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What happens in a circuit that has a capacitor in parallel with a diode?

  1. Feb 5, 2012 #1
    I attached a PNG of the simple schematic. I know that the cuircuit should shift the waveform thats coming from the source. What i dont understand is why or by how much it will be shifted up.
    Any help in explaining this is appreciated.
     

    Attached Files:

  2. jcsd
  3. Feb 5, 2012 #2

    Can you give some numbers for the waveform? Amplitude, DC bias/offset? Also, that diode is in series, not parallel.
     
  4. Feb 5, 2012 #3
    Model the diode as a resistor in the forward bias, and an open circuit in the reversed bias. Suppose [itex]v(t)[/itex] is the potential, and [itex]i(t)[/itex] is the current through the external circuit, and let [itex]\varepsilon(t)[/itex] be the e.m.f. of the source. Then, the circuit equations become:

    [tex]
    i(t) - \frac{v(t)}{R_d} = C \, \left( \dot{\varepsilon}(t) - \dot{v}(t) \right), \ v(t) < 0
    [/tex]

    [tex]
    i(t) = C \, \left( \dot{\varepsilon}(t) - \dot{v}(t) \right), \ v(t) > 0
    [/tex]

    You also need to supply a connection between [itex]i(t)[/itex], and [itex]v(t)[/itex]. These equations are non-linear, and require further analysis.
     
  5. Feb 5, 2012 #4
    The circuit in the diagram is called "clamper" or "DC restoration".

    Say the input varies from +5V to -5V. Initially the capacitor has no voltage (Vc = 0). Say that the input input is -5V; as the capacitor has no charge, for a very brief moment the diode gets 5V, so it is forward polarized and there is a big surge of current to the capacitor, which charges until it has 4.4V, so the output is -0.6V. At that point the diode cuts, as it can only conduct while it has 0.6V or 0.7V.

    A little time later the input changes to +5V; as the capacitor has 4.4V, the other terminal of the capacitor rises to 9.4V. The diode is reverse polarized, so it doesn't conduct anything. There will be a current to the load, but as long as the load has a large resistance (say 100k or more), the current will be small and the capacitor will keep the 4.4V charge.

    Therefore, if the input of the circuit is a square wave of -Vo to +Vo, the output will be a square wave of -0.6V to 2Vo - 0.6V. That is, the circuit will shift the input up, preserving its shape (**PROVIDED** the load is very light), until the lowest voltage of the output is -0.6V.


    ps: if the load is significant (say 1kΩ) or the frequency of the input is high, then the output will be distorted and the bias will not be enough to reach DC.
     
    Last edited: Feb 5, 2012
  6. Feb 5, 2012 #5
    This imagine will help you understand how the circuit works

    attachment.php?attachmentid=43529&stc=1&d=1328462588.png
     

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