9 Pin to 25 Pin cross cable

9 Pin--------- 25Pin
----------------------

1---Black-----8
2---Red-------3
3---Orange--2
4---Blue------20
5---Green----7
6---Brown---6
7---Yellow--4
8---Violet----5
9---Grey-----22

Stability of Biased Differetial Protection in case of through fault

Transformer is a vital equipment in power system. Being a costly equipment its protection is considered important. For lower ratings, less than 10MVA, transformers are  mostly radially connected. Therefore over current and earth fault protections are used for protetcion. But for higher ratings where transformers are connected in parallel operation, over current and earth fault protection can lead to unwanted trippings. In these conditions, differential protection is most suitable protection. As the Primary and secondary current transformers are available in same premises, implementation of differential protection can be easily done. 

Transformer Differential protections are biased differential protections, the biasing current is percentage of arithmetic sum of current in primary and secondary winding. Operating current is vector sum of current in primary and secondary winding. We have to take care of transformation ratio and vector groups for stable operation of Transformer Differential Protection. 

Let us consider a simple two winding transformer, having N1 turns on primary and N2 turns on secondary.

For this transformer I1 x N1 = I2 x N2

or I2 =  I1 x (N1/N2)

(here N1 and N2 are proportional to induced e.m.f. or voltage for practical purposes). Using this relation, relay can calculate both side currents in normal condition and identify the faults inside protected zone (ie transformer).

For measuring I1 and I2 in larger transformers, we use CT (Current Transformers). To physically balancing the secondary currents of CT, we can use CTs with mathing ratios. For example for a 400/220kV Transformer, we have to use 220/1 CT on 400kV side and 400/1 CT on 220kV side. The secondary currents of these two CTs would be same under normal conditions.

Now 400/1 CT is standard and easily available in market, but finding a 220/1 CT is difficult. So, we can use interposing CTs in between main CTs and Relays. they are used to exactly match the ratios. So, we can use 300/1 CT instead of 220/1 and correct the difference by using interposing CT. Interposing CTs serve the another purpose of filtering zero sequence currents, which will be discussed later.

In Numeric protections used presently, no physical Interposing CTs are installed. In these relays virtual Interposing CTs are used. Relays convert analog signals into digital and use formulas to implement CT ratio matching and Zero sequence filtering.

Important issues for Transformer Differential protection:

The following issues are important for transformer differential protection:

1. Phase shifting due to Transformer vector groups (Click here for more about Transformer vector groups)

2. CT ratio errors

3. CT saturation

4. Tap changing

5. Inrush currents

6. Hysteresis losses

The operation of transformer differential protection is based on diffrential current. Stability of transformer differential protection in case of through fault is important feature. On through faults of high level, CTs may get saturated or they may have ratio errors. The protection should not operate for out-of-zone faults.

The diofferential relay has follwoing features to provide stability in these cases:

1. Initial pick up : There is some minimum current to get the relay tripped. say we use 0.2Amps secondary current to get the relay operated. If the differential current is below this value the relay will not operate irrespective of Bias current. 

2. Biasing : There may be some ratio errors in CTs (upto 5%) and some difference in primary and secondary currents due to tap changing (upto 10%). Biasing is used to avoid false trippings from thses effects. The differential protection uses some percentage of through current for biasing (or restraining). For example, we use tripping slope as 20% and differential & restraining currents in a typical relay are :

Differential current  = |I1+I2|

Restraining current = (|I1|+|I2|)/2

The relay will trip when:

Differential current / Restraining current > 0.2

These days numeric differential relays come with variable slopes. The slope upto certain bias current levels (eg 4x) is 20% and after that slope is 30%. or slope may be hyperbolic, it varies from manufacturer to manufacturer. The last portion of slope is parallel to X-axis (say for differential current > 8x), means relay will operate if differential current exceeds a certain limit irrespective of Bias current. This area of relay characteristics is harmonically unrestrained. Typical operating charactristics of a differential relay with two slopes is shown below, the relay operates when the fault location is above the charactristics line (Shown in blue colour in this figure).

3.Harmonic restraining: When charging the transformer from one side a huge inrush current flows into transformer due to magnetic properties of Iron core. This will cause tripping of differential relay. On the other hand there may be a fault in transformer on first charging, so we can not keep relay out of service. For avoiding tripping during this period, second harmonic restraining is used. Second harmonics are observed during charging of transformers at 10-25% levels. Differential relays use 2nd Harmonics in restraining circuits and fundamental frequency in differential circuits to avoid false trippings.

4. Interposing CTs : Phase shifting due to Transformer vector groups is balanced by using Interposing CTs (ICTs). The ICTs on star side are connected in delta and on Delta side they are connected in Star. They also eliminate zero sequence components. 

Let us take the case of Yna0 (Star/Star connected auto transformer, voltage 400kV/220kV, ICTs connected in Delta configuration on both sides). The Primary and secondary CTs are connected in Star. The main Differential relay data for a typical transformer is as follows:

W1 CT ratio 500 (HV side CT ratio =500/1)

W2 CT ratio 1000 (IV side CT ration =1000/1)

W1 Interposing CT multiplier 1.09 (W1 Ct ratio / HV side Full load current)

W1 Interposing CT connection Yd1 (For Zero sequence current filtering)

W2 Interposing CT multiplier 1.20 (W2 CT ratio / HV side Full load current)

W2 Interposing CT connection Yd1 (For Zero sequence current filtering)

87 Initial Setting 0.2xIn (Minimum Differential current for operation)

87 Bias Slope 0.2x (Bias slope = Idiff/Ibias)

We may assume one normal load condition. The typical currents should be:

Primary IL1=0.40A @ 000deg Secondary IL1=0.22A @ 180deg

Primary IL2=0.40A @ 240deg Secondary IL2=0.22A @ 60deg

Primary IL3=0.40A @ 120deg Secondary IL3=0.22A @ 300deg

In case of single phase to earth fault outside protected zone, the current of that phase on both sides will rise by same ratio. Say R phase on 400kV side 2.2A and on 220kV side 4A. Relay will not operate in this condition due to absence of Idiff (for simplicity we will leave load current now).

Primary IL1=2.20A @ 000deg Secondary IL1=4.00A @ 180deg

Primary IL2=0.00A @ 240deg Secondary IL2=0.00A @ 60deg

Primary IL3=0.00A @ 120deg Secondary IL3=0.00A @ 300deg

But in actual practice for the same fault current as described in above para, due to presence of Zero sequence currents, the primary and secondary currents will not match. For L1-E fault on 220kV side values may be as follows (actual values would be somewhat in between these two cases, for simplicity of calculation we are taking two cases on the extreme):

Primary IL1=2.93A @ 0deg Secondary IL1=4A @ 180deg

Primary IL2=1.47A @ 180deg Secondary IL2=0A @ 0deg

Primary IL3=1.47A @ 180deg Secondary IL3=0A @ 0deg

So for above values, relay should not operate. The calculation for this condition is as below:

Primary side:

ICT output a = ICT multiplier x (A-C) /√3

=1.09x(2.93+1.47)/1.732

=2.77@0deg

ICT output b = ICT multiplier x (B-A) /√3

=1.09x(-1.47-2.93)/1.732

=2.77@180deg

ICT output c = ICT multiplier x (C-B) /√3

=1.09x(-1.47+1.47)/1.732

=0.00@0deg

Secondary side:

ICT output a = ICT multiplier x (A-C) /√3

=1.20x(-4.00-0.00)/1.732

=2.77@180deg

ICT output b = ICT multiplier x (B-A) /√3

=1.20x(0.00+4.00)/1.732

=2.77@0deg

ICT output c = ICT multiplier x (C-B) /√3

=1.20x(0.00-0.00)/1.732

=0.00@0deg

Relay operates when |I1+I2| > bias slope x (|I1|+|I2|)/2

R-phase

LHS = |I1+I2| = 2.77@0deg + 2.77@180deg =0

RHS = bias slope x (|I1|+|I2|)/2 =0.2x(2.77+2.77)/2 = 0.554

Y-phase

LHS = |I1+I2| = 2.77@0deg - 2.77@180deg =0

RHS = bias slope x (|I1|+|I2|)/2 =0.2x(2.77+2.77)/2 = 0.554

B-phase

LHS = |I1+I2| = 0@0deg + 0@180deg =0

RHS = bias slope x (|I1|+|I2|)/2 =0.2x(0+0)/2 = 0.0 (in this case |I1+I2| > Is condition will apply, i.e. relay would operate when |I1+I2| > 0.2)

Now it can been seen that in all phases Idiff is zero in all cases which indicates stable condition. So, so relay should not operate. 

During actual faults, we can analyze behaviour of differential relay from disturbance record files. (click here for more about disturbance record file). One sample disturbance record file of actual fault in transformer (inside protected zone) is given below:

In this file:
W1 Ia, W1 Ib, W1 Ic : HV side -R phase, Y phase, B phase  currents
W2 Ia, W2 Ib, W2 Ic : LV side -R phase, Y phase, B phase  currents
W1 Ict Ia, W1 Ict Ib, W1 Ict Ic : HV side interposing CTs output
W2 Ict Ia, W2 Ict Ib, W2 Ict Ic : LV side interposing CTs output 

In this case differential current was in Y-phase so bottom three channels are shown only for Y-phase.

I-Diff : Y phase diffrential current
I-Bias : Y phase Biasing current