Mechanical sending units, fuel system


Old 02-14-05, 07:51 PM
Visiting Guest
Posts: n/a
Mechanical sending units, fuel system

Im trying to find out how exactly a mechanical sending unit works, pertaining to the fuel system. Ive tried numerous sites and have come up with next to nothing. Any help is greatly appreciated.
Sponsored Links
Old 02-14-05, 08:27 PM
Join Date: Sep 2004
Posts: 132
Fuel tank sending units work either through a potentiometer or a rheostat (same thing... one is a club, the other is a finely crafted handgun). Some work as positive co-efficent (high resistance = high fuel level) and some are negative co-efficient (high resistance = low fuel level).

Newer vehicles have a lot of additional circuitry involved in fuel level reporting and make some of the old tests unreliable.
Old 02-14-05, 09:49 PM
Visiting Guest
Posts: n/a
I really appreciate your help, but I really dont understand what you just typed. Im only in my first year studying Automotive Technology in High School and all I need to know is just basically how it operates. Thank you for your help.
Old 02-14-05, 10:52 PM
Join Date: Sep 2004
Posts: 132
One more try... this should be explained in whatever textbook you were given... been a long time since my first year of apprenticeship but there were a couple of good books that pertain to basic automotives even as we know the trade today.

In the most basic fuel guage system, there will be the power supply (battery voltage, a guage, a constant voltage regulator and a float on an arm atached to a variable resistance located in the gas tank. As the float arm moves up and down, the resistance of the circuit will change... this will cause the gauge to move up and down in sympathy with the in tank sensor..

To get a better grasp on how this (and other circuits) work, I can only suggest seeking help on how to decipher voltage drops. While it may seem complicated, electricity only needs some applied logic to be fully understood.
Old 02-15-05, 04:42 AM
the_tow_guy's Avatar
Group Moderator
Join Date: Feb 2001
Location: SW Fla USA
Posts: 11,567
Here's a simple non-electrical analogy:

Attach a garden hose to a sprinkler and turn on the faucet all the way. The water will shoot up about 10 feet, for example. Now put a little light pressure on the hose with your foot and you'll see that the water now only goes up 5 feet or so. Next, apply your full weight to the hose and the sprinkler goes down to a small dribble.

The water coming out of the faucet is your voltage and the sprinkler is your gas guage. When the tank is full and there is no resistance on the sender unit (potentiometer/rheostat) the sprinkler shows "full" (10 foot spray). When there is some resistance (your foot on the hose resisting the flow of water) the spray only goes half way (half tank of gas). When the tank is empty the resistance is very high (resistance to the water flowing through the hose analogus to the electricity flowing through the guage circuit) because your weight is stopping almost all of the water.

Get it?

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off

Thread Tools
Search this Thread
Display Modes