0
\$\begingroup\$

I'm calculating the energy losses of using AC to power rectified (DC) devices.

Rectified loads have capacitor filters that demand sharp peaks of current. This increases conduction losses across the entire AC electrical installation.

We can model all rectified loads in an installation as one large rectifier, for simplicity:

  • The current in this rectifier may be 20 A (or ~5 kW).

  • The ripple may be 5% (Vr = 16 V).

  • The capacitor will then be around 0.01 F:

    C = I / (2f * Vr) = 20 A / (2 * 50 Hz * 16 V) = 0.01 F

In a classical AC load, the RMS current is the average current. But in a rectified load, the RMS current demanded is much higher than the average current, because of the sharp peaks. This causes conduction losses.

To calculate the RMS current powering this ~5kW rectifier, I did:

  • The capacitive reactance is:

    Xc = 1 / (2 * π * f * C) = 0.3 Ω

  • The RMS current is:

    I_rms = V_rms / Xc = 230 V / 0.3 Ω = 770 A

Which is obviously wrong!

Can you help me understand what I should have done differently?

\$\endgroup\$
3
  • \$\begingroup\$ In a typical AC load, the RMS current would be the average current: 20 A <-- you can't connect a load that requires AC to a DC supply. Also, your calculation for the RMS current of a capacitor on a bridge rectifier fed from AC is incorrect because, once the capacitor reaches peak voltage, the bridge diodes get reverse biased and will only need to provide a a fraction of the current to keep the capacitor topped-up. \$\endgroup\$
    – Andy aka
    Commented Feb 15, 2023 at 17:39
  • 1
    \$\begingroup\$ Not sure what exactly your question is, but try reading on power factor correction, which is mandatory in all but the least power hungry products. \$\endgroup\$
    – tobalt
    Commented Feb 15, 2023 at 17:40
  • \$\begingroup\$ Thank you dear strangers! I edited the question to make it more clear, and have since received a response that made me realize I was asking the wrong question. Thank you, once again! \$\endgroup\$
    – user332999
    Commented Feb 16, 2023 at 3:29

1 Answer 1

3
\$\begingroup\$

Your assumption about AC/DC PSUs have large current spikes needs some looking into.

First, at startup, AC/DC PSUs will limit inrush current, usually with a thermistor. Otherwise they would blow their fuse at startup (if not fry the rectifier too.)

Second, AC/DC PSUs devices don't discharge their capacitors completely from cycle to cycle. Instead, there will be a ripple current at the top of each cycle that 'tops off' the capacitors just before the peak. This current isn't instantaneous either, but rides the slower (near-peak) dV/dt part of the AC cycle.

Your concern about the peak rectifier currents isn't misplaced though: they can be quite large for a big (~1kW) supply with large bulk capacitance. Which leads me to...

Third, in more recent times larger devices (greater than 60W) are required or are encouraged to have some sort of power factor correction on their inputs. This mitigates the effect of peak-by-peak charging input capacitors, and thus presents a cleaner, less spiky load to the grid.

Basically, the PFC circuit is a boost-type converter interposed between the rectifier and bulk capacitor, which supplies the cap charging current locally throughout the AC cycle rather than just at the AC peak. PFC white paper from MPS: https://media.monolithicpower.com/mps_cms_document/2/0/2020-power-factor-correction-_pfc__r1.0.pdf

That said, there is some argument to using a local DC grid (12V or 48V, say). This is an emerging trend in data centers, but applies to microgrids as well. It allows better integration of a UPS, avoiding the additional DC-to-AC conversions (and losses therein) of an AC-based system.

This isn't an issue with AC-DC power factor per se, it's just basic losses inherent in any conversion step. Fewer conversions = less loss.

\$\endgroup\$
2
  • \$\begingroup\$ Thank you very, very much! I was clearly lost in trying to calculate power losses that way, when PFC can easily restore the quality of the signal and reduce losses. Thank you dear stranger! \$\endgroup\$
    – user332999
    Commented Feb 16, 2023 at 3:28
  • \$\begingroup\$ De nada, amigo. \$\endgroup\$ Commented Feb 16, 2023 at 4:50

Your Answer

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