Hello to all, making measurements on an ac/dc power supply (230Vac -> 24Vdc) I noticed that the cos phi is 0.56. The active power is 65W and the reactive power is about the same value. Now, I'm sure my load for power supply is 59W at 24Vdc (resistive load) How would you compute the efficiency of the power supply, i.e. power of the load / active power or power of the load / (active+reactive power)? Obviously there is a big difference. The current absorbed by the network is the sum of the two components related to active and reactive power, then I would suggest the latter calculation . Thank you. Greetings. Andrea

What is the computation FOR? If you're sizing a fuse or circuit breaker, it might be useful to know the input in VA terms, as well as in real-power terms.

I have to deal with a certain number of loads feeded by that kind of power supply, so the amount of reactive power affects the choice of circuit breakers and power consumption from the wall.

I'm seriously thinking to replace the power supplies with others that have a better cos fi (0,96 vs. 0,56) to deal with less current from the wall (and also less harmonics)

I've done this, I obtain 0,72A @ 230Vac as input current. I'll use this data to size breakers and wiring. Using a different power supply, with a good active pfc, the current drops t o 0,37A @ 230Vac for the same system - the second power supply is obviously more expensive but I'm wondering if reactive power will be figured out as a cost by the customer that know current consumption

Fuses are actual rated with the I²t parameter, so you really want to know the power supply peak current.

In a simple capacitor input power supply, current is drawn only just before the positive or negative voltage peak. The half cycle is 180 degrees long, but the actual conduction angle might be as short as 18 degrees i.e 1 ms at 50 Hz.

However, the total charge for this half cycle must be transferred during the conduction angle, i.e in this case 10 times the average current.

Assuming a resistive load drawing 10 Arms (2.3 kW@230 V), the fuse heating would be 1 A²s during each half cycle. However, if the same charge needs to be transferred during 18 degrees (1 ms), the peak current would be 100 A and the fuse heating 10 A²s, thus blowing the fuse long before 2.3 kW power.

Putting such peaky loads on all three phases would put the _sum_ of all three phases into the neutral wire, since the current peaks do not overlap. The neutral wire heating would be up to 9 times the heating of an individual phase conductor.

Only with overlapping currents will some (or all) phase currents cancel out in the neutral conductor.

The conduction angle depends on the source impedance of the feeding electric network and the size of the capacitor in the power supply. In practice, conduction angles would be slightly longer than 1/10 of the half cycle time.

You really need to measure the peak current or conduction angle using a current transformer and oscilloscope.

In the EU, power supplies greater than 75 W needs to have a good power factor, requiring some PFC circuitry. It appears that your supply has been designed to be just below that limit and hence does not need a PFC.

At least in EU, buy one that is rated above 75 W and the power factor is OK.

I think you have an instrumentation problem. Read Phil Allison's early response to you for understanding.

If your input power is 65W and your input VA is 230*.72, then your power factor is 65/166 or about .4 which does not agree with your initial post of .56 (Cosine whatever).

Additionally, using the PFC, the output power is 65W and your input power is .37*230 (85W) so the efficiency is now 76% (maybe).

One or more of your instruments may not be capable of measuring what you need. Use caution when making measurements. Check your instruments' specifications to see if they can handle the waveform factors. Do you have an oscilloscope?

** FORGET the term "reactive power " - it is meaningless for your situation where there is NO reactive current.
** You are obsessed with serious miscomprehensions.

If you have 65W load with a *power factor* of 0.56 that load is 65/0.56 =

116VA.

So the RMS current draw is just under 1 amp at 120VAC.

Inrush surge current will be the thing that gets you with such multiple loads and you do not even know what it is.

Not for efficiency. That's (real power out)/(real power in)

Power factor is a different issue. That's (real power)/(volt-amps), where both measurements are made at the same point. Volt-amps (and real power for that matter) may or may not include the harmonic components, depending on how you define power factor*.

*Something that bites old school power guys on occasion when they try to use capacitors to fix a 'power factor' problem that is a non linear load harmonic issue.

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Paul Hovnanian mailto:Paul@Hovnanian.com
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