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I have a 2k ohm sensor that fits properly but I can not find a proper 10k that fits a M14 hole for the water temperature on my car.

The gauge I am using only reads the NTC 10k thermistor so I was wondering if it was possible to convert the resistance. Bit new to the thermistor world.

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  • \$\begingroup\$ I think this issue with this generally is that you would need some sort of power source for the conversion circuitry, which would result in a solution that would not be an exact replacement for your NTC. It would maybe-possibly-theoretically be possible to make som sort of circuitry that converts the 2k thermistor to a 10k thermistor that draws power from the measuring circuit itself, but that would require intense engineering and knowledge and speaclist construction that is so far beyond the scope of this problem as to be effectively impossible. \$\endgroup\$ – BeB00 Oct 17 at 8:45
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    \$\begingroup\$ The actual answer to this problem, as stated by @Spehro Pefhany, is no. It will be 100% impossible for you to find anything that can do that conversion. However, you could look at it like this: Someone made the 10k sensor that was there before, so all you need to do is find out how to get one of those. This is vastly simpler and less expensive than trying to engineer some insane conversion circuit. \$\endgroup\$ – BeB00 Oct 17 at 8:47
  • \$\begingroup\$ What is "the gauge you are using" and do you have a link to its datasheet or manual? \$\endgroup\$ – Brian Drummond Oct 17 at 12:45
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Assuming the resistance-vs-temperature curves are proportional between the two thermistors, it is possible to make a ground-referenced 2k sensor behave like a ground-referenced 10k sensor. It just requires a bit of trickery with current mirrors.

schematic

simulate this circuit – Schematic created using CircuitLab

OA1 keeps the voltage across the 2k thermistor (R3) the same as the terminal voltage of the sensing circuit represented by V1. This causes a certain amount of current to flow through R2 and R3.

OA2 keeps the voltages across R1 and R2 the same. Since R1 is 5× the value of R2, this means that only 1/5 the current is required through R1, controlled by Q3.

Q4 mirrors the current through Q3 (these need to be matched transistors, kept at the same temperature, etc.). So, whatever voltage the sensing circuit applies to Q4, it behaves like R3, except with 5× the resistance.

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    \$\begingroup\$ Wow cool attempt .Sure I would beat it into shape for production .But this is not a commercial site .It should inspire the asker to implement something .+1 from me .This post in my opinion should get more votes . \$\endgroup\$ – Autistic Oct 17 at 10:32
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    \$\begingroup\$ Interesting idea... It can be added to the "bag of tricks" for creating "virtual elements (resistors)". Only, before that, it should be explained in simple words. The resistor R3 (2 k) serves only as a "shaping element" that, according to Ohm's law, "produces" current proportional to the voltage across it with a ratio of 2 k. Then this current is scaled x 0.2 and is used as a replacement of the original current. So, the idea is to replace the original current by a "scaled current" (at the same voltage). It is implemented by a "scaled functional current source"... \$\endgroup\$ – Circuit fantasist Oct 17 at 14:01
  • \$\begingroup\$ … So, the whole circuit acts as a 2-terminal "scaled voltage-to-current converter"... that can be considered as a "10 k virtual resistor". Another idea would be, instead of fully replacing the current with a scaled one, to decrease the current through the original 2 k resistor by adding a "helping current"... or by inserting an "opposing voltage" in series with the voltage drop across the resistor. These circuit solutions would be specific implementations of Miller's idea. \$\endgroup\$ – Circuit fantasist Oct 17 at 14:01
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Simpler idea (thanks, Circuit fantasist), although this version requires that the 2k thermistor be "floating" with respect to ground. See my other answer if that isn't possible.

schematic

simulate this circuit – Schematic created using CircuitLab

By reducing the voltage across the actual thermistor to 1/5 of the applied voltage, it draws 1/5 the current from the measuring circuit, and therefore appears to have 5× the resistance.

The current through R1 and R3, while nonzero, will be negligible (less than 1% of the thermistor current).

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    \$\begingroup\$ Note to OP. You need to find an op-amp that will work over the range of voltages you need with the power supply you have available. \$\endgroup\$ – Transistor Oct 17 at 14:39
  • \$\begingroup\$ Dave, This is just amazing! I came up with exactly the same circuit today that I scribbled on a piece of paper while walking in the park and thinking (I just had not put the compensating resistor R4). The idea for it came to me from the op-amp gyrator circuit (simulated inductor). This can also be called "imperfect bootstrapping". I also began preparing an answer... but later I hesitated. Now I will explain why ... \$\endgroup\$ – Circuit fantasist Oct 17 at 19:50
  • \$\begingroup\$ I realized something new for me. Really, the 2 k sensor resistance is artificially increased 5 times by inserting (subtracting) 1/5V1 in series. But I think this "magic" is in action only when V1 (and accordingly, VOA = 1/5V1) varies... and V1 "sees" 10 k resistance. But here V1 is constant... and 1_5V1 stays constant when R2 (thermistor resistance) varies. My doubt is that when R2 varies, its resistance is not modified (virtually increased) by this trick. Miller effect needs "dynamic voltage" but in this arrangement, it is static. Am I right? \$\endgroup\$ – Circuit fantasist Oct 17 at 20:07
  • \$\begingroup\$ @Circuitfantasist: No, my circuit works even as V1 varies. The voltage drop across R2 is always 1/5 of V1, regardless of the actual values of V1 or R2. And therefore, the current is also 1/5 of what it would be if R2 were connected to ground. Therefore, the effective resistance is 5x. \$\endgroup\$ – Dave Tweed Oct 17 at 21:16
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    \$\begingroup\$ @Circuitfantasist: I make no assumptions about how the sensing circuit works. It could either hold the voltage constant and measure the current, hold the current constant and measure the voltage, or some combination of the two (e.g., resistive divider). \$\endgroup\$ – Dave Tweed Oct 18 at 2:08
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Okay, inspired by Dave (+1), a generic answer that works if one side of the sensor is grounded is to use a Generalized Impedance Converter (GIC), a configuration much beloved by my old professor Adel Sedra. It will work even if the sensor is energized by AC.

schematic

simulate this circuit – Schematic created using CircuitLab

As should be obvious from the configuration, the same voltage appears across the 2K sensor as appears across the simulated 10K sensor, so self-heating will be 5x worse.

The impedance as seen at the left port is

\$Z_{in} = \frac{Z_1 Z_3 Z_5}{Z_2 Z_4}\$, in the case of the above circuit R5*5.

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    \$\begingroup\$ This is mostly for general interest. Adding an active circuit is a bit extreme. Since 10K thermistors are cheap and plentiful it would seem easier to simply adapt a thermistor to whatever the mechanical requirements are. Much easier to say if one has access to a lathe or two, I suppose. \$\endgroup\$ – Spehro Pefhany Oct 17 at 17:45
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    \$\begingroup\$ Probably a drill press is the only machine tool needed if one can find a bolt of the apropriate size. \$\endgroup\$ – Jasen Oct 17 at 21:56
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    \$\begingroup\$ It would be a great challenge to find (maybe in another question) an intuitive explanation of this mess of resistors and op-amps... \$\endgroup\$ – Circuit fantasist Oct 18 at 8:50
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Generally speaking, no. If you made it read correctly at one temperature it would be very inaccurate at other temperatures.

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  • \$\begingroup\$ That was my fear. Looks like I am back to trying to make a 10k sensor fit. \$\endgroup\$ – Pete Sother Oct 17 at 3:57
  • \$\begingroup\$ Depending on how the gauge reads the sensor, it might be possible just by lowering the pullup resistance. Which gauge are you using? \$\endgroup\$ – Bruce Abbott Oct 17 at 4:49
  • \$\begingroup\$ I would probably advise against trying to modify the OEM circuitry of your car unless you have a pretty good grasp of EE, which I'm not sure is the case here. Why cant you just find a replacement for what was there before? It must exist (even if there wasnt something there before, this hole is M14 for a reason). \$\endgroup\$ – BeB00 Oct 17 at 8:49
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    \$\begingroup\$ This is an aftermarket gauge. a poor quality little one at that. I choose it because I was able to 3d print a housing that fit on one of the expansion slots. the OEM one is old and reads hot or not basically. 80s cars.... \$\endgroup\$ – Pete Sother Oct 17 at 13:15

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