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Need better full wave RMS Voltage Formula?

Discussion in 'Electronic Design' started by Hawker, Mar 27, 2007.

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  1. Hawker

    Hawker Guest

    I have been using a formula to compute RMS voltage for full wave
    rectified and filtered power for eons. I don't know where it came from,
    or if it is even accurate, but the numbers I get are usually in the ball
    park.

    It seems to work great if the input capacitor is sufficiently large that
    the ripple voltage is small. But if you make the capacitor to small the
    numbers are negative, and hence not valid. Even if you have no capacitor
    there is an RMS DC voltage. I googled and googled and could not find a
    better formula. Does anyone have one?

    What I am using is.
    F=Frequency (120hz in us)
    C = capacitor value
    V = Peek Voltage
    R = Load Resistor

    Ripple Voltage = (F^-1 / (2*SQRT(3)*V) ) / (RC)
    VDC = (1-((2F)^-1 / (RC))) * V

    This works great for large caps and/or light loads but is not accurate
    for small caps or large loads (you get negative numbers)

    Obviously I need a better formula. Can anyone help?

    Thanx
    Hawker
     
  2. Tom Bruhns

    Tom Bruhns Guest


    If you have access to a computer while you're doing your calcs, I'd
    recommend using Spice to look at the problem. LTSpice/SwitcherCAD is
    free, and easily up to the task.

    If you insist on calculating things "by hand," then consider that
    there are two time periods: the first is when the capacitor is
    charging from the source, and the second is when it is discharging
    through the resistive load. The first will be close to a segment of a
    sinewave, offset by the voltage drop in the rectifier diodes, which
    can be significant, and also modified by the drop in the resistance of
    the source (transformer winding resistance, mainly)--which should be
    fairly small and often is ignored. The second is an exponential
    decay, determined just by C and R. One problem is that it's common to
    assume the exponential decay starts at the peak voltage, but with
    small capacitance, that's not true, and not even a good
    approximation. The output voltage will follow the sinusoid beyond the
    peak voltage. The point at which the circuit transitions from the
    first region to the second is when the negative slope of the sinusoid
    becomes greater than the slope of the exponential decay: V/RC.

    You can put all that together and come up with a pretty accurate
    estimate of the waveform, and find the average DC from that, but it's
    a whole lot easier to just let Spice do it for you. Then it's easy to
    add in transformer winding resistance, leakage inductance, the
    inductance of filter inductors you specifically add, etc.

    Cheers,
    Tom
     
  3. Jim Thompson

    Jim Thompson Guest

    See....

    Newsgroups: alt.binaries.schematics.electronic
    Subject: From S.E.D - Need better full wave RMS Voltage Formula -
    RectifierRipple.pdf
    Message-ID: <>

    for how this was calculated prior to Spice.

    ...Jim Thompson
     
  4. Jim Thompson

    Jim Thompson Guest

    There's no closed-form solution. It requires iteration (that listing
    was from an ancient magnetic-card-reader calculator). Spice is an
    iterative solver.

    You can do pretty well graphically... sine wave and a straight edge
    ;-)

    ...Jim Thompson
     
  5. Phil Allison

    Phil Allison Guest

    "Hawker"

    ** You need to carry out a sanity check.

    A filter electro should not be subjected to large, 120 Hz ripple voltages -
    ie 50% of the peak supply or more.

    Bad for the cap as the max ripple current rating is likely to be exceeded.

    Eg: for a 1000uf, 50 volt cap with 35 volts of peak ripple,

    I = C dv/dt = 5 amps.



    ....... Phil
     
  6. Hawker

    Hawker Guest

    Wow thanx, but a bit more than I can sort though. I was not able to grok
    your notes. Sorry.
    Seem this should be able to be dispelled down to a single formula.
    Given Frequency*2, RMS input AC voltage, capacitance and load (in amps
    or resistance) what is the RMS DC voltage assuming full wave rectified.
    The reason SPICE won't cut it is I have a bunch of dependant formulas in
    several configurations. I need to put the formula into a spreadsheet
    with the rest of the data to come up with slope formula for a program
    given this and every other set of input data that is part of this
    monstrosity.

    thanx
    Hawker
     
  7. Tom Bruhns

    Tom Bruhns Guest

    So...put the whole thing into SPICE. Or feed SPICE parametric data
    for just this part of it.

    How much is a closed-form formula worth to you? Would a reasonably
    compact iterative solution with a few input parameters work? (e.g.,
    frequency, diode drop, capacitance, load resistance-or-current,
    whether full or half-wave rectification. Assumptions: constant diode
    drop when forward biased, no transformer/mains resistance [a "stiff"
    voltage driving the rectifier], ...) What exactly do you need for
    output? Average DC voltage? RMS output voltage? Peak output
    voltage? Peak-to-peak ripple voltage? What accuracy?

    Cheers,
    Tom
     
  8. Hawker

    Hawker Guest

    Well in essence this client is giving me a rectified AC voltage and has
    a small filter cap on it. I need to PWM a large DC load for a constant
    VA drive with it so I need to compute, given the load, what the RMS DC
    will be so I can figure out duty cycle. I know the input voltage, load
    and capacitor. Given that the input voltage will be "universal" or
    85-265VAC before the full wave rectifier the error in voltage drop will
    me minimal compared to the line voltage and can be ignored (or assumed
    ..7V). We can assume in infinitely low impedance source for this
    calculation. An accuracy of a few percent (5%?) should be good enough.
    It would be nice to know ripple but only RMS is really required. the
    only parameter not in your request is line frequency which could be 50
    or 60 hz (100 or 120 for full wave). The device will always be full wave.

    In the end I may end up just figuring this out my measuring. The issue
    is the source could be 60 or 50hz and my 50Hz supply can't give me the
    current I need to test this. Hopefully his will.
     
  9. Tom Bruhns

    Tom Bruhns Guest

    Hmmm...well, it wasn't much of a request on my part, more on yours I
    think! I did say that frequency would be an input parameter. I can
    see my way clear to doing the calculation, assuming zero source
    impedance and zero drop in the diodes. You didn't say how much it was
    worth to you...

    If you're thinking of measuring some scenarios, I'll again ask, why
    not just use Spice? LTSpice will give you the RMS of a waveform with
    a mouse click, and setting up to do a bunch of calcs like you need is
    very easy. There's always enough current available in a Spice
    source. ;-) Also, if you are thinking of measuring, there's no
    reason you can't scale things to low current; a given R*C value will
    give you the same answer, regardless of what R is, given the
    assumption of zero source resistance and zero or constant diode drop
    -- at least for the case of an RC discharge. A constant current
    discharge will give the same answer if you scale I proportionally to
    C, so 10A and 10000uF will give the same answer as 10mA and 10uF --
    because dV/dt is the same in both cases. The zero source impedance
    will always keep the cap at the value of the input sinusoid until the
    sinusoid drops faster than the load is discharging the cap, and then
    the cap goes into either an exponential or a linear discharge ramp.
    Because of diode drops and finite source impedance, that won't be
    perfectly true, but for the voltage range you mentioned, it should
    give you a result easily within 5%.

    Cheers,
    Tom
     
  10. Try 'calculus', at least that what it was called when I attended High
    School.

    Have fun.

    Stanislaw.
     
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