Instrument Security Factor (ISF) of Current Transformer

Instrument security Factor (ISF) is used for specifying current limit, by which current will be delivered by a current transfprmer (CT) during fault conditions. 

ISF is used for metering class CTs. For protection class CTs, there is similar term called Accuracy Limit Factor (ALF).

Metering instruments like energy meters, transducers etc. are meant for measuring electrical quantities with high accuracy under normal operation. Measurement under fault condition is not required and insignificant due to short duration of faults. Metering instruments are designed to carry current upto a certain level. 

For example, an energy meter used with CT of 1000/1A ratio carries maximum 1A under full load. Therefore its input current carrying paths needs to be designed for current upto 150 to 200% of rated current (say 2A continuous) . For fault conditions, which will prevail for short duration, it should withstand upto 5A or 10A. But in case of severe fault of 40kA or 50kA on primary side, secondary current may go upto 40A or 50A. If meter is to be designed for such heavy currents its size and connections will be bulky. Its cost will go up and accuracy will go down. Therefore, CT core is designed to saturate at such heavy currents and maximum secondary current is specified as ISF.

ISF is ratio of rated instrument limit primary current to rated primary current. If ISF is specified as 5 for a 1000/1A CT, then CT metering core will saturate at 5000A primary and current in secondary will not go above 5A. Sometimes, auxiliary reactors are used to achieve ISF limit.

Equivalent circuit of CT:

I_P = Primary current

I_S = Secondary current

V_S = Secondary voltage

Z_E = Exciting impedance

I_E = Exciting current

R_S = Secondary resistance

X_L = Leakage reactance

Z_B = Burden impedance 

Effect of burden on ISF: ISF is specified for a certain burden. If actual burden is different from rated burden, ISF will change as below:

ISF at actual burden = ISF at rated burden x (Rated burden / Actual burden)

* Internal secondary winding resistance of CT to be added for calculating burden.

For example, for a CT with rated burden 20VA and ISF 5, if actual burden is 10VA, ISF will be =5 x (20/10) =10. I.e. maximum secondary current will be 10A.

Resistance, Reactance and Impedance

Resistance: When electric current flows through a material there will always some opposition to this flow. Resistance (R) is measure of opposition to flow of electrical current. Resistance is measured in ohm (Ω). The resistance depends on resistivity (ρ), length (l) and area (a) of material:

1. Length of material: Resistance is directly proportional to length of material. That’s why poor voltage conditions are observed in remote locations.
2. Area of material: Resistance is inversely proportional to the area of material. That’s why we use thicker wires for heavy duty appliances.
3. Resistivity of material: Resistivity is fundamental property of material by which it opposes flow of electric current. Resistivity of some of common materials at 20⁰C is given below:

Resistance will cause energy loss, which is equal to I²R. For transfer of electrical power from source to appliance, low losses are expected, therefore materials with lower resistivity are used as conductors like copper, aluminum etc. In some cases, these losses may be intentional. Like in heaters, I²R losses should be higher. Therefore, we use material with high resistivity like Nichrome etc. Rubber, Air and PVC are used for insulation material due to higher resistivity.

Resistance of a material is practically considered constant over a short working range of temperature. However, it varies with temperature due to change in resistivity.  For example resistivity of copper at 21⁰C will be 1.68 x 10^-8 x (1 + 0.00404) Ωm.

Resistance will be same for DC as well as AC currents.   

Reactance (X): In AC systems, voltage and current are continuously varying due to sine wave form. Due to this, in addition to resistance, one more component opposes the flow of current called reactance (X). Reactance have two components, Capacitive reactance (Xc) and inductive reactance (XL).  

Xc will be due to capacitance (C) in the circuit and XL will be due to inductance (L) in the circuit. 

For Parallel RLC circuit, same voltage will be applied across three elements:

Current in Capacitance (IXc) leads voltage by 90⁰ and current in Inductance (IXL) lags voltage by 90⁰, Threrefore, IXc and IXL are 180⁰ apart. Total current due to reactance (IX) will be difference of IXc and IXL. Being parallel connected circuit, total reactance (X) of the circuit will be 

For series RLC circuit same current will flow through all the three elements. Voltage across three elements will be different in magnitude and angle. In this case magnitude of voltage across these components can be higher than source voltage magnitude (VAC).

Reactance of the circuit will be

Impedance (Z): Impedance is combined effect of resistance (R) and total reactance (X). Currents flowing through resistance and reactance are 90⁰ apart. 

For parallel RLC circuit Impedance will be 

For Series RLC circuit impedance will be

For example: If in above circuit of Parallel connected RLC:

V=100V

R = 10 Ω

C = 100 μF

L = 10 mH

f = 50 Hz

Current through resistor

IR = 100/10 = 10A @0⁰

Xc = 1/ (2 x 3.1428 x 50 x 100 x 10^-6)

Xc = 31.82 Ω

Current through capacitor

IXc = 100/31.82 = 3.142A @90⁰

XL = 2 x 3.1428 x 50 x 10 x 10^-3

XL = 3.14 Ω

Current through inductor

IXL = 100/3.14 = 31.85A @-90⁰

Total reactance X will be

X = 1/(1/3.14 - 1/31.82) = 3.48 Ω

IX = 100/3.48 = 28.7A @-90⁰

Or 

IX = IXL - IXc = 31.85 - 3.142 = 28.7A

Z = 1/[√{(1/R)² + (1/X)²}]

Z = 1/[√{(1/10)² + (1/3.48)²)}] = 3.29 Ω

Or calculation of impedance by caculating current first:

IZ = √(IR² + IX²)

IZ = √(10² + 28.7²) = 30.39 A

Z = V/IZ = 100/30.39 = 3.29 Ω

From the above it can be seen that current through inductor is 31.85A which is higher than current supplied by source.

Now, if same components are connected in series, the impedance of circuit will be as below:

V=100V

R = 10 Ω

C = 100 μF

L = 10 mH

f = 50 Hz

Xc = 1/ (2 x 3.1428 x 50 x 100 x 10^-6)

Xc = 31.82 Ω

XL = 2 x 3.1428 x 50 x 10 x 10^-3

XL = 3.14 Ω

Total reactance X will be

X = Xc - XL = 31.82 - 3.14 = 28.68 Ω

Z = √(R² + X²) 

Z = √(10² + 28.68²) = 30.37 Ω

Current through circuit = V/Z = 100/30.37 = 3.29 A

Voltage across resistor = IR = 3.29 x 10 = 32.9V @0⁰

Voltage across capacitor = IXc = 3.29 x 31.82 = 104.7V @-90⁰

Voltage across inductor = IXL = 3.29 x 3.14 = 10.3V @90⁰

From the above it can be seen that voltage across capacitor is 104.7V wich is higher than source voltage of 100V. Therefore, while working with RLC circuits, precuations must be taken to avoid electric shock from higher voltages in the circuit.

R-X diagram for impedance relays

R-X plane:

Impedance relays are widely used as distance relays for protection of transmission lines / feeders. Various manufacturers use different algorithms in their relays for optimum performance. R-X diagrams are used to illustrate characteristics of relay. From these diagrams protection engineers can easily understand relay behaviour during fault conditions and visualize difference in characteristics of relays of different model / manufacturer. Relay testing using secondary injection is also based on verification of operating characteristics on R-X diagrams. For example, chrematistics of some of the manufacturers are as below:

In above diagrams, horizontal axis represents Resistance and Vertical axis represents Reactance. Here, -ve value of resistance means power import, +ve value means power export, -ve value of reactance means capacitance and +ve value of reactance means inductance. Lines / feeder are have inductive behaviour in case of fault coonditions, therefore their reactance will be +ve and for forward direction fault, power will injected into fault from relay location, therefore resitance will also be positive. Therefore, Lines / feeders have their line representation in first quadrant based on per unit positive sequence reactance (X₁) and positive sequence resistance (R₁). We can represent feeder on R-X plane by calculating total X₁ and R₁ of Transmission Line / Feeder (Click here for more about line parameters).  

Zone settings parameters in relays also have R and X value (or Z and φ) for each zone. For quadrilateral characteristics resistive reach, R value for left and right blinders is also specified. From these relay setting parameters R-X diagrams can be plotted.

Importance of RMS value in electrical system

In Alternating Current (AC) system magnitude of electrical quantities are expressed in root mean square (rms) values. 

In Direct Current (DC) system, magnitude of current or voltage is constant over the period of time. We can specify that 5A current is flowing or 12V potential difference is existing between two points. For example, DC 5A is shown below, Red being positive and black being negative. In this image, one AC signal is also shown in Blue colour.

The question is: at which point we should specify the current as it is continuously varying and changing its sign from +ve to -ve and -ve to +ve in each cycle. One of the approach may be specifying peak value:

The problem with peak value is that, in actual, it appears for a fraction of time in a cycle (two time in a cycle, one in +ve and another in -ve). for rest of the cycle, magnitude is less than peak value. If we compoare 5A DC and 5A AC (peak value), DC current will have more power over a period of time. That is how concept of RMS value came. It is value of AC signal, for which same value DC signal dissipates the same power in a resistor. 

That means heat produced by 5A DC over a period of time "T" and by 5A AC over time "T" shall be same. Therefore, power of both AC and DC signals is same.

For understanding rms value, let us calculate actual energy dissipated in heat for one cycle of AC. First divide AC cycle in "n" parts on time scale. Calculate energy of each part by formula I²Rt and add them up over the period of one cycle.  

In above image energy for one cycle will be I₁²Rt₁ + I₂²Rt₂ + I₃²Rt₃ +…e I_n²Rt_n 

Energy (over one cycle) E = (I₁²Rt₁ + I₂²Rt₂ + I₃²Rt₃ +…. I_n²Rt_n)

In the above equation t₁ =t₂ =t₃ …=t_n =t

Energy (over one cycle) = (I₁² + I₂² + I₃² +…. I_n²) Rt

Average energy for n^th part

Energy (for n^th part) = (I₁² + I₂² + I₃² +…. I_n²) Rt /n

Power (for n^th part) = E/t = (I₁² + I₂² + I₃² +…. I_n²) R /n        …EQ1

As we know Power P = I²R                                                    …EQ2

From EQ1 and EQ2:

I²R = (I₁² + I₂² + I₃² +…. I_n²) R /n

I² = (I₁² + I₂² + I₃² +…. I_n²)/n

I =  √ {(I₁² + I₂² + I₃² +…. I_n²)/n}

Therefore, RMS value is is the square root of the arithmetic mean of the squares of the values.

In AC system values mentioned are always rms values unless otherwise specified. For exmple in AC system:

• 200V means 200V RMS value
• 10A means 10A RMS value
• 200Vpeak means 200V peak value
• 10Apeak means 10A peak value

RMS value for common waveforms is as below:

• Sine wave: RMS value = Peak value / √2
• Square wave: RMS value = Peak value
• Saw tooth wave / triangle wave: RMS value = Peak value / √3

Peak value is significant for insulation calculations,

What if CT secondary is open circuit?

CT secondary shall always be kept short circuited. What happens if CT is open circuit?

Current Transformer (CT) is an instrument transformer used to reduce or mutiply current by a pre-defined factor. 

It has small number of primary turns and large number of secondary turns. Generally primary winding has only one turn, as primary conductor simply pass through CT core.  Secondary winding has large number of turns (may be 1000, depending on ratio). In the image below, only few turns are shown in CT secondary for simplicity . Primary winding and secondary winding are wound on magnetic core.

From the above, Primary current Ip flows through primary winding. Due to this current magnetic field is developed and magnetic flux φ flows in the core. This alternating flux produces voltage Vs in secondary winding. Normally, Zs is very low (or may be short circuit),  this voltage develops current Is in secondary winding. 

According to Lenz's law:

The current induced in a circuit due to a change in a magnetic field is directed to oppose the change in flux and to exert a mechanical force which opposes the motion.

Therefore, flux produced by secondary current is opposite to flux produced by primary winding. Theroritically, both flux should be same but in oppsite direction. However, in actual, there will be a very small difference due to leakage flux. The resultant flux, difference of flux produced by primary current and flux produced by secondary current, is very small for normal load current. It magnifies in case of higher fault current. CT core is designed to carry resultant flux upto a certain fault level. 

For low value of Zs, CT can be considered as short circuited transformer. Reactance of primary winding will be very small, hence very low voltage drop across primary terminals.

If CT is open circuit, there will be no secondary current and no flux produced by secondary current. Magnetic core will get saturated by high amount of flux produced by primary current. 

In saturated part of waveform there will be no change in flux and due this there will be no e.m.f. induced in secondary. This will result in no current in secondary.

As CT is already staurated in this case, voltgae developed will not be sinusoidal. It will contain harmonics and wave shape will be spiky. The peak value of voltage will be very high in the range of 2-3kV. This higher voltage is dangerous for persons working in panels and equipment. This high voltage can lead to failure of insulation in CT, Cables, relays, terminal blocks etc.

Switch on to fault (SOTF)

 Switch on to fault (SOTF) feature is used to instantaneously isolate the fault in case a line is charged with persisting fault / remote end earth switch closed. Main requirement is to avoid time delayed tripping in this condition.

Normal case when line is charged from both ends: Consider line is protected by distance protection, during fault following happens:

1. Fault in Zone-1: Instantaneous tripping (phase selective is implemented).
2. Fault in Zone-2: Instantaneous tripping with carrier receive from remote end (phase selective is implemented).
3. 3-phase Fault near to bus: Relay will decide direction based on memory voltage.

In case line is charged with persisting fault (remote end still open): Consider line is protected by distance protection, during fault following happens:

1. Fault in Zone-1: Instantaneous tripping (phase selective is implemented). Auto reclose will close the breaker, if implemented.
2. Fault in Zone-2: Delayed tripping with Zone-2 time as no protection will operate from remote end and no carrier will be sent from remote end. Auto reclose will not operate for delayed tripping.
3. 3-phase Fault near to bus: Relay cannot decide direction based on memory voltage as there is no memory voltage.

From above, we need a protection which can instantly clear faults in Zone-2 and 3-phase fault near to bus. Line closure is detected either from binary input wired from CB close switch or from dead line detect based on voltage and currents (Typical setting Voltage <40% and Current <20%). SOTF feature remain activated for ~0.2 seconds after this detection.

Different methods for detecting SOTF conditions are as below, distance relay may use combination of these:

1. Non-directional Zone-2 impedance based SOTF protection: In this case fault impedance is measured regardless of its direction. If it is less than Zone-2 impedance instantaneous 3-pole trip will be issued and auto reclose is blocked.
2. Over current based SOTF protection: In this case current through all phases is monitored. If current in any phase is higher than set value, which is kept ~ 200% of maximum full load current of line, 3-pole trip is issued and auto reclose is blocked

 

CT saturation

CT saturation is one of the phenomena which is important for differential protections. During saturation, core of CT gets saturated due to higher level of flux density than designed value. After this no more change of flux occurs even the primary current is still changing. 

Due to no change in flux secondary current vanishes. We may get secondary current having distorted waveform during CT saturation.

For differential protections, this may lead to differential current for out of Zone faults and relay may give unwanted trip command. Most of the manufacturers compensate CT saturation with different techniques.

 Let us see two cases with and without CT stauration

For case-1 External fault without CT saturation:

Differential current = 1 + 1 + 2 – 4 = 0 kA
Bias current = 1 + 1 + 2 + 4 = 8 kA
Idiff / IBias = 0/8 = 0.0

For case-2 External fault with CT saturation, CT for fourth feeder is staurated and measuring only 1kA current instead of actual 4kA current: 

Differential current = 1 + 1 + 2 – 1 = 3 kA
Bias current = 1 + 1 + 2 + 1 = 5 kA
Idiff / IBias = 3/5 = 0.6

For case-2 we can avoid unwanted relay operation if we keep differential slope setting higher than 0.6

* For simplicity current is considered without any DC component.