Our (old) boat has a fresh water tank whose level was reported by a now broken gauge. For cosmetic reasons I need to replace the broken gauge with a similar form factor (I do not wish to cut the panel about). So I cannot just buy any new tank gauge.
The old gauge is from the Anders Acclaim AM25 series and had been adapted (messily) to act as an Ommeter with a "Euro type" tank sender. I can obtain a physically compatible replacement ammeter (1 milliamp max) and I can change the tank sender to the "US type" (240 ohms empty, 33.5 ohms full).
But I cannot work out what circuit to build such that the gauge will read zero amps when the tank is empty and 1milliamp when the tank is full.
Boat's electrics are a nominal 12v (more like 12.5 when exhausted and 14.4 when charging).
Edit: the replacement ammeter is described as having a coil resistance of 200ohms here http://www.topqualitytools.co.uk/meter-78×60-0-1ma-am25a2/
Edit: Further hunting has produced an alternative compatible gauge, a voltmeter, the AM25F2 (0-75mV DC). Might this be utilised in a solution?
Best Answer
simulate this circuit – Schematic created using CircuitLab
Figure 1. An imbalanced bridge solution. R5 is the assumed resistance of your milliammeter.
I offer this as a rough and ready solution to your question. It is based on the principle of the Wheatstone Bridge which, when the ratios of R1:R2 = R3:R4 are matched will have zero current through R5 and AM1. R3 is set at 240 Ω so that balance occurs when the tank is empty and the meter will read zero.
The values for R2 and R4 were established by messing around with the CircuitLab simulator (which you can do by editing my question - just don't save it) rather than going through all the maths.
To keep the values of R2 and R4 reasonably low and avoid you having to do precise resistor matching we need to drop the bridge voltage down to about 2 V or so. R6 and D1 do this. The forward voltage, Vf, of an LED is reasonably constant over a range of currents so this voltage will remain reasonably constant with variation in battery voltage. The 680 Ω resistor will give about 15 mA. A red, yellow or green LED will give around 2 V at this current. Blue and white would give a higher voltage.
Figure 2. LED I-V curves for various colours of LED. Green will give a 2 V drop at around 15 mA. Source: LEDnique.
Update:
For a 1 mA, 200 Ω meter R2 and R4 should be reduced to about 680 Ω.