First question - What is this resistor network doing? Answered that it's linearizing the thermistor but this isn't what I was looking for. I'm looking for an explanation that describes the relationship between the resistors and the output

Edit (June 23 2015) for Question #2:

I'm doing calculations on the beginning resistor circuit and something's not making sense. I need help calculating or an equation which will give me the voltage for Vout.

Here are the numbers I've been using.

  1. V+ = 12V
  2. RA, RB, RC, Rd = 15kohm for simplicity I'm using all the same amount
  3. Rth (multiple data) = 15k, 9.6k, 2k, 39k

I took measurements from the RA,RB,Rth node (#1) and RB,RC,RD node (#2) and the Rth,RD (Vout #3) node.

For: •

  • Rth = 15k -> #1 = 7.5, #2 = 4.5 and #3 = 6V.
  • Rthh = 9.6k -> #1 = 7.4, #2 = 4.6 and #3 = 6.3V.
  • Rth = 2k -> #1 = 7.2, #2 = 4.8 and #3 = 6.9V.
  • Rth = 39k -> #1 = 7.64, #2 = 4.36 and #3 = 5.24V.

I understand #1 + #2 = V+. But if I do voltage division for #3, it doesn't give me the correct voltage. I'm assuming because Rc is coming into play with Rd (parallel?) but I'm not 100% sure. I've been told the only reason #1+#2 = V+ is because Ra = Rc. I came to that conclusion because I'm trying to find the relationship between all of the resistors.

What am I missing?

Overall circuit task: I am trying to put 12V into the circuit and get 1-5V out of the opamp depending on the amount of light the Thermopile picks up for an outside application. So the temperature could be 30F to 110F (0C-43C).

Circuit 1


2 Answers 2


They're linearizing the thermistor. Basically by putting a resistance in parallel with it, you can force it to be more linear over a certain range of temperatures. If this circuit is going to feed an ADC, linearizing the thermistor is unnecessary and often detrimental.

  • \$\begingroup\$ I see. How would you recommend the circuit to be then? \$\endgroup\$
    – Luke
    Jun 18, 2015 at 13:15
  • \$\begingroup\$ @Luke All you've done is throw up a schematic and asked what a part of it is doing. I have no idea what you need or expect it to do. \$\endgroup\$
    – Matt Young
    Jun 18, 2015 at 13:16
  • \$\begingroup\$ @MattYoung I understand the linearizing into an ADC is unnecessary (just linearize in software), but why could it be detrimental? Thanks. \$\endgroup\$
    – TronicZomB
    Jun 18, 2015 at 13:19
  • \$\begingroup\$ @TronicZomB Linearization costs sensitivity. \$\endgroup\$
    – Matt Young
    Jun 18, 2015 at 14:09
  • \$\begingroup\$ Please see the link I commented onto my post. The resistor (Rd) is in series with Rth and is used when a voltage source is applied. If Rd was in parallel, it would be because of a current source. Or that's what I got from the website. \$\endgroup\$
    – Luke
    Jun 18, 2015 at 18:16

The schematic is for a thermopile IR sensor. The part shown below is normally inside the sensor package, with the values determined by the sensor manufacturer.

enter image description here

Like all thermocouple devices, the thermopile does not measure actual temperature but has a voltage output that is (to first-order accuracy- the relationship is nonlinear and the exact relationship involves both temperatures) proportional to the difference of the temperatures across the junctions. In this case there are probably about 100 thin-film junctions in series.

Since it is only measuring a temperature difference, it requires compensation for the internal temperature at the thermopile 'cold' junctions. This is achieved by an NTC thermistor and resistor network.

The resistor network scales and (roughly) linearizes the very nonlinear (Steinhart-Hart equation) behavior of the thermistor to provide an increase of a similar number of uV/K as the thermopile loses when the ambient temperature increases (probably some mV/K since there are many junctions in series).

Since the thermopile produces voltage and the network provides a ratio, the network inside the sensor must be designed to work at a specific reference/supply voltage.

  • \$\begingroup\$ Unfortunately, I've spoken with the manufacturer and they aren't really helpful for values of Ra, Rb, RC, or Rd. For the thermopile that I'm using, the thermistor and thermopile have separate leads. I appreciate the detailed description about a thermopile. Could you elaborate further on the resistor network? How it functions? Rth and Rd create a voltage divider that changes as temp increases/decreases, but what is the function of Ra, Rb and Rc? \$\endgroup\$
    – Luke
    Jun 18, 2015 at 18:13
  • \$\begingroup\$ As I said it's to linearize and scale. If you have the thermistor specifications you can use the procedure in this Hammatsu data \$\endgroup\$ Jun 18, 2015 at 18:17
  • \$\begingroup\$ Actually I was going through that guide before you brought it up to me. I got stuck on Step 2 and 3. For step two, I was able to get the cooling/heating data but I wasn't sure what to do about the voltage measurements. Do I cool/heat the thermopile then put 12V across it and measure at the temp? Just things I don't know. Step 3 - Technically you'd have Vb and Vc as unknowns after you solve for Voutmin and Voutmax. I went through and solved for the unknown for one equation, plugged into the other, get an answer, then put it back into the first question but got really bad numbers. \$\endgroup\$
    – Luke
    Jun 23, 2015 at 19:47
  • \$\begingroup\$ Step 3 #3. is very oddly worded. Order of magnitude refers to 10^2 or .01 less than. Does that mean Rb is (Rth+Rd)*0.01 or is it (Rth+Rd)-((Rth+Rd)*0.01)? From my heating/cooling measurements I got Rth average value to be 12k and Rd to be 9.6k from step 1. So for step 3, 21.6k*0.01 = 216. Should Rb be 216 ohm or is it 21.6k-216 = 21384ohm. I've tried to contact Hamammatsu but very delayed not helpful responses. That's why I'm on Stack. \$\endgroup\$
    – Luke
    Jun 23, 2015 at 19:50
  • \$\begingroup\$ Order of magnitude is 10:1 \$\endgroup\$ Jun 23, 2015 at 20:11

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