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Homework Help: Op-amp as a comparator (analog-to-digital conversion)

  1. May 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Okay, This time it's an exam question. I am doing some past papers and there are questions about comparators which are not introduced in this course (this course is preliminary course for engineering students who are interested in electrical, electronic or computer engineering). So I guess the instructor wants us to know the logic of a comparator just within the 3 hours examination (in which not all 3 hours are for that particular question...)
    Now I find it difficult... maybe I am not familiar with the concepts yet... But anyway I have tried to solve it...Here are the details and my attempts:

    First, we are given an op-amp circuit for building a radio (the situation is that you are in a lost island which is rich of op-amps and resistors but no logic gates... so are you going to create logic gates AND and OR using just the above materials):

    Q4 a.png
    To find Vx, I use the node method and finally get the value of Vx in terms of VA and VB:

    Vx = - (Rf / Ra) (VA + VB)

    Then, assuming Ra = 2Rf and Vcc = 5 V, we are now relating the unknowns voltage with real values:
    Q4 b.png
    What I get are 0, -2.5, -2.5 and 5 from top to bottom.

    Third, now we are go to transform voltages to logic values:
    Let Vy = - Vx:
    Q4 c.png
    What I write are R1 = Rb, R2 = 3Rb, R3 = 2Rb, R4 = 2Rb, IN1 connects to Vcc, IN2 connects to GND.

    Then, it comes to the comparator question:
    Q4 d.png
    What I am sure about is that Vs+ = +Vcc, Vs- = GND...
    But how to know which V- and V+ should be respectively?

    AND gate in logic means y = ab. How to see this from the op-amp above?

    I know that 5 V is "true" which represents logic value "1"; 0 V is "false" which represents logic value "0".
    So this thing should either output 5 V (Vcc) or 0V (GND)...

    So is V+ = Vy and V- = Vth?

    Now, other than AND gate, we also can create an OR gate using the above materials...
    Q4 e.png
    Since I totally don't understand the logic of 4d, I also don't understand this. But I guess it mentions Vth this way, which shows that Vth is used as a reference for comparing? But why is just changing Vth can be resulted in two different logic gates functions?

    Here comes the final part: to design a 3-input AND gate using no more than 4 op-amps...
    Once again I don't understand the logic, so I don't know what to do.
    Q4 f.png

    2. Relevant equations
    For Q4d, Vout = K (V+ - V-) where K is a huge gain about 105.

    3. The attempt at a solution
    The attempts are written above.
  2. jcsd
  3. May 22, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
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