We know that, the rate of charging of a capacitor depends on resistance (as given by its time constant 'RC').
1. But how is capacitor being charged in HVDC, even if we donot add a resistor?
Since we donot add resistor, time constant RC=0, which means capacitor should charge instantaneously. But in my simulation, it is being charged slowly (took about 8seconds, to reach the DC link voltage).
2.Is there any way to reduce the charging time (not by reducing the capacitor value(which is 78.25 milliFarad),as it was obtained by calculation for my system)?
Usually, the HVDC is started through a pre-insertion resistor. By using this resistor, the charging process (transformer reactance, DC capacitance, and DC cable) can be slowed down and current spike can be avoided. So, maybe it has to do with this pre-insertion resistor.
C=dq/dv, T=1/RC, no capacitor is pure, all are having impurity loss angle delta, means current cannot be lead by 90 degree with respect to applied voltage, hence would be charge depending on time constant (T) equal to 1/RC...
1. From the rectifier side
2. Either your HVDC link simulation model contains a pre-insertion resistor that's not obvious to you, or (more likely) the current inrush is being limited by the rectifier V-I characteristic.
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