Electronics Forum / Basics / July 2009

Considering the heating power of a waveform, for a DC signal the power
dissipated by a load will be:

   P = IE

Where P is the power in watts,
     I is the current through the load in amperes, and
     E is the voltage across the load in volts, DC.

Consider, now, the same load dissipating the same power with a
sinusoidal voltage supplying the current, and it becomes apparent that
since the voltage isn't staying constant, but has peaks and valleys,
then its peaks must be higher than the peak DC in order to cause the
same dissipation in the load.

The RMS value of a signal is defined, roughly, as the square root of the
average of the squares of a set of numbers or quantities.

So, to get the RMS value of a sinusoidal signal, what you could do would
be to heat up a load using a DC supply, measure the temperature of the
load once it had stabilized, then disconnect the DC supply, connect an
AC supply and drive the load until it stabilized at the same
temperature.

Then you'd sample the varying voltage, square each sample, then at the
end of one period add up all the squared samples, divide by the number
of samples and take the square root of that sum.

That'll be the Root-Mean-Squared value of the voltage required to heat
up the load the same as DC and, by contrivance, it'll be numerically the
same as the DC voltage.

Now, if you look at the samples that are being taken and you grab the
two that are the highest in voltage, you'll find that they occur at the
positive and negative _crests_ of the AC signal, and that their value
will be same as the value of the DC signal times the square root of 2,
which is about 1.414

So, we can say:

         |x| peak
    C = ----------
           x RMS      

Where C  is the crest factor of the signal being measured, and
     x is the value of the voltage being measured

For DC, then, since we have no peaks and valleys, we'll have:


         |x| peak     1
    C = ---------- = --- = 1
           x RMS      1

and for a sinusoid:


         |x| peak     1.414
    C = ---------- = ------- = 1.414
           x RMS        1

JF

It refers to the maximum load current crest factor that the UPS can
tolerate. Crest factor here is (peak current) / (RMS current) of the
load. Presumably the UPS can keep making a sine wave up to that 5:1
point.

A resistive load has a current crest factor of 1.414. Some switching
power supplies (the non-PFC kind) are much worse.

Google   UPS crest factor

John

"notme"

** No it does not -   it says " waveform ".

   Go read it again.

** Got nothing to do with the output voltage wave

 -  it has everything to do with the CURRENT draw wave.

The current drawn by loads typically attached to a UPS is anything but sine
wave in shape -  it is more like a stream of short pulses in time with the
voltage peaks and with the same polarity as them.

So, in your example, the maker is saying the UPS will supply continuous
current *peaks* of  up to 5 times the RMS current rating of the UPS.

Eg   Say the UPS is rated at 300 watts at 120 volts, then the RMS current
rating is 2.5 amps.

With a CF of 5, the allowable peak value is 12.5 amps.

....   Phil