0
\$\begingroup\$

I don't understand which form of power, reactive or real, should be dominant in a maximally efficient DC-DC boost converter.

As the circuit charges the inductor, the current and voltage will be out of phase. This seems like a textbook example of reactive power: the load, an inductor, is almost purely reactive and there's a phase shift between current and voltage (the equation for reactive power clearly shows a phase angle, φ, closer to π/2 would make the power more reactive: Q = |S|sin(φ) ). Intuitively, this seems like a good thing - you don't want real power being dissipated in the inductor and producing waste heat.

This seems to conflict with the definitions of real vs reactive power!

According to wikipedia's entry for "AC power", real power is that which results in the net transfer of energy in one direction (eg. into the load), and reactive power is power which results in no net transfer of energy between source and load because the energy is passed back and forth between them. With an ideal DC-DC converter, however, no energy will be passed back from the inductor to the charging circuit because the energy will instead be sent to the converter's output. This completely goes against the definition of reactive power as having a net zero energy transfer.

How can the two ways of analyzing the converter's power be reconciled?

\$\endgroup\$

1 Answer 1

6
\$\begingroup\$

As the circuit charges the inductor, the current and voltage will be out of phase.

No, if you apply a voltage across an inductor you will get a current ramp. You are thinking in terms of AC sinusoidal situations and this is not one of them.

you don't want real power being dissipated in the inductor and producing waste heat.

No, but you want the inductor to store energy in its magnetic field and this, can be seen as real power being forced into the inductor; it so happens that it doesn't turn to heat AND, importantly, can be liberated into the output circuit in the 2nd half of a typical switching cycle of a switching power converter.

How can the two ways of analyzing the converter's power be reconciled?

Go back to basics: \$V = L\dfrac{di}{dt}\$

This can be used to: -

  1. Derive the inductor current and energy used in a switching converter or, in applicably in switching supplies, it can be used to: -
  2. Show that the current through an inductor is 90 degrees lagging the voltage across it when talking about sine wave excitation (inapplicable to switching converters but important in AC analysis of motors, transformers etc..)

It is the former that is helpful in the analysis of switching converters and not the latter.

\$\endgroup\$
2
  • \$\begingroup\$ You're of course totally right about the current being a ramp (so roughly a triangle wave if the voltage is a square wave), but what I was trying to convey about the phase shift is that the current and voltage are shifted (even if they are not sinusoidal - the Fourier transform would require a phase shift between the two) and would thus suggest the power is largely reactive according to the equation for Q. \$\endgroup\$
    – Giesbrecht
    Commented Mar 6, 2019 at 9:19
  • \$\begingroup\$ When it comes to inductors and applying squares waves, there is no phase shift. The current begins ramping immediately on applying the voltage and, upon that voltage becoming 0 volts, the current ramps down with no delay or phase shift. When we talk about phase shifts we are talking about sine waves and sine waves are not applicable to switching converters. You cannot differentiate the power taken "into storage" in the magnetic field of a switching converter and that turned to heat by a resistor. \$\endgroup\$
    – Andy aka
    Commented Mar 6, 2019 at 9:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.