Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Necessary to spec power factor?

Status
Not open for further replies.
T

treez

Guest
Hello,

My company specifies its flashlamps that run on 230VAC in terms of current drawn as measured on a true rms multimeter.........they draw about 25mA on 230VAC.

I told them they need to specify the power factor (since they have mains rectifiers and no pfc stage)

But they tell me this is not necessary, and even though the power factor will be about 0.55, this does not really mean that twice the power is needed to power the lamps, as the extra current supplied is "reactive" current, and this is not dissipated.......or rather , some is dissipated in the mains grid supply wires, but since theyre made of thick copper, not much is dissipated there........................so do we need to specify power factor?


is it really twice the power thats dissipated due to the poor power factor....really twice the amount of oil burned in the power station....?
The lamps are for sale to industrial customers.
 

I presume, flash lamps are not considered as lighting equipment. Then no harmonic current limits have to be kept for equipment below 75 W input power.

Otherwise, the EN 61000-3-2 limits would apply (no minimal power for discharge lamps).

The popular misconcepts about reactive power have been sufficiently discussed at edaboard, I think. "some is dissipated in the mains grid supply wires" is in fact the best guess.
 
  • Like
Reactions: treez

    T

    Points: 2
    Helpful Answer Positive Rating
"is it really twice the power thats dissipated due to the poor power factor....really twice the amount of oil burned in the power station....?
The lamps are for sale to industrial customers".

If the current is in phase with the applied voltage, then the load is resistive and W = I X V applies. If the current and voltage is 90 degrees out of phase, then the product I X V = VA and is imaginary power, which causes currents to flow BUT NO POWER IS DISSIPATED. depending on the phase of your current, there will be some real (in phase) power and some imaginary (out of phase) power. Power generators don't like out of phase currents because the voltage regulators don't work, so there is a spec but as I reckon that your phase angle will be leading it will serve to compensate for all the lagging devices at you premises. :)
Frank
 
  • Like
Reactions: treez

    T

    Points: 2
    Helpful Answer Positive Rating
Thanks.

So mains rectifier bridge followed by smoothing cap gives leading phase angle..........electric motors gives lagging phase angle?
 

Current spikes before peak voltage are definitely lagging as it stores charge in caps.
But this is a highly non-linear low power device. It won't affect anyone's breaker loading, so the information is academic.
 
  • Like
Reactions: treez

    T

    Points: 2
    Helpful Answer Positive Rating
It's capacitive reactive power (leading current) and much more harmonic (distortion) apparent power.
 
  • Like
Reactions: treez

    T

    Points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top