I'm planning on using the LM35 temperature sensor to measure both positive and negative temperatures. The recommended schematic to measure negative temperatures without a negative voltage supply is this:

LM35 negative temperature

I understand that the role of the diodes is to lift/offset the negative pin from GND. What I don't understand is: What's the role of the 18 kOhm resistor?

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    \$\begingroup\$ The 18k is probably because the output is open collector. Why not use the lm135 instead? Those diodes will probably make you lose the initial 0.5c accuracy anyway. \$\endgroup\$ Commented Jan 9, 2014 at 12:10
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    \$\begingroup\$ @Erik: Definitely not, because the tempertaure is indicated by the difference of the + and - outputs. (Which are a bit misleading: for below-0 temperatures + will be more negative than -). \$\endgroup\$ Commented Jan 9, 2014 at 12:17
  • \$\begingroup\$ @ErikFriesen The LM135 is pricier.. \$\endgroup\$
    – m.Alin
    Commented Jan 9, 2014 at 12:29

1 Answer 1


A negative temperature is indicated by the LM35 with an output (marked in your circuit as +) level that is more negative than its ground pin (marked in your circuit as -). But the LM35 has no supply of a voltage that is more negative than its ground pin, hence it needs something to create that negative voltage from. That is the current supplied by the 18k resistor.

  • \$\begingroup\$ So I guess I'll need to buffer the (+) output of the sensor before reading it with a PIC ADC? IIRC the PIC's ADC has a max. input impedance of 2 kOhm.. \$\endgroup\$
    – m.Alin
    Commented Jan 9, 2014 at 12:43
  • \$\begingroup\$ I am not sure. IIRC the input of a PIC A/D is mostly capacity, so you might get by with a longer acquisition time. \$\endgroup\$ Commented Jan 9, 2014 at 12:57
  • \$\begingroup\$ Actually what I'm interested in finding out is if the resistor makes the total output impedance of the sensor to be 18 kOhm or if it's considered as being in parallel with the (+) pin output impedance, so it's total output impedance is much lower than 18 kOhm.. \$\endgroup\$
    – m.Alin
    Commented Jan 9, 2014 at 13:02

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