A 3 phase 240VAC line and I get a max output voltage of 589Vdc.
designing the circuit for 100KVA.
After the three phase rectification my output voltage is 589Vdc, after these I need to place a capacitor branch at dc side. So how to calculate the Capacitor value.
Can any help me how to calculate the capacitor value at dc side of the 3 phase rectification and also if possible with some equations.
You'll need to know the load current because the cap will be supplying that current between the voltage peaks from the bridge.
If you manipulate I = CdV/dT you should be real close.
Things are not adding up already. 240 VAC 3ph rectified should have peak voltage of 340 volts DC not 589.
If you have 589 VDC you are running off of a 416 VAC circuit which could be possible if its a 240/416 Wye system.
At 100 Kw on 589 volts you would be drawing around 170 amps. Without knowing what load you are running a rough guess would be that you would need between 40,000 and 100,000+ uf.
Don't drop your screwdriver on the buss bars when thats charged up!
The sums suggest you are going across phases.
240AC RMS x Sqrt(2) gives you a peak voltage. Then multiply by sqrt (3) gives you your voltage of approx 588V.
Jaguarjoe is correct. You need to know the current being drawn and the suggested equation is a good start.
Actually I am doing the simulation work for Resistance Welding Current Transformer for IBE Robots.
3-Phase Rectification , capacitor branch, Full Bridge Inverter, Transformer and then the rectification. I have designed the circuit. Vrms is 400V,50Hz. Designing for 100KW power.
Now my doubt was how can I calculate the capacitor value in capacitor branch,