1
\$\begingroup\$

I drew a single-phase AC/DC rectifier in PLECS and used bipolar PWM to control it.

enter image description here

Here's my calculation process and related parameters (the duty cycle of Vtri is 0.5):

enter image description here

I think my calculation is correct, but the waveform of the grid current Is is not an AC sinusoidal waveform; it has a DC component.

enter image description here

I wonder what went wrong?

\$\endgroup\$
6
  • \$\begingroup\$ You have a capacitor on the output. Assuming the capacitor has been charged, no further current is needed to charge it. Depending on the load discharging the capacitor will determine what the input current waveform looks like. This is the problem that PFC (power factor correction) circuits solve in real world power supplies. \$\endgroup\$
    – Kartman
    Aug 29, 2022 at 23:26
  • \$\begingroup\$ Yes, the initial capacitance value of the DC side capacitor is 500V. If I want to get the required current curve, do I need to change it to 550V? \$\endgroup\$
    – mikux111
    Aug 29, 2022 at 23:57
  • \$\begingroup\$ Voltage rating isn’t the issue. In your simulation graph the DC voltage across the capacitor and DC current into the capacitor. This will demonstrate the load your bridge rectifier is driving. If your load was purely resistive, then your grid current would be sinusoidal. \$\endgroup\$
    – Kartman
    Aug 30, 2022 at 5:12
  • \$\begingroup\$ Thank you very much for your answer, but rectifiers generally require capacitance to reduce the voltage ripple on the load, right? Without this capacitor, I find the grid current will also be distorted. In addition, I am very sorry for my misexpression. I said that I set the initial voltage of 500 volts for this capacitor, and its initial voltage affects the grid current waveform. Should I set it to 0V or 550V? \$\endgroup\$
    – mikux111
    Aug 30, 2022 at 7:12
  • \$\begingroup\$ @mikux111 Given your equations, this is what I get in LTspice. The bottom plot shows that there is a slight displacement between the voltage and current, but the voltage across the output is sinusoidal, not wavy, like yours, and I had to use the negative value of the angle you've calculated. \$\endgroup\$
    – a concerned citizen
    Aug 30, 2022 at 7:52

0

Reset to default

Your Answer

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

Browse other questions tagged or ask your own question.