*Originally posted by uart *

Actually I've made quite a few power measurements on PC's using **only a multimeter**, and quite a cheap one at that.

Actually I have done this in the past also and got a pretty good result, but it's quite a pain in the butt. As you've said, the power supply runs on 340V DC. If you're on 240V AC, you simply rectify it and you have 340V(240V x sqrt2). If you're on 120V, the voltage, the voltage is doubled then rectified to yield 340v. (120x2xsqrt2). Voltage multiplier takes peak to peak from AC and it only works if input is AC.

So what I did was, shoot the voltage up to 240V AC using a transformer, then rectify it to 340V DC using a 600V rated bridge and 5,000µF worth of capacitor bank. You need to charge the capacitor through a resistor on initial power up or else you'll blow the fuse or worse, you'll blow the bridge rectifier.

Next, you set your computer to 230V(just in case) and connect the two prongs to 340V DC and connect a multimeter in series. You should place a jumper lead across meter probes, then remove it once connection has been established, so you won't push the inrush current through the meter. If you let the inrush current through your meter, it could cost you an expensive 600V CAT III meter fuse($5-10 each).

The trick is that the front end of a normal computer PSU feeds straight into a full wave rectifier (see note) and a cheap multimeter only really measures “proportional rectified average” (pi/sqrt(8) times rectified average current) rather than true RMS current. The upshot of this is that if you measure the AC input current with a cheap multimeter and then divide the result by pi/sqrt(8) (approx 1.11) then you have a very good estimate of the average DC current supplied by the PSU front-end full wave rectifier. As you know the DC working voltage of this rectifier (approx 1.4 time the RMS line voltage) then you can deduce the PSU input power to a reasonable approximation.

Wow that's new to me. I never really understood how non true-RMS DMM's measured voltage. True RMS ammeter and voltmeter both have a little computer that samples the waveform thounsands of times a second and calculate the true RMS current on the fly. Although, input power factor varies(meaning that waveform varies) depending on the load to an extent and I have a feeling that this method will fail to compensate for that.

Before you lampoon this technique remember that even if you use the most accurate equipment on the input power measurements that the figure of 70% efficiency is only a “ball park” figure and it could easily be somewhere between 65% and 80%, so there are inherent approximations no matter which way you do it. BTW I’m not criticizing the 70% figure as I think 70% to 75% is pretty right.. .
I actually measured efficiency just a few minute ago, but don't count me on it.

I connected a dummy load(a long ass extension cord.. lol!) on 5V line and drew 17.8A from 5V bus and my PSU is rated at 33A on 5V.

The current was 17.80A based on 0.5% +3 digit DMM. I *assumed* voltage was 5.0V, because I only had one DMM and if I remove the DMM from current measurement, it alters the series resistance and varies current, so that was not an option.

That gives output of 89W

Line voltage=122.1V

Line current=1.56A

VA=189VA

power=121W

PF=0.64.

89/121=73.6% efficinecy at ~30% total load all on 5V bus.

During this measurement, power factor was 0.64, therefore input current was 1.58A(true RMS) and 189VA.

One thing to remember about multiple computer loads is wattage is sum of each computer's wattage, but not necessarily so for current.

Power factor is down to 0.64 due to harmonic distortion. 1.5A+1.5A+1.5A+1.5A will be less than 6A, because harmonics from multiple computers tends to cancel out the distortion and increase the power factor. The resulstant current is what really counts when you're trying to squeeze in many computers on the same circuit. You'll have to actually connect all the computers you intend to connect on the same circuit through power analyzer to see what kind of current they're actually putting on line. You shouldn't exceed 12A for computer loads on 15A circuit as computer loads are more or less continuous duty and National Electric Code specifies you can only load circuit to 80% of rated current if continuous use.