VA versus watts

[harryagain]

My you're a clown. Have you the faintest idea what you're talking about?

[harryagain]

Except with resistive loads, power factor always means a lot.

[harryagain]

Slide rule, hmm? THAT explains a few things.

[meow2222]

Switching slowness is a fun one, especially if you end up with 2 devices on at once :)

NT

[harryagain]

There seems to be a misprint there. It says 150VA at the top which makes sense and 60VA at the bottom which doesn't.

[harryagain]

Well as it's intended for resitive load (filiment lamps) then it would be.

[The Natural Philosopher]

Oh dear. what is the power factor of a diode an inductor and resistor in series with a capacitor across the resistor, driven by a non sinusoidal alternating voltage then harry?

and even if you can calculate the answer, what use its it?

[The Natural Philosopher]

Theres a lot of magic smoke in them transistor things..

[The Natural Philosopher]

Do you actually know what an 'electronic transformer' is, harry?

I thought not.

[The Other Mike]

Harry you are no fun. I don't know Watts up with you, someone stolen one of your Joules?

[tony sayer]

In article , The Other Mike scribeth thus

Big ole boy by the look of it, those mass rating in tons then?..

[John Rumm]

and what kind of load are we talking about here?

So why are you wasting words?

[John Rumm]

Funny, we told harry that ages ago, and he is still off on one.

[John Rumm]

I don't follow the logic there...

A SMPSU without PFC on its input may well have a non unity power factor (0.6 - 0.8 (leading) seems quite common on older PC PSUs for example).

The PF on the input will remain pretty much the same regardless of the nature of the load. A resistive load will not substantially change the PF of the load presented by the PSU itself.

[harryagain]

As you don't give any values no-one can say. But it is a simple rectification/smoothing circuit indended to produce some sort of DC output from AC (with the resistor as a load possibly) You might have found it in a mains radio/amplifier in days of yore. But more likely with a rectifier bridge, not one diode. Or even a thermionic twin rectifier valve.

The power factor on the DC side would be neither here nor there (or unity to be pedantic) if there was no ripple. On the AC side it would be whatever it was, (likely near unity).

If the AC is non sinusoidal it is likely originating from some electronic switching device. Possibly even a lighting "transformer". But Power Factor is impertant whatever the waveform. Even DC with a ripple. (Which is actually still AC.)

[harryagain]

The rating of all electrical equipment/cables/wires/motors etc is only limted by heat losses caused by inherent inefficiencies.

[meow2222]

pf on the transformer secondary is fairly low, the diode(s) only conduct near the peaks. Mains side pf isnt 1 either.

the diodes

nope

[harryagain]

I don't thinkyou have a clue what you're talkinf about.

A diode is not a switching device.

Do you not know what AC stands for?

[harryagain]

I take it that's a "no" then?

[harryagain]

Transformer heatloss is (A) resistive. Heat loss is I x I x R. (Is quareed R) And (B) iron losses in the core.

The iron losse are pretty much constant regardless of load. The resistive losses vary with current (load).

This is where elctronic devices can win out, the iron losses are much reduced. or even zero.