by pebe » Sat Apr 27, 2013 1:00 pm
Hi Ron,
Here is the circuit. Basically, it passes a small current through the 30ohm sensor, P1. The resulting voltage across it is fed to the input, pin2, of IC1. The output of the op-amp switches on the transistor which shunts R4 and R5, effectively varying the current drawn from ‘S’ pin of the gauge. The figures in blue indicate the voltages as P1 is varied from zero to its maximum resistance.
R1 and D1 & D2 form a stable 1.4V supply which is fed to R2 and P1. Varying the sensor from min to max resistance gives a voltage of from zero to 0.68V into pin2 of the op-amp IC1. The low value of 1.4V for the supply means that only 40mA max flows through the sensor.
The output of IC1 controls TR1 so that its current shunts R4 and R5 until the voltage that is tapped off P2 and fed into pin3, equals the voltage fed in from P1. With P1 at max, the voltage at pin ‘S’ of the gauge is 4.3V, and that gives the same result as connecting a 90ohm resistor between ‘S’ and ‘-‘ of the gauge.
D3 is fitted in series with the negative supply to the gauge. That is because it is not possible to get zero volts between collector and emitter of TR1, even when it is switched hard on. It will not effect the operation of the gauge because it is the difference in current through its two internal coils – not the supply voltage – which operates the needle.
To set it up, adjust P1 for max resistance and then adjust P2 until the voltage between ‘S’ and ‘-‘ of the gauge is 4.3V.
- Attachments
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- Fuel Gauge.GIF (6.15 KiB) Viewed 115240 times