Project: Auxilliary gas gauge for a motorcycle

Discussion in 'The Completed Projects Collection' started by Jack_K, Nov 2, 2009.

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1. Jack_K Thread Starter Active Member

May 13, 2009
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I wanted to add an extra gas gauge to my motorcycle because the existing gauge is very small and not very accurate. I selected a gauge used on 2000 to 2003 Harley Davidson cruisers. I picked this because it should be rugged enough for motorcycle use and I got a really good deal on ebay. This gauge needs 12 volts and ground plus a connection from the fuel level sending unit of the motorcycle.

Unfortunately the HD fuel level sending unit has a different resistance range (10 – 70 ohms) than my Stratoliner’s sending unit (10 – 210 ohms). That means the HD gauge will not read accurately (actually, not even close). The HD gauge can be “fooled” by an electronic circuit that reduces the voltage going to the gauge.

The gauges are not quite linear but we can consider them to be so. That means we can use the slope-intercept equation y=mx+b to calculate the change in voltage required.

I was able to measure the voltage across the motorcycle's gas gauge sending unit. I obtained the following voltages across the sending unit corresponding to the gauge reading:

Ohms Resistor voltage
10 -- 0.53 Full
88.8 -- 3.60 Half tank indicated
214 -- 5.90 Empty (low fuel light on)

Putting resistors to ground, in series with the HD gauge, produced the following voltages:
Ohms Resistor voltage
10 --- 0.40 Full
20 --- 1.33
30 --- 1.92
40 --- 2.52
50 --- 2.94
60 --- 3.24
70 --- 3.50 EMPTY

It seems like it is pretty close to linear. A little hard to tell but close enough.
Using gauge 2 as x and gauge 1 as y --

y = mx + b, m is the change in y divided by the change in x.

m = (0.4 – 3.50) / (0.53 – 5.9) ~ .57728

To find b we will plug in a value for x and y. Let's use the first point --

y = .577x + b
0.4 = .577 * .53 + b
b = 0.094v

The amplifier needs to have a gain of .577 and you need to add 0.094V to the input.
The equation would be Vresistor2 = .577 * Vresistor1 + 0.094V

I used an L2722 op-amp because Sgt Wookie pointed out that it can drive the output almost all the way to ground and I needed a 0.40 volt output.

Adjust R1/R2 to get 0.577 times V1 at pin 7 of the op-amp. Adjust R8 to get 0.094 volt at the junction of R4, R7, R8, and the diode. Actually, the diode can be omitted.

This circuit turned out to work beautifully and is actually more accurate than the existing gauge.

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Last edited: Nov 3, 2009
2. SgtWookie Expert

Jul 17, 2007
22,183
1,728
Nice job, Jack.

The amount of current necessary to be sunk from the new gas gauge eliminated most common opamps from consideration. The L2722 is an active product that was readily available, compact, and inexpensive. It's not rail-to-rail, but met the requirements of the project.