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This sounds like an automotive question, but its really not.

I have a engine temperature sensor (presumably a thermistor, as it reduces resistance with increasing temperature). I need to mate this to a temperature gauge, but the gauge is calibrated for a different sensor with a different temp. response curve.

temp/ohm curves are known for both sensors (see graph) enter image description here

For various reasons I can't replace the engine sensor. Also the gauge is a sealed unit, so lets assume I can't modify the gauge internals.

Is there anyway I can insert a circuit between the sensor and gauge, so the gauge 'thinks' its connected to the 'correct' sensor? Preferably a circuit of passive components as I don't have too much electronics experience.

Initially I assumed I might be able to do this by putting a suitable resistor in parallel with the sensor, but any resistor like that will only 'flatten' the response curve and as you see, response curve needs to be 'stretched'.

A couple of other factors to consider.

  1. I don't need super accuracy; +/- 3 deg C is fine
  2. I only this to work well in the range 85-105 deg C

Is this possible in principle and what is the general approach?

EDIT: More info

The red curve is given by: resistance(ohms)=114+6014*e^(T*-0.0396), where T=temp in deg C

Measured resistances from the existing engine sensor (blue line) is below. I would hope there is enough data there to predict the remaining values up to 105degC (do themistors follow a standard curve?)

enter image description here

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Use a microcontroller to measure the thermistor, then program a lookup table and make it control an electronic potmeter. Intermediate values can be calculated by interpolation.

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  • \$\begingroup\$ probably the most general and flexible approach to this kind of problem. But as I said, my electronics expertise is pretty basic. I was hoping for a solution, that perhaps, involved a couple of well placed resistors. But I may have to resort to learning about microcontrollers if ee.se doesn't produce a simpler solution. \$\endgroup\$ – Ken Apr 20 '13 at 21:52
  • \$\begingroup\$ Entry level microcontroller evaluation boards are not too expensive and many to choose from. A really easy option is arduino.cc For this application the smallest one will be sufficient, and they are all very easy to program from a PC. \$\endgroup\$ – jippie Apr 20 '13 at 21:59
  • \$\begingroup\$ looks like I'll have to go with your micro-controller route. do you know of any online resources for the circuit you describe (or similar). I'm a programmer by trade, so coding won't be a problem. It just knowing which bits to get and how to hook them up. \$\endgroup\$ – Ken Apr 22 '13 at 10:31
  • \$\begingroup\$ No I don't have a good example at hand. If you are unfamiliar with microcontrollers and want to make a quick start, check out an Arduino Uno compatible board (if you search around, you can get them for nice prices). Advantage is you don't need anything else but a USB cable to start fiddling with it. That would be step 1. Then to find a proper digital potmeter, I'd advice to join us in chat: chat.stackexchange.com/rooms/15/electrical-engineering We may be able to guide you a bit. \$\endgroup\$ – jippie Apr 22 '13 at 17:23
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If the existing engine sensor were the red graph then I'd say this can't be done with a few simple components. But because it's the other way round and you only want accuracy in a portion of the temperature range then this is achievable.

However, your graphs do not give enough technical detail about the resistance of either gauges in the 85ºC to 105ºC range to calculate a circuit so to answer your question, YES there is a way to do what you want BUT, NO it's impossible to do this accurately with the data you have provided.

A more in-depth look at the resistances of both sensors from 75ºC to 120ºC is required. Can you provide this information?

EDIT - Addition You now have data for the engine sensor albeit a little limited. You obviously have spreadsheet capabilites and this means you are nearly there. Given the resistance formula you have for the red-curve try creating another table of numbers that puts this resistance in parallel with another fixed value resistance, Rf. You do this by Rf.Rt/ (Rf + Rt) where Rf is the fixed value and Rt is the thermistor value.

See what experimenting with different values for Rf does to the new numbers - how far off the engine values do you get?

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  • \$\begingroup\$ I've added hard data to the question, maybe that will help?! \$\endgroup\$ – Ken Apr 20 '13 at 19:48
  • \$\begingroup\$ @Ken if you don't have data above 98ºC how can you judge whether a solution meets your spec (+/-3ºC). I can't guess the data but maybe you can by using a spreadsheet and following the trend mathematically. If you can do that then quite possibly two resistors will do what you need. \$\endgroup\$ – Andy aka Apr 20 '13 at 20:20
  • \$\begingroup\$ I kind-of assumed that thermistors followed a predictable curve, that if you have some of the data points you can reverse engineer the mathematical curve for the device. I'll pull the sensor out and try to get some readings at higher temps \$\endgroup\$ – Ken Apr 20 '13 at 20:45
  • \$\begingroup\$ setting up a spreadsheet to model a resistor in parallel was the first thing I tried (using 1/a=1/b+1/c), but whatever value resistor I tried tended to 'flatten' out the blue curve (reduce it's range), when in fact I need the 'expand' the blue curve to match the red one. \$\endgroup\$ – Ken Apr 20 '13 at 20:50
  • \$\begingroup\$ @Ken I see what you mean... It is the other way round. OK it's not gonna be cured by a simple resistor thing. To get around this is non-trivial compared to just a couple of resistors. I thought you were trying things the other way round. Sorry - it's more complex and requires at least an op-amp and some way of calibrating the gauge against a current fed into it. \$\endgroup\$ – Andy aka Apr 20 '13 at 22:42
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If i understand your question correctly you want to go from a lower resistanse to a higher resistance. That may be doable wth a series + parallel resistor.

I bet your thermistor follows an exponential curve (try plotting the log and see if you get a straight line). That shouuld make it easy to extrapolate the missing values.

I didn't solve the problem for you, as you sound like you want to conquer this using your own skills -rigth?

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  • \$\begingroup\$ as mentioned in a earlier comment, a parallel resistor will only tend to flatten curve (move it further away from the shape of the red curve), I need some way to 'add' curvature, and a resistor in series only serves to shift the curve up or down \$\endgroup\$ – Ken Apr 21 '13 at 9:24
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First your formula for the red curve makes no sense. However,using the curves it appears that you could get a good approximation of the red curve by using four of the thermistors corresponding to the blue curve in series. Using four multiplies the slope of the blue curve by four and makes it closer to the red curve.

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  • \$\begingroup\$ The formula for the red curve was found on website related to that gauge, It's the only source I have... Unfortunately the location of the thermistor in the engine will only allow for one thermistor. \$\endgroup\$ – Ken Apr 21 '13 at 9:24

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