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Single Phase Full wave controlled Bridge Rectifier RL Load RMS V & I

  1. Jan 5, 2015 #1
    Hi,

    For a single phase full wave fully controlled bridge rectifier carrying a RL load, how is the RMS voltage and current calculated? I have researched textbooks but can only seem to find RMS equations for Resistive loads and I am not sure whether the negative load voltage will affect the value of the RMS calculation?
     
  2. jcsd
  3. Jan 5, 2015 #2
    If the inductance is considered as high enough to produce a constant ripple free current, am I correct in thinking that the RMS current will simply be the same as the Mean DC current value, worked out by dividing the average dc voltage value by the circuit resistance for an RL Load?
     
  4. Jan 6, 2015 #3

    donpacino

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    http://en.wikipedia.org/wiki/Root_mean_square

    RMS is simply the quadratic mean of a signal. If the signal is a sine wave, simply take the amplitude of the voltage or current and divide it by the square root of 2
     
  5. Jan 6, 2015 #4
    Hi, many thanks for the reply. I understand that this is the case for a full sine wave, but in the case of a rectifier where the wave is not complete, surely the value of the RMS current will not stay as sqrt(2)?
     
  6. Jan 6, 2015 #5

    jim hardy

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    See if this old Hammond appnote is of any help..
    http://www.hammondmfg.com/pdf/5c007.pdf

    You specified that the inductor is large enough to 'smooth' the current
    so you're into a wave that's the sum of a DC component and an AC component
    and will have to be solved by integration...... unless you're lucky enough to find an example already worked out.
    Try a search on "RMS of non-sinusoids"

    Filter capacitor as in the Hammond note complicates things.
    If you're not including one of those, then i believe your intuition is correct - DC amps where there's no ripple must equal RMS amps. That's by definition isn't it - RMS gives same heating value as DC ?

    I used to work out those problems by simulation in a Basic program using finite difference in maybe 100 steps per line cycle. That'd give me a feel for what was a reasonable approximation.

    Am i on the right question here?
    RMS current through the transformer winding may not be the same as through its load because Crest Factors come into play. A peaky waveform has higher RMS than a smooth sinusoid.

    old jim
     
  7. Jan 6, 2015 #6

    davenn

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    Thanks Jim

    nice pdf :)

    Dave
     
  8. Jan 6, 2015 #7
    I am assuming by "fully controlled" you are using a 4 thyristor (SCR) bridge - with phase angle control? Then you need to include the firing angle in the calculation - and the filtering in the inductor is dependent on the firing angles etc.. If the DC load is purely restive, and the source is ideal ( no inductive or capacitive elements) then the V out is rectified sine ( it will not look like DC as there is no smoothing or filtering) - the math is simple - however to calculate with RL load it is more complicated- I would simulate in LTSPICE.
    IMO - no such thing as ripple free DC in this case except when there is 0 load current (;-)
     
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