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Clarification on RMS and frequency

Discussion in 'Electronic Basics' started by RR, Dec 4, 2005.

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  1. RR

    RR Guest

    Hi,

    As I understand it, RMS is calculated on the area under the sine or cosine
    wave for the AC supply and the peak voltage.

    It happens to work out at 0.707 of peak, as I understand, for mains supply.

    The frequency (50Hz or 60Hz, 100Hz, or 1Hz) makes no difference to this if
    the peak is unchanging. Is that correct?

    I've attempted to follow the calculations here:
    http://www.alpharubicon.com/altenergy/understandingAC.htm
    and the frequency seems to always reduce to a factor of 0.5.

    So, why does an appliance designed for 240volts 50Hz care whether you supply
    240volts 60Hz or 240volts 100Hz, for that matter?

    Or does merely depend on the type of appliance? I can see how a motor that
    relies on the frequency would care, but some appliances should not care
    less.

    What have I missed?

    tia,
    RR
     
  2. Bob

    Bob Guest

    RMS is easy to understand once you've learned what it's supposed to
    represent.

    The purpose of the RMS value is to assign a single value to a time-varying
    waveform, and that value is the value that would produce the same power,
    into resistor.

    For example, if you apply a steady 2 volts to a 5 ohm resistor then that
    resistor will dissipate 4/5 watts, because P=V^2/R. Square the voltage,
    divide by the resistance, and you have the power.

    What if you have a time-varying voltage into that 5 ohm resistor and you
    wanted to find the "effective" voltage (i.e., a single value for the voltage
    that would result if it were a steady voltage being applied)? You could find
    the effective voltage two ways:

    FIrst, apply the varying voltage to the resistor, and then:

    Measure the power being dissipated (maybe using a calorimeter), then
    rearrange P=V^2/R and solve for V.

    or

    By knowing what the varying waveform was (mathematically or graphically),
    you first square the waveform (each point on the voltage vs. time curve),
    then find the 'mean' of the squared waveform, and finally take the square
    root of that mean. The reason you need to square the voltage vs. time
    waveform, again, is because the power is proportional to the square of the
    voltage. This 'root of the mean of the square' technique will give you the
    effective voltage -- aka, the RMS voltage (root, mean, square).

    As for your second question, the frequency matters, to some appliances, only
    if they have some frequency-sensitive components:

    motors
    transformers
    capacitors
    inductors
    and so on.

    Bob
     
  3. John G

    John G Guest

    An appliance with NO inductance or capacitance will not care. i.e. a
    simple heater or a simple lamp.

    But most other appliances are more complicated and then raising or in
    fact lowering the frequency will have undesirable effects.
     
  4. RR

    RR Guest

    I've read lots about RMS and "almost" understood. I think your description
    was the best I've read. Thanks.
    So, that covers just about everything except an electric bar heater (incl
    kettles, ovens, etc.), as most appliances would have one or more of those
    items.

    Switch mode power supplies (power packs) have inductors and capacitor and
    transformers, but they are typically "smart" enough to deal with a range of
    inputs. Correct?

    cheers,
    RR
     
  5. Pooh Bear

    Pooh Bear Guest

    With regard to frequency, if the appliance has a conventional line frequency
    power transformer, the magnetising current is frequency related. If the
    transformer in a 60Hz piece of equipment wasn't also designed to operate
    correctly at 50Hz too it may have problems ( the core may indeed saturate and
    the transformer burn out - this tends to be a marginal issue and may even take
    hours to happen ).

    Graham
     
  6. Pooh Bear

    Pooh Bear Guest

    Yes. Switch mode supplies don't care about the line frequency.

    Graham
     
  7. RR

    RR Guest

    Thanks for that.

    I'm in the opposite situation. My appliances are designed for 50Hz
    (Australia) and my backup generator is outputting 51.9 HZ.

    I need to increase it's voltage output (it's only giving me 215V instead of
    230-240V) and that will increase the frequency to about 54-55Hz. (It doesn't
    have an AVR).

    Do power transformers and motors get damaged if they have a frequency fed in
    that's a little too high?

    I was told that low frequency is a problem but slightly higher frequency is
    OK.

    tia,
    RR
     
  8. John G

    John G Guest

    I think my signature line may be pertinent.

    A few cycles here or there will not generally matter except if there is
    a speed dependant motor eg. a synchronous clock ( not common to-day)
     
  9. Jasen Betts

    Jasen Betts Guest

    the clock runs fast as do any other synchronous motors etc...

    some devices aren't effected. (lightbulbs, radiant heaters, etc...)
    transformers designed for 60Hz run hot at 50Hz, going the other way is less
    troublesome.

    Bye.
    Jasen
     
  10. Andrew Holme

    Andrew Holme Guest

    Some equipment is designed to operate on anything between 50 and 60Hz, to
    suit both the US and European standards. For example, I have a wireless set
    in front of me with a plate screwed to the front which says: "48 - 62 ~" In
    the US, this would run off 110V at 60Hz, but I'm feeding it 110V at 50Hz
    here in the UK, and it works fine - although the transfomer does hum a bit.

    As a general rule, if you're not sure about the specification of an
    appliance, I think higher frequency would be safer because it produces a
    lower magnetising current in the transformer.

    Unless the equipment uses the mains frequency for speed control or timing, I
    can't see how a few cycles here or there would make a lot of difference.
     
  11. RR

    RR Guest

    Lots of different problems, but all related to similar goals.

    We're moving to a large block of land, and thinking of possibly not
    connecting to the grid.

    So, solar and wind power into batteries then back out through inverter(s).
    Hence my question titled "Switch between power sources".

    But, you can't run washing machines from batteries (at least, it doesn't
    seem very practical).

    So, that's where the generator comes in....

    Of course, a few cycles ain't going to matter, but we might be talking 10%
    above the appliance's rating (55Hz instead of 50Hz). Hence, my concern and
    question.

    Pooh Bear said that being under-frequency will tend to burn out things. But
    what about over frequency?

    cheers,
    RR
     
  12. Pooh Bear

    Pooh Bear Guest

    Yup, basically that's right. You should have no trouble.

    Graham
     
  13. Bob Myers

    Bob Myers Guest

    Sort of, but you'll get a better idea of what it's really all about simply
    by going through what "RMS" stands for in the first place.

    The problem is one of determining the "effective" voltage, current, or
    whatever of an alternating source; in other words - and to use the most
    popular example - if I pass an AC current through a resistor, how much
    power is dissipated in that resistor? How can I compare AC to DC in
    this sense?

    You clearly can't use the peak voltage or current - the waveform isn't
    at the peak but for an instant, so obviously doing a power or some
    other such calculation based on that value would be wrong. The next
    idea would probably be to try to find the "average" value of the
    waveform, but that winds up even worse - if you average any "pure"
    AC (meaning that it is symmetrical, regardless of the form of the wave,
    and spends as much time above zero as below), you get a result of
    exactly zero. That's obviously not right, either, since the resistor DOES
    heat up.

    So instead, we start by squaring the function that describes the AC
    waveform; if you square such a wave, everything winds up above
    zero, right? Then find the average, or mean, of the squared waveform
    (so now you have a constant value), and to correct for the squaring
    operation you did in the first place, find the square root of that average
    value.

    In short, you are finding the (square) Root of the Mean of the Square.
    "RMS," see?

    It happens to work out to 0.707 peak for a pure sinusoid; other
    waveforms wind up with different fractions of the peak value, and
    this is really only a shortcut which can be applied to those examples
    where one of these "regular" waveforms is in question. But the
    above process - square the wave, average it, and take the square
    root of the result - applies to all.

    Obviously, frequency does not enter into this at all - RMS is 0.707
    of peak for ANY sinusoid, regardless of frequency.
    The final result is just a numeric value - it has no frequency (that
    is the result of averaging the squared waveform - an average has
    no frequency, since it's a constant value, right?). You seem to get
    into some unexpected frequencies during all of this because the
    squared waveform is itself a periodic wave with a frequency 2X
    the original (at least for anything but a 50% duty-cycle square
    wave).

    When an appliance "cares" about the line frequency, one of two
    things are generally at issue:

    1. The appliance relies on the frequency of the line in order to
    run at the proper speed - e.g., the motor in an electric clock.

    2. The appliance contains components (typically, transformers or
    similar magnetics) which are designed to operate at one frequency,
    and will be less efficient at others.

    Bob M.
     
  14. John G

    John G Guest


    Bob,
    Where you a trained teacher?
    Thats one of the best replies I have seen here to a very basic question.

    Sometimes the posters get carried away with their own imagined expertise
    and launch off into long explanations that are technically way over the
    head of the original poster and sometimes not even accurate.
     
  15. Bob Myers

    Bob Myers Guest

    First of all, thanks for the nice comment - but no, I was never exactly
    trained
    as a teacher, other than on-the-job experience. I spent several years as a
    part-time electronics instructor at a local community college, and my job
    today often involves tutorial sorts of presentations. It's the simplest
    stuff
    that can often be the most challenging to try to get across, though, that's
    for
    sure.
    Unfortunately, I'd have to agree with you there - too often, we DO seem
    to forget that this is sci.electronics.BASICS, not
    sci.prove-your-understanding
    -of-electronics-to-the-Nth-decimal-place. That sort of thing is often
    counter-productive, anyway - far better to get an almost-intuitive,
    "back-of-the-envelope" grasp on what's going on than to shoot for the
    most precise, state-of-the-art answer possible first time out.

    Bob M.
     
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