Chapter 1 EMI

MeasurementsMeasurements

REVIEW OF UNIT I and UNIT II REVIEW OF UNIT I and UNIT II

Learning Objectives

• To introduce basic components of measurement system

• To understand the basic properties of measurement systems

Basic components in a measurement system

Amplification and Conditioning

INSTRUMENTATION CHARACTERISTICS

• Shows the performance of instruments to be used.

• Divided into two categories: static and dynamic characteristics.

• Static characteristics refer to the comparison between steady output and ideal output when the input is constant.

• Dynamic characteristics refer to the comparison between instrument output and ideal output when the input changes.

STATIC CHARACTERISTICS

1. ACCURACY

– Accuracy is the ability of an instrument to show the exact reading.

– Always related to the extent of the wrong reading/non accuracy.

– Normally shown in percentage of error which of the full scale reading percentage.

2. PRECISION

• An equipment which is precise is not necessarily accurate.

• Defined as the capability of an instrument to show the same reading when used each time (reproducibility of the instrument).



Example :X : resultCentre circle : true value

XXXXXXXXXX

XXXXXX

X X

X

x xHigh accuracy, high precision

Low accuracy, high precision

Low accuracy, low precision

Accuracy vs Precision

High Precision, but low accuracy.

There is a systematic error.

Accuracy vs Precision (Cont)

High accuracy means that the mean is close to the true value, while high precision means that the standard deviation σ is small.

3. TOLERANCE

• Closely related to accuracy of an equipment where the accuracy of an equipment is sometimes referred to in the form of tolerance limit.

• Defined as the maximum error expected in an instrument.

• Explains the maximum deviation of an output component at a certain value.


4. RANGE OF SPAN

• Defined as the range of reading between minimum value and maximum value for the measurement of an instrument.

• Has a positive value e.g..:The range of span of an instrument which has a reading range of –100°C to 100 °C is 200 °C.


5. BIAS• Constant error which occurs during the

measurement of an instrument.• This error is usually rectified through

calibration.Example :A weighing scale always gives a bias reading. This equipment always gives a reading of 1 kg even without any load applied. Therefore, if A with a weight of 70 kg weighs himself, the given reading would be 71 kg. This would indicate that there is a constant bias of 1 kg to be corrected.


6. LINEARITY• Maximum deviation from linear relation

between input and output.• The output of an instrument has to be linearly

proportionate to the measured quantity.• Normally shown in the form of full scale

percentage (% fs). • The graph shows the output reading of an

instrument when a few input readings are entered.

• Linearity = maximum deviation from the reading of x and the straight line.


Linearity

OutputReadings

Measured Quantity

7. SENSIVITY• Defined as the ratio of change in output towards

the change in input at a steady state condition.• Sensitivity (K) = Δθο Δθi

Δθο : change in output; Δθi : change in input

Example 1: The resistance value of a Platinum Resistance Thermometer changes when the temperature increases. Therefore, the unit of sensitivity for this equipment is Ohm/°C.


Sensitivity

Variation of the physical variables

Most sensitive


Example 2:Pressure sensor A with a value of 2 bar caused a deviation of 10 degrees. Therefore, the sensitivity of the equipment is 5 degrees/bar.

• Sensitivity of the whole system is (k) = k1 x k2 x k3 x .. x kn

k1 k2 k3θiθo

STATIC CHARACTERISTICSExample:Consider a measuring system consisting of a transducer, amplifier and a recorder, with sensitivity for each equipment given below:

Transducer sensitivity 0.2 mV/°CAmplifier gain 2.0 V/mVRecorder sensitivity 5.0 mV/V

Therefore,Sensitivity of the whole system:

(k) = k1 x k2 x k3 k = 0.2 mV x 2.0 V x 5.0 mV °C mV Vk = 2.0 mV/°C

Example :The output of a platinum resistance thermometer (RTD) is as follows:

Calculate the sensitivity of the equipment.Answer :

Draw an input versus output graph. From that graph, the sensitivity is the slope of the graph.K = Δθο graph = (400-200) ohm = 2 ohm/°C Δθi slope (200-100) °C

Input(°C) Output(Ohm)0 0

100 200200 400300 600400 800

8. DEAD SPACE / DEAD BAND

• Defined as the range of input reading when there is no change in output (unresponsive system).

Dead Space

OutputReading

MeasuredVariables

- +


9. RESOLUTION

• The smallest change in input reading that can be traced accurately.

• Given in the form ‘% of full scale (% fs)’.

• Available in digital instrumentation.


10. THRESHOLD

• When the reading of an input is increased from zero, the input reading will reach a certain value before change occurs in the output.

• The minimum limit of the input reading is ‘threshold’.


DYNAMIC CHARACTERISTICS

• Explains the behaviour system of instruments system when the input signal is changed.

• Depends on a few standard input signals such as ‘step input’, ‘ramp input’ dan ‘sine-wave input’.

Response time

One would like to have a measurement system with fast response.

In other words, the effect of the measurement system on the measurement should be as small as possible.

EXAMPLE OF DYNAMIC CHARACTERISTICS

Response from a 2nd order instrument:Output

100%

90%

10%

trTime

EXAMPLE OF DYNAMIC CHARACTERISTICS

Response from a 2nd order instrument:1. Rise Time ( tr )• Time taken for the output to rise from

10% to 90 % of the steady state value.2. Settling time (ts)• Time taken for output to reach a

steady state value.

Measuring Instruments: ammeter, voltmeter, ohmmeter.You must be able to calculate currents and voltages in circuits that contain “real” measuring instruments.

You know how to calculate the current in this circuit:

Measuring Instruments: Ammeter

V

R

.VI = RIf you don’t know V or R, you can measure I with an ammeter.

.VI = R+rTo minimize error the ammeter resistance r should very small.

Any ammeter has a resistance r.Any ammeter has a resistance r. The current you measure is

r

Example: an ammeter of resistance 10 m is used to measure the current through a 10 resistor in series with a 3 V battery that has an internal resistance of 0.5 . What is the percent error caused by the nonzero resistance of the ammeter?

V=3 V

R=10

r=0.5

Actual current: VI = R+r

3I = 10+0.5

I = 0.286 A = 286 mAYou might see the symbol

used instead of V.

V=3 V

R=10

r=0.5

Current with ammeter:

A

VI = R+r+R

3I = 10+0.5+0.01

I = 0.285 A = 285 mA RA

0.286-0.285% Error = 1000.286

% Error = 0.3 %

A Galvanometer

When a current is passed through a coil connected to a needle, the coil experiences a torque and deflects.

An ammeter is based on a galvanometer. OK, everything is electronic these days, but the principles here still apply.

For now, all you need to know is that the deflection of the galvanometer needle is proportional to the current in the coil (red).

A typical galvanometer has a resistance of a few tens of ohms.

Hold it right there. Didn’t you say an ammeter must have a very small resistance. Is there a physics mistake in there somewhere?

A galvanometer-based ammeter uses a galvanometer and a shunt, connected in parallel:

AI

GRG

RSHUNTIG

ISHUNT

I

Everything inside the blue box is the ammeter.

The resistance of the ammeter is

A G SHUNT

1 1 1R R R

G SHUNTA

G SHUNT

R RR R R

Homework hint: “the galvanometer reads 1A full scale” means a current of IG=1A produces a full-scale deflection of the galvanometer needle. The needle deflection is proportional to the current IG.

If you want the ammeter shown to read 5A full scale, then the selected RSHUNT must result in IG=1A when I=5A. In that case, what are ISHUNT and VAB (=VSHUNT)?

GRG

RSHUNTIG

ISHUNT

I

Homework hint: “the galvanometer reads 1A full scale” means a current of IG=1A produces a full-scale deflection of the galvanometer needle. The needle deflection is proportional to the current IG.


A B

“the galvanometer reads 1A full scale” means a current of IG=1A produces a full-scale deflection of the galvanometer needle. The needle deflection is proportional to the current IG.


A galvanometer-based ammeter uses a galvanometer and a shunt, connected in parallel:

GRG

RSIG

IS

I

A G S

1 1 1R R R

Example: what shunt resistance is required for an ammeter to have a resistance of 10 m, if the galvanometer resistance is 60 ?

S A G

1 1 1R R R

G A

SG A

60 .01R RR 0.010 R -R 60-.01

The shunt resistance is chosen so that IG does not exceed the maximum current for the galvanometer and so that the effective resistance of the ammeter is very small.

(actually 0.010002 )

To achieve such a small resistance, the shunt is probably a large-diameter wire or solid piece of metal.

G A

SG A

60 .01R RR 0.010 R -R 60-.01

Measuring Instruments: Voltmeter

You can measure a voltage by placing a galvanometer in parallel with the circuit component across which you wish to measure the potential difference.

V=3 V

R=10

r=0.5

GRG

a bVab=?

Example: an galvanometer of resistance 60 is used to measure the voltage drop across a 10 k resistor in series with a 6 V battery and a 5 k resistor (neglect the internal resistance of the battery). What is the percent error caused by the nonzero resistance of the galvanometer?First calculate the actual voltage drop.

V=6 V

R1=10 k

R2=5 k

a b 3

eq 1 2R R +R =15 10

-33

eq

V 6 VI 0.4 10 AR 15 10

-3 3abV =IR 0.4 10 10 10 4 V

The measurement is made with the galvanometer.

V=6 V

R1=10 k

R2=5 k

GRG=60

a b

60 and 10 k resistors in parallel are equivalent to an 59.6 resistor. The total equivalent resistance is 5059.6 , so 1.19x10-3 A of current flows from the battery.

I=1.19 mA

The voltage drop from a to b is then measured to be6-(1.19x10-3)(5000)=0.07 V.

The percent error is.

4-.07% Error = 100=98%4

To reduce the percent error, the device being used as a voltmeter must have a very large resistance, so a voltmeter can be made from galvanometer in series with a large resistance.

V GRG

RSer

Everything inside the blue box is the voltmeter.

a b

Vab

a b

Va

b

Homework hints: “the galvanometer reads 1A full scale” would mean a current of IG=1A would produce a full-scale deflection of the galvanometer needle.

If you want the voltmeter shown to read 10V full scale, then the selected RSer must result in IG=1A when Vab=10V.

“the galvanometer reads 1A full scale” would mean a current of IG=1A would produce a full-scale deflection of the galvanometer needle. If you want the voltmeter shown to read 10V full scale, then the selected RSer must result in IG=1A when Vab=10V.

Example: a voltmeter of resistance 100 k is used to measure the voltage drop across a 10 k resistor in series with a 6 V battery and a 5 k resistor (neglect the internal resistance of the battery). What is the percent error caused by the nonzero resistance of the voltmeter?We already calculated the actual voltage drop (3 slides back).

V=6 V

R1=10 k

R2=5 k

a b

-3 3abV =IR 0.4 10 10 10 4 V

The measurement is now made with the voltmeter.

V=6 V

R1=10 k

R2=5 k

VRV=100 k

a b

100 k and 10 k resistors in parallel are equivalent to an 9090 resistor. The total equivalent resistance is 14090 , so 4.26x10-4 A of current flows from the battery.

I=.426 mA

The voltage drop from a to b is then measured to be6-(4.26x10-4)(5000)=3.9 V.

The percent error is.

4-3.9% Error = 100=2.5%4Not great, but much better. Larger Rser is needed for high accuracy.

An ohmmeter measures resistance. An ohmmeter is made from a galvanometer, a series resistance, and a battery.

GRG

RSer

R=?

The ohmmeter is connected in parallel with the unknown resistance with external power off. The ohmmeter battery causes current to flow, and Ohm’s law is used to determine the unknown resistance.

Measuring Instruments: Ohmmeter

Everything inside the blue box is the ohmmeter.

To measure a really small resistance, an ohmmeter won’t work.

Solution: four-point probe.

V

A

Measure current and voltage separately, apply Ohm’s law.

Chapter 1 EMI

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