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UNIVERSITI TUN HUSSEIN ONN
MALAYSIA
FACULTY OF ELECTRICAL AND ELECTRONIC
ENGINEERING
BEF 20903
BEF 23903
ELECTRICAL MEASUREMENTS
High AC Cur rent Measurements
Using Current Transformer
Written By Dr. Zainal Alam Haron
Dept of Electrical Power Engineering
Universiti Tun Hussein Onn Malaysia
Date
Version First draft
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Contents
1. Introduction
2. Types of Current Transformers3. 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
(greater than 100 A)
X
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
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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.
Figure 3. Equivalent circuit of a current transformer.
where
Vp : Primary voltage
Ip : Primary currentRp : 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 : Rsistance of external burden
Np:Ns Is
ZL= RL+ jX LEs
Ip
Im
Io
Ie
Ep
IsXp
Re Xm
XsRs
Rp
Vp Vs
Ideal
CT
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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. I m
and Ie are the reactive and loss components of the exiting current. The resulting
simplified equivalent circuit is shown in Figure 4.
Figure 4. Simplified equivalent circuit of a current transformer.
Np:Ns
ZL= RL+ jX LEs
Ip
Ep
Is
Im
Io
Ie
Re Xm
Vp
Is
Vs
Ideal
CT
Rs
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Equivalent Circuit Of CT Referred To The Secondary Side
Figure 5 shows a simplified equivalent current transformer diagram converted to the
secondary side.
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.
p
s
tN
N
K
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;
ZL= RL+ jX L
Ip
Im
Io
Ie
Re XeEs
Is
Vs
Note: Rshas been assumed negligible here.
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s
p
actI
IK
iii. 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.
t
p
s
s
p
nom KN
N
ratedI
ratedIK
)(
)(
iv. Ratio correction factor (R.C.F.) - This is equal to the transformation
ratio divided by nominal ratio.
nom
act
K
KFCR ..
Thus,
nomact KFCRK .. and
act
nomK
FCRK
..
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:
sts
p
s
p IKIN
NI
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
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primary current is 800 A. Calculate the maximum secondary current and secondary
voltage across the ammeter.
Figure 6. Pictorial view of bar-type current transformer referred to in the Worked
Example above.
Solution
Secondary current:
AN
NII
s
p
ps 5160
1800
Voltage across ammeter:
0.12.055 AAss
RRIV 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 currentis 280 A, calculate:
1. The secondary current
2. The voltage across the secondary terminals
3. The voltage drop across the primary.
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Figure 7.
Solution
(a) The secondary current
80
1
400
5
p
s
s
p
I
I
N
N
5.380
1280
p
s
psN
NII A
(b) The voltage across the secondary terminals
2.42.15.3 sss RIV V
(c) The voltage across the primary terminals
5.52400
52.4
s
p
spN
NVV mV
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 CTs 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 Esand Epare 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 Esby an angle s. The
secondary current is now transferred to the primary side by reversing I sand multiplied
by the turns ratio KT. The total current flows through the primary Ipis then vector sum
of KTIsand Io.
Is-Secondary Current
Es- Secondary induced emf
Ip- primary Current
Ep- primary induced emf
KT- turns ratio = numbers of
secondary turns/number of primary
turns
Io- Excitation Current
Im- magnetizing component of Io
Iw- core loss component of Io
m- main flux.
Figure 9. Phasor diagram of a current transformer.
Im
Ip
Ie Io
KTIs
-Ep
Es
Is
O
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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:
osTos
p
s
s IIKIIN
NI
'
Figure 10 shows a vector representation of the three currents in the equivalent
diagram.
Figure 10. Vector representation of the three currents in a CT.
IpIs = KTIs
Io
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Hence, the percentage current error in the primary current is
%100
)(
)()(%
actualI
actualImeasuredI
p
pp
%100
p
psT
I
IIK
%100
A
AN
K
KK
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 %100%100'%
p
psN
p
ps
I
IIK
I
II
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
TN KK 400
5
2000
Ip= 2000 A
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Is= 4.9 A
Current error %22000
20005400%100
p
psT
I
IIK
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, 105
50
NK
Number of primary turns, Np= 3
We require 45
20
AK
Therefore,3
104
saN
giving Nsa= 2 turns
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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,
ps II '
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)
Figure 12. Equivalent circuit of ideal current transformer connected to a 1 burden.
Figure 13. Simplified equivalent circuit of non-ideal current transformer connected
with a 1 burden.
1000A:5A Is
ZL= 1 Es
Ip
Im
Io
Ie
Ep
Is
1000A:5A Is
ZL= 1 ES
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Power factor of exciting current (Io),
p.f. = 0.4 = cos
Therefore,
o42.664.0cos 1
and
oooo
58.2342.669090
Figure 14. Phasor diagram showing phase relationships between Io, Ie, and Im.
Turns ratio
2001000
A5
A
NK
Current flowing in secondary, Is= 5 A, p.f. = 1. Therefore, reflected secondary
current is
AA5
A10005
1000' AIKI
sNs
IeIo
Im= 80 A
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Figure 15. Phasor diagram showing phase relationships between Ip, Io, and Is.
From Figure 15, primary current
22' cossin oosp IIII
22 58.23cos158.23sin11000 oo
A4.1000
Therefore, actual turns ratio
8.2005
4.1000
s
p
AI
IK
Ratio error,
%04.08.200
8.200200%
actual
actualmeasured
K
KKe Ans.
Is
Ip
Io
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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.
Figure 16. Simplified equivalent circuit of a current transformer.
N
p
s
T KN
NK 400
1
400
20005400'
s
p
s
s IN
NI A
Np:Ns
ZLEs
Ip
Ep
Is
Im
Io
Ie
Re Xm
Vp
Is
Vs
Ideal
CT
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Figure 15. Phasor diagram showing phase relationships between Ip, Io, and Is.
Y- component of primary-referred secondary current,
176628cos2000cos'' oss IyI A
X- component of primary-referred secondary current,
9.93828sin2000sin'' oss IxI A
X- component of excitation current, Im= 90 A
Y- component of excitation current, Ie= 40 A
Total x-component of primary current,
1029909.938)()( ' msp IxIxI A
Total y-component of primary current,
1806401766)()( '
esp IyIyI A
Therefore, actual primary current
Is
Ip
Io
= 28
o
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207918061029)()( 2222 yIxII ppp A Ans.
Percentage current error
)(
)()(%
actualI
actualImeasuredI
p
pp
p
ps
I
II
'
%8.3%1002079
20792000
Therefore,
%8.3% Ans.
How to Reduce Errors in Current Transformer
In current transformer design, the core characteristics must be carefully selected
because excitation current Ioessentially 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 currenttransformer while current is flowing in the primary circuit. If the secondary is
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accidentally opened, the primary current Ipcontinues to flow unchanged because the
impedance of the primary is negligible compared to that of the electrical load. The
line current thus becomes the excitingcurrent 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 satfor most of the time.
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
ip
sat
Ndtd
e
3000 V
secondary
voltage
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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
afterthe secondary circuit is again closed.
To facilitate maintenance of ammeter instrumentation, short-circuiting switches are
often installed in parallel with the CTs secondary winding, to be closed whenever the
ammeter is removed for service (see Figure 18).
Figure 18. Short-circuit switch allows ammeter to be removed from an active current
transformer.
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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
Step 2. Select Ammeter.
Solution
Measurement of larger currents requires the use of a
current transformer. Standard practice is to use a 5-A full-
scale ammeter with the appropriate current transformer.
Ammeters so used are calibrated in accordance with the
selected transformer.
Step 1. Calculate Current
pfV
PI
0.8240
20,000
A104I Therefore,
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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.
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 asomewhat higher current, a
150:5 current transformer is
therefore selected. The
ammeter is a 5-A meter with
its scale calibrated from 0 to150 A. Single-phase
power line
NL
AC ammeter
Current
transformer
Step 4: Draw the connection diagram.
The ammeter is connected to the line, through the current
transformer, as in the figure.
Figure 20.
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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
http://1.bp.blogspot.com/-KDqgan_Tqx4/Tb_Orvezo0I/AAAAAAAAAQE/5USXfVeN0qk/s1600/Bushing+Type+CT+%28BCT%29.JPGhttp://2.bp.blogspot.com/-SffsYcQlgUg/Tb_Om4HbV5I/AAAAAAAAAQA/pa2DEciE-gI/s1600/Bar+Type+CT.jpg -
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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
http://4.bp.blogspot.com/-Ep63lXsmEK8/Tb_O3YxM3RI/AAAAAAAAAQM/-MU74OzeyK8/s1600/wound+type+CT.gif -
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A CT for operation on a 110 kV grid
Outdoor CTs
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 ClampsThe 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)
http://en.wikipedia.org/wiki/Current_transformerhttp://en.wikipedia.org/wiki/Current_transformer -
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(b)
Figure 24. Schematic of an analogue AC current clamp meter.
2. Current Clamp with Voltage Output
(a)
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(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:
5.2200.0
5.0
s
o
o
I
V
R
Figure 26
VResistiveshunt
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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 MeterThe 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.
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Figure 29. Block diagram of a digital AC current clamp meter.
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-
7/26/2019 High Current Measurements
39/39