Protection Riddle No.60 -Role of resistor in high impedance REF
What is the reason of using resistor in high impedance relay?

#1
Sat, December 12th, 2009 - 14:31
The main role of external resistor in high impedance REF circuit is expansion of available protection setting range for suitable setting selection and increasing of protection stability against out of zone earth fault which can be caused CT saturation regarding probability variation wide of used current transformers characteristics. Indeed the prime task in REF high impedance calculating settings is to calculate the value of the stabilizing resistor Rstab and stability factor K. A stabilizing resistor is required to ensure through fault stability when one of the secondary CT’s saturates while the others do not. The requirements can be expressed as:

VS = IS.Rstab  and
VS > K.If .(Rct + 2Rl + RB)

where:
VS = stability voltage setting
VK = CT knee point voltage
K = relay stability factor
IS = relay current setting
Rct = CT winding resistance
Rl = CT secondary lead resistance
RB = resistance of any other components in
the relay circuit
Rstab = stabilising resistor

For example, assume:

VK = 97 V
Rct = 3.7 Ω
Rl = 0.057 Ω

Starting with the desired operating time, the VK/VS ratio and K factor can be found.
An operating of 40ms (2 cycles at 50Hz) is usually acceptable, and hence, from submitted relay manufacturer we can write:

VK/VS = 4
K = 0.5

The maximum phase fault current can be estimated by assuming the source impedance to be zero, so it is limited only by transformer impedance to 5250A, or 10A secondary after taking account of the ratio compensation. Hence the stability voltage can be calculated as

VS = 0.5 x 10( 3.7 + 2 x 0.057) = 19.07V

Hence, Calculated VK = 4 x 19.07 = 76.28V However,

Actual VK = 91V and
VK/VS = 4.77

Thus from above Figure, with K = 0.5, the protection is unstable.
By adopting an iterative procedure for values of VK/VS and K, a final acceptable result of VK/VS = 4.55, K = 0.6, is obtained. This results in an operating time of 40ms.
The required earth fault setting current Iop is 250A (assume the maximum earth fault current is limited by the
earthing resistor to 1000A). The chosen E/F CT has an exciting current Ie of 1%, and hence using the equation:
Iop = CT ratio x (IS + nIe)
where:
n = no of CT’s in parallel (=4)
IS = 0.377, use 0.38 nearest settable value.
The stabilising resistance Rstab can be calculated as 60.21Ω.

Reference: ALSTOM - Protective Relay Application Guide (PRAG)

#2
Fri, September 28th, 2012 - 13:02
I am not able to follow the following para extracted from your riddle no 60The required earth fault setting current Iop is 250A (assume the maximum earth fault current is limited by the
earthing resistor to 1000A). The chosen E/F CT has an exciting current Ie of 1%, and hence using the equation:
Iop = CT ratio x (IS + nIe)
where:
n = no of CT’s in parallel (=4)
IS = 0.377, use 0.38 nearest settable value.
The stabilising resistance Rstab can be calculated as 60.21Ω.
can u please explain

#3
Sat, September 29th, 2012 - 08:40
CT ratio = 600/1=600

Ie = 0.01 A

n = 4 ( 3 ph+1 N )

250 = 600 x ( Is + 0.04)

Is= 0.37666

#4
Tue, July 1st, 2014 - 05:43
Hai, is it the same calculation for voltage operated relay?

#5
Tue, July 1st, 2014 - 08:47
Yes, generally relay setting could be applied based on percentage of CT rated secondary current. For REF with voltage setting, same calculated percentage should be applied.

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