fuel sensor to ADC, need help

I'm doing 8051 project of Motorbike meter, troubling with fuel sensor, my bike having 100 ohm fuel float sensor, that varies from 93 ohm to 5 ohm, (93 ohm for empty & 05 for full), i need 0 to 5 volt variation to ADC, means for 93 ohm i need 0 volts & for 5 ohm i need 5 volts,,, plz help me im in final stage of project, this is only the remaining thing

You do not want to simply put 5v on the sensor. 5v/5 ohms is an amp or 5watts which will heat the resistor in a gas tank.
So you have two choices: use .5v and either have fewer bits (you do not need 256 to 2048 (8 to 12 bits) resolution for gas level or use .1v and a op amp to amplify it up to 4.5v full scale.

how to do that, i cant getting that, any schematics????? plz
and also originally they have used 12v through galvanometer (fuel level indicator), means for empty 6V across float sensor, for full 0.5 v, voltage across sensor will varie with respect to battery voltage

OK, ill do as u said, ill take 1v range, and use non inverting amplifier, but that is not a problem, i need to fed op amp with 0 to 1 volts using float sensor rit, i cant do 0 volts with float sensor, max of 0.5 volts ( with good variation range) op is 2.5 v, is there any solutions???

got it thanks ,,
by designing this one**broken link removed**

but proble remains,,,,,,, for the 50 percent of 93 ohm resistance i have to get 2.5v, but its giving me 3.3V,, its a big difference. it gives false values,
means if tank capacity is 10L at 50% 93oh resistence it have to show 5L, but by this designe it will shows 6.6Ltrs,, how to rectify it??????

if this false change is constant example : instead of 2.5 ----> 3.3 . use a voltage divider with ratio r1/r1+r2 = 2.5/3.3

hope that helps , good luck

how......
i cant get it

This is an easy problem.

It sounds like Vdd is 5V, right?
Connect a 500 ohm resistor from +5v to the gas tank sensor. This is a divider.
When empty R is 500 + .5 and full it is 500 + 93 ohms.
The current is 5v/500.5= 1ma when full and 5v/593= 8.4ma when empty.

The voltage at the sensor full is .5 ohms * 1ma = .04 volts.
The voltage at the sensor empty is 93 ohms * 8.4ma = .78 volts

If you use the 8051s ADC in 8 bit mode the are 256 steps from 0 to 5v so each step is 5v/256 = .019v

The voltage from the divider to the ADC is .78v to .04v which is a range of .74v. Even through the voltage range is small it is still .74v/.019v = 39 ADC steps.

Why do you care if you are not using all 256 steps over the 5v range from the ADC? A gas gauge with 39 steps is enough. It is 2.5% per step. I doubt your digital gas gauge has 6 digits. Maybe it has 8 bars. Then 8 steps is enough.

One resistor and some easy code. You can start with .06v for empty and .7v for full and have the MCU store any value less than .06 when empty and more than .7 when full, so the empty and full readings are self calibrated and accurate.

- - - Updated - - -

Make R2 small so the gain is less......

finally done it, getting full 0 to 5 volts, for 50% 2.5XX volts... i needed full values form 0 to 5 v (256bits) bcoz im showing values in liters.

thanX all

by the way, which op-amp is better for this purpose, vcc +15(max), and input is 5uV to 70nV and op should 0.0XX to 5V, i've designed it with LT1006, is it good?