enter image description here

In the diagram above I want to know the value and the rating of the capacitor that should be used for good rectification

The Line to Line voltage is 400 , and by relation
Vdc =(3√2 VL)/pi=(3√2 x 400)/pi = 540 V

The ballast load has the power rating of 32KW .

This gives me the current of 55A dc . How can I translate this to the capacitance value ?

One other Question . The generator has 3 phase wires and one neutral but the rectifier takes in three phase wires . Should I connect the neutral to this rectifier ? probably at the anode side of lower cathodes ?

Reference Link

Thank you

  • 2
    \$\begingroup\$ Don't connect neutral. \$\endgroup\$ Commented Feb 6, 2016 at 22:10
  • \$\begingroup\$ what's the speed of the PWM? \$\endgroup\$ Commented Feb 6, 2016 at 22:11
  • \$\begingroup\$ How much ripple do you want? \$\endgroup\$
    – user16222
    Commented Feb 6, 2016 at 22:59
  • \$\begingroup\$ @JonRB I dont know about ripples ! . What are the advantages of having none ! . Isnt this system dependent ! \$\endgroup\$ Commented Feb 6, 2016 at 23:14
  • \$\begingroup\$ @jasen the duty cycle varies and since my pic controller has 24mHz crystal , this is the highest frequency of the pulse train I can generate ! \$\endgroup\$ Commented Feb 6, 2016 at 23:16

4 Answers 4


One of the big advantages with 3-phase supply in this situation is that even with zero capacitance you get a good DC supply with no drops to zero volts.

3-phase rectification

Figure 1. Three-phases, half-wave rectification and full-wave rectification with DC value superimposed. Source: Wikipedia. Click image for high-res view.

For a three-phase full-wave diode rectifier, the ideal, no-load average output voltage is

$$ V_\mathrm {dc}=V_\mathrm {av}=\frac{3{\sqrt 3}V_\mathrm {peak}}{\pi} $$

Adding capacitance may help with power glitches and transients. You probably don't need any.

You can't connect neutral in this situation. Neither the DC+ or DC- are at neutral potential (voltage).


For worst case (100% duty cycle),

$$ C = \frac{I\ t}{\Delta V}, \ $$

where \$C\$ is the capacitance in farads, \$I\$ is the DC load current in amperes, \$t\$ is the period of the rectified AC in seconds, and \$ \Delta V\$ is the allowable ripple across \$C\$ in volts.

Leave neutral disconnected.


I want to know the value and the rating of the capacitor

For the voltage rating of your capacitor you should consider using 1000 volt capacitors.

There will be inductance in the wires from your 3 phase electrical utility supply. Each situation is different as to how much inductance will be present.

The inductance along with your capacitor form a resonant circuit.

When you first turn on the disconnect (circuit breaker) to your system, there will be an inrush of current to charge up the capacitor. The inductance will continue to drive current into your capacitor resulting in the capacitor voltage temporarily overshooting your calculated 540 volts.

The amount of overshoot is dependent upon the inductance and capacitance, which are at this time undefined.

In the following model, I show a switch between rectifiers and capacitor because it was simpler to model than a 3 phase circuit breaker turning on.

enter image description here


Transistor is correct in saying that you dont need capacitance across the 3 phase bridge .You should have some capacitance for your PWM stuff .If you dont you will have current pulses from the PWM going down your three lines with EMC ramifications .Sizing your cap is just like you would do on a buck convertor .


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.