CT’s connection topology: cases

The location of the CT’s is critical to ensure the proper operation of the active filter. The CT’s are the “eyes” of the filter and it will react in accordance with the information supplied by them.

The location of the CT’s must always be in closed loop configuration. This means that the CT’s must see the load current and the filter current.

In some cases, summation CT’s might be needed to fulfil the closed loop requirement.

Typical circuit topologies and adequate CT’s location are described hereafter in the following order:

Case 1: Global compensation – one feeding transformer.

Case 2: Individual compensation – one feeding transformer.

Case 3: Global compensation – transformer busbar not accessible.

Case 4: Two independent feeding transformers.

Case 5: Back up generator.

Please bear in mind that the active filter always needs 3 CT’s: one per phase.

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This is the most frequent configuration: one transformer feeds several non-linear loads. The active filter is installed in central position and filters the combined harmonic currents.

This configuration and the proper location of the CT’s is represented hereafter.

Figure 6.1. Global compensation – one feeding transformer.

The connection of the CT’s to the active filter must be as represented herafter:

PQF LOAD LOAD LOAD

L1 L2 L3

Load side Supply side

K L

k l

K L

k l

K L

k l

PQF X5.2

X5.3 X5.4 X5.5 X5.6

X5.1

L1 L2 L3

Figure 6.2. CT’s connection to the active filter.

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In the example hereafter, the active filter PQF is connected to compensate Load 1 only. It does not see load 2.

Figure 6.3. Individual compensation – one feeding transformer

The connection of the 3 CT’s to the active filter is described in 6.4.1.

LOAD 2

LOAD 1 PQF

K = P1, L = P2, k = S1, l = S2

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Figure 6.4. Transformer busbar with no access: single-line diagram

For this configuration, three CT’s (one per phase) have to be installed on side A et on side B (in total, 6 CT’s). Those CT’s will then feed 3 summation CT’s (one per phase) that are connected to the active filter. This CT topology is represented in figure 6.5.

LOADS (Side A)

LOADS

(Side B) PQF

Figure 6.5. Transformer busbar with no access: CT connection (to be done for each phase)

The CT’s installed in each phase of side A et B (CT1 and CT2) must be identical (X / 5) and feed a summation CT whose secondary is 5A (5+5/5A).

The summation CT is then connected to the active filter in accordance with chapter 6.4.1.

A total of 3 summation CT’s (one per phase) must be used. The CT ratio to be programmed in the filter is: 2X / 5.

The CT – summator – PQF connection is represented here below. This has to be done for each phase.

LOADS

(Side A) LOADS

(Side B)

PQF

Summation CT (one per phase) Primary 1: 5 A Primary 2: 5A Secondary: 5A CT 1 (one per

phase) Primary: X

Secondary: 5A CT 2 (one per

phase) Primary: X Secondary: 5A

Figure 6.6. Connection between CT1, CT2 , the summation CT and PQF for one phase.

6

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Two independent transformers (the tie is normally open) feeds two different set of loads. One active filter is fitted on each LV busbar.

This system may however also work in degraded mode: the tie is closed and only one transformer feeds the whole LV system.

By connecting the CT’s as described hereafter, it is still possible to filter properly the harmonics and to correct the power factor.

PQF PQF

T1 T2

S1, k S2, l

S1, k S2, l

P1 P2 P1 P2

S1

S2

k l

PQF

Side A Side B

P1, K

P2, L

P1, K P2, L

Figure 6.8. Two independent transformers: CT connection for one phase.

For each phase, 3 CT’s must be installed: - one to measure I1 - one to measure I2 - one to measure I0.

Those CT’s must be identical: X/5 A.

CT I1 and CT I0 feed a summation CT which is connected to PQF1.

CT I2 and CT I0 feed a summation CT which is connected to PQF2.

Those summation CT’s must be 5+5 / 5 A.

Condition 1: the tie is open.

PQF1 sees I1 and PQF2 sees I2 (I0 = 0). The two transformers work independently and the total current to be compensated is I1 + I2.

Condition 2: the tie is closed but both transformers feed the loads.

In this configuration, PQF1 sees (I1-I0) and PQF2 sees (I2+I0). The total current seen by the two filters is I1 + I2.

S1, k

S2, l

P1 P2 P1 P2

S1 S2 S1, k

S2, l

S1 k

S2 l

P1 P2 P1 P2

S1 S2

I1 I2

I0

I’1-I’0 PQF 1

I’2+I’0 PQF 2

k l k l

T1 T2

P2, L

P1, K P1, K

P2, L P1

K

P2 L

Condition 3: the tie is closed but only one transformer feeds the loads (degraded mode).

If only T1 feeds the loads with the tie closed, PQF1 sees (I1-I0) and PQF2 sees I0 (I2 is zero). If only T2 feeds the load, I1 will be zero.

The above described connection must be done for each phase. The CT ratio to be programmed in the filter is: 2X/5.

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Many installations are fitted with back up generators to ensure the proper operation of the installation in case of mains power outage.

A typical configuration is given here below.

Figure 6.9. Back-up generator: typical single-line diagram

The CT connection must be such that the active filter works whatever the type of supply: generators or transformer-MV network.

For each phase, one CT is installed in the transformer feeding and one in the G

LOAD PQF

Figure 6.10. Back-up generator: CT connection (for one phase)