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# pickup of differential current

## pickup of differential current

(OP)
Hi....
Can somebody explain me (math or by words)...
If we have differential protection of transformer 132/33kV, YND5 with differential protection setting 0.3Ibase (where Ibase is 137A), with CT ratio on HV side of 200/1, and on LV side of 800/1.

Pick up value of differential on HV side is (0.3*137)/200= 0.205A -> this is for 3ph fault.
Pick up value on HV side for 1ph fault is 0.205*1.5=0.3075A

Question is....why it is 1ph fault 1.5times more than for 3ph fault (on Y side)?
Same for delta (LV) side, why is there factor sqrt3 between 1ph and 3ph fault?
Thanks

### RE: pickup of differential current

It looks like with those settings, there is a zero sequence filter on the Y side, a delta simulation on the D side

On the Y side, a single phase injection has 1/3 +ve sequence, 1/3 -ve sequence, and 1/3 zero sequence.

So 2/3 of what you inject into the filter leaves - hence you need to inject 1.5 × pickup. If you inject balanced three phase current, there is no zero sequence, so what goes in comes out.

On the D5 side, it looks like the relay is doing a delta connection simulation - which is (Phase 1 - Phase 2)/SQRT(3) - phase 1 and phase 2 vary depending on the transformer arrangement.

If you inject a single phase, phase 1 is the injected current, phase 2 is zero, so the equation is ("injected current" - 0)/SQRT(3) - so the output is SQRT(3) less than what was injected

When you inject balanced currents, phase 1 and phase 2 have the same magnitude, but 120 degrees apart.

If you enter this into the equation - the number out the other side will have the same magnitude, but with a 30 deg shift - so no change in magnitude from what was injected.

Note that other relays may do a delta simulation on the Y side and no filter on the D side - so it will change depending on the design.

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