UNIVERSITI TUN HUSSEIN ONN MALAYSIA FACULTY OF ELECTRICAL AND ELECTRONIC ENGINEERING
BEF 20903 BEF 23903 ELECTRICAL MEASUREMENTS
High AC Current Measurements Using Current Transformer
Written By
Date Version
Dr. Zainal Alam Haron Dept of Electrical Power Engineering Universiti Tun Hussein Onn Malaysia First draft
Contents 1. Introduction 2. Types of Current Transformers 3. Equivalent Circuit of a current transformer 4. Current Transformer Ratios 5. Phasor Diagram of Current Transformer 6. Errors in current transformer 7. Phase angle error 8. Methods to minimize errors 9. Types of current transformer construction 10. Clamp meter
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LEARNING OUTCOMES After completing this module you will be able to:
1. Explain the principle of operation, construction and use of current transformers to measure AC currents,
2. Interpret the different current transformers at suppliers catalog,
3. Specify the different kinds of current transformers.
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1.0 INTRODUCTION High AC currents can be measured with shunts under special circumstances, but most often such direct connections to AC lines are extremely dangerous to humans. One way to get around the hazard is to use current transformers (CT) that isolate AC line voltages and reduce input current by a specified ratio. A current transformer produces a scaled down replica of the input quantity to the accuracy expected for the particular measurement. The common laws for transformers are valid for current transformers.
Tasks of Current Transformers The main tasks of a current transformer are: 1. To transform currents from a usually high value to a value easy to handle for measuring instruments. 2. To insulate the metering circuit from the primary high voltage system. 3. To provide possibilities of standardizing the measuring instruments to a few rated currents.
Advantages of current transformers
1. The measuring instruments can be placed for away from the high voltage side by connecting long wires to the current transformer. This ensures the safety of instruments as well as the operator.
2. A current transformer can be used to extend the range of current measuring instruments like ammeters.
3. The power loss in current transformers is very small as compared to power loss due to the resistance of shunts.
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Comparison Between Resistive Shunt And Current Transformer
Table 1; Shunt versus current transformer trade-off Advantage
Shunt
Current transformer
Meter cost
X
Meter can handle higher currents
X
(greater than 100 A) Meter power consumption
X
Fewer accuracy issues (saturation, phase response at high-power factors)
X
2.0 TYPES OF CURRENT TRANSFORMERS Based on the construction, two types of CTs can be identified. 1)
Clamp-on CT This is a C.T., in which the core can be opened with the help of a clamp and the conductor (whose current is to measured) can be inserted into the core. This conductor acts as a primary winding. The secondary winding is wound on the laminated core. A low range ammeter is connected across the secondary, which measures current of the conductor. It is a portable instrument, which can be used in laboratories.
Figure 1. Clamp-on CT 5
2)
Bar type CT A bar type CT has a circular ring type core over which secondary winding is wound, across which a direct reading ampere meter is connected. When a bar conductor or a Bus bar whose current, is to be measured is inserted in to the ring, the ampere meter reads the current (Figure 2).
Figure 2. Bar type CT
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3.0 EQUIVALENT CIRCUIT OF A CURRENT TRANSFORMER For quantitative analysis, a current transformer can be represented by the equivalent circuit of Figure 3.
Ip
Rp
Xp
Is’
Rs
Np:Ns
Xs
Is
Io Ie Vp
Re
Im Xm Ep
Es
ZL = RL + jX L
Ideal CT Figure 3. Equivalent circuit of a current transformer.
where Vp
:
Primary voltage
Ip
:
Primary current
Rp
:
Primary winding resistance
Xp
:
Primary winding reactance
Io
:
Exciting current
Ie
:
Core loss current
Re
:
Equivalent core loss resistance
Im
:
Magnetizing current
Xm
:
Magnetizing reactance
Ep
:
Primary winding induced voltage
p, s
:
Primary and Secondary windings
:
Flux surrounding the windings
Is
:
Secondary winding current
Es
:
Secondary induced emf
Rs
:
Secondary winding resistance
Xs
:
Secondary winding reactance
RL
:
Résistance of external burden 7
Vs
XL
:
Reactance of external burden
KT
:
Turn ratio = Ns/Np
It can be seen from the equivalent circuit diagram shown in Figure 3 that the primary current contains two components:
An exciting current, which magnetizes the core and supplies the eddy current and hysteresis losses, etc.
A remaining primary current component, which is available for transformation to secondary current in the inverse ratio of turns.
Simplified equivalent circuit The equivalent diagram in Figure 3 comprises all quantities necessary for error calculations. The primary internal voltage drop does not affect the exciting current and the errors. Therefore the primary internal impedance is not indicated in the diagram.
The secondary internal impedance, however, must be taken into account, but only the winding resistance Rs. The leakage reactance is negligible where continuous ring cores and uniformly distributed secondary windings are concerned. The exciting impedance is represented by an inductive reactance in parallel with a resistance. Im and Ie are the reactive and loss components of the exiting current. The resulting simplified equivalent circuit is shown in Figure 4.
Is’
Ip
Rs
Np:Ns
Is
Io Ie Vp
Re
Im Xm
Ep
Es
ZL = RL + jX L
Ideal CT Figure 4. Simplified equivalent circuit of a current transformer. 8
Vs
Equivalent Circuit Of CT Referred To The Secondary Side Figure 5 shows a simplified equivalent current transformer diagram converted to the secondary side. Ip’
Is Io’ Ie’
Es
Im’
Re’
Xe’
ZL = RL + jX L
Vs
Note: Rs has been assumed negligible here.
Figure 5. Simplified equivalent circuit of current transformer referred to the secondary side.
Current Transformer Ratios The transformer ratios for a CT can be defined as follows: i.
Turns ratio - This is the ratio of the turns of the transformer windings.
Kt
Ns Np
where Ns = No. of turns in secondary winding Np = No. of turns in primary winding
ii.
Transformation (Actual) ratio--This is the ratio of primary winding current (Ip) to the secondary winding current (Is) of the transformer;
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K act
iii.
Ip Is
Nominal ratio - This is the ratio of the rated primary winding current (Ip(rated)) to the rated secondary winding current (Is(rated)) of the transformer. K nom
iv.
I p (rated ) I s (rated )
Ns Kt Np
Ratio correction factor (R.C.F.) - This is equal to the transformation ratio divided by nominal ratio.
R.C.F
K act K nom
Thus,
K act R.C.F K nom
and
K nom
R.C.F K act
Note If the exciting current Io could be neglected the transformer should reproduce the primary current without errors and the following equation should apply to the primary and secondary currents:
Ip
Ns I s Kt I s Np
In reality, however, it is not possible to neglect the exciting current.
Worked Example 1 A bar-type current transformer which has 1 turn on its primary and 160 turns on its secondary is to be used with a standard range of ammeters that have an internal resistance of 0.2 . The ammeter is required to give a full-scale deflection when the 10
Worked Example
A bar-type current transformer which has 1 turn on its primary and 60 turns on its secondary is to be used with a standard range of mmeters that have an internal resistance of 0.2Ω’s. The ammeter is current is 800 A.when Calculate maximum secondary equired to give aprimary full scale deflection the the primary current is 800current and secondary Amps. Calculate voltage the maximum current and secondary across the secondary ammeter. oltage across the ammeter.
Figure 6. Pictorial view of bar-type current transformer referred to in the Worked Example above.
Solution Secondary current:
Np I s I p Ns
1 800 5A 160
Voltage across ammeter: Vs I s R A 5 R A 5 0.2 1.0 volts
Worked Example 2 A current transformer has a rating of 50 VA, 400 A/5 A, 36 kV, 50 Hz. It is connected into an a.c. line having a line-to-neutral voltage of 14.4 kV. The ammeters, relays and connecting wires on the secondary side possess a total impedance (burden) of 1.2 . If the transmission line current is 280 A, calculate: 1. The secondary current 2. The voltage across the secondary terminals 3. The voltage drop across the primary. 11
Figure 7.
Solution (a)
The secondary current
Np Ns
Is 5 1 I p 400 80
Is I p
(b)
Ns 1 280 3.5 A Np 80
The voltage across the secondary terminals Vs I s Rs 3.5 1.2 4.2 V
(c)
The voltage across the primary terminals V p Vs
Np Ns
4.2
5 52.5 mV 400
Exercise A toroidal transformer has a ratio of 1000 A/5 A. The line conductor carries a current of 600 A. (a) Calculate the voltage across the secondary winding if the ammeter has an impedance of 0.15 . (b) Calculate the voltage drop the transformer produces on the line conductor. (c) If the primary is looped four times through the toroidal opening, calculate the new current ratio.
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Figure 8.
[Answer: 0.45 V; 2.24 mV; 250 A/5 A]
Changing The Transformer Turns Ratio Relatively large changes in a current transformers turns ratio can be achieved by modifying the primary turns through the CT’s window where one primary turn is equal to one pass and more than one pass through the window results in the electrical ratio being modified.
So for example, a current transformer with a relationship of say, 300/5A can be converted to another of 150/5A or even 100/5A by passing the main primary conductor through its interior window two or three times as shown. This allows a higher value current transformer to provide the maximum output current for the ammeter when used on smaller primary current lines.
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PHASOR DIAGRAM OF CURRENT TRANSFORMER The vector diagram for the circuit shown in Figure 4 is shown in Figure 9 Here flux has been taken as reference. EMF Es and Ep lags behind the flux by 90o. The magnitude of the phasors Es and Ep are proportional to secondary and primary turns. The excitation current Io is made up of two components, namely, Im and Ie. The secondary current Io lags behind the secondary induced emf Es by an angle s. The secondary current is now transferred to the primary side by reversing Is and multiplied by the turns ratio KT. The total current flows through the primary Ip is then vector sum of KT Is and Io.
Ip
-Ep
Is-Secondary Current KTIs
Es - Secondary induced emf
Ie
Ip - primary Current Ep - primary induced emf
Io
KT - turns ratio = numbers of O
Im
secondary turns/number of primary turns Io - Excitation Current Im - magnetizing component of Io Iw - core loss component of Io Φm - main flux.
Is Es
Figure 9. Phasor diagram of a current transformer.
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ERRORS IN CURRENT TRANSFORMER Current transformers suffer from two errors:
1. Current ratio (or turns ratio) error 2. Phase angle error
1. The Current Ratio Error Figure 3 shows that not all the primary current passes through the secondary circuit. Part of it is consumed by the core, which means that the primary current is not reproduced exactly. The relation between the currents will in this case be:
I s'
Ns I s I o KT I s I o Np
Figure 10 shows a vector representation of the three currents in the equivalent diagram.
Io
Is’ = KTIs
Ip
Figure 10. Vector representation of the three currents in a CT.
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Hence, the percentage current error in the primary current is
%
I p (measured ) I p (actual) I p (actual) KT I s I p Ip
100 %
100 %
KN KA 100 % KA
According to the definition above, the current error is positive if the secondary current is too high, and vice versa.
Current error is an error that arises when the current value of the actual transformation ratio is not equal to rated transformation ratio.
Current error %
I s ' I p Ip
100%
KN Is I p Ip
KN = rated transformation ratio Ip = actual primary current Is = actual secondary current
Worked Example 3 In case of a 2000/5A class 1 5VA current transformer
KN
2000 400 K T 5
Ip = 2000 A
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100%
Is = 4.9 A
Current error
KT I s I p Ip
100%
400 5 2000 2% 2000
Worked Example 4 An application requires a 20:5 CT ratio, but only a 50:5 CT is available. Given that the number of primary turns (Np) is 3, determine the number of secondary turns that need to be added so that a 20:5 actual ratio will be obtained.
Solution Given:
Nameplate transformation ratio, K N
Number of primary turns, Np = 3
20 4 5
We require
KA
Therefore,
4
giving
Nsa = 2 turns
10 N sa 3
17
50 10 5
2. Phase Angle Error For a ideal current transformer the angle between the primary and reversed secondary current vector is zero. But for an actual current transformer there is always a difference in phase between two due to the fact that primary current has to supply the component of the exiting current. The angle between the secondary current phasor reversed (Is’) and the primary current (Ip) (see Figure 8) is termed as Phase Angle Error; that is,
I s' I p The phase angle error is usually expressed in minutes, and if the reversed current phasor leads the primary current phasor then the phase angle error is defined as positive; otherwise it is taken as negative. It will be seen that with a moderately inductive burden, resulting in Is’ and Io approximately in phase, there will be little phase error and the exciting component will result almost entirely in ratio error. A reduction of the secondary winding by one or two turns is often used to compensate for this.
Figure 11. Picture of current CTs in a switchyard.
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Worked Example 5 The exciting current of a transformer (of ratio 1000A:5A) with a burden of 1 is 1 A at a p.f. = 0.4. Calculate actual transformation ratio and ratio error.
Solution Given :
Exciting current, Io = 1 A, p.f = 0.4 Turns ratio, KT = NS/NP =1000/5 = 200 Burden, ZL = 1 (resistive)
Is
1000A:5A
ZL = 1
ES
Figure 12. Equivalent circuit of ideal current transformer connected to a 1 burden.
Is’
Ip
1000A:5A
Is
Io Ie
Im Ep
Es
ZL = 1
Figure 13. Simplified equivalent circuit of non-ideal current transformer connected with a 1 burden.
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Power factor of exciting current (Io), p.f. = 0.4 = cos
Therefore,
cos1 0.4 66.42 o
and
90 o 90 o 66.42 o 23.58 o
Io
Ie α
Φ Im = 80 A Figure 14. Phasor diagram showing phase relationships between Io, Ie, and Im.
Turns ratio
KN
1000 A 200 5A
Current flowing in secondary, Is = 5 A, p.f. = 1. Therefore, reflected secondary current is
I s' K N I s
1000 A 5 A 1000 A 5A
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Ip
Is’ Io α Φ
Figure 15. Phasor diagram showing phase relationships between Ip, Io, and Is’.
From Figure 15, primary current
Ip
I
' s
I o sin
I
cos
2
2
o
1000 1sin 23.58 1cos 23.58 2
o
o 2
1000.4 A
Therefore, actual turns ratio
KA
Ip Is
1000 .4 200 .8 5
Ratio error,
%e
K measured K actual 200 200.8 0.04% Ans. K actual 200.8
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Worked Example 6 A CT has a single turn primary and 400 secondary turns. The magnetizing curent is 90 A while core loss current is 40 A. Secondary circuit phase angle is 28o. Calculate the actual primary current and ratio error when secondary current carries 5 A current.
Solution The information given is only sufficient to be applied to the simplified equivalent circuit of a current transformer shown in Figure x.
Is’
Ip
Is
Np:Ns
Io Ie Vp
Re
Im Ep
Xm
Es
Ideal CT
Figure 16. Simplified equivalent circuit of a current transformer.
KT
I s'
N s 400 400 K N Np 1
Ns I s 400 5 2000 A Np
22
ZL
Vs
Ip
Is’ = 28o Io α Φ
Figure 15. Phasor diagram showing phase relationships between Ip, Io, and Is’. Y- component of primary-referred secondary current, I s' y I s' cos 2000 cos 28 o 1766 A
X- component of primary-referred secondary current, I s' x I s' sin 2000 sin 28 o 938.9 A
X- component of excitation current, Im = 90 A
Y- component of excitation current, Ie = 40 A
Total x-component of primary current,
I p ( x) I s' ( x) I m 938.9 90 1029 A
Total y-component of primary current,
I p ( y ) I s' ( y ) I e 1766 40 1806 A
Therefore, actual primary current
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I p I p ( x) 2 I p ( y ) 2 1029 2 1806 2 2079 A Ans.
Percentage current error
%
I p (measured ) I p (actual) I p (actual)
I s' I p Ip 2000 2079 100 % 3.8% 2079
Therefore,
% 3.8% Ans.
How to Reduce Errors in Current Transformer In current transformer design, the core characteristics must be carefully selected because excitation current Io essentially subtracts from the metered current and affects the ratio and phase angle of the output current. The higher the exciting current or core loss the larger the error. It is desirable to reduce these errors, for better performance. For achieving minimum error in current transformer, one can follow the following, 1.
Using a core of high permeability, low hysteresis loss magnetic materials, and large cross section. The number of joints in the core should be minimum to minimize the air gaps and the reluctance. Three materials are preferred for making cores: i.
Cold rolled grain oriented (CRGO) silicon steel
ii.
Hot rolled grain oriented (HRGO) silicon steel
iii.
Nickel Iron alloys.
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The grain-oriented materials are preferred as their grains are oriented in the direction so as to provide unidirection to the magnetic field, usually rectangular (Figure 160a) or a ring type (Figure 16b) shape is made. In ring shape, the joints may be eliminated and due to orientation of the grains, flux is always along the grains and so the reluctance as minimum.
Figure 16. Current transformer cores. (a) rectangular core; (b) ring core.
The various alloys used for making cores are
(i) Silicon steel (4% silicon) (ii) Mumetal (76% nickel) (iii) Permendur (50% cobalt) (iv) Hypernik (50% nickel) 2.
Keeping the rated burden to the nearer value of the actual burden.
3.
Ensuring minimum length of flux path and increasing cross – sectional area of the core, minimizing joint of the core.
4.
Lowering the secondary internal impedance.
Secondary voltage of open-circuit CT Every precaution must be taken to never open the secondary circuit of a current transformer while current is flowing in the primary circuit. If the secondary is 25
accidentally opened, the primary current Ip continues to flow unchanged because the impedance of the primary is negligible compared to that of the electrical load. The line current thus becomes the exciting current of the transformer because there is no further bucking effect due to the secondary ampere-turns. Because the line current may be 100 to 200 times greater than the normal exciting current, the flux in the core reaches peaks much higher than normal. The flux is so large that the core is totally saturated for the greater part of every cycle. Referring to Figure 17, as the primary current rises and falls during the first half cycle, flux in the core also rises and falls, but remains at a fixed saturation level sat for most of the time. ip sat
e
d N dt
3000 V
secondary voltage
Figure 17. Primary current, flux, and secondary voltage when a CT is open-circuited.
The same thing happens during the second half-cycle. During these saturation intervals, the induced voltage across the secondary winding is negligible because the flux changes very little. However, during the unsaturated intervals, the flux changes at an extremely hig rate, inducing voltage peaks of several thousand volts across the 26
open-circuited secondary. This is a dangerous situation because an unsuspecting operator could easily receive a bad shock. The voltage is particularly high in current transformers having ratings above 50 VA.
Thus, for reasons of safety, if a meter or relay in the secondary circuit of a CT has to be disconnected, we must first short-circuit the secondary winding and then remove the component. Short-circuiting a current transformer does no harm because the primary current remains unchanged and the secondary current can be greater than that determined by the turns ratio. The short-circuit across the secondary may be removed after the secondary circuit is again closed.
To facilitate maintenance of ammeter instrumentation, short-circuiting switches are often installed in parallel with the CT’s secondary winding, to be closed whenever the ammeter is removed for service (see Figure 18). To facilitate maintenance of ammeter instrumentation, short-circuiting switches are often installed in parallel with the CT's secondary winding, to be closed whenever the ammeter is removed for service (see figure).
Note:
Though it may seem strange to intentionally short-circuit a power system component, it is Short-circuit switch allows perfectly proper and quite Figure 18. Short-circuit switch allows to be removed ammeter toammeter be removed from an from an active current necessary when working with active current transformer current transformers. transformer. circuit.
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Worked Example Current is to be measured in a single-phase line which supplies a 240-V, 20 kW load with a 0.8 power factor. Select an appropriate ammeter and current transformer. Direct reading ammeters are available with full-scale reading ranging from 2 to 20 A.
Figure 19
Solution
Step 1. Calculate Current
I
P V pf
20,000 240 0.8
Therefore,
I 104 A Step 2. Select Ammeter.
Measurement of larger currents requires the use of a current transformer. Standard practice is to use a 5-A fullscale ammeter with the appropriate current transformer. Ammeters so used are calibrated in accordance with the selected transformer. 28
Step 3
Select Current Transformer Because the current is greater than 20 A, a current transformer is required. A transformer is chosen which can accommodate a somewhat higher current, a 150:5 current transformer is therefore selected. The ammeter is a 5-A meter with its scale calibrated from 0 to 150 A.
N
L Current transformer
AC ammeter Single-phase power line
Step 4: Draw the connection diagram.
Figure 20.
The ammeter is connected to the line, through the current transformer, as in the figure.
TYPES OF CURRENT TRANSFORMER CONSTRUCTION Below are the types of current transformers in the manner of construction and applications.
1.
Bar-type Current Transformer A Bar-type current transformer is one that has a fixed and straight single primary winding turn passing through the magnetic circuit. The primary winding and secondary winding(s) are insulated from each other and from the core(s) and are assembled as an integral structure.
2.
Bushing-type Current Transformer A bushing-type current transformer is one that has an annular core and a secondary winding insulated from and permanently assembled on the core but has no primary winding and no insulation for a primary winding. This type of current transformer is for use with a fully insulated conductor as the primary winding. A bushing type current transformer usually is used in equipment where the primary conductor is a component pan of other apparatus.
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3.
Window-type Current Transformer A window-type current transformer is one that has a secondary winding insulated from- and permanently assembled on the core, but has no primary winding as an integral part of the structure. Primary insulation is provided in the window, through which one turn of the line conductor can be passed to provide the primary winding.
4.
Wound-type Current Transformer A wound-type current transformer is one that has a primary winding consisting of one or more turns mechanically encircling the core or cores. The primary winding(s) and secondary winding(s) are insulated from each other and from the core(s) and are assembled as an integral structure.
Bar Type CT
Bushing Type CT
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Window Type CT
Wound Type CT Figure 21: Type of current transformers
Outdoor CTs Current Transformer could either be: 1. Indoor current transformer: One that, because of its construction, must be protected from the weather. 2. Outdoor current transformer: One of weather-resistant construction, suitable for service without additional protection from the weather.
In a typical arrangement in outdoor HV CTs, the secondary is wound on bushing type insulated core. The prinary is mounted in insulator bushing insulation around the primary (see Figure 22)
Figure 22
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Outdoor CTs
A CT for operation on a 110 kV grid
Figure 23. Pictures of different outdoor current transformers of various constructions.
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CLAMP METER A clamp meter is an electrical tester that combines a basic digital multimeter with a current sensor. A common form of current sensor comprises a split ring made of ferrite or soft iron and hinged at one one to form a pair of jaws that can be opened to clamp around the conductor whose current is to be measured. A wire coil is wound round one or both halves, forming one winding of a current transformer. The conductor around which it is clamped forms the other winding. This allows properties of the electric current in the conductor to be measured, without having to make physical contact with it, or to disconnect it for insertion through the probe. Types Of Current Clamps The output of the current clamp can be read by any AC ammeter whose input impedance is compatible with the specifications of the current clamp. Current clamps are also available that convert the current input signals into a voltage signal for measurement by a voltmeter.
Some current clamps incorporate a rectifier circuit whose output is a DC voltage that is proportional to the average current being measured. Such clamps facilitate the use of strip chart recorders for obtaining real time trends of current loads. To obtain a TRUE RMS output, a DC-to-RMS converter is attached to output of the dc current clamp.
1.
Current Clamp with Current Output
(a) 33
•
Clamp–on current probes or “current clamps” enable you to measure currents without breaking the electric circuit.
•
Current clamps offer a safe, cost–effective, simple and accurate way to take current measurements.
(b)
Figure 24. Schematic of an analogue AC current clamp meter.
2.
Current Clamp with Voltage Output
(a)
34
measure currents without breaking the electric circuit. •
Current clamps offer a safe, cost–effective, simple and accurate way to take current measurements.
Resistive shunt
V
(b)
Figure 25. Schematic of an analogue AC current clamp meter.
The Output Voltage and “Burden Resistor” The output voltage (Vo) should be set as low as practically possible to minimize the insertion loss. Assuming 0.5 V is the optimum secondary output voltage in a circuit and the output current is 20 A, a 1:100 ratio transformer will yield a secondary current of
≅ 200 mA. Per Figure 26, the burden resistor should be:
Ro
Vo 0 .5 2 .5 I s 0.200
Figure 26
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Current Clamp incorporating a rectifier and filter circuit
Figure 27. Current clamp incorporating a rectifier and filter circuit to convert the ac current signal into a voltage signal.
Figure 28. Picture of an analogue current clamp.
Digital AC Current Clamp Meter The most common application nowadays is the use of a current probe with a digital multimeter. The probe output is connected to a DMM set on the AC current range to handle the probe output.
36
Figure 29. Block diagram of a digital AC current clamp meter.
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