High Current Measurements

7/26/2019 High Current Measurements

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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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High Current Measurements