CURRENT MEASUREMENT SENSORS

Experiment 3

Time : 3 hoursEquipment3 x multimetersACS712 Hall current sensor AC1005 CT sensor Dual DC bench power supply 3A240VAC variac12V transformer3x 12v 21/5W bulbs

Preliminary•Read up on Hall effect so you understand what you are seeing in part 1 of the experiment.•Answer the first 3 questions.•Read the data sheet for the ACS712 current sensor.•Read up on Current transformers (CT) so you understand what you are seeing in part 2 of the experiment.•Read the data sheet for the AC1005 CT.

IntroductionIn this experiment a commercial Hall effect current sensing IC will be used to monitor both AC and DC currents. A current transformer (CT) will then be used to measure AC currents.

TheoryThe motion of a charged particle moving in an electric and magnetic field is governed by the Lorentz force which is given by

F = q E + v × B( ) (1)

The movement of a charged particle in a conductor or semiconductor, under the influence of an electric field, can be considered as a random motion with a net drift. The average drift velocity v in the direction of the current flow is proportional to the applied electric field such that

v = µE (2)

where µ is called the mobility of the charged particles, which for a semiconductor, can either be negatively charged (electrons) or positively charged (holes). If there are n charged particles per unit volume, then the current density, J is defined as

J = nqv (3)

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It follows from equation (2) that

J = nqµE ≡ σE (4)

where σ is the conductivity of the semiconductor.

If the semiconductor is placed in a transverse magnetic field , then from equation (1), the magnetic force component will act to shift the charged particles to the surface. This drift of the charged particles, transversely to their direction of motion, sets up an electric field, the Hall field, to counteract the deflection. At equilibrium, the force due to the Hall field cancels the magnetic force.

Q1 Show that the Hall field can be written as

EH = −R J × B( ) (5)

where R is the Hall coefficient.

R =1nq

(6)

The geometry of the hall sensor is shown in figure 1.

VH

J

B

t

w

y

x

z

Figure 1. The Hall effect.

J = Ji and B = Bk (7)

Q2 Show that

R =EH

JB (8)

where EH is the amplitude of the Hall field.

Q3 Hence show that

VH = −RW

⎛⎝⎜

⎞⎠⎟IB (9)

where VH is the Hall voltage across the semiconductor established by the Hall field, and I is the external current supplied to the semiconductor.

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Current Transformers

http://commons.wikimedia.org/wiki/File:Leistungsschalter-110KV.jpg

Figure 2. The current transformer.

Current transformers are a simple method of measuring high AC currents whilst being completely electrically isolated from the circuit being measured. This can be very important especially when measuring the current on high voltage circuits. In figure 4 above we see the current to be measured (I) is flowing in the large conductor. It is passed through a circular magnetic core which concentrates the magnetic flux produced around the current carrying conductor. The conductor essentially forms the primary winding of a transformer, which in this case has only 1 turn. A secondary winding is placed around the magnetic core. This winding normally has many turns (maybe 1000). The voltage produced on the secondary of a voltage transformer can be calculated by

Vsec = VprimNsec

Nprim

In the case of a CT we do not know, nor are we interested in the primary voltage. Remember the primary voltage would be the voltage dropped across the part of the conductor going through the CT. This would normally be really small. In the case of the CT we are interested in knowing the current flowing through the primary conductor. The secondary current is just proportional to the inverse of the ratio we used to calculate the secondary voltage. That is

Isec = I primNprim

Nsec

=1Nsec

Remember the primary has only 1 turn (Nprim=1). If we now measure the secondary current then we can of course calculate what the primary current is by multiplying it by Nsec. We could use a ammeter to measure the secondary current but mostly we would prefer to have a voltage that represents the value of primary current. This voltage could be used as an input to a computer or another circuit etc. To turn the secondary current into a voltage we simply pass it through a resistor and measure the voltage drop across it. This resistor is called the burden of the CT.

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Burden resistor (R Ohms)

V = I/Nsec x R

I/Nsec

Figure 3. The burden resistor.

A small value of burden resistor will not impede the secondary current very much, but it will only develop a small voltage across it. A small burden will give the most linear output with respect to the primary current. A large burden will give a much larger output voltage but due to the greater opposition to secondary current it will give a less linear output. Look at the data sheet for the AC1005 CT at the back of this document to see the linearity versus burden.

WARNING: Care must be taken when using CTs to ensure the secondary is not open circuited. Remember the secondary is trying to produce a current that is controlled by the primary current. If the secondary has a high resistance burden then the secondary voltage will rise higher to keep the same secondary current flowing. If the burden is an infinitely high resistance then the secondary voltage will theoretically become infinitely high. Of course in reality it will just become very high. This could potentially be dangerous for the transformer or surrounding equipment and personnel. The transformer you are using today will not produce dangerous voltages.

The other problem that can occur with a CT is when the primary current reaches a higher value for which the CT is designed. Normally the primary conductor is a heavy cable so it may not be affected. However higher than design currents may produce magnetic fields in the CT that saturate the core material. In this case the secondary current will be less than expected and the output voltage will also less than expected. We now have the situation where a dangerously high current may be flowing in the primary circuit, but the equipment monitoring it may see that it is less than the real current. Always ensure the CT can handle (without saturation) more than the highest possible current.

EXPERIMENT

Hall Effect SensorsIn this part of the experiment you will be using a semiconductor hall sensor, the ACS712.

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Figure 4. The ACS712 Hall effect current sensor and test board.

Current in

Current out Vcc (8-12v)

GND

Output

PSU (8-12v)

PSU

(12v

max

)

.

Figure 5. DC Current sensor test circuit.

(a) DC current and power measurementConnect 2 x 21W bulbs in parallel to form a dummy load which should be placed in series with the current sensor and an ammeter as shown in figure 5 (If the power supply has a digital ammeter, this may be used). Start with the DC bench power supply set initially to 0V. A voltmeter should be placed across the bulb load. Another voltmeter should monitor the output voltage from the current sensor. A power supply should be used to supply approx 8V to the current sensor PCB. Record (i)voltage across the load, (ii)current measured by the ammeter and (iii)the voltage output of the current sensor. Increase the voltage of the power supply connected to the dummy load until a current of 0.2A flows in the load. Repeat the measurements and continue to increase the current in steps of 0.2A until 3.0A is reached.

Tabulate your results. Plot the sensor voltage versus load current and comment on the relationship. Refer to the data sheet to back your findings. Calculate and plot the power in the load versus supply voltage.

Q5 Plot and comment on the relationship between the power in the load versus the supply voltage.

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(b) AC current and power measurement

Current in

Current out Vcc (8-12v)

GND

Output

PSU (8-12v)

240vVariac

240 - 12vTransformer

Figure 6. AC Current sensor test circuit.

Repeat part (a) using the 12VAC transformer connected to the mains via a variac. The variac can then be used to allow voltages of approx 0 to 12V to be delivered to the dummy load. Examine the current sensor output on a CRO and AC volt meter..

A similar analysis to part (a) should be undertaken except the sensor output should be measured on the CRO. Record both maximum and minimum values from the screen. Calculate the RMS value for each measurement (Your CRO may be able to do this).

Q6 If the load were inductive explain what you would observe when measuring this circuit with an AC voltage applied.

Current TransformersIn this part of the experiment you will be using a small current transformer, the AC1005. Figure 6 shows the test box. The current to be measured is connected to the primary terminals and will flow solely through the heavy wire passing through the centre of the CT. The secondary of the CT is available at the secondary terminals. There is no burden connected across the secondary. You will need to provide this.

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Figure 7. Current transformer test box.

(c) AC current measurementSet up the circuit shown in figure 6. Insert a burden resistance of 100Ω. Increase the variac in stages such that the current through the load increases in steps of 0.2A up to a maximum of 3A. Record and plot the voltage at the secondary of the CT versus primary current. Change the burden resistance to 10KΩ and repeat the measurements. Comment on your results.

Primary

240 - 12vTransformer

Secondary

Rburden

240vVariac

Figure 8. Test circuit for CT measurements.

Disconnect the primary wire through the CT and pass it through the centre of the CT a second time and reconnect it. You now have a 2 turn primary. Repeat the measurement from above, using the 10KΩ burden and comment on the results.

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IP+IP+

IP–IP–

IP

5GND

2

4

1

3ACS712

7

8+5 V

VIOUTVOUT

6FILTER

VCC

CBYP0.1 µF

CF1 nF

Application 1. The ACS712 outputs an analog signal, VOUT . that varies linearly with the uni- or bi-directional AC or DC primary sampled current, IP , within the range specified. CF is recommended for noise management, with values that depend on the application.

ACS712

DescriptionThe Allegro® ACS712 provides economical and precise solutions for AC or DC current sensing in industrial, commercial, and communications systems. The device package allows for easy implementation by the customer. Typical applications include motor control, load detection and management, switch-mode power supplies, and overcurrent fault protection. The device is not intended for automotive applications.

The device consists of a precise, low-offset, linear Hall circuit with a copper conduction path located near the surface of the die. Applied current flowing through this copper conduction path generates a magnetic field which the Hall IC converts into a proportional voltage. Device accuracy is optimized through the close proximity of the magnetic signal to the Hall transducer. A precise, proportional voltage is provided by the low-offset, chopper-stabilized BiCMOS Hall IC, which is programmed for accuracy after packaging.

The output of the device has a positive slope (>VIOUT(Q)) when an increasing current flows through the primary copper conduction path (from pins 1 and 2, to pins 3 and 4), which is the path used for current sampling. The internal resistance of this conductive path is 1.2 m� typical, providing low power loss. The thickness of the copper conductor allows survival of

ACS712-DS, Rev. 14

Features and Benefitsƒ Low-noise analog signal pathƒ Device bandwidth is set via the new FILTER pinƒ 5 �s output rise time in response to step input currentƒ 80 kHz bandwidthƒ Total output error 1.5% at TA = 25°Cƒ Small footprint, low-profile SOIC8 packageƒ 1.2 m� internal conductor resistanceƒ 2.1 kVRMS minimum isolation voltage from pins 1-4 to pins 5-8ƒ 5.0 V, single supply operationƒ 66 to 185 mV/A output sensitivityƒ Output voltage proportional to AC or DC currentsƒ Factory-trimmed for accuracyƒ Extremely stable output offset voltageƒ Nearly zero magnetic hysteresisƒ Ratiometric output from supply voltage

Fully Integrated, Hall Effect-Based Linear Current Sensor IC with 2.1 kVRMS Isolation and a Low-Resistance Current Conductor

Continued on the next page…

Approximate Scale 1:1

Package: 8 Lead SOIC (suffix LC)

Typical Application

TÜV AmericaCertificate Number:U8V 06 05 54214 010

Go to the course website to see the full data sheethttp://maxwell.me.gu.edu.au/sok/ees/resources/ACS712.pdf

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AC1005 • 5 Amp Current Transformer

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Notes:1) Unless requested, the terminating resistor and the one-turn

primary are not supplied2) Pin 3: Normally for mechanical support only but will be

used on center tapped designs

Low Cost 50/60Hz Current TransformersApplications• Sensing Overload Current• Ground fault detection• Metering• Analog to Digital Circuits

Electrical Specifications @ 20°C ambient

0.01

0.1

1

10

100

Out

put V

olta

ge in

Vol

ts

1 10 100 Input Current in Amps

Output Volts vs Input CurrentFor various ohmic loads

50100200

500

1K

2K5K10K

Infinity 10

100

1000

10000

Load

Res

ista

nce

RL

in o

hms

0 100 200 300 400 Percent ratio error in %

%RE vs RL at Rated primary current(AC1005)

0.1

1

10

100

Sec

RM

S Ex

citin

g Vo

ltage

in V

olts

0.01 0.1 1 10 100 1000 Sec RMS Exciting current in mA

Typical Excitation Curve(AC1005~AC1020)

I - TurnPrimary 1/8W

100 ohms

2

1

0.5V5A 5mA RL

Dimensions

7.62

15.24

11.12

5±1mm

TALEMA INDIAW/Y

AC1005

Ø9.502 3 1

23.8

0

23.80

1.75

Ø 0.813

Germany: Int.+49 89 - 841 00-0 • Ireland: Int.+35 374 - 954 8666 • Czech Rep: Int.+420 37 - 744 9303 • India: Int.+91 427 - 244 1325 http://www.talema-nuvotem.com

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