Reading this datasheet of the LM94022, chapter 8.3, where is shown the following example:

Application example of the LM94022

I don't know how they get the formulas. I more agree to achieve something like this:

Example that makes more sense

taken from this answer.

What am I missing to get to the same formulas of the application example?

  • 2
    \$\begingroup\$ I agree with your derivation; note that if comparator Vout is not 4.1V then it becomes a little more difficult to solve. \$\endgroup\$ – Peter Smith Mar 31 '17 at 14:15


simulate this circuit – Schematic created using CircuitLab

Basic Thevenin stuff. VREF is what you want. VREF = I3*R2. Plug in the equations to make I3, and solve. Put in VOUT = 0 for one threshold and VOUT = V+ to find the other.


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