This document discusses current transformers (CTs), including their construction, theory of operation, polarity, equivalent models, accuracy, saturation, burdens, parallel operation, and residual magnetism. It describes the main types of CT construction, how primary current creates magnetic flux sensed by the secondary winding, and factors that influence accuracy such as burdens, saturation, and exciting current.

1. Current Transformers
CT Basics
– Construction
– Theory of Operation
– Polarity
– Equivalent Model
– Open Circuit Voltage
– Accuracy
CT Transient Performance
CT Construction
Bar-Type
– A fixed insulated straight conductor that is a
single primary turn passing through a core
assembly with a permanently fixed secondary
winding.
Bushing Type
– A secondary winding insulated from and
permanently assembled on an annular core with
no primary winding or insulation for a primary
winding.

2. CT Construction
Window Type
– A secondary winding insulated from and
permanently assembled on the core with no
primary winding but with complete insulation
for a primary winding.
Wound Type
– A primary and secondary winding insulated
from each other consisting of one or more turns
encircling the core. Constructed as multi-ratio
CTs by the use of taps on the secondary
winding.
Theory of Operation
Op
Ip - Primary Current
Ip
Np - Primary Winding Turns
Is - Secondary Current
Np Es - emf Induced
φp - Magnetic Flux Due to Primary Ip
φs - Magnetic Flux Due to Secondary Is
φe - Magnetic Flux Due to Exciting Ie
Oe Zb - Burden Impedance
Is
Zb Vs Ns
Os

3. Theory of Operation
When a time varying current Ip flows, a
magnetomotive force Fp (mmf) is developed by:
Fp = Ip * Np.
The primary mmf creates a magnetic flux Φp in
the core given by: Φp = Fp/Rm where Rm is the
core reactance.
The direction of Φp is determined by the right
hand rule.
Φp Links the secondary winding, inducing an
electromotive force Es (emf), resulting in a
secondary current Is flowing through burden Zb.
Theory of Operation
The magnetomotive force Fs developed is due to Is and
given by: Fs = Is * Ns.
Fs creates an opposing flux Φs to Φp by Lenz’s law
resulting in a magnetic flux Φe in the core by: Φe = Φp -
Φs.
Φe is sufficient to maintain Es at a given Ip and load
impedance.
Since the magnetic flux is proportional to the mmf we get:
– Fe = Fp - Fs or
– Ie * Np = Ip * Np - Is * Ns dividing by Ns
– Ie * Np/Ns = Ip * Np/Ns - Is
– Is = Ip * Np/Ns if Ie is small

4. Polarity
The CT primary and secondary terminal is physically
marked with a polarity.
The marking indicates the instantaneous direction of the
secondary current in relation to the primary current.
When current flows in at the marked primary, current is
flowing out of the marked secondary:
PRI. SEC.
Hint: Direction of the secondary current can
determined as if the two polarity terminals formed
a continuous circuit
Equivalent Model
The transformation of current induces errors.
Some energy form the primary winding is used to:
– Establish magnetic flux in the core.
– Change the direction of the magnetic flux in the core
named hysteresis losses.
– Generate heat due to eddy currents.
– Establish leakage flux.
To account for losses a fictitious component is
introduced, the exciting current Ie.

5. Equivalent Model
Ip/n Is
1:n
Ie/n
Zpn**2 Zs
Ic/n Im/n
Zb
Rcn**2 Xmn**2
The primary current Ip is stepped down in magnitude by 1:n through a
no-loss transformer.
– Zpn**2 - primary winding impedance
– Zs - secondary winding impedance
– Rcn**2 - hysteresis and eddy current losses referred to the secondary
– Xmn**2 - magnetic reactance accounting for losses to establish flux
referred to the secondary
Equivalent Model
Ip/n Is
1:n
Ie/n
Rs
Ic/n Im/n
Es Us Zb
Rcn**2 Xmn**2
If the secondary winding is uniformly distributed on the core, Zs is
resistive = Rs.
The voltage drop across the primary winding is negligible to the source
voltage to which it is connected and does not effect current flow,
Zp/n**2 = 0.
The secondary current is reduced by the shunting current of the
exciting branch. The greater Ie the less accurate Is represents Ip.

6. I
Equivalent model
Ie' Ip'
Vector relationship from equivalent model
– Is is reference.
Is
– Voltage across burden is: Us = Is * Zb.
– emf induced is: Es = Us + Is * Rs = Is * (Zb + Rs)
– The angle α between Us and Is defines the burden
power factor
– Treating the exciting reactance as linear, Im lags
Es by 90
– The magnetic flux Φe from Im lags Es by 90
– The Vectorial sum Im and Ic define Ie
– Ip is the Vectorial sum of Is and Ie
Oe
– Is is less then Ip by a delta I and with a phase
Es
IsXb
angle error of β.
Us IsRb Ie' Ic'
Im'
Open Circuit Voltage
Open secondary causes Φs to go to zero.
Ip drives the core to saturation each half cycle.
The action of Ip changing from maximum to zero back to
maximum causes Φp to change from saturation in one
direction to its saturated value in the opposite direction.
The rapid rise of Φp induces high voltage spikes in the
secondary winding.
A formula for peak voltage derived from CT tests is:
Vpeak = 35 × Zb × Ip / n
.
Tests have shown values ranging fron 500 to 11,000 volts.

7. Accuracy
Definition: Ability to reproduce the primary current in
secondary amperes in both wave-shape and magnitude.
ANSI/IEEE C57.13 designates two rating classes C or T
describing capability.
– C: the ratio can be calculated, leakage flux is negligible
due to uniform distribution of secondary winding.
– T: the ratio must be determined by test, leakage flux is
appreciable due to undistributed windings.
Designations are followed by a terminal voltage rating that
the CT can deliver to a standard burden at 20 times rated
secondary current without exceeding 10% ratio correction.
– Voltage classes are 100, 200, 400, 800
Accuracy
The burdens are in ohms and at a .5 pf.
– Standard burdens are B-1, B-2, B-4, B-8
Example:
– C800: 800 V/ 5 A * 20 = 8
– If current is lower the ohmic burden can be higher in
proportion.
– Accuracy applies to the full winding. If a lower tap of a
multiratio CT is used the voltage capability must be
reduced proportionally.

8. CT Transient Performance
CT Saturation
CT Burden
CT Parallel Operation
Residual Magnetism
CT Saturation
Type C class CTs performance can be calculated from the
excitation characteristics.
– The excitation curve specifies the relationship of the
exciting current Ie to exciting voltage Vs.
– The point of the curve where the tangent is at a 45 to
the abscissa is called the knee.
Above the knee is the saturated region where the
change in Ie no longer results in an appreciable
change to Es.
Below the knee is the unsaturated region where Ie is
negligible.

9. CT Saturation
Type T class CTs performance must be determined from
test curves of primary to secondary current at standard
burdens.
Factors influencing the threshold of saturation under
steady state are Zb and Is.
– Transformer operation is given by:
Es = 4.44 fNAB max substitute Φe = AB max
Φe = Es / 4.44 fN where
Es = Is( Zb + Rs)
– Decrease Zb or Is (through the turns ratio) will limit Es
and thus Φe.
CT Saturation
Steady State Analysis
– Saturation by the AC component is avoided by selecting the proper turns
ratio, decreasing the burden, or choosing the proper CT accuracy class.
– Criteria: The product of Is and Zb does not exceed the saturation or knee
point voltage of the CT.
– Procedure:
Determine the secondary current(Ip/n) from the primary fault current
at the desired turns ratio.
Determine the total secondary burden, Zt = Zb + Zlead + Zct.
Calculate the required secondary voltage, Vs = (Ip/n) * Zt.
Determine the secondary excitation current Ie required for the value
of Vs from the excitation characteristic curve.
Determine the approximate burden current Is by subtracting Ie from
Ip/n.
Check the effective ratio with the desired ratio to see if the
performance is within the intended accuracy.

10. CT Saturation
Steady State Example: CT ratio 500:5 C100, If = 12000 A
– Ip/n = 12000/100 = 120
– Zct = 80*.005 = .4
– Zlead is 200 feet of full circuit run of #10 Awg,
Zl = e**(.232(10)-2.32)*200/1000 = .2
– Relay burden of .15
– Zt = .4 + .2 +.15 = .75
– Vs = 120 * .75 = 90
– Vs of 90 is approximately Ie = 18
– Is = 120 - 18 = 102
– Effective ratio is 12000/102 = 117.6 or 588.23:5
CT Saturation
Steady state analysis
– An alternate procedure to check CT performance is to use Ks the
saturation factor and a criteria of Ks = Vk/Vs. Where Vk is the
effective knee point voltage of the CT from the excitation
characteristic curve and Vs is the voltage across the CT secondary.
Determine Vk from the excitation characteristic curve and CT
ratio.
Calculate Vs from (Ip/n)*Zt.
Determine Ks and check criteria.

11. CT Saturation
DC Saturation
– When a fault occurs the current usually contains a DC component.
The total flux required to produce the offset drives the CT into
saturation.
– Flux reaches a saturation during the positive cycle and the exciting
reactance decreases shunting the primary current thus distorting the
secondary.
– During the negative cycle of the primary current, the core becomes
unsaturated.
– As the DC component decays the negative cycle of the primary
current and flux become greater and the core eventually runs out of
saturation during a complete cycle returning to steady state.
– The DC component time constant is the X/R ratio of the primary
circuit.
CT Saturation
DC Saturation Analysis
– If saturation is to be avoided the secondary voltage requirement Vs must
be (1 +X/R) times the voltage required for the AC component.
– Example: CT 2000:5 C800 tap 1500:5, If = 19349 A with a source
impedance of 81 deg.
Ip/n = 19349/300 = 64.5
Zct = 300 * .0025 = .75, Zlead = .2, Zb = .15
Zt = .75 + .2 +.15 = 1.1
X/R = tan(81) = 6.31
Vs = (6.311 +1) * 64.5 * 1.1 = 518.9
Vs of 518.9 is approximately Ie = 20
Is = 64.5 - 20 = 44.5
Effective ratio is 19349/44.5 = 434 or 2175:5
– Due to 1500:5 tap the voltage must be reduced by the proportional amount
giving the CT a 600 volt rating.

12. CT Burden
Burden is defined as the total impedance of the secondary
circuit. This includes:
– CT winding resistance
– Leads resistance from CT to relay and back
– Impedance of the connected relays
CT winding resistance
– Higher ratio CTs (3000:5) resistance is .0025 Ω/turn
– Lower ratio CTs (300:5) resistance is .005 Ohms/turn
Resistance of leads is based on the AWG gage given by:
.232G−2.32
Ω / 1000' = e
residual relays. Note: Most microprocessor relays
calculate I residual.
CT Burden
Relay Impedance
– Microprocessor based relay almost negligible
– Electromechanical given in VA @ rated I
Burden is also influenced by CT connection and type of
fault.
– Example: Y-connected CTs
3Ph fault is balanced, current through the phase
relay is the only burden.
SLG fault is unbalanced, current must go through
the phase and

13. Parallel Operation
Parallel interconnection increases the burden seen
by each individual relay.
– The increase is dependent upon the type of connection,
number of transformers, and distribution of current
between transformers.
Parallel interconnection can be used to supply a
high burden when low ratios are required.
– Primaries are connected in series and secondaries in
parallel.
Residual Magnetism
(Remanence)
Caused by current interruption and the magnetizing force
becoming zero while the flux density in the core is
operating at a high level. Occurs when:
– Direct current is passed through a winding
– Application of a high overcurrent interrupted at a peak magnitude
Effect of residual magnetism is the accuracy of the
secondary when the CT is next energized. How effects:
– The flux changes start at the remnant value near saturation
distorting the waveform.
– The primary current is required for the excitation reducing the
secondary output and increasing the CT error.