Pressure Sensor Circuit - advice/critiques appreciated!

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rockrockmcrock

Joined Jan 6, 2012
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Hi guys,

I'm a (modern) electronics newbie - but am slowly learning the basics for this project. Loads of peeps have already been very helpful on helping me get started, but I still seem to have loads to learn, so any advice is hugely appreciated.

The problem:
I want to retain some old bi-metallic gauges in one of my classic cars (got to keep it original on the dash right?). These older gauges are "bi-metallic electronic". However modern gauges work in a different way (i.e. potentiometer based). This means that modern 'sensors' for gauges are mostly 'reversed' in the way they work (more on this later). Of course without the right signal from the sensor the gauge will not work. Is there a simple circuit to make modern sensors work with old bi-metallic gauges?

How bi-metallic gauges and sensors work:
All gauges are 10v. The gauges work by a 'sensor' which acts as a variable resistor and limits the current on the gauge circuit in line with pressure. The current heats a coil wrapped around a bi-metallic strip in the gauge with responds by moving the needle on the gauge. Low pressure = high resistance (240 ohms) => no heating of coil & gauge needle reads low, high pressure = low resistance (33 ohms) => coil is at it's hottest & gauge needle reads high.

How modern sensors work:
Most sensors are 10v. Modern sensors work the opposite way round but over roughly the same range. In other words, Low pressure = low resistance (33 ohms) => gauge needle reads low, high pressure = high resistance (240 ohms) => gauge needle reads high. If you put this modern sensor on the old bi-metallic gauge setup the reading's would be reversed (i.e. the needle would read high when at low pressure and low when at high pressure.

What the project circuit would need to do:
Work from a 12v car battery, produce a 10v regulated sources and then reverse the modern sensors output - i.e. when the sensor gives high resistance, allow more current to flow on gauge circuit and visa versa.

The Current Project Circuit
So far (with a lot of help from others - thanks H!) I have come up with this circuit - but will it work?:


Here's my understanding of the above circuit and how it works (I might be sooo wrong though!):
  • Voltage regulator produces 10v & up to 1A (easily enough) at 2 & 6 for 10v rated gauge and RSensor
  • Voltage divider (2,3,4) effectively inverts RSensor output, by producing a voltage at 3 which is proportional to RSensor resistance (i.e. increasing RSensor resistance, increases voltage at 3). RTrimmer can be used to vary the sensitivity for the op amp (ie "high-low" voltage change delta - is that correct?)
  • No current is carried across the Op Amp (3,5), this is the need for the second circuit 6,7,8,9 - the voltage at 3 is simply used to proportionally vary the current across 6,7,8,9.
  • The Op Amp triggers the NPN transistor to control the current and the feedback loop 8 to the Op Amp provides the reference input to the Op Amp (comparison of 8,3).
  • To ensure the right current in 6,7,8,9, RSense is used, such that current across RSense (8,9) must equal current across Gauge (6,7). Does this mean I need to know the resistance of the Gauge (Load) though to be accurate?
OK, assuming I'm not completely wrong about the above - any advice the following hugely appreciated:
  1. NPN Transistor needs to carry the current and be triggered by a low current - hence a high gain, 400mA+ specification is ok?
  2. The Op Amp - there seems to be loads of variables to consider here....totally confused at present, but slogging on! All help appreciated!
  3. Any improvements to the circuit

Thanks again guys and to all in the community (internet and real-world) who have helped on this already. This seems to be a very widespread issue in the classic car world, but no one has found an easy solution that I can see - if we can get this solved then the classic car community would be in your debt!

RRMR
 

Dodgydave

Joined Jun 22, 2012
11,303
You mentioned the fatal subject THE CAR.......


You have a constant current regulator in R sense, the current is set by the voltage at pin 3 of the op amp, as the voltage rises and falls at pin 3, the current in R sense follows it and the transistor drives the gauge with the same current.

Your posted question will probably be closedown,
 
Last edited:

bertus

Joined Apr 5, 2008
22,277
Hello,

I am closing this thread as it violates AAC policy and/or safety issues.

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