If I filter the output of a rectifier using an LC circuit on the output, the input source would have a current draw with a square wave shape. I know that a lot of commercial transformers like sinusoids. Is this considered bad power factor since square wave are kind of out of phase with sinusoids or are the two unrelated.

  • \$\begingroup\$ Yes. The EU mandated switched mode psus over 100W (needs fact check) must have power factor correction. \$\endgroup\$
    – Kartman
    Apr 12, 2021 at 6:23
  • \$\begingroup\$ @Simon See my answer more quantitative ... Relevant to any "system" ... \$\endgroup\$
    – Antonio51
    Nov 3, 2021 at 12:57

3 Answers 3


It's not that 'a lot of commercial transformers like sinusoid'. It's simply that every transformer has issues (usually heating or vibration) with frequencies over the one it is designed for. So a 50Hz transformer would have to handle 100Hz, 150Hz, 200Hz and so on.

The EN 61xxx standard (in EU, the US would probably have some FCC related stuff) mandate a limit on the shape of the current from the equipment powered by mains (harmonic distortion testing).

Since the Fourier theorem holds for periodic signals, there is a limit on the upper harmonics of the current signal: with a purely sinusoidal current the upper harmonics would be 0, but even a rectifier bridge will alter that (since the diodes conducts only at 0.7V).

During the test you literally do a fast fourier transform of the current signal, with some good precision (IIRC you need at least 16 bit of resolution) and compare the upper harmonics with the fundamental and the standard limits: it's a pass/fail process. Buy the standard or look around for the relevant figures.

The power factor is strictly related and is a summarization of the harmonic test: it is in fact equal to the cos-phi of linear reactive loads, and some nasty integral in the general form (the definition is easily found on internet). So a bad power factor not necessarily indicates harmonic distortion (if it's only due to reactive load); severe harmonic distortion however causes a bad power factor (due to the way the PF is calculated). An reactive-only PF is not strictly bad for the transformer, is more an efficiency issue for the transmission network (reflected power causes unnecessary heating), and is handled by other kinds of standards.

Diode rectifiers are somewhat bad, but most of the time they pass the test; the worst load is a phase controlled load (like TRIAC dimmers) since the straight current edge has a huge harmonic content.

Switched regulators fall in between: they pull current in pulses of about 10-250kHz and that makes a bad PF. For small power supplies you simply put an inductor on the input of the supply, to smooth the current (old ATX supplies actually used a fluorescent tube ballast coil for that!). More powerful or better one use an active PFC circuitry which is quite complex to explain but can reach even PF 0.99

  • \$\begingroup\$ active PFC circuitry is "just" a boost converter programmed to keep current proportional to voltage, right? \$\endgroup\$
    – user253751
    Apr 12, 2021 at 8:55
  • \$\begingroup\$ This is interesting and a ton of useful information! So would this mean that the square wave shape's Fourier series components operating at higher frequencies (upper harmonics) are what the transformers struggle to handle? \$\endgroup\$
    – Simon
    Apr 12, 2021 at 19:00
  • \$\begingroup\$ user253751, yes, it's "just" that. The "just" part is that if you don't have a dedicate IC usually you have to use a DSP to do it right, simple no? \$\endgroup\$ Apr 13, 2021 at 6:35
  • \$\begingroup\$ For the OP: yes, it's the high frequency component that overheats the laminate with eddy currents. As a rule of thumb iron transformers can't handle more than 1kHz (core datasheet specifies the details). But the EN is mostly concerned with EMC altough it influences both \$\endgroup\$ Apr 13, 2021 at 6:37

Power factor depends on the shape of the current I(t) that the load draws from the voltage source V(t).

Let's assume V(t) is a perfect sine wave shape at a certain fixed frequency f.

If the load has: D,R,L,C components, than the load is not linear due to the presence of diodes D.

I(t), the current drawn by the load, will not be linear.

I(t) will have, in general, components at frequencies: f, 2f, 3f, 4f, etc.

That's not good for the PFC of your circuit.

To answer your question:

Yes, the power factor will be less than 1 and will be considered a "bad power factor".

Well, maybe not that bad.

  • \$\begingroup\$ If it's a square wave, there are no even frequencies. There will be if it's a half-wave rectifier. \$\endgroup\$ Apr 12, 2021 at 8:30
  • 1
    \$\begingroup\$ I indeed wrote "...in general..." \$\endgroup\$ Apr 12, 2021 at 8:42
  • \$\begingroup\$ Ah, you're right. I missed that. \$\endgroup\$ Apr 12, 2021 at 8:43

By "definition", Power factor is :

PF = Active power on Load / Apparent input power. (because it impact "transformers").

Active power on load is mean(V(Vload) * I(Iload)) ( W ) on one period.

Input "apparent power" is rms(Vinput) * rms(Current input) ( VA ).

Example of use

0- Checking full wave on resistor PF = 1

enter image description here

1- in SCR circuit (half wave) : delay= 5 ms, PF = ~ 0.495

enter image description here

2- in triac circuit (full wave) : delay= 5 ms, PF = ~ 0.700 (off coarse, better then half wave)

enter image description here

3- in triac circuit (full wave) : delay= 7.5 ms, PF = ~ 0.297 (off coarse)

enter image description here

4- in SCR circuit (half wave, RLD load) : delay= 5 ms, PF = ~ 0.348

enter image description here


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