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Comparator Circuits

  1. Feb 13, 2014 #1
    1. The problem statement, all variables and given/known data
    We have to answer questions from a particular circuit (comparator) in our textbook for homework. The circuit is attached to this thread. I haven't started the questions yet as I don't even understand the circuit! Based on my notes I think my textbook is wrong and would appreciate it if someone could glance at the attachment and confirm or deny the case.

    2. Relevant equations
    Open loop gain = A (V+ - V-) where V+ is the non-inverting input, V- is the inverting input of the op-amp and A is the amplification factor.

    3. The attempt at a solution
    Where it says next to the diagram "suppose we want the reference voltage to be 4V. This would mean that the LED would light when the input voltage is equal to or greater than 4V". This is what I think is wrong:

    This means V+ (non-inverting input) is fixed at 4V. If Vin (which is V-) is equal to or greater than this then according to the open loop gain equation, the output voltage would be negative. This means that the anode of the diode is negative which means no current will flow. From my class notes it says the anode must be positive for a diode to light.

    So either my class notes are wrong or the textbook is wrong? Huge thanks to anyone that can help me with this!

    Attached Files:

  2. jcsd
  3. Feb 13, 2014 #2


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    Staff: Mentor

    You are correct and the book is wrong. The comparator output goes high when the + input is higher than the - input. Or said differently, the output is high when the - input is below the + input.
  4. Feb 13, 2014 #3
    Phew! Thought I was missing something. Thank you berkeman!
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