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I have a NTC thermistor It has zero-power resistance of 5kΩ at 25°C. I will use general voltage dividing and reading but I have 2 questions.

  • How to calculate the series resistance needed when using 10V supply ?

  • What changes whether the series resistance is at the upper leg or the lower leg?

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Let's have a voltage divider with 10VDC input

schematic

simulate this circuit – Schematic created using CircuitLab

If R1 is the NTC and R2 is an ordinary resistor, Vout increases when the temperature increases. If R2 is the NTC and R1 is an ordinary resistor, Vout decreases when the temperature increases.

Generally Vout = Vin*R2/(R1+R2) if we think the output current =0. If there's some substantial load, the formula is more complex.

The hard part: How we solve the temperature if we know the Vout and the other resistor? It's quite complex calculation and we must have the curve how the resistance of the NTC depends on the temperature

NOTE: Vout = 0V does not mean 0 degrees and Vout changes not 1V per a degree. The latter is called also "Vout has non-linear dependence on the themperature"

In the past very complex circuits were developed to make the dependence between the voltage and the temperature simpler. An example:

schematic

simulate this circuit

The selection of the components requires a strong math. But it's possible to achieve the following:

  • the voltmeter scale has zero degrees in the wanted position
  • the needle in the voltmeter turns nearly linearly as the temperature changes. This means the scale has near equally wide degrees.
  • the calibration for an individual NTC is possible by having some resistors as trimpots.
  • By having R1 as a trimpot, the aging of the battery could be compensated (needs also a switch and a fixed resistor in place of the NTC during the calibration)

This is a typical bridge for resistor-like physical sensors. The exact calculations are beyond the scope of this answer.

The operational amplifier circuits increase the possiblities and reduce a little the complexity of the calculations. The bridge can be replaced by a differential amplifier.

A computer (input via an ADC) moves all difficulties to the software. It does not remove the need of strong math, but in software the temperature-ohm-dependency can be taken into the account exactly. This is the used way also in the digital multimeters that have a temperature measuring range for a specific probe.

Your other resistor: nearly 5 kOhm, for example 4,7 kOhm is a good choice because

  • it makes the voltage change big per a degree near 25 degrees and the dependence is quite linear, too, but not exactly linear
  • at 25 degrees you have 5 volts room to both directions

Read other articles in this site (there are several) and find some application notes and circuit diagrams.

ADDENDUM due the comment: Self-heating is possible to compensate or reduce to be non-disturbing.

Reduction: Switch the Vin off for most of the time or make the thermal contact to the ambience more effective

Compensation: by hard math. If the self heating does not cause substantial heat generating or heat consuming extra processes, you can assume a known heat source in a linear medium. This requires also the time been taken into the account.

I recommend the reduction. Measure once per a second a 10 ms period and keep the Vin off for 990 ms. The self heating is reduced 99%.

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  • \$\begingroup\$ @user107577 added a hint against self heating. \$\endgroup\$ – user287001 Feb 11 '17 at 6:05
  • \$\begingroup\$ Many thanks for self reduction reducing method User2870001. Sorry i couldn't select it as an answer. \$\endgroup\$ – user107577 Feb 11 '17 at 6:20
  • \$\begingroup\$ @user107577 only to know If you had selected my story by accident, you could allways flag it and ask the peer to remove the selection. At least one of my answers has been selected as the best and thatafter someone has proved it fake. The flagging is for giving the signal to the peer. \$\endgroup\$ – user287001 Feb 11 '17 at 6:32
  • \$\begingroup\$ I dont have enough reputation if i had i could. Ok friend i selected as answer. Thanks again. \$\endgroup\$ – user107577 Feb 11 '17 at 6:37
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To maximize the resolution, match the thermistor value at the operating temperature.

Usually it's best to ground the thermistor so that you are not running your supply (typically it is the ADC reference) out where it can pick up noise.

You will get a lot of self-heating with a 10V supply (with most thermistors, especially in still air). That's 5mW. Take care that they do not specify the dissipation constant in some unrealistic situation (eg. flowing water at 1m/s) if you are operating in something like still air.

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  • \$\begingroup\$ Match the resistance in the middle of the operating range for maximum resolution then. Minimize voltage across the thermistor to minimize heating. 1.4mW/°C - if that is valid for your conditions you will get 3.5°C rise. If you use 5V rather than 10V the rise would be 1/4 or < 1°C. You can throw away some of the resolution of the ADC maybe and use a higher value resistor, really up to you depending on requirements. \$\endgroup\$ – Spehro Pefhany Feb 11 '17 at 5:02
  • \$\begingroup\$ Dissipation is Vtherm^2/Rtherm so if the thermistor can change through the value = Rseries that will be worst-case (and Vtherm = Vsupply/2). Otherwise it will be at maximum thermistor resistance. Self heating is power/dissipation coefficient. \$\endgroup\$ – Spehro Pefhany Feb 11 '17 at 5:33
  • \$\begingroup\$ With 5K series the worst case is at Rth = 5K, 5V across thermistor, 5V across resistor (matched load impedance to source impedance of thevenin equivalent for maximum power transfer- maybe you have studied this?). So 5mW and 3.5°C. \$\endgroup\$ – Spehro Pefhany Feb 11 '17 at 5:43
  • \$\begingroup\$ +1 for matching at the operating temperature, but ideally it's the centre of the operating range, getting reasonable linearity for +/- 10C from match. However self heating might drive OP to keep the current down and use a larger resistor than that. \$\endgroup\$ – Neil_UK Feb 11 '17 at 5:47
  • \$\begingroup\$ 100 ohms and 5K with 10V supply so 5.1K and about 2mA, and power in thermistor is I^2*R = 0.4mW. \$\endgroup\$ – Spehro Pefhany Feb 11 '17 at 5:50
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Often the series resistor matches the NTC resistance at room temperature to maintain maximum linearity. It doesn't matter much whether it's in the top or bottom position. Typically it's in the bottom, but it may depend on how you are interfacing it to the rest of the system, and how you are programming it.

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  • \$\begingroup\$ Thanks for response sir. You advice to use 5k resistance What do you mean by lineraity? I will directly send voltage to 16 bit adc and for calculation connection to lower lag seems easy. Any other critical factor you may say to take care . \$\endgroup\$ – user107577 Feb 11 '17 at 3:10
  • \$\begingroup\$ What I meant by linearity is that if the other resistor is matched, you will see the voltage change by the largest amount when the temperature changes, so therefore get the best resolution. Perhaps I should have said resolution instead of linearity. If the other resistor is very large or small, the resulting divider voltage will cover a smaller portion of the full scale range of the input. \$\endgroup\$ – AngeloQ Feb 11 '17 at 13:29

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