What does this diode do??

Discussion in 'General Electronics Chat' started by lucy_b14, Feb 27, 2009.

  1. lucy_b14

    Thread Starter New Member

    Feb 15, 2009
    2
    0
    Hi

    I came across the attached circuit in a paper. I assume D1 rectifies half the input wave (since the current source and capacitor in parallel are representative of a piezoelectric source) but I'm not sure of the purpose of diode D2.

    Could anyone please explain the effect of D2?

    Thanks!
     
  2. davebee

    Well-Known Member

    Oct 22, 2008
    539
    46
  3. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
    2,040
    287
  4. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    Don't you need two capacitors for a voltage doubler?
     
  5. awright

    Well-Known Member

    Jul 5, 2006
    84
    7
    You have two. The PZT device acts as a capacitor. A perfect current source has infinite resistance, so does not short the PZT capacitor.

    awright
     
  6. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    OK so I'm being thick here, please explain in more detail how this works?

    The capacitor in the piezo element is in parallel with the supply and the D1/D2 junction, not in series. A series cap provides the charge pump function inherent in the Cockcroft multiplier described by others

    Label the top of the current source A and the bottom B.

    If A is positive, relative to B D1 conducts, D2 blocks and C is charged accordingly.
    If A then becomes negative D1 now blocks, cutting off C from discharge, and D2 conducts, providing a path for the current.

    Thus C is maintaned at the positive voltage on A.

    If the piezo element is now triggered, generating a positive voltage at A this will be added to the voltage already on C.

    If the piezo element is now triggered, generating a negative voltage at A, D1 again blocks, thus the voltage on C is unaffected.

    So this circuit appear to me to be a circuit to add a positive peizo generated voltage to any impressed on the output capacitor by the rest of the circuit.

    Not quite a multiplier.
     
    Last edited: Feb 28, 2009
  7. b.shahvir

    Active Member

    Jan 6, 2009
    444
    0
    I think its a clipper circuit.
     
  8. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    6,357
    718
    There is a mechanical delay with the piezo, compared to the time to charge the capacitor. Exact delay depends on size of the element for the resonant frequency.

    I'm in the same boat as studiot, though the constant current source may be tripping me up in quick mental visualization.
     
  9. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    Note also that piezo devices are essentially voltage devices. They are not capable of supplying much current. So all analyses will depend upon the size of the capacitors.
     
  10. b.shahvir

    Active Member

    Jan 6, 2009
    444
    0
    But what is the practical use of this circuit then ?
     
  11. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    Good question, to which we would all like the answer.
     
Loading...