Full Bridge Rectifier

Discussion in 'Homework Help' started by Jess_88, Aug 9, 2012.

  1. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    Hey guys,

    I have been tasked with designing a full bridge rectifier, with given input voltage, power, frequency input and %ripple factor.

    I am fining it very difficult to get equations for a full wave rectifier with a smoothing capacitor.

    Can someone direct me to or post some important equation relative to this circuit?

    From what I understand, there should be some generic formula relating input voltage and current to load voltage current and load resistance. Everywhere I go,I seem to get different equations.

    Thanks so much guys
     
  2. daviddeakin

    Active Member

    Aug 6, 2009
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    Haven't we been here before?
    http://forum.allaboutcircuits.com/showthread.php?t=73053&highlight=ideal

    Rectification is a nonlinear process, and for a real system there may be no analytical solution to the problem. In other words, no neat formula. Somewhere you have to state some simplifications or approximations.

    For example, if you assume ideal components then you might make the following assumptions:
    • The capacitor will be charged to the peak AC voltage during each half cycle, Vac*sqt(2).
    • The ripple voltage is symmetrical- perhaps a neat triangle wave.
    • The ripple factor is small, so you can assume the discharging current between each 'top up' is constant, so that C=Q/V and Q=It can be applied easily.

    With these assumptions you should be able to derive most of the relevant formulae yourself. The only really tricky one is power factor...
     
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  3. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
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    ok, now I understand. Thanks

    One more thing.
    Is the peak instantaneous voltage applied across a diode the same as the maximum repetitive peak reverse voltage? I'm having some trouble fining out what this actually means... Same for the peak instantaneous voltage applied across a capacitor.

    I'm assuming the peak "instantaneous" voltage applied across a capacitor is just the peak voltage across the load (Vac*sqt(2))???
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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  5. daviddeakin

    Active Member

    Aug 6, 2009
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    In a bridge rectifier, yes. Each diode has to withstand a peak reverse voltage of Vac*sqt(2) when it is not conducting.

    Yes, I assume that is what is intended.
     
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  6. #12

    Expert

    Nov 30, 2010
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    I feel a need to interfere....

    When the capacitor is charged and the sine wave is in its "other" polarity, the diodes suffer double the peak voltage of the sine wave. In a full wave bridge, 2 diodes in series share this reverse voltage stress.

    After the pulse of charging current finishes entering the filter capacitor, the rest of the wave shape on the capacitor is a decreasing exponential based on Vo=dV (e^-t/RC) but in reality, it is very close to a linear ramp because the load is almost a constant current draining the capacitor for a few milliseconds. The equasion to describe the voltage drop is: (radical 2) C Er F = I where Er means Voltage ripple, peak to peak. This equasion is very useful in power supply design because it gives you the voltage drop on the capacitor between pulses. That lowest voltage is what you have left to supply your regulator,
     
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  7. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
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    ok so the wording here is confusing me a little.
    When you say "the sin wave is in the "other" polarity", do you mean when the input voltage source is in its negative half cycle?
    "The diodes suffer double the peak voltage of the sin wave...2 diodes share the reverse voltage stress"
    Because the voltage is shared, is the voltage aross each diode just the peak voltage???
    And for claification, the instantaniouse voltage across the capacitor is the peak voltage from the voltage source?
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I'm confused as well. In the case of the centre-tapped transformer full-wave rectifier with two diodes, each diode has a PIV of approximately 2Vm [ignoring the conducting diode drop].

    In the bridge rectifier case the two non-conducting diodes may be thought of as being in parallel with the transformer secondary potential - less one diode drop.
     
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  9. Jess_88

    Thread Starter Member

    Apr 29, 2011
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    Yeah and in my case, i'm not required to account for a voltage drop across each diode :\
     
  10. daviddeakin

    Active Member

    Aug 6, 2009
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    In a two phase (2 diode, centre tapped) rectifier, the diodes must withstand twice the peak AC voltage.

    In a 4-diode bridge rectifier they only have to withstand the peak AC voltage, because there is always one diode clamping one of the transformer windings to ground (ignoring diode drop).
     
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  11. #12

    Expert

    Nov 30, 2010
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    Yes.
    I didn't think about the center-tapped configuration because nobody used that word before I posted but there are still 2 diodes in each conductive event and they share the stress of the reverse voltage when the voltage wave changes to its other polarity.
     
    Last edited: Aug 11, 2012
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  12. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I think the confusion arose over exactly what voltage the two non-conducting diodes were "sharing" in the bridge configuration. At any instant each non-conducting diode is effectively in "series" with a conducting diode with their total impressed voltage being the secondary voltage. If the conducting diode is viewed as an ideal unipolar switch then the non-conducting diode simply sustains the full secondary voltage - which is what daviddeakin stated earlier.
     
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