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My ASUS motherboard has one thermistor connector that takes a 10 kΩ thermistor.

I want to connect 2 or 3 thermistors to this connector but only the one with the lowest resistance to be read by the motherboard.

Is this possible? If yes, then what would be the best approach? So far I tried larger thermistors in parallel but that is not accurate.

The chip is an NCT6796D.

schematic

simulate this circuit – Schematic created using CircuitLab

schematic

simulate this circuit

I am thinking to try an Arduino.
I'm trying to figure out how to make a potentiometer with it.
Maybe this will work:

schematic

simulate this circuit

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    \$\begingroup\$ Sorry, but generally, questions asking if something is possible are not very good; because everything is possible unless some restriction prevents it, such as laws of physics. \$\endgroup\$
    – Justme
    Commented Sep 9, 2022 at 4:18
  • \$\begingroup\$ It depends on what sort of circuit is being used to read the thermistor. If there is a current being passed through it to read a voltage, then this Q/A can be modified to provide a solution to picking the smallest voltage. If OTOH the circuit is using the thermistor as the feedback R in an RC oscillator to count frequency, or in a Wheatstone bridge, then it's maybe impossible, certainly more complicated. Post the schematic of the measuring circuit. \$\endgroup\$
    – Neil_UK
    Commented Sep 9, 2022 at 4:33
  • \$\begingroup\$ Now the additional requirement is that it must be the best way. That is also very subjective, and you don't define what best means to you. Likely you don't even want the best solution but just a good enough solution. But please understand that ways to do it depends on how the motherboard uses the thermistor, and it is currently unknown. Do you have the schematics of the motherboard, or can you reverse-engineer how the motherboard uses the thermistor? \$\endgroup\$
    – Justme
    Commented Sep 9, 2022 at 5:13
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    \$\begingroup\$ When you say “lowest” do you mean lowest temperature or lowest resistance? These are commonly NTC thermistors so those two conditions are opposite \$\endgroup\$
    – Bryan
    Commented Sep 9, 2022 at 5:22
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    \$\begingroup\$ Not possible in any simple way. \$\endgroup\$
    – Drew
    Commented Sep 9, 2022 at 6:21

2 Answers 2

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This is a variation of the answer from Neil_UK, where I (ab-)use a LM339 comparator with open collector outputs. This way no diodes are needed and the sense voltage can be as low as ~0.2V.

There is a separate reference IC just in case the mainboard reference cannot provide the additional current.

schematic

simulate this circuit – Schematic created using CircuitLab

As Neil_UK mentioned in a comment, using a comparator this way has a high risk of oscillations and stability problems. I soldered this circuit and indeed it needs a capacitor (here C3) at the output as mentioned in figure 11 of the LM339 datasheet from ST. The output low voltage is clipping at 30 mV.
It is possible to keep the NTC on the mainboard.

I simulated the circuit with LTSpice. The different NTC values are replaced by two sine generators with different frequencies and amplitudes. The output (green track) shows the min(a,b) function.

Analog_UMin

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  • \$\begingroup\$ Can I power the lm339 with 2v from the MCP1501? Can I use LM4040 instead of MCP1501? \$\endgroup\$
    – Chris
    Commented Sep 10, 2022 at 23:06
  • \$\begingroup\$ @Chris No, LM339 will probably not work with 2V in this circuit. It needs some headroom (1.5V) between maximum input voltage and VCC. So 3.3 V may be just sufficient. Yes, the shunt regulator LM4040 will work as well. It was indeed my first idea, but I was too lazy to create a schematic symbol for it. \$\endgroup\$
    – Jens
    Commented Sep 10, 2022 at 23:17
  • \$\begingroup\$ I'm trying to understand how lm339 works. Does it output the voltage that is larger or the voltage that is smaller? \$\endgroup\$
    – Chris
    Commented Sep 10, 2022 at 23:21
  • \$\begingroup\$ @Chris The datasheet shows a simplified inner circuit. There is only an NPN at the output, that can sink current. No source current comes out of it. In some aspects like an OpAmp with a missing upper output transistor. \$\endgroup\$
    – Jens
    Commented Sep 10, 2022 at 23:28
  • \$\begingroup\$ I see the NPN transistor. I guess there should be no worries to power it at a voltage higher than 2V. \$\endgroup\$
    – Chris
    Commented Sep 11, 2022 at 0:28
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It uses a lot of components, but I think it fulfills your requirement. Or maybe it's not a lot - one diode, one resistor and 1/4 of an opamp package per thermistor.

This only works if the motherboard uses an ADC to measure the voltage. They often measure chip voltages, and the OP's circuit is capable of generating a voltage, so that's a reasonable supposition. However, it's still possible to measure that thermistor in other ways, perhaps charging and discharging a capacitor and measuring timing, which this circuit won't handle.

schematic

simulate this circuit – Schematic created using CircuitLab

Each thermistor generates a voltage with its corresponding resistor, reproducing the voltage that the motherboard expects to see.

Only the lowest voltage channel will control the output voltage through its diode. Opamp feedback removes the effect of the diode voltage drop, rather like it's used in a precision rectifier circuit. Thermistors generating a higher voltage will have their op-amp output at positive rail, with their diode reverse biassed.

You'll notice I've not specified the voltage rails for these opamps. If the thermistor voltage never drops below a volt or thereabouts, and you use a rail2rail or single supply opamp that can drive to close to its negative supply, then you will get away with powering it from GND as the negative rail. Diode leakage is not important in this circuit, so you could use schottky diodes instead of silicon to get down to 0.5 V output with a R2R opamp. Otherwise you'll need a negative supply. Positive rail needs to be at least Vref, or more depending on how high the output voltage is required to go.

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  • \$\begingroup\$ Note that, if the source impedance can be rather high, the op-amps can be removed, just use the diodes from a much larger (100s k, megs?) pullup to each thermistor divider; completely passive. Conversely, if a low impedance is required, a forth amp should buffer the output, after the pull-up. (Less than 5k seems unlikely to be needed, given the original case; again, excepting for those possible weird use cases.) \$\endgroup\$
    – Tim Williams
    Commented Sep 9, 2022 at 9:36
  • \$\begingroup\$ @TimWilliams I'd need to see a circuit. How do you handle the diode drops without introducing a large temperature offset into the reading? \$\endgroup\$
    – Neil_UK
    Commented Sep 9, 2022 at 9:38
  • \$\begingroup\$ Nevermind -- I see VREF in OP's case is 2.048V, which leaves little room even for schottky diodes. Can be an option at higher voltages, but such is not the case here. \$\endgroup\$
    – Tim Williams
    Commented Sep 9, 2022 at 9:44
  • \$\begingroup\$ Thank you for the answer! I'm not sure if this will work to my situation. Like I said "my motherboard have one connector" and the thermistor connects to that. The 10k resistor is soldered to MB. If can't easily get the Vref then this will not work. \$\endgroup\$
    – Chris
    Commented Sep 9, 2022 at 16:10
  • \$\begingroup\$ It does not matter if it uses many components. I'm not going to commercially sell it. There will be one pcb therefore 5 component or 50 does not make too much difference. \$\endgroup\$
    – Chris
    Commented Sep 9, 2022 at 16:39

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