Analog Electrical Devices and Measurements

2141-375 Measurement and Instrumentation

Analog Devices: Current Measurements

θsin ILBFBLIF

=×=rrr

Force on a conductor

F B

I A conductor is placed in a uniform magnetic field B T, at an angle of θ. The current flow in the conductor is IA. Force exerted on the conductor can be calculated from

θ

Analog Devices: Current Measurements

ILBkx =

Force on a conductor

B

I

With no current flowing through the conductor, the spring will be at its outstretched length. As current flows through the conductor, the spring will stretch and developed force required to balance the electromagnetic force.

IFs = kx

F = IBL

k = spring constant, x = the total distance moved by the spring and θ is 90o

xBLkI =

D’Arsonval or PMMC Instrument

Important parts of PMMC Instrument

•Permanent magnet with two soft-iron poles• Moving coil • Controlling or restoring spring

Torque Equation and Scale

When a current I flows through a one-turn coil in a magnetic field , a force exerted on each side of the coil

L = the length of coil perpendicular to the paper

F = IBLF = IBLF = IBL

Since the force acts on each side of the coil, the total force for a coil of N turns is

D

F = NIBL

The force on each side acts at a coil diameter D, producing a deflecting torque

TD = NIBLD

Torque Equation and Scale

The controlling torque exerted by the spiral springs is proportional to the angle of deflection of the pointer:

Since all quantities except θ and I are constant for any given instrument, the deflection angle is

θ = CI

TC = KθWhere K = the spring constant. For a given deflection, the controlling and deflecting torques are equal

BLIND = Kθ

Therefore the pointer deflection is always proportional to the coil current. Consequently, the scale of the instrument is linear.

Galvanometer

• Galvanometer is essentially a PMMC instrument designed to be sensitive to extremely current levels.

• The simplest galvanometer is a very sensitive instrument with the type of center-zero scale, therefore the pointer can be deflected to either right or left of the zero position.

• The current sensitivity is stated in µA/mm

DC Ammeter

PMMC instrument

Ammetershunt

Rs

Coil resistance

Rm

Im VmI = Is + Im Is

Rs

I

Shunt resistance

s

mms I

RIR =m

mms II

RIR−

=

ssmm RIRI =sm VV =

• An ammeter is always connected in series with a circuit.

• The internal resistance should be very low

• The pointer can be deflected by a very small current

• Extension of ranges of ammeter can be achieved by connecting a very low shunt resistor

Example: An ammeter has a PMMC instrument with a coil resistance of Rm = 99 Ω and FSD current of 0.1 mA. Determine the total current passing through the ammeter at (a) FSD, (b) 0.5 FSD, and (c) 0.25 FSD.

Known: FSD of Im Rm and Rs

Solution:

DC Ammeter

2.5510I = Im + Is (mA)

2.4754.959.9Is (mA)

0.0250.050.1Im (mA)

DC Ammeter: Multirange

B

AE

C

D

Rs1

Rs2

Rs3

Rs4

Rm

• A make-before-break must be used so that instrument is not left without a shunt in parallel to prevent a large current flow through ammeter.

Multirange ammeter using switch shunts

Make-before break switch

B

AE

C

D

DC Ammeter: Ayrton shunt Rm

R1 R2 R3

VSIm

I Is Is

Im

B

C

D

AI

+

- An Ayrton shunt used with an ammeter consists of several series-connected resistors all connected in parallel with the PMMC instrument. Range change is effected by switch between resistor junctions

Rm

R1 R2 R3

VSIm

I Is Is

Im

I

B

C

DA

+

-

R1 + R2 + R3 in parallel with Rm

R1 + R2 in parallel with Rm + R3

Example: A PMMC instrument has FSD of 100 µA and a coil resistance of 1kΩ. Calculate the required shunt resistance value to convert the instrument into an ammeter with (a) FSD = 100 mAand (b) FSD = 1 A.

Known: FSD of Im Rm

Solution:

DC Ammeter

Example: A PMMC instrument has a three-resistor Ayrton shunt connected across it to make an ammeter. The resistance values are R1 = 0.05 Ω, R2 = 0.45 Ω, and R3 = 4.5 Ω . The meter has Rm = 1 kΩ and FSD = 50 µA. Calculate the three ranges of the ammeter.

Known: FSD of Im Rm R1 R2 and R3

Solution:

DC Ammeter

DC Voltmeter

PMMC instrument

Series resistance or “multiplier”

mm

s RIVR −=mmsm RIRIV +=

Given V = Range

V

RmRs

Im

V

Coil resistance

Multiplierresistance

mm

s RIRangeR −=

The reciprocal of full scale current is the voltmeter sensitivity (kΩ/V)

The total voltmeter resistance = Sensitivity X Range

• An ammeter is always connected across or parallel with the points in a circuit at which the voltage is to be measured.• The internal resistance should be very high

DC Voltmeter: Multirange

( )RRIV mm +=

• Multirange voltmeter using switched multiplier resistors

Meter resistance

Rm

R1

R2

R3

V

Multiplierresistors

R1 R2 R3Rm

V

• Multirange voltmeter using series-connected multiplier resistor

Where R can be R1, R2, or R3

( )RRIV mm +=

Where R can be R1, R1 + R2, or R1 + R2 + R3

Example: A PMMC instrument with FSD of 50 µA and a coil resistance of 1700 Ω is to be used as a voltmeter with ranges of 10 V, 50 V, and 100 V. Calculate the required values of multiplier resistor for the circuit (a) and (b)

Known: FSD of Im Rm

Solution:

DC Ammeter

Meter resistance

Rm

R1

R2

R3

V

Multiplierresistors

R1 R2 R3Rm

V

(a) (b)

Ohmmeter: Voltmeter-ammeter method

V

A

VS

+

-

Rx

I

IV Ix

V

+

-

VVS

+

-

Rx

I

A+ -VA

+

-

VxV

-

Pro and con:•Simple and theoretical oriented•Requires two meter and calculations•Subject to error: Voltage drop in ammeter (Fig. (a))

Current in voltmeter (Fig. (b))

Fig. (a) Fig. (b)

Measured Rx:

meas xR R≈

measx A A

xV VV VR R

I I I+

= = = +

if Vx>>VA

Therefore this circuit is suitable for measure large resistance

Measured Rx: meas 1 /x

x V V x

RV VRI I I I I

= = =+ +

meas xR R≈if Ix>>IV

Therefore this circuit is suitable for measure small resistance

Ohmmeter: Series Connection

•Voltmeter-ammeter method is rarely used in practical applications (mostly used in Laboratory)•Ohmmeter uses only one meter by keeping one parameter constant

Example: series ohmmeter

15k

0∞

2550

75

100 µA

45k 5k

0

Infinity

Rx

Resistance tobe measured

Standardresistance

RmMeter

resistance

Meter

Battery

VS

R1

Basic series ohmmeter consisting of a PMMC and a series-connected standard resistor (R1). When the ohmmeter terminals are shorted (Rx = 0) meter full scale defection occurs. At half scale defection Rx = R1 + Rm, and at zero defection the terminals are open-circuited.

Basic series ohmmeter Ohmmeter scale

Nonlinear scale

1s

x mVR R RI

= − −

Loading Effect: Voltage Measurement

LinearCircuit V

a

b

Vab Rm VVab Rm

Rth

Vth

a

bUndisturbed condition: Rm = ∞

ab u thV V V= =

Measured condition: Rm ≠ ∞m

ab m thm th

RV V VR R

= =+

Measurement error: 100m u

u

V VerrorV−

= ×

11 /m u

th m

V VR R

=+General equation:

%100/1

1%100 ×+

−=×+

−=thmthm

th

RRRRR

Therefore, in practice, to get the acceptable results, we must have Rm ≥ 10 Rth (error ~ 9%)

Loading Effect

Circuit before measurement

10 V

R1100kΩ

R2100kΩ

5 V

5 V V

V

10 V

100kΩ

100kΩ

6.7 V

3.3 V 100kΩ

10 V

100kΩ

100kΩ

6 V

4 V 200kΩ

Circuit under measurement

V 3.3V 10100//100100

100//100=

+=measV

V 0.4V 10100//200100

100//200=

+=measV

V

10 V

100kΩ

100kΩ

5.2 V

4.8 V 1000kΩ

V 8.4V 10100//1000100

100//1000=

+=measV

Example Find the voltage reading and % error of each reading obtained with a voltmeter on (i) 5 V range, (ii) 10 V range and (iii) 30 V range, if the instrument has a 20 kΩ/V sensitivity, an accuracy 1% of full scale deflection and the meter is connected across Rb

Loading Effect

SOLUTION The voltage drop across Rb with out the voltmeter connection

On the 5 V range

V 550k 5k 45

k 5=×

+=

+= V

RRRV

ba

bb

kΩ 100 V 5 kΩ 20range =×=×= SRm

kΩ 4.76 k 5 k 100k 5 k 100=

+×

=+

=bm

bmeq RR

RRR

The voltmeter reading is

V 4.782 50k 76.4 k 54

k .764=

+=

+= V

RRR

Veqa

eqb

50 V

Ra45kΩ

Rb5kΩ

45 V

5 V

Loading Effect

Error of the measurement is the combination of the loading effect and the meter error

The loading error = 4.782 - 5 = -0. 218 V

The meter error = ± 5 x = ± 0.05 V1100

∴% of error on the 5 V range:%36.5100

V 5V 05.0V 218.0

±=×±−

=

± 6.10± 0.35± 0.3-0.054.9530

± 4.40± 0.22± 0.1-0.124.8810± 5.36± 0.27± 0.05-0.224.785

% errorTotal error (V)

Meter error (V)

Loading error (V)

Vb .

(V)Range

(V)

Loading Effect: Current Measurement

LinearCircuit A

a

b

Vab Rm

I

AVab Rm

Rth

Vth

a

b

I

Undisturbed condition: Rm = 0

Measured condition: Rm ≠ 0

Measurement error:

General equation:

Therefore, in practice, to get the acceptable results, we must have Rm ≤ Rth /10 (error ~ 9%)

ththu RVII /==

( )mththm RRVII +== /

( )thmum RRII /1/ +=

100×−

=u

um

IIIerror

%100/1

1%100 ×+

−=×+

−=mththm

m

RRRRR

AC Voltmeter: PMMC Based

D W

Waveform Amplitude Average RMS

A

A

A

A

A

A

0

πA

πA2

AWDD+

0

0

AWDD+

A

2A

2A2A

3A

AC Voltmeter: PMMC Based• Basic PMMC instrument is polarized, therefore its terminals must be identified as + and -.• PMMC instrument can not response quite well with the frequency 50 Hz or higher, So the pointer will settle at the average value of the current flowing through the moving coil: average-responding meter.

Full-wave Rectifier Voltmeter

RmD1

D2

D3

D4

Rs

Multiplierresistors

VpVrms

Vav

•Using 4 diodes

•On positive cycle, D1 and D4 are forward-biased, while D2 and D3 are reverse-biased

•On negative cycle, D2 and D3 are forward-biased, while D1 and D4 are reverse-biased

•The scale is calibrated for pure sine with the scale factor of 1.11 (A/√2 / 2A/π)

Example: A PMMC instrument has FSD of 100 µA and a coil resistance of 1kΩ is to be employed as an ac voltmeter with FSD = 100 V (rms). Silicon diodes are used in the full-bridge rectifier circuit (a) calculate the multiplier resistance value required, (b) the position of the pointer when the rmsinput is 75 V and (c) the sensitivity fo the voltmeter

Known: FSD of Im Rm

Solution:

AC Voltmeter: PMMC Based

AC Voltmeter: PMMC BasedHalf-wave Rectifier Voltmeter

Rs

RSH

D1

D2

RmVp

Vrms

Vav

•On positive cycle, D1 is forward-biased, while D2 is reverse-biased

•On negative cycle, D2 is forward-biased, while D1 is reverse-biased

•The shunt resistor RSH is connected to be able to measure the relative large current.

•The scale is calibrated for pure sine with the scale factor of 2.22 (A/√2 / A/π)