capacitance = Iload/(Vrip*Hz) = 10/(5*50) = 0.04F
The system is simulated in proteus and i get a perfect 230V max voltage output.
Is the simulation just doing the most ideal or did i do some wrong calculations?
Further I am in doubt how to add the load to the system.
I can do constant power and constant current and constant resistance. I don't know what the best choice is in this situation.
The system is simulated in proteus and i get a perfect 230V max voltage output.
Is the simulation just doing the most ideal or did i do some wrong calculations?
Usually you should have an idea what you want to simulate....
Set up the simulation most close to the real circuit.
If this real circuit has constant current behaviour, then use a constant current load ... and so on....
I now build a three phase diode brigde circuit. When i for example put 20A load to it the current at the AC side is about 3.7kA for one phase?
When i do this with a one phase circuit bridge the load is just 20A like the DC current.
What is the relation between the AC current and the current at the DC side?
This is the circuit:
Currents...
It's not that simple ... especially in this case...
Look at the currents with a scope.
Klaus
Added:
The load is a constant 20A current. All the time exactly 20A.
But the load in the line is 1/3 of time positive constant 20A current and 1/3 of time negative constant 20A current.
The Amperemeter shows RMS current. Since with RMS current calculation all currents are squared it does not matter whether it is positive or negative.
So it's 1/3 negative and 1/3 positive makes 2/3 of time x 20A constant current.
We have 2/3 of duty cycle.
RMS current is sqrt(2/3) x constant current = sqrt(2/3) × 20A = 0.8165 x 20A = 16.323A