The circuit is wired as follows

vcc --- R1 (1M ohm) ---|Thermistor --- Ground

The Thermistor has a resistance of 1M ohm when it is at 25 C. I am around that temperature now and indeed my ohm meter shows it has a resistance around 1M. I am seeing voltage drops of about 1.1V across the resistors but VCC is 3.3V. This blows my mind but I really don't know enough fundamentals to understand how this is the case.

If I switch the voltmeter out for something like an analog pin on an arduino I get very noisy input which does NOT match the voltmeter. On the arduino I am sampling about once a second.

One last point is that the probe I am using is this probe. It seems that grounding the metal sheath might help reduce the noise I am not yet certain.

Improvements to noise. It seems I could benefit from using a capacitor and multisampling to improve the quality of the signal. Information related to my Microcontroller

  • \$\begingroup\$ What is the input impedance of your voltmeter? \$\endgroup\$
    – nanofarad
    Dec 28, 2018 at 22:30
  • \$\begingroup\$ Hint: what is the equivalent resistance of your voltmeter? \$\endgroup\$
    – The Photon
    Dec 28, 2018 at 22:30
  • \$\begingroup\$ Klein tools mm300 is the voltmeter I am looking it up now \$\endgroup\$
    – flips
    Dec 28, 2018 at 22:33
  • \$\begingroup\$ Is there a circuit I can make to detect my voltmeters impedance? I ask because unfortunately this information is not listed in the manual \$\endgroup\$
    – flips
    Dec 28, 2018 at 22:36
  • \$\begingroup\$ Many DVMs have an input impedance of 10 Megohms - try calculating the voltage across the resistor with a 10 Meg in parallel to verify that value. Or, knowing the voltage, calculate the equivalent resistance of your meter and the resistor in parallel. \$\endgroup\$ Dec 28, 2018 at 22:46

2 Answers 2



simulate this circuit – Schematic created using CircuitLab

Figure 1. Circuit (drawn using the built-in CircuitLab schematic editor).

If your multimeter DC voltage range has an input of 1 MΩ which is typical - although 10 MΩ is also common - then you have 1 MΩ in parallel with the 1 MΩ thermistor giving 500 kΩ for the parallel combination. With R1 as shown that creates a 3:1 voltage divider so 1.1 V would be the expected meter reading.

  • \$\begingroup\$ Great! Alright this makes sense for the voltmeter reading. How about for the analog reading on the micro controller? \$\endgroup\$
    – flips
    Dec 28, 2018 at 23:44

You should determine whether your voltmeter is disturbing the behavior of the circuit, due to a low equivalent resistance.

Try this: put two 1 megohm resistors in series between 3.3 V and ground. Use your voltmeter to measure the power supply voltage (should be 3.3 V) and then measure the voltage across each resistor. These voltages should be about 1.65 V each, and should add up to be equal to the measured supply voltage. If not, then the voltmeter is disturbing the circuit behavior.

The input impedance of an Arduino A/D pin is probably going to behave even worse than your voltmeter. You will probably need to add a unity-gain buffer amplifier between the thermistor circuit and the Arduino. Another option would be to use a thermistor with a much lower resistance, say 10 kilohm or less.

  • \$\begingroup\$ For the arduino so I can use something like what is found here? analogictips.com/… I have some lm 358m that should work for this purpose thanks! \$\endgroup\$
    – flips
    Dec 29, 2018 at 0:22
  • \$\begingroup\$ Yes, that's the right kind of circuit. An LM358 might work for you, but note that the maximum output voltage is about 3.3 V for a supply voltage of 5.0 V. Also, the input bias current is large enough that it could introduce an error of 100 mV or so. Can you live with those restrictions? If not, look for a "rail-to-rail input/output" op amp with a lower input bias current. \$\endgroup\$ Dec 29, 2018 at 20:10
  • \$\begingroup\$ I am supplying it 5.0V and the signal coming in to the non-inverting input of the op amp is .95V the signal going out is about 1.44V. I am flabbergasted and will make a proper diagram of my circuit to share so I can get more meaningful feedback \$\endgroup\$
    – flips
    Dec 30, 2018 at 12:46
  • \$\begingroup\$ If you are using your voltmeter to measure the input voltage then your measurement is almost certainly incorrect. \$\endgroup\$ Dec 30, 2018 at 13:01
  • \$\begingroup\$ crap its the same thing again im such an idiot yes! sorry I keep forgetting about its resistance \$\endgroup\$
    – flips
    Dec 30, 2018 at 13:31

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