurement sensor decreases as the level increases.

The diodes in line with the LED's are all 1N4148 diodes. I did not include current resistors inline with the LED. You will need to calculate the required resistors yourself.

View attachment 15826

Will that work? He said the fuel sensor rheostat (variable resistor) has one side grounded internally and only the one lead available, so I don't think he can rewire it so as to have both terminals available to use as you have drawn.

Lefty

browningbuck · Dec 5, 2007

yes it is internally grounded.... thanks for bringing that up didnt know that that would have an issue THANK YOU LEFTY!!!!

SPDCHK · Dec 5, 2007

I must have missed the one wire part.

I actually also have the resistor value wrong as well. It should be 8.62 Ω. (Good luck finding one with that value in any shop

).
8.62 Ω should give 0.7V drop across the resistor when the tank is empty and at fuel sensor measuring 132 Ω, the volt drop over the 8.62 Ω resistor should be exactly 5.0V.

Again, it's just an example.

browningbuck · Dec 5, 2007

ahhh its starting to make sense a little to me... but i dont under stand what wiring a 8.62 ohm resistor in line with ground would do for me, but if it will make all this work i can get a resistor in there , but it is atill just going to common ground.

as for the two resistors that it is using for ref, do those change now that you have changed the value of the one coming off the signal

superbrew · Dec 5, 2007

It's dificult for everyone that is just starting out. I am also just learning the basics.

The resistor values for the ref do not change, they are just a voltage divider setting up a 5V reference.
If you do not understand what a voltage divider is, try googling it, it will help you understand alot of what this circuit is doing.

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