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# [SOLVED]Why SMPS is said to have 'Low power factor' instead of 'High harmonics content'?

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#### Rusdy

##### Newbie level 3
Hi experts out there,

Can anyone please give me links or good reading material, as per "Why SMPS is said to have 'Low power factor' instead of 'High harmonics content'?"

This question has been bugging me for awhile, as from what I know from my Electrical 101, 'power factor' is a factor of real power / reactive (or capacitive) power. Basically, current is sloshing around the circuit and the supply, without being used (therefore, the power waveform goes to negative). For example, an ideal inductance will have average (Real) power of zero because of the negative content.

If that is the definition, why SMPS is said to have low power factor? As far as I know, power waveform of SMPS has no negative content at all (i.e. power x current is always positive). Am I missing something?

I quote from Wikipedia:
"The current drawn from the mains supply by this rectifier circuit occurs in short pulses around the AC voltage peaks. These pulses have significant high frequency energy which reduces the power factor."

What the???? It is true on the first sentence quoted, but the conclusion of "reduces power factor"?????

That is harmonics problem, not power factor. Therefore, any attempt to correct it, should not be called Power Factor Correction (PFC), but 'Harmonic Correction' (HC) :?:.

PS:
1. I don't have a degree in Power Electronics, so this is probably obvious to some (but not me).
2. This guy (Milton Dall) on this website (link here) shares the same sentiment with me.

Power factor (displacement) relates to a lagging or leading current (relative to the phase voltage) 90 degrees out of phase is zero power factor, but total power factor must include harmonics, a switch mode psu has a slightly leading power factor due to the way the input caps are charged (a non PFC one) but is rich in harmonics and is therefore said to have a low power factor (even though the phase of the 50Hz I component is nearly in phase with the voltage) this is because the total watts drawn divided by the total VA drawn gives a low number (e.g. 0.65) this is the total or absolute power factor. The displacement power factor refers only to the phase shift between fundamental V & I (50/60Hz) - for linear loads the phase displacement is related to the total power factor, but not for complex or non-linear loads. Hope this helps, Regards, Orson Cart.

Rusdy and atripathi

Points: 2

### Rusdy

Points: 2
Adding to what Orson said: For non sinusoidal loads PF= (I(fundamental)/I(total))*DPF ; Where DPF is displacement power factor. For SMPS, due to high THD, I(fundamental) to T(total) ratio is poor and hence the low PF.

Rusdy

### Rusdy

Points: 2
Thanks Orson!

total power factor must include harmonics
Ah, no wonder, my definition of 'power factor' is way oversimplified (I should have done Electrical 401!!). Anyhoo, I've just found out a great layman explanation from **broken link removed**. I still don't fully understand (yet) how harmonics can contribute low power factor (hence, SMPS DOES consume more current than it should at the input), but the math is there! Oh well, too hard basket...

Actually, after thinking hard about it, it makes sense. As the voltage only has fundamental frequency (50/60Hz), and the current has harmonics, 3rd, 5th, so on; the harmonics component (of the current) multiplied with voltage component (which is does not exist), equals to zero: i.e. harmonics in current contributes zero real power.

The website I just linked above sums it nicely:

Hence where harmonics are present in the load current (but not in the voltage) the harmonic currents contribute nothing to the power in the load but do reduce the power factor.

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I would say that
the harmonic currents contribute nothing to the power in the load
is not true, imagine a square wave of current in phase with the phase voltage, the current is in phase with the volts and the instantaneous VxI at any point is positive. If you replace the square wave with a fundamental sine wave of the same magnitude as that in the square wave, i.e remove all the harmonics, presumably you will have less power power being drawn. Orson Cart.

I would say that "the harmonic currents contribute nothing to the power in the load" is not true.
The literature is clearly referring to a sine voltage. In so far, the statement is correct.

You should also consider, how power factor is usually calculated. For the contribution of harmonic currents, a sinusoidal voltage is assumed, despite of possibly present voltage harmonics. More detailed standards (e.g. EMI related) specify a maximum level of individual voltage hanrmonics to be kept.

Rusdy

Points: 2