# Three Phase Full Wave Bridge Rectifier Current Calculation

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#### maxplanck

##### Newbie level 6 Hello friends,

How can I calculate a 3-Phase Full Wave Bridge Rectifier Circuit currents? I added an image. Isolation transformer is connect Delta-Delta. I want to calculate other currents according to I load.

(

I d1 = I d2 = I d3 = DC diode currents,

I s1 = I s2 = I s3 = Transformer Secondary RMS currents,

I p1 = I p2 = I p3 = Transformer Primary RMS currents,

)

Thank you,

Regards. Last edited:

#### e_bettio

##### Newbie level 3 Generally speaking, the AC current into a six-pulse bridge is equal to 0.82 of the DC current. So in your diagram:
Is1 = 0.82*Iload. Iload is a DC (average) current, Is1 is an rms current.
If you want to be a bit more scientific about it, sketch the currents.
Id1 is equal to Iload 1/3 of the time, Id4 (as I will call the current through the bottom diode on the same leg as d1) is equal to -Iload for 1/3 of the time. Is1 is the sum of Id1 and Id4.
So Is1, when divided into segments of 30 degrees (twelve segments per cycle) will have the following values:
(I will use Il instead of Iload)

0, Il, Il, Il, Il, 0, 0, -Il, -Il, -Il, -Il, 0.
Now if you want the rms of this, you take the 'r'oot of the 'm'ean of the 's'quare of the waveform, i.e. you first do the 's' (square), then the 'm' (mean), then the 'r' (root).

So, let's do it:
's' (square each segment):
0, Il^2, Il^2, Il^2, Il^2, 0, 0, Il^2, Il^2, Il^2, Il^2, 0
'm' (take the mean of the entire cycle):
(Il^2)*8/12
'r' (take the square root):
Il*sqrt(8/12) = Il*0.82.

The primary side currents depend on the transformer ratio. If the transformer ratio is 1:1, then the primary side currents are equal to the secondary side currents.

Sorry for my wordy explanation, hope it helps. It is much easier to describe with diagrams, but I'm in bed now so I'm not going to bother. Let me know if it doesn't make sense and I will sketch something.

Eric

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