I have an old oil pressure gauge and sensor from a 1970 Land Rover. The gauge still works but the internal pressure plate in the sensor failed so I cut it open to see how it works. I've set it up on my bench with a power supply and it works with the gauge when I apply finger pressure to it. But I haven't been able to figure out how to measure the resistance and voltage drop as the pressure changes. The numbers on my DMM flash up and down as if I'm trying to measure a/c. Any suggestions on how to measure this?

 

I think most of these gauges work by mechanically producing a PWM waveform that results in an average voltage delivered to the heating element that is wrapped around one leg of the strip. This results in an average temperature in the bimetallic strip that turns the gauge dial (basically, the gauge is a temperature gauge that has a pressure-to-temperature converter in it). The pressure bends a spring to make an electrical contact and the heater bends it back to break contact. The more the pressure, the more the spring is bent by it and the more heat it takes to bend it back far enough to break contact. As the metal cools, it again makes contact.

So I would expect a DMM to get very confused by things.

I think the resistor is probably some kind of calibration resistor, since it looks like it's in parallel with the strip. But that's just a guess (in more ways than one).

 

I think the resistor is probably some kind of calibration resistor, since it looks like it's in parallel with the strip. But that's just a guess (in more ways than one).

Thanks to your photo and WBahn I now have some idea of how old pressure sensors work!

If the resistor is actually in parallel with the strip heating wire, which I suspect it is, then you may find that the resistance of the resistor is in the region of 200 ohms and this is wired in parallel with the bimetal heating wire (with resistance probably around 25 ohms) connected in series with the bimetal switch.

So if you read the resistance across the terminals you should just see the resistance of the resistor but if you close the bimetal contact (by hand) you should see it drop to around 22 ohms R1//R2 = R1xR2/(R1+R2).

With no pressure the resistance will be just that of the resistor, when the pressure goes up it will switch the heating wire in parallel until the bimetal disconnects it again. The more the pressure, the more the heating element is switched into the circuit so the average resistance is lower.