See the system examples of the LM35 temperature sensor datasheet on page 16, see picture. This should be used for negative temperatures. However, the voltage change is 10 mV/C and it supports -55 C minimum, resulting in -550 mV. A diode (1N914) has a typical drop of 600 or 700 mV .... why are two diodes used, and not one?

(I play to use a 1N4001 or 1N4148 since I have plenty of them).

enter image description here

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    \$\begingroup\$ A "typical" drop is irrelevant here. You need to ask yourself: what is the minimum diode forward voltage when passing the minimum current from the sensor? \$\endgroup\$ Commented Feb 7, 2019 at 13:14
  • \$\begingroup\$ Yes, just checked, for an 1N4148 the minimum is around 0.4-0.5V, which could be too less, thanks! It also answers my question that I can use 1N4148 instead. \$\endgroup\$ Commented Feb 7, 2019 at 13:17

1 Answer 1


The whole point about this circuit is that the 18 kohm resistor is tied to a negative voltage relative to Vout- and this ensures that the LM35 can produce a signal that reflects negative temperatures. Without the bias resistor to a negative supply rail (relative to Vout-), the LM35 only works from +1 or 2 degC.

2 diodes are used because despite what people simplistically call "the standard volt drop of 0.5 volts to 0.9 volts for a range of diodes", the actual volt-drop depends on the current through the diode and, if it is low (as per the case of this circuit - circa micro amps) then you'll probably need two diode drops to produce something in excess of 0.55 volts.

Here's an extrapolation of what a 1N4148 diode might produce at 1uA of current: -

enter image description here

Given that an LM35 might only draw about 10 uA at -55 degC, you can imagine each diode dropping maybe 0.4 volts hence two diodes are needed.


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