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Voltage trippler ?

Discussion in 'Electronic Basics' started by Guest, Aug 7, 2003.

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  1. Guest

    Guest Guest

  2. Lets call the left output terminal zero volts to simplify the
    description of other voltages. And I will name the caps from left ot
    right, C1, C2, C3, likewise the diodes, D1, D2, D3.

    The right end of C1 is charged to 1X (one times) the negative half
    cycle peak voltage of the AC by the conduction of the D1 when the
    bottom of the transformer winding is negative (remember I have defined
    the top end of the transformer winding as zero volts). It retains
    most of this voltage throughout the rest of the AC cycle.

    C2 sees the full AC voltage from the transformer. During the positive
    half cycles, the D2 conducts, limiting the voltage at the right end of
    C2 to be no more positive than the negative half cycle peak voltage on
    the right end of C1. This charges C2 to the full peak to peak voltage
    of the AC cycle (left end at positive peak while the right end is held
    at the negative peak stored on C1) or 2X the half cycle peak voltage.

    Then on the negative half cycles, the right end of C2 swings from 1X
    the negative half cycle voltage (that it got from C1) to a full cycle
    peak to peak voltage more negative than that (3X the half cycle
    voltage) and that negative peak is pushed into the right end of C3
    through D3. The left end of C3 is still at the 1X negative half cycle
    voltage that was forced into C1.

    So C1 stores a DC voltage of 1X, but C2 and C3 store 2X times the half
    cycle peak voltage. C1 and C3 have their voltages stacked like
    batteries to produce the output voltage of 3X, while C2 bobs up and
    down on the AC waveform pumping charge out of C1 and into C3. C1
    should be twice as big a capacitor as C2 and C3 (since it supplies not
    only load current, but also the charging current for C3 via C2),
    though it needs only half the voltage rating.

    How was that?
     
  3. Guest

    Guest Guest

    http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/voldoub.html#c3
    D2 is the middle one, it conducts during the -ve cycle
     
  4. The middle diode becomes forward biased when the bottom of the
    secondary is most positive.
     
  5. This is a version of a charge pump.

    The easiest way to understand these things is to download a free spice
    simulator and try it out, probing various places in the circuit. You could
    also build one, of course, and probe it with an oscilloscope, which would be
    even better.

    They are confusing circuits, which depend on the one way conduction of the
    diodes to push up the charge level of each stage, thus causing each
    sucessive stage to oscillate at a higher voltage.

    Note that the one is hyperphysics actually lowers the charge at each stage,
    so the left side is higher voltage than the right side. They are usually
    drawn the other way, so left is lower voltage than right.

    As you probably know, the power isn't increased by this configuration (its
    actually decreased by losses in the diodes) so you aren't getting something
    for nothing. They have lousy regulation and current output, in general.

    Regards,
    Bob Monsen
     
  6. Robert Monsen wrote:
    (snip)
    Sorry, but it is not correct to equate more positive with higher
    voltage. These circuits are equally good at making the end that is
    furthest from the transformer, more positive or more negative, just by
    flipping each diode and capacitor. If you get zapped by a big
    negative voltage you would never again think of it as a lower voltage.
     
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