Help with clipping circuit

Discussion in 'General Electronics Chat' started by ronaldocavalcante, Mar 20, 2012.

  1. ronaldocavalcante

    ronaldocavalcante Thread Starter New Member

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    Hi, all. I'm from Bahia-Brasil and this is my first post.

    Please, I need help to calculate currents across two ideal diodes D1 (direct) and D2 (reverse) on clipping circuit: (see circuit)

    Main source: Vs = 7.5 peak (symmetrical triangle wave) with resistor (series) of 5.6 kohm

    Diode D1 with DC source +3.25 V on katode
    Diode D2 with DC source -3 V on anode
    RL = 5.6 Kohm

    I used electronics Workbench but the currents was wrong.

    thanks for help

    ronaldo

    Attached Files:

  2. bertus

    bertus Administrator Staff Member

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    ronaldocavalcante likes this.
  3. ronaldocavalcante

    ronaldocavalcante Thread Starter New Member

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    Hi, Bertus

    I read the site and can't find the currents across D1 e D2, yet.

    The V-out is OK, but the currents is always wrong.

    ronaldo
  4. Ron H

    Ron H E-book Developer

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    Are both resistors 5.6kΩ?
    This looks like homework to me. We don't give answers to homework.
    Show us your work and the answers you got, and maybe we can help you.
  5. ronaldocavalcante

    ronaldocavalcante Thread Starter New Member

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    Yes, both are 5.6K.

    I don't want the result! I'd like a hint to help me solve the question.

    The Vout is simple: V1 and V2 (positive and negative) + 0,7V (diode). In this case (ideal diode) level of the clipping voltage, V1 and V2.

    But to determine currents I tried thevenin and superposition without success.

    Steps: open V2, short in V1 determines Vo1 and IR1a using Vin

    Open V1, short in V2 determines Vo2 and IR1b using V1.

    Vo = Vo1+Vo2 and IR1=IR1a+IR1b.

    When I use the simulation (EWB) Vo is OK and currents not.


    I repeat that not need results, but help the way to try solve the problem.

    However, tahks for your attention.

    regards

    ronaldo
  6. ronaldocavalcante

    ronaldocavalcante Thread Starter New Member

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    Ok.
    Tomorrow I'll make this experience. I've got a scope 60MHz and 2 true rms multimeter.

    As I don't have a function generator, I'll try find here a triangle wave oscilator circuit.

    Maybe my calculation error is found.

    thanks
    Last edited: Mar 22, 2012
  7. Ron H

    Ron H E-book Developer

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    Convert Vs and the two resistors to the Thevenin equivalent. Then the solution will be simple.

    Is Vs ±7.5V? You said 7.5V peak. I'm assuming that means 15V p-p.
  8. ronaldocavalcante

    ronaldocavalcante Thread Starter New Member

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    Solved!!!

    It was my problem. Caculation of decimals was wrong
    Also, I forgot that the function generator of EWB generates the RMS and oscilloscope uses peak voltage.

    thanks for all

    ronaldo
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