T=1/RC,

Yes. If you can afford to charge very slowly, take a big ressistor value. Then you can have low inrush current, and be able to use low power resistor, as the peak power will be E^2/R.  (E is the input DC source voltage).

In case you need to charge quickly, then take R=E/Imax, Imax is the maximal inrush current that the input DC source can support.  Again the resistor should support a peak power of E^2/R. You will see this value could be quite high. In practice, you can use one with a little bit smaller power rating.  

Thank you Mr. Wang. This makes sense.

Just to clarify you, I am pre-charging the inverter DC link via grid before starting PWM and therefore the Grid power goes in to B-6 Rectifier (Formed via FWD of IGBTs) and then it charges the DC link.

In this case, from my view, the limitation comes via Capacitor RMS current withstand capability i.e, Imax. Thus, if I want to calculate resistor and power then,

Take R=Vm/Imax where Vm is maximum grid voltage i.e. 1.4142*Vrms and Imax = Maximum capacitor RMS current withstand capability

After that power rating can be selected as,

Peak power = (Vm^2)/R. Where Vm is peak voltage of the AC grid.

Is my concept is correct? Please verify this method. I will be grateful and obliged.:)

Thanks and Regards

Aalok

Yes. Your approach looks correct to me.

I guess you verifyed that Imax of your capacitor is less that Imax of your input diodes.

In case you want to charge even faster, have in mind that the capacitor could support an inrush current larger than Maximum capacitor RMS current withstand capability.

yes you are right again Mr. Wang. The specified maximum capacitor RMS current is a continuous rating. However, the larger current withstanding capability will merely compromise on the service life of my inverter which is not desirable. Therefore, I consider maximum capacitor RMS current as upper limit.

I also varified that my diode is having much larger current withstand capability as I am working on MW range inverter.

After the calculation, I will again post here to crosscheck with estimated precharging time as per 1/RC to close the question loop.

1) choose precharge time as you like (I suggest 1-3 sec) and calculate resistor value keeping in mind that it takes 3 time constant to fully charge the capacitors. 

2) To size resistor wattage calculate the energy absorbed by the resistor (1/4 Cv2). Don't mind peak power but consider peak voltage and energy: these are the two parameters you have to match in the resistor data sheet.

There are resistors that are specifically made for precharge circuits. Best of all are high voltage ceramic carbon resistor from HVR as they are compact and can sustain high energy and high voltage. Second choice are wirewound resistors; don't use thick film, thet cannot sustain high energy pulse 

HEllo Guid,

Thank you for your insightful answer.. I got your point.

E^2/R  and  R=E/Im

Hello Guid,

Just one doubt. For capacitor energy, it will be 1/2C*(V^2). For 560 V AC input via bridge rectifier, it will be 560*sqrt(2) and it gives me energy of 3.3 kJ. This will be maximum energy in capacitor possible. So, Do I have to choose this much energy handling capacitor?

Can anyone suggest me the method  to calculate Precharge Resistor for a 48V Battery with Pure Resistive load.

please find the image

Hello Deepak, can u share the source of this document ? Would like to understand the problem to correlate more.