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V_in / V_out for a simple didoe/op-amp circuit

  1. Jan 21, 2007 #1


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    1. The problem statement, all variables and given/known data
    Question attatched in image file.

    2. Relevant equations

    3. The attempt at a solution

    A1.b) Input will be a sine wave 6Vpp . When the input goes positive, D1 conducts and short circuits the ‘first’ resistor. D2 doesn’t conduct (it acts as an open circuit) and so Vout = voltage across R in parallel = 3V.

    When the input goes negative D1 doesn’t conduct and D2 does. Therefore, the voltage goes through the ‘first’ resistor but the second resistor in parallel is short circuited. Therefore Vout = 0v since the diode is ideal.

    A1.C) The output from the op-amp will be = to vin . -1 (negative feedback) .When vin goes positive, D conducts and Vout = -0.7V

    When Vin goes negative D acts as an open circuit, so the Vout = 1.5V


    Obviously the answers above somewhat incomplete since a graph of v_in/v_out is also needed, but i just wanted someone to check if i have got the right idea.


    Attached Files:

    Last edited: Jan 21, 2007
  2. jcsd
  3. Jan 22, 2007 #2


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    Staff: Mentor

    Nice job. Your solutions look complete and correct to me.
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