# ups units

Jim Lux James.P.Lux at jpl.nasa.gov
Sat Jun 7 12:26:51 EDT 2003

```First off...
VA is not watts... Watts is active power ( integral of instantaneous voltage
* instantaneous current). VA is RMS volts * RMS amps.

If you hook up an RMS ammeter and measure the current draw, multiply that by
the RMS voltage, you'll actually be measuring VA (Volt-Amps)

If, and only if, the load is resistive so the current and voltage waveforms
are "in phase" and identical, will VA=Watts.  For all other cases (where PC
power supplies happen to be), the current isn't necessarily in phase with
the voltage, nor is the current waveform a nice sinusoid.

The ratio between the "watts" and the "VA" is called the power factor.  So,
if you have a device that runs at 120V, and draws 10 amps, it has a VA of
1200.  But, maybe the current is out of phase a bit, or non-sinusoidal, so
if you actually measure the power (i.e. how fast the wheel spins on the
meter), (maybe you put the load in a sealed, insulated box and measure the
temperature rise?)... you find that it dissipates, say 1000 Watts.  The
power factor, in this case, would be 1000/1200 or 0.83 (which, by the way,
is a typical kind of pf for a switching power supply).

The other thing to bear in mind is the "Watt" number advertised for a power
supply (i.e. a 400W power supply)  may not bear much connection to how much
power is actually consumed or supplied.

The nameplate must (in a regulatory sense) give you the RMS current and RMS
voltage, so you can get VA. It will also usually give Watts.  Depending on
the marking requirements, it might give power factor, as well. (The accuracy
and reliability of nameplate markings is whole 'nother story... Unless
something catches fire, or gets potentially blamed for a disaster, it's
unlikely that the stuff actually gets checked.. certainly not on a "per each
unit manufactured" basis)
---------
Now we come to the other big problem.. inrush current.
The nameplate ratings are for the "steady state" situation.  Many devices,
particularly those with large capacitors inside (i.e. power supplies) draw s
everal times their steady state current when power is first applied.  (AC
Motors also draw 3-4 times their running current when starting, too).  The
circuitbreaker in your panel doesn't trip because it has a "inverse
time/current" characteristic.. A short overload won't trip it, but a
sustained one will (remember that the purpose of the circuit breaker
("overcurrent protection" or OCP) is to keep the place from burning down;
NOT to protect your equipment.  The short inrush might lead to a "light
blink" as the load pulls down the supply voltage a bit, but won't usually
trip the breaker. In fuses, this is the difference between "fast blow" and
"slow blow".. although there are literally dozens of curves available.  For
circuit breakers, they actually have ones with knobs to set the curve, used
in industrial applications (where you're turning on 10 kW of lights at a
shot, or starting a 50 HP compressor, etc.)

Your UPS, on the other hand, probably does overcurrent protection with a
very fast acting current sensor. The OCP on the load side of the UPS is
designed to protect the UPS (i.e. don't fry the output transistors). I
suspect that this is a difference between the inexpensive UPS's and the "big
iron room full of batteries" UPSs.. The former is really designed to run one
PC (and it's associated fairly small inrush), the latter is designed with
slow trip and some overhead in the design.

----- Original Message -----
From: "Glen Kaukola" <glen at mail.cert.ucr.edu>
To: "Beowulf" <beowulf at beowulf.org>
Sent: Thursday, June 05, 2003 12:39 PM
Subject: ups units

> I contacted the company we purchased the Cyberpower units from, and he
> in turn contacted Cyberpower.  Cyberpower informed him that there
> shouldn't be more than one computer plugged into each UPS unit, and said
> something which I didn't understand about 12 amps, 125 volts, and 400 X
> 2 watts being more than 1500AV.  It all sounded rather fishy to me, so I
> was hoping that somebody on this list could clear things up for me.

12A * 125V (which sounds like a worst case, to me.. 125 is really high for a
line voltage) = 1500 VA... That's probably their max output spec..

Consider your 400W power supplies.. If they had 0.8 power factor , the
current each will draw is (400/110) / 0.8 = 4.5 Amps
A 2:1 inrush would be perfectly reasonable...

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