If you help to find out the right formula to calculate Balance Resistor and its power value( R1 and R4 ) as shown in Pic. because i have seen every one is using different different formuals.

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Data given is Vdc min=400Vdc Vdc max=735Vdc C6=C5=100uF/450V , 167uA leakage Current


2 Answers 2


Ignoring implementation specifics downstream and looking at balancing only:

The current through the resistors should be at least 10x the capacitor leakage current to provide a stiff division of the input voltage (i.e. the capacitor leakage current is insignificant relative to the resistor current).

Calculate this at \$Vin\$ (min), so 400v/1.67mA yields close to 240k\$\Omega\$, so two 120k\$\Omega\$ resistors will do the trick.

Power in each device is calculated at \$V_in\$ (max), so from \$V^2/R\$ I get 4.5W for each resistor.

I would probably use 10W devices.

Note that the resistance is high enough that it should not affect normal downstream performance (but that is application specific which I am not trying to answer)


This situation depends on the capacitor tolerances as well as balancing resistors. A common tolerance on electrolytic capacitors is -20% / +80% and this means C5 could be 180 uF and C6 could be 80uF. Under these circumstances at initial power up, C6 will receive approximately 70% of the peak voltage supplied from the bridge.

This means, on the face of it, that 450 V capacitors are unsuitable when Vdc is as high as 735V because 70% of 735V is 515V. This situation is only alleviated after about 3 RC time constants and, if R is 100 kohm then 3 time constants are 30 seconds. Can one of the capacitors survive that long?

If the capacitor tolerance is +/-20% then C5 could be 120 uF and C6 could be 80 uF. Now, at initial power up, C6 will receive 60% of the supply voltage of 735V - this is 441 volts and quite close to the limit.


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