By theory, a bridge diode and a capacitor bank can convert an AC sine wave into a DC signal, but the source current would appear spikey, having a power factor much smaller than unity.

In order to correct the distortion and phase shift of the source current, a power factor correction (PFC) stage is usually used in a power supply. The topology of this stage is usually a boost converter.

I am trying to understand why a boost converter is used to fix this issue. Also, what is the expected waveform out of this boost converter? Is it going to be a say, 400 V, AC signal (for 110 V input)?


2 Answers 2


There are other ways to correct power factor. For example, current can be shunted into mains to make up the difference from an ideal sine wave. Such a converter cycles current into and out of the supply alternately, so its supply voltage changes little (it only needs a DC bus capacitor, and draws slightly more power on average to make up its losses), and this can be used for industrial whole-plant power factor correction, for general loads from inductive motors to peaky rectifiers.

For individual supplies, sinusoidal current must be drawn. Boost is the cheapest way to implement this.

The alternative topologies are buck, forward, flyback, SEPIC, and Ćuk. Well, and some other more speculative or less suitable things.

Of these, Ćuk inverts the voltage, making any subsequent converter problematic (its controller will be referenced to negative, so needs independent DC control power to run it).

Buck and forward can only draw variable input current while Vin > Vout, which will not be true over the entire waveform, even if we make Vout fairly small. And we certainly don't want a large step-down ratio, lest current be obnoxiously high even at fairly modest power levels (consider a 240W supply dropping 120V 2A down to 24V 10A!).

That leaves boost, flyback and SEPIC. And of these, boost is the simplest and most efficient, lacking other considerations. It has the restriction that Vout > Vin(pk). So, for a typical 90-265V AC input, a 400V DC output is chosen, or around there.

Notice the converter's output goes into a filter capacitor, and its output current varies from zero, to a peak of twice the load current. So the output ripple goes as Vr = Io / (2πFC). This is a small-ripple approximation; a more convincing argument is based on energy, as the capacitor must source the energy that's missing from the line as its voltage crosses through zero. Indeed, there is some harmonic distortion, as the capacitor discharges deeper, a constant-power load will increase current draw; conversely, the capacitor's energy drops quadratically (i.e., E ~ CV2).

Flyback generally performs worse, due to the discontinuous currents seen on both primary and secondary; it has the advantage of isolation, so in an application where secondary side mains ripple is acceptable, it might be preferred -- a single converter power supply. I think some LED lighting supplies use this (LEDs have quite low impedance, so are very sensitive to the ripple voltage; this is solved with a current regulator circuit, which costs a minor amount of efficiency.)

SEPIC is not isolated, and requires more effort than boost, but it does have the advantage that Vout can be higher or lower than Vin. Thus, if you had an application that required say 200V DC supply, you could still do that with the 90-265VAC input range.

Honorable mention to the other Ćuk design:
however its requirement for bidirectional switches makes gate drive complicated, and I suppose it's patent protected so probably not getting much use outside of its inventor/owner(s).


The boost converter can take in power in a waveform which approximately looks like a resistive load. The output is usually DC and slightly higher that what simply rectified AC would be. That DC can then directly be used by the normal power supply section, like it was just directly rectified AC.


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