Sample Calculation for Differential Relays

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.



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













At -% Tap






At Tap on +%









Ibias = IHV+ILV


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













At -% Tap






At Tap on +%









Ibias = IHV+ILV


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













At -% Tap






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.


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



Sample Calculation for Differential Relays

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