Crest factor?

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

--
Oops...

I didn\'t answer your main question, sorry.

If it\'s supposed to output 120VRMS and has a crest factor of up to 5:1,
then we rearrange:


     |x| peak =  C * xRMS = 5 * 120VRMS = 600V peak


Which means that if the load is resistive and can stand off +/- 600V
peak, the output of the UPS will be able to deliver as much power into
the load as if the load were being driven with a sinusoid. 

Another caveat is that the output waveform from the UPS will be anything
_but_ sinusoidal.

           
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

--
Yup, that\'s a way better explanation than mine.

Kinda like: "For a crest factor of five, if you\'ve got an audio amp
pushing a sinusoid through an 8 ohm loudspeaker and you suddenly switch
in 4 more 8 ohm loudspeakers in parallel, the output of the amp will
continue to be sinusoidal for a while.

JF

"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