Sample calculation-for-differential-relays

2.1 Differential Relay Settings Calculations MiCOM P63X
Relay Type: MICOM P63X
Required Data
Ratings of the Power and Current Transformers
MVA
Voltage Ratio
Rated Voltage in kV (HV Side)
Rated Voltage in kV (LV Side)
Vector Group
CT Ratio (HV Side)
CT HV Side Vector Group
CT Ratio (LV Side)
CT LV Side Vector Group
Minimum Tap = - %
Maximum Tap = +%
Rated Current (HV Side) = MVA / (√3*kV)
Current on CT Secondary (HV) = Rated Current (HV Side)/CT Ratio
Required Ratio Compensation = 1/ Current on CT secondary (HV)
Rated Current (LV Side) = MVA / (√3*kV)
Current on CT Secondary (LV) = Rated Current (LV Side)/CT Ratio

Required Ratio Compensation = 1/ Current on CT Secondary (LV)
Relay current on LV side I2 = Current on CT secondary LV side / Ratio compensation for LV side
Calculations for OLTC tap setting -% and +%
Full load Current (- % or +% HV Side) = MVA / (√3*kV)
Relay Current for HV side I1= Current on Ct secondary (HV)/ Ratio compensation for HV side
I1 is calculated for extreme transformer taps
The differential current is = I1-I2
Bias current I bias = (I1+I2)/2
Differential current I dif and Ibias is calculated foe extreme Transformer taps
Therefore the operating current of the relay
If Ibias < 2
= Is + m1* I bias
If Ibias > 2
= m2*( Ibias – 2)+ m1 * 2
Where m1 and m2 are the slope of relay char
Pick up setting is chosen such that the Differential current at worst tap condition should not be more than
90% of operating current for better stability
Substation Name : 220 kV GSS Debari
Relay Type: MICOM P633
Transformer Name: CROMPTON
MVA = 100

Voltage Ratio = 220/132/11
Rated Voltage in kV (HV Side) = 220
Rated Voltage in kV (LV Side) = 132
Vector Group = YNA0d11
CT Ratio (HV Side) = 300/1
CT HV Side Vector Group = Star/Star
CT Ratio (LV Side) = 500/1
CT LV Side Vector Group = Star/Star
Minimum Tap = -15
Maximum Tap = +10
=100*10^6/ (√3*220*10^3)
=262.4319 A
= 262.4319/300
=0.875 A
Required Ratio Compensation = 1/ 0.875
= 1.143 A
=100*10^6/ (√3*132*10^3)
=437.38 A

=437.38 /500
= 0.875 A
=1.143 A
Calculations for OLTC tap setting -10%
Full load Current for 220 kV %( HV Side) Winding at -15% = MVA / (√3*0.9*kV)
=100*10^6/ (√3*220*0.85*10^3)
=308.743 A
= 308.743 /300
=1.029 A
As the adopted ratio correction is 1.143 A the current to relay bias terminal
= 1.029 * 1.43
=1.176 A
Hence the differential current is = 1.176 – 1
= 0. 176 A
Bias current = (I1+I2)/2
= (1.176+1)/2
= 1.088 A
[Since the bias current is less than 1.5 A the slope will be within 30%]
Therefore the operating current of the relay will be = Is + (0.3* I bias)

= 0.2 + (0.3 * 1.088)
= 0.526 A
Differential current at worst tap condition should not be more than 90% of operating current for better
stability. Here it s less than 90% so the setting is acceptable.
Calculations for OLTC tap setting +10%
Full load Current for 220 kV %( HV Side) Winding at +10% = MVA / (√3*1.15*kV)
=100*10^6/ (√3*220*1.10*10^3)
=238.54A
= 238.54 /300
=0.795 A
= 0.795* 1.143
=0.909A
Hence the differential current is = 1 - 0.909
= 0.091 A
= (0.909 +1)/2
= 0.955 A
= 0.2 + (0.3 * 0.93)
= 0.48 A

2.2 Procedure for Relay Setting of Transformer Differential Relay MBCH
Data Required
MVA Rating
Voltage ratio
Vector group
CT ratio on HV Side
Winding connection of CT on HV side
ICT ratio on HV Side
Winding connection of ICT on HV side
CT ratio on LV Side
Winding connection of CT on LV side
Transformer Tap
Min -% OLTC Tap and max % OLTC tap
At Normal tap
Rated current (HV Side) = MVA / (3*Rated voltage (HV side))
Rated current (LV Side) = MVA / (3*Rated voltage (LV side))
If CT is Star/Star
Current on CT Secondary (HV) = Rated Current (HV Side) / CT Ratio (HV Side)
If CT is Star/Delta the current shall be multiplied by 3.

Current on the secondary of ICT for Star/Star = Current on CT Secondary (HV Side) / ICT Ratio (HV
Side)
For Star/Delta the current shall be multiplied by 3.
The same procedure is followed for getting the relay current on LV side also.
Idiff = Relay Current = Difference between the HV and LV Current (Current at the secondary of ICT’s)
To make sure that the relay does not operate when the Transformer taps are changed the currents are
calculated at extreme taps.
At Tap on -%
Voltage corresponding to the above tap,
Full load current = MVA / 3 * Voltage corresponding to –ve tap.
At Tap on +%
Full load current = MVA / 3 * Voltage corresponding to +ve tap.
Currents at the CT Secondary and ICT Secondary are calculated by using the above procedure. For
extreme +ve and extreme –ve tap.
Relay current I1= the difference between the HV and LV Currents for extreme +ive tap.
I2= The difference between the HV and LV Currents for extreme +ve tap.
MBCH has an adjustable basic threshold setting of 10% to 50% current I selectable in 10% steps.
Dual Slope – 20% Slope upto In.
- 80% Slope for Current > In.
Relay operating current = Pickup setting + Bias Setting * Bias current

Bias Current = (I1+I2)/2
I Operating 1A = Pickup setting + 0.20 * Bias Current less than 1A +0.8*current above 1A
I Difference = I1- I2
Operating current at extreme taps is calculated with the same procedure.
The pick up setting is acceptable if the I operating is less than the bias current at extreme taps.
Differential Relay Settings Calculations
Substation Name: 220 kV GSS Debari
Relay Type: MBCH
Ratings of the Power and Current Transformer
MVA = 50
Vector Group = YNa0d1
CT Ratio (HV Side) = 150
CT Ratio (LV Side) = 300
CT Ratio (HV Side) = 150
CT Ratio (LV Side) = 300

=50*10^6/ (√3*220*10^3)
= 131.22 A
=50*10^6/ (√3*132*10^3)
= 218.6 A
= 131.22/150
= 0.87 A
=218.6/300
= 0.73 A
Interposing CT (HV Side) = 0.87/0.58
Vector Group Interposing CT (HV Side) = Star/Delta
Interposing CT (LV Side) = 0.73/0.58
Vector Group Interposing CT (LV Side) = Star/Delta
Current on ICT Secondary (HV) = √3*(Current on CT Secondary (HV)/ Interposing CT
(HV Side))
=√3*(0.87/0.58)
=1.01
Current on ICT Secondary (LV) = √3*(Current on CT Secondary (LV)/ Interposing CT
(LV Side))
=√3*(0.73/0.58)

=1.00
Difference between HV side and LV side currents = Current on ICT Secondary (HV) -
Current on ICT Secondary (LV)
= 1.01-1.00
=0.01 A
=50*10^6/ (√3*220*0.88*10^3)
= 149.11 A
= 149.11 /150
=0.99 A
Current on ICT Secondary (HV) = √3*(Current on CT Secondary (HV)/ Interposing CT (HV Side))
=√3*(0.99 /(0.87/0.58))
=1.15 A
Difference between HV side and LV side currents = Current on ICT Secondary (HV) - Current on ICT
Secondary (LV)
= 1.15 – 1.00
= 0.15 A
I bias = (I1+I2)/2
= (1.00 + 1.15)/2
=1.075 A
Consider the setting of Is = 0.1

The I Operating Current = 0.1 + (0.2 * 1) + (0.8*(1.075-1)
= 0.1 + 0.2 + 0.8*0.075
=0. 36A
At tap of -12% with full load the relay require a current of 0.36 A at Is = 0.1
Since I1-I2 = 0.15 which is less than relay operating value, relay will not operate.
So a setting Is = 0.1 is acceptable.
=50*10^6/ (√3*220*1.05*10^3)
=124.97 A
= 124.97 /150
=0.83 A
Current on ICT Secondary (HV) = √3*(Current on CT Secondary (HV)/ Interposing CT (HV Side))
=√3*(0.83 /(0.87/0.58))
=0.96 A
Current on ICT Secondary
= 1.00-0.96
= 0.04 A
I bias = (I1+I2)/2
= (0.96+1.00)/2

=0.98 A
Consider the setting of Is = 0.1
The I Operating Current = 0.1 + (0.2 * 0.98)
= 0.1 + 0.196
= 0.296 A
At tap of +5% with full load the relay require a current of 0.29 A at Is = 0.1
Since I1-I2 = 0.04 which is less than relay operating value, relay will not operate.

2.3 Procedure for Relay Setting of Transformer Differential Relay KBCH
Data Required
MVA Rating
Voltage ratio
Vector group
HV voltage
LV voltage
Transformer percentage impedance:
Transformer vector group:
OLTC Tap: +% OLTC Tap: -%
CT ratio and winding configuration
HV side
LV side
At Normal tap
HV Side full load current = MVA / 3*kV
Current on CT Secondary Ict (HV) = Rated Current (HV Side) / CT Ratio (HV Side)
N1=Required ratio compensation = 1/ Ict sec
Assuming Relay current = 1A
N1 is set on the relay
Current on HV side = Current on CT secondary (HV)/ N1

At Normal tap low voltage side
LV Side full load current = MVA / (3*kV)
Current on CT Secondary Ict (LV) = Rated Current (LV Side) / CT Ratio (LV Side)
N2=Required ratio compensation = 1/ Ict sec (LV)
Current on LV side = Current on CT secondary (LV)/N2
At -% Tap
HV side current = MVA / (3*kV
Where kV is the voltage corresponding to -% tap on HV side
With ratio compensation
Current on CT Secondary =Ict sec /N1
At Tap on +%
HV side current = MVA / (3*kV )
Where kV is the voltage corresponding to +% tap on HV side
Current on CT Secondary =Ict sec /N1
Differential current at extreme taps
At -% tap =Idiff1 =IHV-ILV
At +% tap =Idiff2 =IHV-ILV
Ibias = IHV+ILV /2

Operating current of Relay
I Operating = Pickup setting + 0.20 * Bias Current less than 1A +0.8*current above 1A
Relay Operating current current is calculated using the above equation.
For extreme taps the Idiff and Ibias current are calculated., the pick up setting is chosen such that it will
not operate for extreme taps
Substation Name: 220 kV GSS Nagaur
Relay Type: KBCH (Areva)
Transformer Name: Transformer 2
MVA = 100
Voltage Ratio = 220/132 kV
Minimum Tap = -15
Maximum Tap = +10
=100*10^6/ (√3*220*10^3)

=262.43 A
= 262.43/400
=0.66 A
=1.52
=100*10^6/ (√3*132*10^3)
=437.39 A
=437.39/600
= 0.73 A
=1.37 A
=100*10^6/ (√3*220*0.85*10^3)
=308.74 A
= 308.74 /400
= 0.77 A

= 0.77 * 1.52
=1.18 A
= 0.18 A
= (1.18+1)/2
= 1.09 A
= 0.1 + (0.2 * 1.09)
= 0.37 A
stability. Here it is less than 90% so the setting is acceptable.
=100*10^6/ (√3*220*1.1*10^3)
= 237.26 A
= 238.57 /400
= 0.60 A

= 0.60 * 1.52
= 0.82 A
Hence the differential current is = 1 – 0.82
= 0.18 A
= (0.82 +1)/2
= 0.91 A
= 0.1 + (0.2 * 0.91)
= 0.28 A

2.4 Procedure for Relay Setting of Transformer Differential Relay DTH 31/32
Data Required
MVA Rating
Voltage ratio
Vector group
CT ratio on HV Side
CT ratio on LV Side
Transformer Tap
Min -% or OLTC Tap max +%
Rated current (HV Side) = MVA / (3*Rated voltage (HV side))
Rated current (LV Side) = MVA / (3*Rated voltage (LV side))
If CT is Star/Star
Current on CT Secondary (HV) = Rated Current (HV Side) / CT Ratio (HV Side)
If CT is Star/Delta the current shall be multiplied by 3.
Current on the secondary of ICT for Star/Star = Current on CT Secondary (HV Side) / ICT Ratio (HV
Side)

For Star/Delta the current shall be multiplied by 3.
The same procedure is followed for getting the relay current on LV side also.
Idiff = Relay Current = Difference between the HV and LV Current (Current at the secondary of ICT’s)
To make sure that the relay does not operate when the Transformer taps are changed the currents are
calculated at extreme taps.
At Tap on -%
At Tap on +%
Currents at the CT Secondary and ICT Secondary are calculated by using the above procedure. For
extreme +ive and extreme –ive tap.
Relay current I1= The difference between the HV and LV Currents for extreme +ive tap.
I2= The difference between the HV and LV Currents for extreme +ive tap.
DTH 31 relay has a fixed pick up setting and variable single slope bias setting.
For setting the Bias Setting
The operating current of DTH 31 is given by the equation

Relay operating current = Pickup setting + Bias Setting * Bias current
The pickup setting in DTH 31 = 0.15 (constant)
Bias Current = (I1+I2)/2
Operating Current at normal tap with Bias setting Is = 0.15(or) 0.3
I Operating 1A = 0.15 + 0.15 * Bias Current
I Difference 1A = I1- I2
Operating Current at extreme Minimum tap
I Operating 2A.
I Difference 2A = I1 – I2.
Operating Current at extreme Maximum tap
I Operating 3A.
I difference 3A = I1 – I2.
In each of the above cases IOperating current >IDifference + Tolerance
If the above is true a setting of 0.15 for bias is selected. In case the above is not true. Similar calculation is
carried out for the next Bias Setting.
Bhinmal Differential Relay Settings Calculations
Relay Type: DTH31
Station Name : 220 kV GSS Modak

Transformer Name: NGEF
MVA = 100
Rated Voltage in kV (HV Side) = 220 kV
Rated Voltage in kV (LV Side) = 132 KV
Vector Group = YNyn0d1
ICT Ratio (HV Side) = 7.3/5
ICT Ratio (LV Side) = 3.6/2.89
=100*10^6/ (√3*220*10^3)
=262.4319 A
=100*10^6/ (√3*132*10^3)
=437.39 A
= 262.4319/300
=0.874 A

=437.39/600
= 0.728 A
Interposing CT (HV Side) =
= 7.3/5
Vector Group Interposing CT (HV Side) = Star/Delta
Vector Group Interposing CT (LV Side) = Star/Delta
(HV Side))
=√3*(0.656/1.128)
=1.03 A
(LV Side))
=√3*(0.728/1.261)
= 0.99 A
Difference between HV side and LV side currents = Current on ICT Secondary (HV) –
= 1.03 – .99
=0.04

=100*10^6/ (√3*220*0.85*10^3)
=308.743 A
= 308.74 /300
=1.029 A
Current on ICT Secondary (HV) = √3*(Current on CT Secondary (HV)/ Interposing CT (HV
Side))
=√3*1.029=1.22 A
Current on ICT Secondary (LV) = 1.00 A
= 1.22- 1.00
= 0.22 A
Ibias = ( I1 + I2 ) /2
= ( 1.22+ 1 ) /2
= 1.11 A
Operating Current = Bias setting + Slope 1 * I bias
= 0.15 +0.15*1.11
= 0.3165 A
As Id= 0.22, Operating current = 0.31, Hence stable

=100*10^6/ (√3*220*1.10*10^3)
= 238.57A
= 228.20 /300
= 0.795 A
Side))
=0.9434
Current on ICT Secondary (LV) = 1.00 A
=1- 0.9434
= 0.056 A
At Tap of -10 % I bias = (I1+I2)/2
= (1+0.9434)/2 A =0 .97 A
Consider a bias setting of Is = 0.30
The I Operating Current = 0.15 + Is * I bias
= 0.15 + 0.15* 0.97
=0.29 A
At tap of -10% with full load the relay require a current 0.47 A at Is = 0.30.

Since I1-I2 = 0.056 which is less than the operating value, relay will not operate.
2.5 Differential Relay Settings Calculations RET 670
Differential Relay Settings Calculation Procedure
Relay Type: RET 670
Required Data
MVA
Voltage Ratio
Vector Group
CT Ratio (HV Side)
CT Ratio (LV Side)
CT resistance value (Rct)
Lead resistance value (Rl)
Minimum Tap = - %
Maximum Tap = +%

Relay current on LV side I2 = Current on CT secondary LV side / Ratio compensation for LV side
Relay Current for HV side I1= Current on Ct secondary (HV)/ Ratio compensation for HV side
Bias current I bias = (I1+I2)/2
Differential current I dif and Ibias is calculated foe extreme Transformer taps
= Is + m1* I bias for current less than (4.Iref) + m2* Ibias for current greater than (4 Iref)
Where m1 and m2 are the slope of relay char
90% of operating current for better stability..
Sample Calculation for ABB RET 670
A) Biased Differential
Substation Name : 220 kV GSS Rajwest
Relay Type: ABB RET 670

Transformer Name: TELK
MVA = 315
Minimum Tap = -10
Maximum Tap = +15
=315*10^6/ (√3*400*10^3)
=454.66 A
= 454.66/1000
=0.454 A
= 2.199 A

=315*10^6/ (√3*220*10^3)
=826.66 A
=826.66/1000
= 0.826 A
=1.209 A
Compensated current on LV side=√3*0.826/1.209
=315*10^6/ (√3*400*0.85*10^3)
=534.89 A
= 534.89 /1000
=0.534 A
Inter posing CT (HV) winding = √3*Current on CT sec (HV)/Ratio correction factor
=√3*0.534/2.199
=0.421 A
Hence the differential current is = ABS (0.421– 1.183)
= 0. 762 A

= (0.421+1.183)/2
= 0.802 A
Therefore the operating current of the relay will be = Is + (1.5* M1)+M2*Ibias
= 0.1 + (0.4 * 1.5)*0.8*0.697
= 1.25 A
=315*10^6/ (√3*400*1.1*10^3)
=413.33 A
= 413.33 /1000
=0.413 A
Interposing HV winding current=√3*0.413/2.199
=0.325
I bias = (interposing CT HV current side+ Interposed LV side CT current side)/2
= (0.325+1.183)/2
= 0.754 A
Hence the differential current is = 1.183-0.325
= 0.858 A

Therefore the operating current of the relay will be = Is + (m1+m2)* I bias
= 0.1 + (0.4 +0.8) * 0.754
= 1.01 A
The operating value is larger than the differential value hence the system is Stable for the MAX tap
condition.
B) High Impedance Differential
Substation Name : 220 kV GSS Rajwest
Relay Type: ABB RET 670
Transformer Name: TELK
MVA = 315
% Impedance of the transformer=0.1198
CT resistance (Rct) =4Ω
Lead Resistance value=1.5 Ω

=315*10^6/ (√3*400*10^3)
=454.66 A
Maximum fault current contributed =Rated current /Percentage Impedance
=454.66/0.1198
=3795.15 A
Maximum fault current contributed secondary=3795.15/1000
= 3.795 A
Operating Voltage =3.795*(4+2*1.5)
=26.57V
Rstab = 250Ω
Stabilizing resistance value (Rstab) to be set as per the operating voltage range given in the technical
reference manual table provided by the manufacturer.

2.6 Procedure for Relay setting of Transformer Differential Relay ABB RADSB
Substation : 220kV GSS Dausa
Tranformer Name : CGL
Relay Type: RADSB [ABB]
Data Required
MVA Rating
Voltage ratio
Vector group
CT ratio on HV Side
CT ratio on LV Side
Transformer Tap
Min -% OLTC Tap and max % OLTC tap
Calculation

Interposing CT (HV Side) = Current on CT Secondary (HV) / ICT ratio on HV side
Interposing CT (LV Side) = Current on CT Secondary (LV) /0.578
(HV Side))
(LV Side))
Calculations for OLTC tap setting -%
Current on ICT Secondary (HV) in A
= √3*(Current on CT Secondary (HV)/ Interposing CT
(HVSide))
Calculations for OLTC tap setting +%

Current on ICT Secondary (HV) in A = √3*(Current on CT Secondary (HV)/ Interposing
CT (HVSide))
Relay setting range available : Operating value can be Settable 20, 25, 35, or 50% of In
Id = (I1-I2)
Ibias = ( I1+I2)/2
Operating current = 0.25+ Slope1 setting * I bias
Id should be less than I operating current
The setting shall be selected in such a way that the relay shall not operate for the extreme tap
positions
Relay Type: RADSB [ABB]
Substation : 220kV GSS Dausa
Tranformer Name : CGL
Transformer Name: CGL
MVA = 100
Voltage Ratio = 220/132kV
Vector Group = YnYa0

ICT Ratio (HV Side) = 4.37/2.88
ICT Ratio (LV Side) = 3.64/2.88
=100*10^6/ (√3*220*10^3)
=262.4319 A
=100*10^6/ (√3*132*10^3)
= 437.39A
= 262.4319/(300/5)
= 4.37 A
=437.39/(600/5)
= 3.64 A
Interposing CT (HV Side) = 4.37/2.88
= 1.517
=1.263

Current on ICT Secondary (HV) = (Current on CT Secondary (HV)*Interposing CT
(HV Side))
=4.37/(4.37 /2.88)
= 2.88 A
(LV Side))
=3.64/(3.64/1.263)
=2.88 A
= 4.9891 – 4.9916
= 0.0019606708 A
=100*10^6/ (√3*220*0.85*10^3)
=308.74 A
= 308.74/(300/5)
=5.15 A
Current on ICT Secondary (HV) = (Current on CT Secondary (HV)/ Interposing CT (HV side )
=5.15/(4.862 /1.517)

=3.39 A
Current on ICT Secondary (LV) =(Current on CT Secondary (HV)/ Interposing CT (HV side)
= 2.88 A
= 3.39-2.88
= 0.51 A
I bias = ( I1+I2)/2 = (3.39+2.88)/2 = 3.34
Operating current = 0.25+0.2*3.34 = 0.88 A
Id = 0.51 is less than I operating = 0.88, hence stable
=100*10^6/ (√3*220*1.1*10^3)
= 238.57 A
= 238.57 /(300/5)
=3.98 A
Side))
=3.98/(3.954 / 1.517)
=2.62 A

Current on ICT Secondary (LV) = 2.88
= 2.88-2.62
= 0.26
I bias =( I1+I2) /2
= ( 2.88+2.62) /2 = 2.75 A
I operating = 0.25 + 0.2* 2.75 = 0.8 A
Id= 0.26 is less than I operating = 0.8 A, hence stable

2.7 Procedure for Relay Setting of Transformer Differential Relay GE T60
Data Required
MVA Rating
Voltage ratio
Vector group
HV voltage
LV voltage
HV side
LV side
At Normal tap

Current onLV side = Current on CT secondary (LV)/N2
At -% Tap
Current on CT Secondary = Rated Current (HV Side) / CT Ratio (HV Side)
Current on CT Secondary =Ratio of compensation * CT secondary current
At Tap on +%
Ibias = IHV+ILV

I Operating = Pickup setting + 0.3 * Bias Current for Bias current less than Break Point
I Operating = Pickup setting + 0.3 * Bias Current + 0.8 (Bias current – Break point) for Bias current greater
than Break point.
not operate for extreme taps.
Substation Name: 220 kV GSS Kushkhera
Relay Type: GE T60

Differential Current Setting = 0.25 A
Break Point = 2
Slope1 = 0.3
Slope 2= 0.8
Transformer Name: IMP
MVA = 100
Minimum Tap = -15
Maximum Tap = +10
=100*10^6/ (√3*220*10^3)
=262.43 A
= 262.43/400
=0.66 A

=1.52
=100*10^6/ (√3*132*10^3)
=437.39 A
=437.39/500
= 0.874 A
=1.143 A
Calculations for OLTC at Nominal tap setting
Full load Current for 220 kV %( HV Side) Winding = MVA / (√3*0.9*kV)
=100*10^6/ (√3*220*10^3)
= 262.43 A
= 262.43 /400
= 0.656 A
= 0.656 * 1.52

Compensated current internally = 0.9972 A
= 0.0027A
Bias current = Max (I1, I2)
= (0.9972, 1)
= 1 A
[Since the bias current is less than Break point (2) the slope will be within 30%]
= 0.25 + (0.3* 0.9972)
= 0.549 A
=100*10^6/ (√3*220*0.85*10^3)
= 308.74 A
= 308.74 /400
= 0.772 A
As the adopted compensation is 1.33 A the current to relay bias terminal
= 0.772 * 1.33

Compensated current internally =1.029 A
Hence the differential current is = 1.029 – 0.87
= 0.15 A
Bias current = Maximum of (I1, I2)
= Maximum of (1.029, 0.87)
= 1.029 A
= 0.25 + (0.3 *1.029)
= 0.558 A
=100*10^6/ (√3*220*1.1*10^3)
= 238.57A
= 238.57 /400
= 0.5964 A
= 0.596 * 1.52

Hence the differential current is = 1-0.905
= 0.095 A
Bias current = Maximum of (I1, I2)
= Maximum of (0.905, 1)
= 1 A
= 0.25 + (0.3 * 1)
= 0.55 A

2.8 Procedure for Relay Setting of Transformer Differential Relay CSC326
Data Required
MVA Rating
Voltage ratio
Vector group
HV voltage
LV voltage
HV side
LV side
At Normal tap

I Operating = Pickup setting + 0 * Bias Current if less than 0.656
I Operating = Pickup setting + (0.3 * Bias Current -0.656) if greater than 0.656
I Operating = Pickup setting + 0.7 * (Bias Current-4) + 0.3*4
Differential protection characteristics for transformers:
Substation Name: 220 kV GSS Bharathpur
Relay Type: CSC326

MVA = 100
Minimum Tap = -15
Maximum Tap = +10
=100*10^6/ (√3*220*10^3)
=262.43 A
= 262.43/400
=0.66 A
=1.52
=100*10^6/ (√3*132*10^3)
=437.39 A

=437.39/500
= 0.8747 A
Required Ratio Compensation = 1/0.8747
= 1.143 A
Calculations for OLTC Nominal tap setting:
Full load Current for 220 kV %( HV Side) Winding at = MVA / (√3*kV)
=100*10^6/ (√3*220*10^3*)
= 262.43 A
= 262.43 /400
= 0.656 A
= 0.656 * 1.52
= 0.997 A
Compensated current internally on HV =0.997 A
=100*10^6/ (√3*132*10^3)
=437.39 A

=437.39/500
= 0.8747 A
= 1.143 A
Compensated current on LV = 0.8747 * 1.143
= 0.9993
Hence the differential current is = (0.9973 – 0.9993 )
= 0.002 A
= (0.656+0.8747)/2
= 0.765 A
[Since the bias current is less than 4 A and greater than 0.656 the slope will be within 30%]
Therefore the operating current of the relay will be = Is + (0.3* ( I bias-0.656)
= 0.2 + (0.3 *0.109)
= 0.232 A

=100*10^6/ (√3*220*0.85*10^3)
= 308.74 A
= 308.74 /400
= 0.772 A
= 0.772 * 1.52
= 1.18 A
Hence the differential current is = (1.18– 1.0 )
= 0.18 A
= (1.18+1)/2
= 1.09 A
= 0.2 + (0.3 *0.434)
= 0.33 A

=100*10^6/ (√3*220*1.1*10^3)
= 238.5 A
= 238.574 /400
= 0.596 A
As the adopted ratio correction is 1 A the current to relay bias terminal
= 0.596 * 1.52
Differential current = I1- I2 = 1-0.905 = 0.095 A
= (0.905 +1)/2
= 0.952 A
= 0.2 + (0.3*(0.952- 0.656))
= 0.288 A

stability. Here it is less than 90% so the setting is acceptable
2.9 Procedure for Relay Setting of Transformer Differential Relay DUOBIAS
Data Required
MVA Rating
Voltage ratio
Vector group
HV voltage
LV voltage
HV side
LV side
At Normal tap

I Operating = Pickup setting + 0 * Bias Current if less than 0.656
I Operating = Pickup setting + (0.3 * Bias Current -0.656) if greater than 0.656
I Operating = Pickup setting + 0.7 * (Bias Current-4) + 0.3*4
Differential protection characteristics for transformers:

Substation Name: 220 kV GSS Neemrana
Relay Type: DUOBIAS
MVA = 100
Minimum Tap = -15
Maximum Tap = +10
=100*10^6/ (√3*220*10^3)
=262.43 A

= 262.43/400
=0.66 A
=1.52
=100*10^6/ (√3*132*10^3)
=437.39 A
=437.39/500
= 0.8747 A
= 1.143 A
Calculations for OLTC Nominal tap setting:
Full load Current for 220 kV ( HV Side) Winding = MVA / (√3*kV)
=100*10^6/ (√3*220*10^3*)
= 262.43 A
= 262.43 /400
= 0.656 A
= 0.656 * 1.52* √3

= 1.727 A
Compensated current internally on HV = 1.727 A
=100*10^6/ (√3*132*10^3)
=437.39 A
=437.39/500
= 0.8747 A
= 1.143 A
Compensated current on LV = 0.8747 * 1.143
= 1
Hence the differential current is = (1 – 1 )
= 0 A
= (1+1)/2
= 1 A
Therefore the operating current of the relay will be = Is + (0.3* ( I bias)
= 0.2 + (0.3*1)
= 0.5A

=100*10^6/ (√3*220*0.85*10^3)
= 308.743 A
= 308.743 /400
= 0.771 A
= 0.771 * 1.52
= 0.888 A
=100*10^6/ (√3*132*10^3)
=437.39 A
=437.39/500
= 0.8747 A
= 1.143 A
Compensated current on LV = 0.874* 1.143

= 0.998 A
Hence the differential current is = (1.0-0.888)
= 0.12 A
= (0.888+1)/2
= 0.944 A
= 0.2 + (0.3 *0.28)
= 0.284 A
Id= 0.12 is less than I operating current = 0.284 A, hence Stable
=100*10^6/ (√3*220*1.1*10^3)
= 238.5 A
= 238.5 /400
= 0.56 A
As the adopted ratio correction is 1 A the current to relay bias terminal
= 0.56 * 1.52

Compensated current internally on LV = 0.851*1.143
= 1 A
Hence the differential current is = 1-0.8512
= 0.1488 A
= (1 +0.1488)/2
= 0.574 A
[Since the bias current is less than 0.656 the slope will be zero.
= 0.2 + (0.3*(0.574))
= 0.2A

2.10 Procedure for Relay Setting of Transformer Differential Relay
7UT61X/7UT51X
Data Required
MVA Rating
Voltage ratio
Vector group
HV voltage
LV voltage
HV side
LV side
At Normal tap

Current onLV side = Current on CT secondary (LV)/N2
At -% Tap
Current on CT Secondary = Rated Current (HV Side) / CT Ratio (HV Side)
At Tap on +%

Ibias = IHV+ILV
I Operating = Pickup setting + 0.25 * Bias Current if less than 2.5A
I Operating = Pickup setting + 0.25 * Bias Current + 0.5 ( Bias current above 2.5 A )
Relay Operating current is calculated using the above equation.
not operate for extreme taps
Substation Name: 220 kV GSS Nagaur
Relay Type: SIPROTECH 7UT61X
Transformer Name: BHEL -II
MVA = 100

Minimum Tap = -15
Maximum Tap = +10
=100*10^6/ (√3*220*10^3)
=262.43 A
= 262.43/400
=0.66 A
=1.52
=100*10^6/ (√3*132*10^3)
=437.39 A
=437.39/500
= 0.73 A
=1.14 A

=100*10^6/ (√3*220*0.85*10^3)
= 308.74 A
= 308.74 /400
= 0.77 A
= 0.77 * 1.52
Compensated current internally =1.18 A
= 0.18 A
Bias current = (I1+I2)
= (1.18 +1)
= 2.18 A
= 0.2 + (0.25 *2.18)
= 0.74 A
Differential current at w tap condition should not be more than 90% of operating current for better
=100*10^6/ (√3*220*1.1*10^3)

= 238.5 A
= 238.5 /400
= 0.60 A
= 0.60 * 1.14
Hence the differential current is = 1 – 0.68
= 0.32 A
Bias current = (I1+I2)
= (0.68 +1)
= 1.68 A
= 0.2 + (0.25 * 1.68)
= 0.62 A

2.11 Differential Relay Settings Calculations GE SR745
Relay Type: GE SR745
Required Data
MVA
Voltage Ratio
Vector Group
CT Ratio (HV Side)
CT Ratio (LV Side)
Minimum Tap = - %
Maximum Tap = +%
Break point

Relay current on LV side I2 = Current on CT secondary LV side * Ratio compensation for LV side
Relay Current for HV side I1= Current on Ct secondary (HV)* Ratio compensation for HV side
Bias current I bias = Maximum of(I1,I2)
Differential current I dif andIbias is calculated foe extreme Transformer taps
= Is + Slope1* I bias for I bias current less than Break point And Is + Slope1*Break point + Slope2* (Ibias-
Break point) for I biascurrent greater than Break point.
90% of operating current for better stability.

Sample Calculation of GE SR 745
Substation Name : 220 kV GSS Khinwasar
Relay Type: GE SR745
Transformer Name: EMCO
MVA = 100
Voltage Ratio = 220/132
Minimum Tap = -15
Maximum Tap = +10
Break point=2
Is=0.2, m1=25%, m2=80%
=100*10^6/ (√3*220*10^3)

=262.4319 A
= 262.4319/400
=0.65607 A
= 1.5242 A
=100*10^6/ (√3*132*10^3)
=437.38 A
=437.38 /500
= 0.875 A
=1.143 A
= 0.875 * 1.143
I1=1 A
=100*10^6/ (√3*220*0.85*10^3)
=308.743 A

= 308.743 /400
=0.7719 A
= 0.7719 * 1.5242
I2=1.176 A
Hence the differential current is = (I1-I2) =1.176 – 1
= 0. 176 A
Bias currentIbias = Max(I1,I2)
= (1.176,1)
= 1.176 A
[Since the bias current is less than Break point the slope will be within 25%]
= 0.2 + (0.25 * 1.176)
= 0.494 A
=100*10^6/ (√3*220*1.1*10^3)
=238.574A
= 238.574 /400

=0.5964 A
= 0.5964* 1.5242
I2=0.909 A
Hence the differential current is= (I1-I2) = 1 - 0.909
= 0.091 A
Bias current Ibias = Max(I1,I2)
= (1, 0.909)
= 1 A
[Since the bias current is less than Break point the slope will be within 25%]
= 0.2 + (0.25 * 1)
= 0.45 A

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