Say I have a boost converter (picture below) that accepts a constant DC input and outputs a constant DC signal. Does this system have a power factor (PF) that is smaller than 1 (because the circuit has inductors and capacitors that would induce imaginary power)?

If PF is indeed not 1, how should I calculate input power? Should I integrate \$V_i \cdot I_i\$ over time?

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    \$\begingroup\$ PF is used for AC sources, not DC. But yes integration of VI(t) is necessary to compute average power in each cycle (s) or time intervals \$\endgroup\$ Aug 30, 2022 at 3:13
  • \$\begingroup\$ Do you mean efficiency? \$\endgroup\$
    – winny
    Aug 31, 2022 at 15:01
  • \$\begingroup\$ "Power Factor," FYI, is mainly something that the power company worries about. They measure, and they bill you for the real power that you use in your home or business, but the price that you pay is supposed to also cover the cost of the \$I^2R\$ power loss in their transmission lines. If you connect a hugely reactive load (i.e., low power factor) then you'll be drawing lots of current—causing lots of \$I^2R\$ loss in the transmission lines—but you won't be using much real power that they can bill you for. \$\endgroup\$ Aug 31, 2022 at 16:47
  • \$\begingroup\$ You dont need to worry about the power factor, you need to worry about the harmonics. \$\endgroup\$
    – Miss Mulan
    Sep 7, 2022 at 21:40

1 Answer 1


Yes, but not very usefully. You're more concerned about ripple current by itself, not in relation to DC capacity (which is what the power factor tells you).

[Edit] To be clear, you can trivially ignore imaginary power by simply taking P = Vdc * Idc. This makes DC uniquely easy to measure the power flow of. (You will also see this definition used below.)

You can use an aperiodic definition for PF = S / P, where S is Vrms*Irms, for each frequency component, then RMS summed over all frequency components (apply Parseval's theorem on S(f)), and P is the average power over all frequency components (apply Parseval's theorem to v(t) * i(t)). (I think? I haven't worked it out in frequency domain in a while, I'm afraid.)

Or in time domain, obtain Vrms and Irms, multiply them, then divide by avg(v(t) * i(t)). A SPICE version of which is shown on my website here:

A somewhat more useful measure might be to consider the reactive power in the switching inductor, relative to output (DC) power. This is roughly equivalent to the ripple fraction (inductor current Ipp / Idc), so is another way to express it; but one which can simply be divided by inductor Q* to get inductor loss, which is handy.

*Assuming same Q for all harmonics, or some equivalent value, which is... kind of an odd thing to have, but I guess would work. Anyway, the fundamental generally dominates, so it's not too bad to use Q at that frequency.

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    \$\begingroup\$ What does PF have to do with a DC-DC application? If OP meant "efficiency", it's better to ask for clarifications, first. \$\endgroup\$ Aug 31, 2022 at 8:17
  • \$\begingroup\$ I don't know why they want it, but assuming the question means what it says -- a definition does indeed exist, so that seems sufficient reason to give it. \$\endgroup\$ Aug 31, 2022 at 14:09
  • \$\begingroup\$ PF relates to the displacement of the fundamental harmonic and the presence of the harmonics related to the fundamental. Since this is a DC-DC application, there cannot be any fundamental, or harmonics related to it, so PF is irrelevant. Maybe this is what OP needs to understand. \$\endgroup\$ Aug 31, 2022 at 14:37
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    \$\begingroup\$ PF also relates to anharmonic content. Indeed it doesn't matter that the other components be harmonic in the standard case; merely that they are orthogonal to the fundamental, over the integration period. Thus, we can generalize this to a fundamental of 0Hz and "harmonics" at all other frequencies :) \$\endgroup\$ Aug 31, 2022 at 14:43
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    \$\begingroup\$ Mathematically speaking, just integrate over all time. Practically speaking, just integrate over the lowest components, such that variation in the resulting measurement is low enough not to mind. Engineers don't care about perfection, good enough will do. :) In this particular case, the converter switching period, more or less, will do a fine job. \$\endgroup\$ Aug 31, 2022 at 14:50

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