Here's
one (comparatively) readable application note on power factor correction for switching power supplies, courtesy of Toshiba. It briefly mentions the use of
passive PFC circuits (basically sticking an inductor in the front end of the
power supply), and goes into a reasonable amount of detail on various active PFC schemes. The active devices used are typically MOSFETs of some description, though there's no inherent reason other more specialized switching transistors (perhaps IGBTs?) could be used in odd situations.
As to why peaky
current causes a poor
power factor, that's a
bit tricky to explain without delving into more calculus than I care to
tackle without either earning a salary or college credit for the work. However, I think I might be able to show the basic concept in a simplified form without (much) explicit calculus, and only a moderate amount of
hand waving.
A few reminders and definitions, first.
Ohm's Law: V=IR,
voltage =
current times resistance; and the fact that
power (in watts) is the product of the
current and the
voltage.
Power, in terms of physics, is the rate of energy consumption or transfer; so for a given rate of work (such as light production), the
power going into the
fixture, averaged over time, must be some constant value.
RMS values of measurements (voltages, currents,
etc.) are defined as being the square root (the "R") of the mean or average (the "M") of the square (the "S") of the instantaneous measurement. For a DC signal, this works out to be the measurement itself, ignoring signs--its absolute value, if you like. For other waveforms that vary periodically, it's a
bit more complicated to work out.
Power factor is defined as the ratio of actual
power (well, the average actual
power, if it varies cyclically for a periodic AC wave) and apparent
power; and apparent
power is the product of the
RMS value of the applied
voltage and the
RMS value of the
current.
(As to why
RMS values are used and defined as they are, that's because they have the really handy property that, for purely resistive loads, you can apply
Ohm's Law and the "normal" definition of
power being volts times amps,
etc. to the
RMS values and obtain useful averaged results, results that directly correspond to those of DC circuits, without having to analyze everything at every
point in the
voltage and
current waveforms. It reduces the analysis along the waveform as a whole to an analysis of a single value.)
DC Circuit (power factor = 1.0)
View attachment 22344
For the DC case, everything is very straightforward. The
voltage is constant, and the
current is constant, with their ratio being determined by the DC resistance of the
widget, and the
power is constant, with the average
power always being equal to the (unchanging) instantaneous
power. For this example, it is 1 volt, 1 amp, a 1
ohm load (per
Ohm's law), and hence 1
watt of
power dissipated.
Pseudo-AC Square Wave Circuit (power factor=1.0)
View attachment 22345
Now, instead, let's consider a
square wave, sort-of-AC situation. The
voltage switches between +1V and -1V periodically. The
RMS value of the
voltage, the square root (the "R") of the mean or average (the "M") of the square (the "S") of the signal's instantaneous value, is pretty obviously 1V: the
voltage is always either +1V or -1V, and either +1 or -1 squared is 1. The mean of 1, no varying, is likewise 1, and the square root of that is 1 V
rms. Similarly, the
current in our 1
ohm load also varies identically between +1A and -1A, and we see 1 A
rms. The actual
power consumed at any given
point in time is exactly 1W, and the apparent power--the product of the
RMS voltage and the
RMS current--is also 1W, so the
power factor is 1.0.
The Interesting Case (power factor < 1.0)
View attachment 22346
Finally, for the example that we're really discussing. Let's change the load so it's not a purely resistive load, but instead something that consumes
current for only part of our cycle. I'm arbitrarily saying the first half of the
square wave, but that detail doesn't really matter. The average (actual)
power should be the same as before, so we need to use twice the power--two watts--for the half of the time the
current flows, to make up for the zero watts it consumes the other half of the time. In order for that to work, the
current during the "on" part of the cycle must be 2A rather than 1A. Computing the
RMS current, we can see that half the time the square of the
current measurement will be 4 and the other half of the time it will be 0, so the mean of the square of the measurement is 2, and the
current is 1.41 A
rms (i.e. the square root of 2). The apparent
power here is 1.41W and the actual
power only 1W, giving a
power factor of 1/1.41 or about 71%.
For a realistic scenario of a non-PFC switching
power supply, where the
voltage is a (nominal)
sine wave and the
current a sort of-but-not-really
square wave sort of thing, it requires performing integration to compute the
RMS values explicitly. Because of that, I'll just wave my hands a
bit and say the same basic concept holds. Intuitively (if this stuff can be said to be intuitive) the increase from the squaring part of the
RMS computation will outweigh the corresponding decrease in the mean part of the
RMS computation.
Power factor can be thought of as a sort of metric for how much a load deviates from being purely resistive--whether that's because it is (partly) reactive or capacitive, or has other non-linear characteristics. If a
widget has a high
power factor, it behaves like a good
resistor and therefore the
current and
voltage waveforms are in
phase and have similar waveforms, and for that reason will not lead to unexpectedly high
neutral line currents.