1
\$\begingroup\$

I was studying about 3-phase power supplies. After going through several searches, i found this formula of three phase bridge rectifier's output voltage:

Vdc = ( 3 x sqrt(2)/pi ) x Vrms

Vdc = 1.35 x Vrms

So, in the figure given below, should I take 200 Vac as a phase voltage or line voltage. Is it Rms or peak voltage. what should i put in the equation.

three phase bridge rectifier

Improve this question
\$\endgroup\$
  • 2
    \$\begingroup\$ Since no Neutral is used. It seems to be delta wiring, thus delta voltage \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 8 '18 at 15:31
1
\$\begingroup\$

AC voltages that are stated simply as "nnn VAC" are assumed to be RMS voltages unless otherwise specified. Three-phase voltages that are stated that way are assumed to be line-to-line voltages unless otherwise specified. The circuit shown shows no neutral connection, so that also indicates the line-to-line voltage is the voltage of interest. The source could be delta or wye with no neutral connection to the circuit. Either way, the line-to-line voltage is what the circuit uses.

Improve this answer
\$\endgroup\$
  • \$\begingroup\$ @ Charles Coewie. Thank you for answering me sir, but could you tell me the output voltage of this circuit and the correct formula for the calculation \$\endgroup\$ – Z. Miller Mar 10 '18 at 2:53
  • 1
    \$\begingroup\$ I believe the formula you presented in the question is correct for the unfiltered output of a three phase rectifier neglecting the forward drops of the diodes. Each phase contributes to the output voltage the peak of the waveform and the portion of the waveform +/- 30 degrees from the peak. In other words the portion of the waveform that starts at 60 degrees and ends at 120 degrees. The minimum voltage is sin(60) X sqrt(2) X input and the maximum is sqrt(2) X input. The average is given by the formula that you have. I am too tired to derive that or find the derivation tonight. Can you do it? \$\endgroup\$ – Charles Cowie Mar 10 '18 at 4:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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