I have a fuel gauge that is mismatched to the sending unit on a hotrod. It is ready about twice the fuel it should, so I put together a voltage divider using a dual cement 5 watt resistor. The center or common leg goes to the fuel gauge +terminal, one of the legs goes to the 12v lead, and one of the legs goes to ground. The thing works perfectly except that the resistor gets hot as heck, and if I leave it on will burn the resistor up.

The resistor I am using is a dual .22 Ohm. I re-purposed it out of an Onkyo/Integra that was busted. The actual resistance is exactly .218 and .217 Ohms as measured on my micro-ohmeter.

The Red wire=12VDC, white/red wire is the 6VDC, black wire GND.

 

Use higher resistor values to reduce the heating.

 

Yes, your .44 ohm (total) resistor is dissipating about 330W from the 12V.
No wonder it's a little warm.

You could perhaps use a voltage regulator to get the 6V instead.

 

So that's what is happening...

The sending unit is a potentiometer, basically. I'm not sure how much power the gauge needs.

Do you think a pair of 1/2 watt 2.2 KOhm resistors would provide the current needed to make the gauge work?

 

Probably not.
Measure the gauge resistance.

 

I was trying to pull this off with parts I had laying around. The receiver had an abundant supply of .22 dual cement resistors for instance. I tried making a voltage divider using 470KOhm 1/2 watt resistors and it wasn't enough to move the gauge needle.

What if I put a resistor on the White/red lead? Wouldn't that lower the current and the demand?

 

You need to know the gauge resistance to determine how high a resistance the resistors can have.
Design by random selection of part values usually doesn't work very well.

 

Here are the values of the 3 terminals on the gauge:

+ to GND = 181.6 Ohms
+ to Sending unit = 48.9
Sending unit to GND = 133.3

The sending unit on the tank to the ground is 106.8 Ohms with 1/3 tank of gas.

 

Here is the wiring diagram:

 

I would put everything back to normal. Empty tank to your best guess.... R for empty level....I would want at least a gallon left in tank.
Install a potentiometer in series with the grey terminal back to the gauge. Adjust pot for R reading on gauge. Replace pot with fixed wattage resistor.

 

Unfortunately the higher resistance tells the gauge there is more fuel, so that is backwards. The only way to do this is reducing the input voltage.

 

Try putting a 47-50 ohm 1W resistor between +12V and POS.

Edit: Take two 100Ω ½W resistors and wire them in parallel. Insert in series from +12V into POS on the gauge,

 

Try the two resistor voltage divider with two 100Ω, 1W resistors and see how that works.
You might have to tweak one of the values to get a proper empty and full readings.

 

Thanks a million you all! What a great resource this has turned out to be! Here are a couple of thing I might need clarity on. Just to be sure in my own mind:

Try putting a 47-50 ohm 1W resistor between +12V and POS.

Edit: Take two 100Ω ½W resistors and wire them in parallel. Insert in series from +12V into POS on the gauge,

Let me get clarity on this one. Does this 50Ω1W ( or these if parallel 100's1/2W) resistor go between the 12V and my voltage divider red wire?

Like this:
12VDC -> 50Ω with total 1W rating -> red wire (POS side of my .22Ω dual cement resistor voltage divider)

...to reduce the current the divider dissipates.

Try the two resistor voltage divider with two 100Ω, 1W resistors and see how that works.
You might have to tweak one of the values to get a proper empty and full readings.

I don't have any 1 watt resistors but I think I may have four 220Ω 1/2 watt resistors so I could parallel two pairs to make two 110Ω 1W.


I kind of like the idea of using my voltage divider a little better since I have about a half dozen of those dual .22Ω cement resistors, and it makes a nice little package. My problem was I wasn't limiting the current. I know that now. If I reduce the current with 50Ω before the divider circuit, this should work. Does that sound right you guys?

 

....................
I don't have any 1 watt resistors but I think I may have four 220Ω 1/2 watt resistors so I could parallel two pairs to make two 110Ω 1W.

I kind of like the idea of using my voltage divider a little better since I have about a half dozen of those dual .22Ω cement resistors, and it makes a nice little package. My problem was I wasn't limiting the current. I know that now. If I reduce the current with 50Ω before the divider circuit, this should work. Does that sound right you guys?

Yes you can parallel the resistors.
But my resistors were to replace your voltage divider, not supplement it.

You can't add resistors in series with your low value cermet resistors, since then you won't have the proper voltage division anymore.
You need to brush up on Ohm's law.

 

I found this handy voltage divider calculator.

http://www.raltron.com/cust/tools/voltage_divider.asp

I can easily make a 4V voltage divider with two of my dual .22 cement resistors stuck together. The far ends of those resistors read .44Ω, and either outside leg to the middle is .22Ω. If I glue a pair of the resistors together and wire with .22 and .44Ω, I can turn 12VDC into 4VDC.

Yes you can parallel the resistors.
But my resistors were to replace your voltage divider, not supplement it.

You can't add resistors in series with your low value cermet resistors, since then you won't have the proper voltage division anymore.
You need to brush up on Ohm's law.

I knew that you were creating a 6V divider for a 12V circuit. I was actually referring to MrChips' response. I was simply wanting to keep using the divider I made and reducing current going into the hot side of it. Check my post real fast so you can see what I mean...

 

My solution is to get rid of the voltage divider entirely.

Disconnect the cable going to the POS terminal of the gauge.
Connect the cable you removed to one end of the 50Ω 1W resistor.
Connect the other end of the 50Ω 1W resistor to the POS terminal of the gauge.

You will find it handy to use some quick connectors and a short length of electrical cable, exactly like what you just disconnected.

The 50Ω 1W resistor can be made by putting two 100Ω ½W resistors in parallel.

 

For everyone's info, here is a schematic that shows just what is going on in a sender/gauge circuit.

Lowering the voltage at the "Wiring Connector" will lower the gauge reading (make it go toward empty).

Adding passive resistance at the circuit to the "Tank Unit" will increase the gauge ready (make it go toward full).

My gauge said my tank was full when I knew for a fact it was only 1/3 full. To repair that, I am reducing the voltage going into this circuit. My method was to use stuff I had laying around the shop. It just so happened that I had a dead receiver with a bunch of 5 watt dual cement resistors. The problem is that they are all .22 ohms per leg. this made for a perfect little compact 1/2 divider circuit to make 6VDC out of 12VDC. The problem was that there was far too much current so the resistor turned into a space heater. I still don't have a definitive answer as of this typing, but my hope is to put a single resistor (50 ohm 1 watt) between the 12V and the input + lead of my .22 ohm x .22 ohm voltage divider.

 

My solution is to get rid of the voltage divider entirely.

Disconnect the cable going to the POS terminal of the gauge.
Connect the cable you removed to one end of the 50Ω 1W resistor.
Connect the other end of the 50Ω 1W resistor to the POS terminal of the gauge.

You will find it handy to use some quick connectors and a short length of electrical cable, exactly like what you just disconnected.

The 50Ω 1W resistor can be made by putting two 100Ω ½W resistors in parallel.

We were typing at the same time. If you could glance at my post above, you will see a diagram of how the fuel sender to gauge circuit works. Reducing current to the gauge input won't lower the voltage to the gauge so I don't think that is gonna fly. Am I wrong?

 

Would I lie to you?
You have four 220Ω resistors. Why don't you try it? What have you got to lose?

Make sure you twist all four resistors together in parallel.