POTENTIOMETERS AND INSTRUMENT TRANSFORMERS

 

D.C & A.C Potentiometers

An instrument that precisely measures an electromotive force (emf) or a voltage by opposing to it a known potential drop established by passing a definite current through a resistor of known characteristics. (A three-terminal resistive voltage divider is sometimes also called a potentiometer.) There are two ways of accomplishing this balance: (1) the current I may be held at a fixed value and the resistance R across which the IR drop is opposed to the unknown may be varied; (2) current may be varied across a fixed resistance to achieve the needed IR drop.

The essential features of a general-purpose constant-current instrument are shown in the illustration. The value of the current is first fixed to match an IR drop to the emf of a reference standard cell. With the standard-cell dial set to read the emf of the reference cell, and the galvanometer (balance detector) in position G1, the resistance of the supply branch of the circuit is adjusted until the IR drop in 10 steps of the coarse dial plus the set portion of the standard-cell dial balances the known reference emf, indicated by a null reading of the galvanometer. This adjustment permits the potentiometer to be read directly in volts. Then, with the galvanometer in position G2, the coarse, intermediate, and slide-wire dials are adjusted until the galvanometer again reads null. If the potentiometer current has not changed, the emf of the unknown can be read directly from the dial settings. There is usually a switching arrangement so that the galvanometer can be quickly shifted between positions 1 and 2 to check that the current has not drifted from its set value.

Circuit diagram of a general-purpose constant-current potentiometer, showing essential features Potentiometer techniques may also be used for current measurement, the unknown current being sent through a known resistance and the IR drop opposed by balancing it at the voltage terminals of the potentiometer. Here, of course, internal heating and consequent resistance change of the current-carrying resistor (shunt) may be a critical factor in measurement accuracy; and the shunt design may require attention to dissipation of heat resulting from its I² R power consumption. Potentiometer techniques have been extended to alternating-voltage measurements, but generally at a reduced accuracy level (usually 0.1% or so). Current is set on an ammeter which must have the same response on ac as on dc, where it may be calibrated with a potentiometer and shunt combination. Balance in opposing an unknown voltage is achieved in one of two ways: (1) a slide-wire and phase-adjustable supply; (2) separate in-phase and quadrature adjustments on slide wires supplied from sources that have a 90° phase difference. Such potentiometers have limited use in magnetic testing.

An instrument that precisely measures an electromotive force (emf) or a voltage by opposing to it a known potential drop established by passing a definite current through a resistor of known characteristics. (A three-terminal resistive voltage divider is sometimes also called a potentiometer.) There are two ways of accomplishing this balance: (1) the current I may be held at a fixed value and the resistance R across which the IR drop is opposed to the unknown may be varied; (2) current may be varied across a fixed resistance to achieve the needed IR drop.

The essential features of a general-purpose constant-current instrument are shown in the illustration. The value of the current is first fixed to match an IR drop to the emf of a reference standard cell. With the standard-cell dial set to read the emf of the reference cell, and the galvanometer (balance detector) in position G1, the resistance of the supply branch of the circuit is adjusted until the IR drop in 10 steps of the coarse dial plus the set portion of the standard-cell dial balances the known reference emf, indicated by a null reading of the galvanometer. This adjustment permits the potentiometer to be read directly in volts. Then, with the galvanometer in position G2, the coarse, intermediate, and slide-wire dials are adjusted until the galvanometer again reads null. If the potentiometer current has not changed, the emf of the unknown can be read directly from the dial settings. There is usually a switching arrangement so that the galvanometer can be quickly shifted between positions 1 and 2 to check that the current has not drifted from its set value.

Potentiometer techniques may also be used for current measurement, the unknown current being sent through a known resistance and the IR drop opposed by balancing it at the voltage terminals of the potentiometer. Here, of course, internal heating and consequent resistance change of the current-carrying resistor (shunt) may be a critical factor in measurement accuracy Potentiometer techniques have been extended to alternating-voltage measurements, but generally at a reduced accuracy level (usually 0.1% or so). Current is set on an ammeter which must have the same response on ac as on dc, where it may be calibrated with a potentiometer and shunt combination. Balance in opposing an unknown voltage is achieved in one of two ways: (1) a slide-wire and phase-adjustable supply; (2) separate in-phase and quadrature adjustments on slide wires supplied from sources that have a 90° phase difference. Such potentiometers have limited use in magnetic testing.

An electrical measuring device used in determining the electromotive force (emf) or voltage by means of the compensation method. When used with calibrated standard resistors, a potentiometer can be employed to measure current, power, and other electrical quantities; when used with the appropriate measuring transducer, it can be used to gauge various non-electrical quantities, such as temperature, pressure, and the composition of gases.

Distinction is made between DC and AC potentiometers. In DC potentiometers, the voltage being measured is compared to the emf of a standard cell. Since at the instant of compensation the current in the circuit of the voltage being measured equals zero, measurements can be made without reductions in this voltage. For this type of potentiometer, accuracy can exceed 0.01 percent. DC potentiometers are categorized as either high-resistance; with a slide-wire resistance ranging from The higher resistance class can measure up to 2 volts (V) and is used in testing highly accurate apparatus. The low-resistance class is used in measuring voltage up to 100 mV. To measure higher voltages, up to 600 V, and to test voltmeters, voltage dividers are connected to potentiometers. Here the voltage drop across one of the resistances of the voltage divider is compensated; this constitutes a known fraction of the total voltage being measured.

In AC potentiometers, the unknown voltage is compared with the voltage drop produced by a current of the same frequency across a known resistance. The voltage being measured is then adjusted both for amplitude and phase. The accuracy of AC potentiometers is of the order of 0.2 percent. In electronic automatic DC and AC potentiometers, the measurements of voltage are carried out automatically. In this case, the compensation of the unknown voltage is achieved with the aid of a servomechanism that moves the slide along the resistor, or rheostat. The servomechanism is actuated by the imbalance of the two voltages, that is, by the difference between the compensating voltage and the voltage that is being compensated. In electronic automatic potentiometers, the results of measurements are read on dial indicators, traced on recorder charts or received as numerical data. The last method makes it possible to input the data directly into a computer. In addition to measurement, electronic automatic potentiometers are also capable of regulating various parameters of industrial processes. In this case, the slide of the rheostat is set in a position that predetermines, for instance, the temperature of the object to be regulated. The voltage imbalance of the potentiometer drives the servomechanism, which then increases or decreases the electric heating or regulates the fuel supply.

A voltage divider with a uniform variation of resistance, a device that allows some fraction of a given voltage to be applied to an electric circuit. In the simplest case, the device consists of a conductor of high resistance equipped with a sliding contact. Such dividers are used in electrical engineering, radio engineering, and measurement technology. They can also be utilized in analog computers and in automation systems, where, for example, they function as sensors for linear or angular displacement.

 

Instrument Transformers Basics

Why instrument transformers?

In power systems, currents and voltages handled are very large.

Direct measurements are not possible with the existing equipment’s. Hence it is required to step down currents and voltages with the help of instrument transformers so that they can be measured with instruments of moderate sizes.

 

Instrument Transformers

Transformers used in conjunction with measuring instruments for measurement purposes are called “Instrument Transformers”.

The instrument used for the measurement of current is called a “Current Transformer” or simply “CT”.

The transformers used for the measurement of voltage are called “Voltage transformer” or “Potential transformer” or simply “PT”.

 

Instrument Transformers:

Fig 1. Indicates the current measurement by a C.T. The current being measured passes through the primary winding and the secondary winding is connected to an ammeter. The C.T. steps down the current to the level of ammeter.

Fig 2. Shows the connection of P.T. for voltage measurement. The primary winding is connected to the voltage being measured and the secondary winding to a voltmeter. The P.T. steps down the voltage to the level of voltmeter.

 

Merits of Instrument Transformers:

1.       Instruments of moderate size are used for metering i.e. 5A for current and 100 to 120 volts for voltage measurements.

2.        Instrument and meters can be standardized so that there is saving in costs. Replacement of damaged instruments is easy.

3.       Single range instruments can be used to cover large current or voltage ranges, when used with suitable multi range instrument transformers.

4.       The metering circuit is isolated from the high voltage power circuits. Hence isolation is not a problem and the safety is assured for the operators

5.       There is low power consumption in metering circuit.

6.       Several instruments can be operated from a single instrument transformer.

Ratios of Instrument Transformer:

Some definitions are:

1.       Transformation ratio: It is the ratio of the magnitude if the primary phasor to secondary phasor.

 

Transformation ratio:

Current Transformer equivalent circuit:

X1 = Primary leakage reactance

R1 = Primary winding resistance

 X2 = Secondary leakage reactance

Z0 = Magnetizing impedance

R2 = Secondary winding resistance

Zb = Secondary load

Note: Normally the leakage fluxes X1 and X2 can be neglected

 

Current transformer, simplified equivalent circuit:

Current Transformer Basics:

Current Transformers (CT’s) can be used for monitoring current or for transforming primary current into reduced secondary current used for meters, relays, control equipment and other instruments. CT’s that transform current isolate the high voltage primary, permit grounding of the secondary, and step-down the magnitude of the measured current to a standard value that can be safely handled by the instrument.

Ratio :The CT ratio is the ratio of primary current input to secondary current output at full load. For example, a CT with a ratio of 300:5 is rated for 300 primary amps at full load and will produce 5 amps of secondary current when 300 amps flow through the primary. If the primary current changes the secondary current output will change accordingly. For example, if 150 amps flow through the 300 amp rated primary the secondary current output will be 2.5 amps (150:300 = 2.5:5).

 

Current Transformer: Cautions:

 

Inspect the physical and mechanical condition of the CT before installation.

Check the connection of the transformer requirements for the instrument or the system requirements before connecting the CT.

Inspect the space between the CT phases, ground and secondary conductor for adequate clearance between the primary and secondary circuitry wiring.

Verify that the shorting device on the CT is properly connected until the CT is ready to be installed. The secondary of the CT must always have a burden (load) connected when not in use. NOTE: A dangerously high secondary voltage can develop with an open-circuited secondary.

Construction of Current Transformer:

Current transformers are constructed in various ways. In one method there are two separate windings on a magnetic steel core. The primary winding consists of a few turns of heavy wire capable of carrying the full load current while the secondary winding consist of many turns of smaller wire with a current carrying capacity of between 5/20 amperes, dependent on the design. This is called the wound type due to its wound primary coil.

Another very common type of construction is the so-called “window,” “through” or donut type current transformer in which the core has an opening through which the conductor carrying the primary load current is passed. This primary conductor constitutes the primary winding of the CT (one pass through the “window” represents a one turn primary), and must be large enough in cross section to carry the maximum current of the load.

Construction of Current Transformer

Circuit connection for current and power measurement using C.T.

Equivalent Circuit of C.T.

Relationship in current transformer:

Fig 1 represents the equivalent circuit and Fig 2 the phasor diagram of a current transformer.

n= turns ratio = (No. of secondary winding turns)/(No. of primary winding turns)

rs = resistance of the secondary winding;

xs = reactance of the secondary winding;

re = resistance of external burden i.e. resistance of meters, current coils etc. including leads;

xe = reactance of external burden i.e. reactance of meters, current coils etc. including leads;

Ep = primary winding induced voltage Es

= secondary winding induced voltage Np

= No. of primary winding turns;

Ns = No. of secondary winding turns;

Vs = Voltage at the secondary winding

terminals; Is = secondary winding current;

Ip = primary winding current;

ϴ = phase angle of transformer;

secondary winding current = phase angle of total burden including impedance of secondarywinding

Now in a well-designed current transformer Io <<nIs. Usually Io is less than 1 percent of IpandIp is, therefore, very nearly equal to nIs.

Eqn. (2) can be written as

As

           I _m = I ₀cosα and I _e= I ₀sinα

 

Phase angle:

The angle by which the secondary current phasor, when reversed, differs in phase from primary current, is known as the “phase angle” of the transformer.

+ve if secondary reversed current leads the primary current

-ve if secondary reversed current lags behind the primary current. The angle between Is and Ip is θ. Therefore, the phase angle is θ.

From the phasor diagram,

Errors in current transformers:

• Turns ratio and transformation ratios are not equal.

• The value of transformation ratio is not constant.

It depends upon:

1.       Magnetizing and loss components of exciting current,

2.       The secondary winding load current and its power

This introduces considerable errors into current measurement

In power measurement it is necessary that the phase of secondary winding current shall be displaced by exactly 180° from that of the Primary current. Here, phase difference is different from 180° by an angle θ. Hence due to C.T. two types of errors are introduced in power measurements.

Due to actual transformation ratio being different from the turn’s ratio.

Due to secondary winding current not being 180° out of phase with the primary winding current.

Problem No.1

Two current transformers of the same nominal ratio 500/5 A, are tested by Silsbee’s method. With the current in the secondary of the transformer adjusted at its rated value, the content in the middle conductor I = 0.05e-j126.9° A expressed with respect to current in the secondary of standard transformer as the reference. It is known that standard transformer has a ratio correction factor (RCF) of 1.0015 and phase error +8’. Find RCF and phase angle error of transformer under test.

 

Problems on CT

1. A current transformer has a single turn primary and 200 turns secondary winding. The secondary winding supplies a current of 5 A to a non-inductive burden of 1 resistance. The requisite flux is set up in the core by an mmf of 80 A. The frequency is 50 Hz and the net cross section of the core is 1000 sq. mm. Calculate the ratio and phase angle of the transformer. Also find the flux density in the core. Neglect the effects of magnetic leakage, iron losses and copper losses.

A current transformer with a bar primary has 300 turns in its secondary winding. The resistance and reactance of secondary circuits are 1.5 and 1.0 respectively including the transformer winding. With 5 A flowing in the secondary winding, the magnetizing mmf is 100 A and the iron losses is 1.2W. Determine the ratio and phase angle error. Solution: Primary winding turns Np= 1;

Secondary winding turns Ns = 300;

Turns ratio = Ns/Np = 300/1 = 300.

In the absence of any information to the contrary we can take nominal ratio to be equal to the turns ratio, or

Kn = n = 300

Problem No. 3

A 100/5 A, 50 Hz CT has a bar primary and a rated secondary burden of 12.5 VA. The secondary winding has 196 turns and a leakage inductance of 0.96 mH. With a purely resistive burden at rated full load, the magnetization mmf is 16 A and the loss excitation requires 12 A. Find the ratio and phase angle errors.

Solution: Secondary burden = 12.5 VA.

Secondary winding current = 5 A

Secondary circuit impedance = 12.5/5² = 0.5 .

Secondary circuit reactance = 2π × 50 × 96 ×10^-3 = 0.3

Phase angle of secondary circuit δ = sin^-1 0.3/0.5 = sin^-1 0.6

Therefore, sin δ = 0.6 and cosδ = √(1)² + (0.6)² = 0.8.

Primary winding turns Np = 1. Secondary winding turns Ns = 196.

Turns ratio n = Ns/Np = 196.

Nominal ratio = Kn = 1000/5 = 200 Magnetizing current

Im = (magnetizing mmf)/(primary winding turns) = 16/1 =16 A

Loss component Ie = (excitation for loss/primary winding turns) . = 12/1 = 12 A.

Potential Transformer Basics

Potential transformers are normally connected across two lines of the circuit in which the voltage is to be measured. Normally they will be connected L-L (line-to-line) or L-G (line-to-ground). A typical connection is as follows:

Relationships in a Potential Transformer:

The theory of a potential transformer is the same as that of a power transformer. The main difference is that the power loading of a P.T. is very small and consequently the exciting current is of the same order as the secondary winding current while in a power transformer the exciting current is a very small fraction of secondary winding load current.

Fig 3.and Fig 4. shows the equivalent circuit and phasor diagram of a potential transformer respectively.

Secondary voltages when referred to primary side are to be multiplied by n. When secondary currents are referred to primary side, they must be divided by n

Actual Transformation ratio

An enlarged concise phasor diagram is shown in Fig. 5.

θ = phase angle of the transformer

= angle between VP and VS reversed

= phase angle of secondary load circuit

= phase angle between IP and VS reversed. Now

oa = VPcosθ

 

From Phasor diagram

The terms in the denominator involving IP and IS are small and, therefore, they can be neglected as compared with nV_S .

Errors in potential transformers

Ratio error (Voltage Error):

The actual ratio of transformation varies with operating condition and the error in secondary voltage may be defined as,

% Ratio Error = Kn – R × 100

                                 R

Phase angle error:

In an ideal voltage transformer, there should not be any phase difference between primary winding voltage and secondary winding voltage reversed. However, in an actual transformer there exists a phase difference between V_P and V_S reversed.

Problem No. 1

A potential transformer, ratio 1000/100 volt, has the following constants:

Primary resistance = 94.5 , secondary resistance=0.86

Primary reactance = 66.2 , total equivalent reactance = 110 No

load current = 0.02 A at 0’4 p.f.

Calculate: (i) Phase angle error at no load

ii) burden in VA at UPF at which phase angle will be Zero.

 

Problem No.2

A potential transformer rated 6900/115 Volts, has 22500 turns in the primary winding and 375 turns in the secondary winding. With 6900 volts applied to the primary and the secondary circuit open circuited, the primary winding current is 0.005A lagging the voltage by 73.7°. With a particular burden connected to the secondary, the primary winding current is 0.0125A lagging the voltage by 53.1°. Primary winding resistance = 1200 , Primary winding reactance = 2000 , secondary winding resistance = 0.4 , secondary winding reactance = 0.7 .

(i)                  Find the secondary current and terminal voltage using the applied primary voltage VP = 6900 + j0 as reference. Find the load burden also.

(ii)                Find the actual transformation ratio and also the phase angle.

If the actual ratio = the nominal ratio under above conditions, what change should be made in the primary turns?

 

Testing of Instrument Transformers

Methods for finding ratio and phase angle errors experimentally are broadly classified into two groups:

1. Absolute method: In these methods the transformer errors are determined interms of constants i.e., resistance, inductance and capacitance of the testing circuit.

2. Comparison method: In these methods, the errors of the transformer under testare compared with those of a standard current transformer whose errors are known.

Each of the two methods can be classified, according to measurement technique employed as

1.       Deflection Method: These methods use the deflections of suitable instruments for measuring quantities related to the phasor under consideration or to their deflection. The required ratio and phase angles are then found out from the magnitudes of deflection. These methods may be made direct reading in some cases.

2.       Null Methods: These methods make use of a network in which the appropriate phasor quantities are balanced against one another. The ratio and phase angle errors are then found out from the impedance elements of the network. The method may be made direct reading in terms of calibrated scales on the adjustable elements in the network.

 

Testing of Current Transformer

There are three methods:

1.       Mutual Inductance method: This is an absolute method using null deflection.

2.       Silsbee’s Method: This is a comparison method. There are two types; deflectionand null.

3.       Arnold’s Method: This is a comparison method involving null techniques.

Silsbee’s Method:

The arrangement for Silsbee’s deflection method is shown in Fig.1. Here the ratio and phase angle of the test transformer ‘X’ are determined in terms of that of a standard transformer ‘S’ having the same nominal ratio.

Procedure:

The two transformers are connected with their primaries in series. An adjustable burden is put in the secondary circuit of the transformer under test.

An ammeter is included in the secondary circuit of the standard transformer so that the current may be set to desired value. W₁ is a wattmeter whose current coil is connected to carry the secondary current of the standard transformer. The current coil of wattmeter W₂ carries a current I which is the difference between the secondary currents of the standard and test transformer. The voltage circuits of wattmeters are supplied in parallel from a phase shifting transformer at a constant voltage V.

The phasor diagram is shown in Fig. 2

1. The phase of the voltage is so adjusted that wattmeter W₁ reads zero. Under these conditions voltage V is in quadrature with current I_ss . The position of voltage phasor for this case is shown as V_q .

The phase of voltage V is shifted through 90° so that it occupies a position V_P and is in phase with I_ss .

Or phase angle of test transformer,

as W2p is very small. Hence if the ratio and phase angle errors of standard transformer are known, we can compute the errors of the test transformer. W2 must be a sensitive instrument. Its current coil may be designed for small values. It is normally designed to carry about 0.25A for testing CTs having a secondary current of 5A.

 

Problem No.1

Two current transformers of the same nominal ratio 500/5 A, are tested by Silsbee’s method. With the current in the secondary of the transformer adjusted at its rated value, the content in the middle conductor I = 0.05e^-j126.9° A expressed with respect to current in the secondary of standard transformer as the reference. It is known that standard transformer has a ratio correction factor (RCF) of 1.0015 and phase error +8’. Find RCF and phase angle error of transformer under test.