2102311 Electrical Measurement and
Instruments (Part II)
¾ Bridge Circuits (DC and AC)
¾ Electronic Instruments (Analog & Digital)
¾ Signal Generators
¾ Frequency and Time Interval Measurements
¾ Introduction to Transducers
อาภรณ ธีรมงคลรัศมี
ตึกไฟฟา 6 ชัน้ หอง 306
Textbook:
-A.D. Helfrick, and W.D. Cooper, “Modern Electronic Instrumentation and Measurement
Techniques” Prentice Hall, 1994.
- D.A. Bell, “Electronic Instrumentation and Measurements”, 2nd ed., Prentice Hell, 1994.
Resistor
Resistor Types
Types
Importance parameters
™Value
™Tolerance
™Power rating
™Temperature coefficient
Type
Values (Ω)
Power rating
(W)
Tolerance (%)
Temperature
coefficient
(ppm/°C)
Wire wound
(power)
10m~3k
3~1k
±1~±10
±30~±300
Wire wound
(precision)
10m~1M
0.1~1
±0.005~±1
±3~±30
Carbon film
1~1M
0.1~3
±2~±10
±100~±200
Metal film
100m~1M
0.1~3
±0.5~±5
±10~±200
Metal film
(precision)
10m~100k
0.1~1
±0.05~±5
±0.4~±10
Metal oxide film
100m~100k
1~10
±2~±10
±200~±500
Data: Transistor technology (10/2000)
picture
™Color codes
™Alphanumeric
Resistor
Resistor Values
Values
Color
Silver
Gold
Digit
-
Multiplier
Tolerance
(%)
Temperature
coefficient
(ppm/°C)
±10
K
10-1
±5
J
-
-
±250
K
10-2
-
-
4 band color codes
Most sig. fig.
of value
Tolerance
Least sig. fig. Multiplier
of value
Ex.
Black
0
100
Brown
1
101
±1
F
±100
H
Red
2
102
±2
G
±50
G
Orange
3
103
-
±15
D
Yellow
4
104
-
±25
F
R = 560 Ω ± 2%
Green
5
105
±0.5
D
±20
E
Blue
6
106
±0.25
C
±10
C
Alphanumeric
Violet
7
107
±0.1
B
±5
B
Gray
8
108
-
±1
A
White
9
109
-
-
Data: Transistor technology (10/2000)
-
±20
M
Green
Blue
Brown
Red
R, K, M, G, and T =
x100, x103 , x106 , x109 , and x1012
Ex. 6M8 = 6.8 x 106 Ω
59Ρ04 = 59.04 Ω
Resistor Values
Commonly available resistance for a fixed resistor
R = x ± %∆x
Tolerance
Nominal value
Ex. 1 kΩ ± 10%
≡
900-1100 Ω
For 10% resistor
10, 12, 15, 18, …
10
12
15
R
E
R ≈ √10n
where E = 6, 12, 24, 96
for 20, 10, 5, 1% tolerance
n = 0, 1, 2, 3, …
For 10% resistor E = 12
n = 0; R = 1.00000…
n = 1; R = 1.21152…
n = 2; R = 1.46779…
n = 3; R = 1.77827…
Resistance
Resistance Measurement
Measurement Techniques
Techniques
Bridge circuit
Voltmeter-ammeter
Substitution
Ohmmeter
Voltmeter-ammeter
V
V
A
A
R
R
Substitution
A
Supply
Unknow
resistance
A
Rx
Supply
Decade resistance
box substituted in
place of the
unknown
Voltmeter-ammeter method
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))
+
+
VA A
I
A
+
IV
+
+
Vx
VS
V V
-
-
-
Ix
I
VS
-
V V
-
Rx
Fig. (b)
Fig. (a)
V V + VA
V
Measured Rx: Rmeas = = x
= Rx + A
I
I
I
if Vx>>VA
Rmeas ≈ Rx
Therefore this circuit is suitable for measure
large resistance
Rx
Measured Rx: Rmeas =
if Ix>>IV
Rx
V
V
=
=
I I x + IV 1 + IV / I x
Rmeas ≈ Rx
Therefore this circuit is suitable for measure
small resistance
Ohmmeter
•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
Battery
k
45
R1
Rm
A
0µ
Basic series ohmmeter
75
10
Vs
− R1 − Rm
I
25
5k
50
0
VS
Meter Infinity
resistance
15k
0
Rx
Rx =
Nonlinear scale
Standard
resistance
∞
Resistance to
be measured
Meter
Ohmmeter scale
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.
Bridge Circuit
Bridge Circuit is a null method, operates on the principle of
comparison. That is a known (standard) value is adjusted until it is
equal to the unknown value.
Bridge Circuit
AC Bridge
DC Bridge
(Resistance)
Wheatstone Bridge
Kelvin Bridge
Megaohm Bridge
Inductance
Capacitance
Maxwell Bridge
Hay Bridge
Owen Bridge
Etc.
Schering Bridge
Frequency
Wien Bridge
Wheatstone Bridge and Balance Condition
Suitable for moderate resistance values: 1 Ω to 10 MΩ
A
Balance condition:
R2
R1
I1
V
No potential difference across the
galvanometer (there is no current through
the galvanometer)
I2
D
B
I4
I3
R3
R4
C
Under this condition: VAD = VAB
I1R1 = I 2 R2
And also VDC = VBC
I3 R3 = I 4 R4
where I1, I2, I3, and I4 are current in resistance
arms respectively, since I1 = I3 and I2 = I4
R1 R2 or
=
R3 R4
R2
Rx = R4 = R3
R1
Example
1Ω
1Ω
1Ω
1Ω
12 V
12 V
1Ω
2Ω
1Ω
(a) Equal resistance
1Ω
2Ω
(b) Proportional resistance
1Ω
10 Ω
10 Ω
12 V
12 V
2Ω
20 Ω
(c) Proportional resistance
2Ω
10 Ω
(d) 2-Volt unbalance
Measurement Errors
1. Limiting error of the known resistors
Using 1st order approximation:
A
R1
V
R2
R2  ∆R1 ∆R2 ∆R3 
Rx = R3 1 ±
±
±

R1 
R1
R2
R3 
2. Insufficient sensitivity of Detector
B
D
 R ± ∆R2 
Rx = ( R3 ± ∆R3 )  2

R
±
∆
R
1 
 1
3. Changes in resistance of the bridge
arms due to the heating effect (I2R) or
temperatures
4. Thermal emf or contact potential in the
bridge circuit
R3
C
Rx
5. Error due to the lead connection
3, 4 and 5 play the important role in the
measurement of low value resistance
Example In the Wheatstone bridge circuit, R3 is a decade resistance with a specified in
accuracy ±0.2% and R1 and R2 = 500 Ω ± 0.1%. If the value of R3 at the null position is
520.4 Ω, determine the possible minimum and maximum value of RX
SOLUTION Apply the error equation
Rx =
R2  ∆R1 ∆R2 ∆R3 
Rx = R3 1 ±
±
±

R1 
R1
R2
R3 
520.4 × 500  0.1 0.1 0.2 
±
±
1 ±
 = 520.4( 1 ± 0.004) = 520.4 ± 0.4%
500
 100 100 100 
Therefore the possible values of R3 are 518.32 to 522.48 Ω
Example A Wheatstone bridge has a ratio arm of 1/100 (R2/R1). At first balance, R3 is
adjusted to 1000.3 Ω. The value of Rx is then changed by the temperature change, the new
value of R3 to achieve the balance condition again is 1002.1 Ω. Find the change of Rx due to
the temperature change.
R2
1
=
×
= 10.003 Ω
1000.3
SOLUTION At first balance:
R1
100
R2
1
= 1002.1×
= 10.021 Ω
After the temperature change: Rx new = R3
R1
100
Rxold = R3
Therefore, the change of Rx due to the temperature change is 0.018 Ω
Sensitivity of Galvanometer
A galvanometer is use to detect an unbalance condition in
Wheatstone bridge. Its sensitivity is governed by: Current sensitivity
(currents per unit defection) and internal resistance.
consider a bridge circuit under a small unbalance condition, and apply circuit
analysis to solve the current through galvanometer
Thévenin Equivalent Circuit
Thévenin Voltage (VTH)
I1
VS
A
VCD = VAC − VAD = I1 R1 − I 2 R2
I2
R1
C
R2
D
G
R3
where I1 =
R4
B
Therefore
V
V
=
and I 2
R1 + R3
R2 + R4
 R1
R2 
−
VTH = VCD = V 

R
+
R
R
+
R
3
2
4 
 1
Sensitivity of Galvanometer (continued)
Thévenin Resistance (RTH)
R1
C
R2
A
R3
D
R4
RTH = R1 // R3 + R2 // R4
B
Completed Circuit
RTH
C
Ig=
G
VTH
VTH
RTH+Rg
Ig =
VTH
RTH + Rg
D
where Ig = the galvanometer current
Rg = the galvanometer resistance
Example 1 Figure below show the schematic diagram of a Wheatstone bridge with values of
the bridge elements. The battery voltage is 5 V and its internal resistance negligible. The
galvanometer has a current sensitivity of 10 mm/µA and an internal resistance of 100 Ω.
Calculate the deflection of the galvanometer caused by the 5-Ω unbalance in arm BC
SOLUTION The bridge circuit is in the small unbalance condition since the value of
resistance in arm BC is 2,005 Ω.
Thévenin Voltage (VTH)
A
1000 Ω
100 Ω
R1
5V
G
D
R3
1000
 100

VTH = VAD − VAC = 5 V × 
−

100
+
200
1000
+
2005


≈ 2.77 mV
R2
C
R4
2005 Ω
200 Ω
Thévenin Resistance (RTH)
B
(a)
100 Ω
C
RTH = 100 // 200 + 1000 // 2005 = 734 Ω
1000 Ω
A
200 Ω
D
2005 Ω
B
Ig =
(b)
RTH= 734 Ω
C
Ig=3.34 µA
VTH
2.77 mV
G
D
(c)
The galvanometer current
Rg= 100 Ω
VTH
2.77 mV
=
= 3.32 µ A
RTH + Rg 734 Ω + 100 Ω
Galvanometer deflection
d = 3.32 µ A ×
10 mm
= 33.2 mm
µA
Example 2 The galvanometer in the previous example is replaced by one with an internal
resistance of 500 Ω and a current sensitivity of 1mm/µA. Assuming that a deflection of 1 mm
can be observed on the galvanometer scale, determine if this new galvanometer is capable
of detecting the 5-Ω unbalance in arm BC
SOLUTION Since the bridge constants have not been changed, the equivalent circuit
is again represented by a Thévenin voltage of 2.77 mV and a Thévenin resistance of
734 Ω. The new galvanometer is now connected to the output terminals, resulting a
galvanometer current.
Ig =
VTH
2.77 mV
=
= 2.24 µ A
RTH + Rg 734 Ω + 500 Ω
The galvanometer deflection therefore equals 2.24 µA x 1 mm/µA = 2.24 mm,
indicating that this galvanometer produces a deflection that can be easily observed.
Example 3 If all resistances in the Example 1 increase by 10 times, and we use the
galvanometer in the Example 2. Assuming that a deflection of 1 mm can be observed on the
galvanometer scale, determine if this new setting can be detected (the 50-Ω unbalance in
arm BC)
SOLUTION
Application of Wheatstone Bridge
Murray/Varrley Loop Short Circuit Fault (Loop Test)
•Loop test can be carried out for the location of either a ground or a short
Power or
circuit fault.
R3
communication cable
X1
R1
short
circuit fault
R2
R4
ground
fault
Short
circuit
fault
Murray Loop Test
Let R = R1+R2
Assume: earth is a
good conductor
X2
At balance condition:
R3 R1
=
R4 R2
 R3 
R1 = R 

R
R
+
 3 4
 R4 
R2 = R 

R
R
+
 3 4
The value of R1 and R2 are used to calculate back into distance.
Murray/Varrley Loop Short Circuit Fault (Loop Test)
Examples of commonly used cables (Approx. R at 20oC)
Wire dia. In mm
Ohms per km.
Meter per ohm
0.32
218.0
4.59
0.40
136.0
7.35
0.50
84.0
11.90
0.63
54.5
18.35
0.90
27.2
36.76
Remark The resistance of copper increases 0.4% for 1oC rise in Temp.
Let R = R1+R2 and define Ratio = R4/R5
R
R1
At balance condition: Ratio = 4 =
R5 R2 + R3
Ratio
R1 =
R + R3
Ratio + 1
R - RatioR3
R2 =
Ratio + 1
X1
R4
R1
R2
R5
X2
R3
Varley Loop Test
Short
circuit
fault
Example Murray loop test is used to locate ground fault in a telephone system. The total
resistance, R = R1+ R2 is measured by Wheatstone bridge, and its value is 300 Ω. The
conditions for Murray loop test are as follows:
R3 = 1000 Ω and R4 = 500 Ω
Find the location of the fault in meter, if the length per Ohm is 36.67 m.
R3
Power or
communication cable
X1
 R3 
1000
R1 = R 
= 200 Ω
 = 300 ×
1000 + 500
 R3 + R4 
R1
R2
R4
X2
Murray Loop Test
SOLUTION
Short
circuit
fault
 R4 
500
R2 = R 
300
=
×
= 100 Ω

1000 + 500
 R3 + R4 
Therefore, the location from the measurement point is 100 Ω× 36.67 m/Ω = 3667 m
Application of Wheatstone Bridge
Unbalance bridge
Consider a bridge circuit which have identical
resistors, R in three arms, and the last arm has the
resistance of R +∆R. if ∆R/R << 1
A
R
R
Thévenin Voltage (VTH)
V
C
D
G
VTH = VCD ≈ V
R
R+∆R
B
RTH = R
VTH=V
Small unbalance
occur by the external
environment
∆R
4R
Thévenin Resistance (RTH)
RTH ≈ R
C
∆R
4R
G
D
This kind of bridge circuit can be found in sensor
applications, where the resistance in one arm is
sensitive to a physical quantity such as pressure,
temperature, strain etc.
5 kΩ
5 kΩ
6V
Rv Output
signal
5 kΩ
R v (kΩ )
Example Circuit in Figure (a) below consists of a resistor Rv which is sensitive to the
temperature change. The plot of R VS Temp. is also shown in Figure (b). Find (a) the
temperature at which the bridge is balance and (b) The output signal at Temperature of
60oC.
6
5
4
3
2
1
0
4.5 kΩ
0
20
40
60
80
100 120
Temp (oC)
(b)
(a)
R3 × R2 5 kΩ× 5 kΩ
SOLUTION (a) at bridge balance, we have Rv =
=
= 5 kΩ
R1
5 kΩ
The value of Rv = 5 kΩ corresponding to the temperature of 80oC in the given plot.
(b) at temperature of 60oC, Rv is read as 4.5 kΩ, thus ∆R = 5 - 4.5 = 0.5 kΩ. We will
use Thévenin equivalent circuit to solve the above problem.
VTH = V
∆R
0.5 kΩ
= 6 V×
= 0.15 V
4R
4 × 5 kΩ
It should be noted that ∆R = 0.5 kΩ in the problem does not satisfy the assumption ∆R/R
<< 1, the exact calculation gives VTH = 0.158 V. However, the above calculation still gives
an acceptable solution.
Low resistance Bridge: Rx < 1 Ω
Effect of connecting lead
R2
V
At point p:
rearrange
R3
G
R1
The effects of the connecting lead and the connecting
terminals are prominent when the value of Rx decreases
to a few Ohms
m
p
n
Rx
Ry
Ry = the resistance of the connecting lead from R3 to
Rx
At point m: Ry is added to the unknown Rx, resulting in too
high and indication of Rx
At point n: Ry is added to R3, therefore the measurement of Rx
will be lower than it should be.
R
Rx + Rnp = ( R3 + Rmp ) 1
R2
R
R
Rx = R3 1 + Rmp 1 − Rnp
R2
R2
Where Rmp and Rnp are the lead resistance
from m to p and n to p, respectively.
The effect of the connecting lead will be
canceled out, if the sum of 2nd and 3rd term is
zero.
Rnp R1
R1
Rmp
R2
− Rnp = 0 or
Rx = R3
R1
R2
Rmp
=
R2
Kelvin Double Bridge: 1 to 0.00001 Ω
Four-Terminal Resistor
Current
terminals
Current
terminals
Voltage
terminals
Voltage
terminals
Four-terminal resistors have current terminals
and potential terminals. The resistance is
defined as that between the potential
terminals, so that contact voltage drops at the
current terminals do not introduce errors.
Four-Terminal Resistor and Kelvin Double Bridge
R3
R2
• r1 causes no effect on the balance condition.
• The effects of r2 and r3 could be minimized, if R1 >>
r2 and Ra >> r3.
• The main error comes from r4, even though this value
is very small.
Rb
G
Ra
R1
r3
r2
r1
Rx
r4
Kelvin Double Bridge: 1 to 0.00001 Ω
2 ratio arms: R1-R2 and Ra-Rb
the connecting lead between m and n: yoke
l
R2
R3
I
m
Rb
V
k
G
p
Vlk =
Ry
Ra
n
R1
o
Eq. (1) = (2) and rearrange:
The balance conditions: Vlk = Vlmp or Vok = Vonp
Rx
R2
V
R1 + R2
(1)
here V = IRlo = I [ R3 + Rx + ( Ra + Rb ) // Ry ]
Vlmp


Ry
Rb 
= I  R3 +
Ra + Rb + Ry 

Rb Ry
R
Rx = R3 1 +
R2 Ra + Rb + Ry
 R1 Ra 
 − 
 R2 Rb 
(2)
Rx = R3
R1
R2
If we set R1/R2 = Ra/Rb, the second term of the right hand side will be zero, the relation
reduce to the well known relation. In summary, The resistance of the yoke has no effect
on the measurement, if the two sets of ratio arms have equal resistance ratios.
High Resistance Measurement
Guard ring technique:
Volume resistance, RV
Surface leakage resistance, Rs
Guard
ring
µA
High
voltage
supply
Is
Iv
V
(a) Circuit that measures insulation volume
resistance in parallel with surface leakage
resistance
Rmeas = Rs // Rv =
V
I s + Iv
Is
High
voltage
supply
µA
Iv
V
Material
under test
Is
(b) Use of guard ring to measure only volume
resistance
Rmeas = Rv =
V
Iv
High Resistance Measurement
Example The Insulation of a metal-sheath electrical cable is tested using 10,000 V supply
and a microammeter. A current of 5 µA is measured when the components are connected
without guard wire. When the circuit is connect with guard wire, the current is 1.5 µA.
Calculate (a) the volume resistance of the cable insulation and (b) the surface leakage
resistance
SOLUTION
(a ) Volume resistance:
IV =1.5 µA
RV =
V 10000 V
=
= 6.7 ×109 Ω
IV
1.5 µA
(b ) Surface leakage resistance:
IV+ IS = 5 µA
RS =
IS = 5 µA – IV = 3.5 µA
V 10000 V
=
= 2.9 × 109 Ω
IS
3.5 µA
MegaOhm Bridge
Just as low-resistance measurements are affected by series lead impedance, highresistance measurements are affected by shunt-leakage resistance.
RA
E
E
G
RC
the guard terminal is connect to a bridge
corner such that the leakage resistances
are placed across bridge arm with low
resistances
R1 // RC ≈ RC
R2 // Rg ≈ Rg
RA
RB
since R1 >> RC
since R2 >> Rg
RB
R2
Rx
G
R1
RC
Rx ≈ RA
RB
RC
Capacitor
Capacitance – the ability of a dielectric to store electrical charge per
unit voltage
Area, A
conductor
Dielectric, εr
thickness, d
A ε 0ε r
C =
d
Typical values pF, nF or µF
Dielectric
Construction
Capacitance
Breakdown,V
Air
Meshed plates
10-400 pF
100 (0.02-in air gap)
Ceramic
Tubular
0.5-1600 pF
500-20,000
Disk
1pF to 1 µF
Aluminum
1-6800 µF
10-450
Tantalum
6-50
500-20,000
Electrolytic
Mica
Stacked sheets
0.047 to 330 µF
10-5000 pF
Paper
Rolled foil
0.001-1 µF
200-1,600
Plastic film
Foil or Metallized
100 pF to 100 µF
50-600
Inductor
Inductance – the ability of a conductor to produce induced voltage
when the current varies.
N turns
L =
A
µo µ r N
2
A
l
µo = 4π×10-7 H/m
l
µr – relative permeability of core material
Ni ferrite:
µr > 200
Mn ferrite:
µr > 2,000
L
Air core inductor
Re
Cd
Equivalent circuit of an RF coil
Iron core inductor
Distributed capacitance Cd
between turns
Quality Factor of Inductor and Capacitor
Equivalent circuit of capacitance
Cp
Cs
Rs
Rp
Parallel equivalent circuit
Series equivalent circuit
Equivalent circuit of Inductance
Ls
Lp
Rs
Rp
Series equivalent circuit
Rs2 + X s2
Rp =
Rs
Rs2 + X s2
Xp =
Xs
Parallel equivalent circuit
Rs =
R p X p2
R +X
2
p
2
p
Xs =
X p R p2
R p2 + X p2
Quality Factor of Inductor and Capacitor
Quality factor of a coil: the ratio of reactance to resistance (frequency
dependent and circuit configuration)
Inductance series circuit:
Q=
Inductance parallel circuit: Q =
X s ω Ls
=
Rs
Rs
Rp
Xp
=
Typical Q ~ 5 – 1000
Rp
ω Lp
Dissipation factor of a capacitor: the ratio of reactance to resistance
(frequency dependent and circuit configuration)
Capacitance parallel circuit:
D=
Capacitance series circuit:
D=
Xp
Rp
=
1
ω C p Rp
Rs
= ω Cs Rs
Xs
Typical D ~ 10-4 – 0.1
Inductor and Capacitor
RP2
LS = 2
⋅ LP
RP + ω 2 L2P
RS
ω 2 L2P
I
RS = 2
⋅
R
P
RP + ω 2 L2P
Q=
I
V
V
RS2 + ω 2 L2S
LP =
⋅ LS
ω 2 L2S
LP
LS
RS2 + ω 2 L2S
RP =
⋅ RS
RS2
RP
V
ω LS
IωLS
RS
V/ωLP
V/RP
θ
Q=
RP
ω LP
θ
I
IRS
CS =
RS =
1+ ω C R
⋅ CP
ω 2CP2 RP2
2
2
P
1
⋅ RP
1 + ω CP2 RP2
I
RS
V
CP
LS
RP
2
D = ω CS RS
I
V
2
P
IRS
I/ωCS
I
VωCP δ
δ
V
CP =
1
⋅ CS
1 + ω 2CS2 RS2
1 + ω 2CS2 RS2
⋅ RS
RP =
ω 2CS2 RS2
D=
V/RP
1
ω CP RP
AC Bridge: Balance Condition
B
Z2
Z1
I1
V
I2
C
D
A
all four arms are considered as impedance
(frequency dependent components)
The detector is an ac responding device:
headphone, ac meter
Source: an ac voltage at desired frequency
Z3
Z4
Z1, Z2, Z3 and Z4 are the impedance of bridge arms
At balance point:
D
I1 =
General Form of the ac Bridge
Complex Form:
Polar Form:
Z1Z4 ( ∠θ1 + ∠θ 4 ) =Z2 Z3 ( ∠θ 2 + ∠θ 3 )
EBA = EBC or I1 Z1 = I 2 Z 2
V
V
and I 2 =
Z1 + Z 3
Z2 + Z4
Z1 Z 4 = Z 2 Z 3
Magnitude balance:
Phase balance:
Z1Z4 =Z2 Z3
∠θ1 + ∠θ 4 =∠θ 2 + ∠θ 3
Example The impedance of the basic ac bridge are given as follows:
Z1 = 100 Ω ∠80o (inductive impedance)
Z3 = 400 ∠30o Ω (inductive impedance)
Z 2 = 250 Ω (pure resistance)
Z 4 = unknown
Determine the constants of the unknown arm.
SOLUTION The first condition for bridge balance requires that
Z4 =
Z 2 Z 3 250 × 400
=
= 1, 000 Ω
Z1
100
The second condition for bridge balance requires that the sum of the phase angles of
opposite arms be equal, therefore
∠θ 4 =∠θ 2 + ∠θ 3 − ∠θ1 = 0 + 30 − 80 = −50o
Hence the unknown impedance Z4 can be written in polar form as
Z 4 = 1, 000 Ω ∠ − 50o
Indicating that we are dealing with a capacitive element, possibly consisting of a
series combination of at resistor and a capacitor.
Example an ac bridge is in balance with the following constants: arm AB, R = 200 Ω
in series with L = 15.9 mH R; arm BC, R = 300 Ω in series with C = 0.265 µF; arm CD,
unknown; arm DA, = 450 Ω. The oscillator frequency is 1 kHz. Find the constants of
arm CD.
B
Z1
I1
V
Z1 = R + jω L = 200 + j100 Ω
I2
C
D
A
SOLUTION
Z2
Z3
Z4
Z 2 = R + 1/ jω C = 300 − j 600 Ω
Z3 = R = 450 Ω
Z 4 = unknown
D
The general equation for bridge balance states that
Z4 =
Z1 Z 4 = Z 2 Z 3
Z 2 Z 3 450 × (200 + j100)
=
= j150 Ω
(300 − j 600)
Z1
This result indicates that Z4 is a pure inductance with an inductive reactance of 150 Ω
at at frequency of 1kHz. Since the inductive reactance XL = 2πfL, we solve for L and
obtain L = 23.9 mH
Comparison Bridge: Capacitance
R1
Vs
Measure an unknown inductance or
capacitance by comparing with it with a known
inductance or capacitance.
R2
At balance point:
D
C3
Rx
R3
Cx
Z1 Z x = Z 2 Z 3
1
1 =R1 ; Z 2 = R2 ; and Z 3 = R3 +
jω C3


1 
1 
R1  Rx +
 = R2  R3 +

j
ω
C
j
ω
C
x 
3 


where Z
Unknown
capacitance
Diagram of Capacitance
Comparison Bridge
Separation of the real and imaginary terms yields:
R2 R3
Rx =
R1
and
C x = C3
Frequency independent
To satisfy both balance conditions, the bridge must contain two variable
elements in its configuration.
R1
R2
Comparison Bridge: Inductance
R1
Vs
Measure an unknown inductance or
capacitance by comparing with it with a known
inductance or capacitance.
R2
At balance point:
D
L3
Lx
R3
Rx
where
Z1 Z x = Z 2 Z 3
Z1 =R1 ; Z 2 = R2 ; and Z3 = R3 + jω L3
Unknown
inductance
Diagram of Inductance
Comparison Bridge
Separation of the real and imaginary terms yields:
R1 ( Rx + jω Lx ) = R2 ( RS + jω LS )
R2 R3
Rx =
R1
and
Lx = L3
Frequency independent
To satisfy both balance conditions, the bridge must contain two variable
elements in its configuration.
R2
R1
Maxwell Bridge
R1
Measure an unknown inductance in terms of
a known capacitance
R2
C1
V
R3
Z x = Z 2 Z3 Y1
At balance point:
D
Lx
Rx
where
Unknown
inductance
Diagram of Maxwell Bridge
1
Z 2 = R2 ; Z3 = R3 ; and Y1 = + jω C1
R1
1

Z x = Rx + jω Lx = R2 R3  + jω C1 
 R1

Separation of the real and imaginary terms yields:
Rx =
R2 R3
R1
and
Lx = R2 R3C1
Frequency independent
Suitable for Medium Q coil (1-10), impractical for high Q coil: since R1 will be very
large.
Hay Bridge
Similar to Maxwell bridge: but R1 series with C1
R1
R2
C1
V
At balance point:
D
where
Lx
R3
Z1 = R1 −
Z1Z x = Z 2 Z3
j
; Z 2 = R2 ; and Z3 = R3
ω C1

1 
+
R
 1
 ( Rx + jω Lx ) = R2 R3
jω C1 

Diagram of Hay Bridge
L
(1)
R1 Rx + x = R2 R3
C1
Lx jRx
−
+ jω Lx R1 = R2 R3
which expands to R1 Rx +
C1 ω C1
Rx
(2)
= ω Lx R1
ω C1
Rx
Unknown
inductance
Solve the above equations simultaneously
Hay Bridge: continues
ωLx
θL
ω 2C12 R1 R2 R3
Rx =
1 + ω 2C12 R12
and
Z
R1
Rx
θC
1
ωC1
Z
Lx =
R2 R3C1
1 + ω 2C12 R12
X L ω Lx
tan θ L =
=
=Q
R
Rx
X
1
tan θ C = C =
R ω C1 R1
tan θ L = tan θ C or Q =
Phasor diagram of arm 4 and 1
Thus, Lx can be rewritten as
For high Q coil (> 10), the term (1/Q)2 can be neglected
1
ω C1 R1
R2 R3C1
Lx =
1 + (1/ Q 2 )
Lx ≈ R2 R3C1
Schering Bridge
C1
Used extensively for the measurement of capacitance
and the quality of capacitor in term of D
R2
R1
V
Z x = Z 2 Z3 Y1
where Z = R ; Z = 1 ; and Y = 1 + jω C
2
2
3
1
jω C3
R1
At balance point:
D
C3
Cx
Rx
Unknown
capacitance
Rx −
 − j  1

j
j
ω
C
= R2 
+

1
ω Cx
ω
C
R
x  1


Diagram of Schering Bridge
which expands to
j
R2C1
jR2
Rx −
=
−
C3 ω C3 R1
ω Cx
Separation of the real and imaginary terms yields:
Rx = R2
C1
C3
and
C x = C3
R1
R2
Schering Bridge: continues
Dissipation factor of a series RC circuit:
D=
Rx
= ω Rx C x
Xx
Dissipation factor tells us about the quality of a capacitor, how close the
phase angle of the capacitor is to the ideal value of 90o
For Schering Bridge:
D = ω Rx Cx = ω R1C1
For Schering Bridge, R1 is a fixed value, the dial of C1 can be calibrated directly in D
at one particular frequency
Wien Bridge
R1
Unknown
Freq.
R2
C1
Measure frequency of the voltage source using series
RC in one arm and parallel RC in the adjoining arm
Z 2 = Z1Z 4 Y3
1
1
Z1 = R1 +
; Z 2 = R2 ; Y3 =
+ jωC3 ; and Z 4 = R4
jωC1
R3
At balance point:
Vs
D
R3
R4
C3
Diagram of Wien Bridge
which expands to R2 =


j   1
R2 =  R1 −
R
j
ω
C
+
 4
3
ω
C
R
1 

 3

RC
R1 R4
jR4
+ jω C3 R1 R4 −
+ 4 3
R3
C1
ω C1 R3
Rearrange Eq. (2) gives
f =
1
2π C1C3 R1 R3
R2 R1 C3
=
+
R4 R3 C1
ω C3 R1 =
(1)
1
ω C1 R3
(2)
In most, Wien Bridge, R1 = R3 and C1 = C3
(1)
R2 = 2 R4
(2)
f =
1
2π RC
Wagner Ground Connection
C
C5
R1
Rw
A
Cw
C6
R2
1
2
C1
D
B
C2
Rx
C3
R3
Wagner ground
Cx
D
C4
Diagram of Wagner ground
One way to control stray capacitances is by
Shielding the arms, reduce the effect of stray
capacitances but cannot eliminate them
completely.
Stray across arm
Cannot eliminate
Wagner ground connection eliminates some
effects of stray capacitances in a bridge circuit
Simultaneous balance of both bridge makes the
point 1 and 2 at the ground potential. (short C1
and C2 to ground, C4 and C5 are eliminated from
detector circuit)
The capacitance across the bridge arms e.g. C6
cannot be eliminated by Wagner ground.
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