My main task is to make a 10k NTC sensor look like a 100k sensor. I have attached the first 3 images below which show the Excel sheet I am working on which includes the temperature (Range of -55 to 300 degrees C) and NTC resistance values (for both 10 and 100k) accordingly. I have then tried to replace Rp1, Rp2 and Rs with some values in order to match the data. I think it will be easier to apply the voltage divider formula to this circuit as shown in the third picture. I want to find the voltage across the "red" resistor as a function of NTC temperature but not sure how to do this or if there is any other easier way to go about this.

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Using \$V_{out} = V_{in} \times Req~/~(R1 + Req)\$, I believe Vin will be 5V, R1 will be 13kohm and Req wil be Req=RP2∥(RS+(RP1∥RNTC)). I am not sure how to work out the values for Rs, Rp1 and Rp2 which will need to be substituted.

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I hope I have made this more clear and would highly appreciate any help with this.

  • 1
    \$\begingroup\$ Which diagram of all the ones you provided you mean? And any resistance value would depend on which resistance range the NTC has which means it depends on NTC type and range of temperatures you intend to measure. \$\endgroup\$
    – Justme
    Jul 21, 2023 at 13:13
  • \$\begingroup\$ Do you mean Rp2? In any case it all depends on what "suitable Vout ratings" you are aiming for and the specification of the NTC \$\endgroup\$ Jul 21, 2023 at 13:56
  • \$\begingroup\$ Is this for temperature sensing on an MCU, by any chance? In that case, there is a simpler solution: don't solve for resistance at all, go directly to temperature. \$\endgroup\$ Jul 21, 2023 at 15:42
  • \$\begingroup\$ Thankyou for the response @Justme , I have made the question more clear now. Hope this helps. \$\endgroup\$
    – Mahnoor
    Jul 21, 2023 at 18:12
  • \$\begingroup\$ Thankyou for the response @Peter Jennings , I have made the question more clear now. Hope this helps. \$\endgroup\$
    – Mahnoor
    Jul 21, 2023 at 18:13

2 Answers 2


Your proposed solution is impossible: from -55 to 300°C, a typical NTC thermistor spans five decades of range.

There exists no linear circuit which can simply shift one decade over.

A fit can be made, over a narrow temperature range, using series-parallel resistors, potentially themselves having some nominal tempco, or potentially requiring negative values (which can be constructed relatively easily with op-amps, assuming a power source is available, and that considerations like noise immunity have been made).

A nonlinear circuit that emulates a resistor based on another, could be constructed, at great expense. You will likely find it several orders of magnitude easier to purchase (or import) the correct component in the first place.

If the sensor is sensitive to voltage instead of resistance, a transfer function can be constructed, though it requires great precision to match the full range: again, the dynamic range is five digits, or about 18 bits. High precision components and well-performing ADC-DACs will be required, as well as an MCU. Fewer bits are required if you don't need resolution over the full range (say, -55 to -20°C reads the same minimum-scale value, and 200-300°C reads the same maximum-scale value; and various steps inward, with resolution in the middle span being comparable to device accuracy).

  • \$\begingroup\$ Hello @Tim Williams, thankyou for your response. Ignoring the -55 to 300°C range, how would I further calculate Vout 'as a function of temperature' using the following Vout values? I would like to try to calculate Vout across any range of NTC temperature. I have been told to take Rp1 and Rp2 as 1*10^12, NTC is 10,000, and Rs values range from 10-100 ohm. This gives V out values (using the voltage divider formula) of 0.1, 0.05, 0.03, 0.03, 0.02, 0.02, 0.01, 0.01, 0.01 and 0.01V. \$\endgroup\$
    – Mahnoor
    Jul 23, 2023 at 15:53

The general voltage divider formula is $$ V_{Z_x} = V_{in}\cdot\frac{Z_x}{Z_t} $$ where \$Z_x\$ is the impedance you are taking the voltage across and \$Z_t\$ is the total impedance.

For two resistors it becomes $$ V_{R_2} = V_{in}\cdot\frac{R_2}{R_1+R_2} $$

To get from the original circuit on the bottom right to the simplified one on the top left you go through the steps shown, combining parallel and series resistances until you have \$R_{P1}\$, \$R_{P2}\$, \$R_S\$ and the NTC resistance all combined into one equivalent resistance.

You can do this one step at a time as shown in the diagrams. The formula for the whole thing is going to be $$ V_{out} = 5V\cdot\frac{R_{eq}}{13k\Omega+R_{eq}} $$ where $$ R_{eq} = R_{P2}\parallel(R_S + (R_{P1}\parallel R_{NTC})) $$

Without knowing what you consider suitable Vout ratings and what the NTC characteristics are there's no way for us to tell you what the unknown resistances \$R_{P1}\$, \$R_{P2}\$ and \$R_S\$ should be.

Two ways you may be able to get a solution are to do a simulation and adjust the values to get the results you want, or put the formulas into a spreadsheet and calculate \$V_{out}\$ with different resistor values.

The datasheet for the NTC thermistor or the source of your schematic may give some clues as to what values to use in that circuit, if you could provide those we may be able to provide a more complete answer. You can also search for articles on thermistor linearization such as this one.

  • \$\begingroup\$ Thankyou for the response @GodJihyo, I have made the question more clear now. Hope this helps. I have provided the NTC data. \$\endgroup\$
    – Mahnoor
    Jul 21, 2023 at 18:15

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