RICHLAND COLLEGE
School of
Engineering Business & Technology
INTC 1307
Instrumentation Test Equipment
Teaching Unit 6
Rev. 0 – W. Slonecker
Rev. 1 – (8/26/2012)– J. Bradbury
AC Bridges
Unit 6: AC Bridges
OBJECTIVES:
1. To explain the operation of an AC Bridge.
2. To review AC concepts of frequency and effects on LRC circuits.
3. To explain the requirements of balancing an AC Bridge.
4. To null an AC Bridge.
AC Bridge Configuration
When the four resistive arms of the basic Wheatstone bridge are replaced by impedances and the
bridge is excited by an AC source, the result is an
AC Bridge. Now, to balance the bridge, two
Z3
conditions must be satisfied, the resistive (R) and
Z1
the reactive components (XC or XL). Once balanced,
the AC Bridge indicates a null.
D
AC bridge circuits are also used for shifting phase,
providing feedback paths for oscillators and
Z2
Z4
amplifiers, filtering out undesired signals, and
measuring the frequency of audio and radio
frequency (rf) signals.
At balance:
|Z| indicates the magnitude of the impedance
Z1   1
Z2   2

Z3   3
Z 4  4
( Z1   1 )( Z 4   4 ) ( Z 2   2 )( Z 3  3 )
Z1 Z 4  Z 2 Z 3
Z1
IA X
Z2
Z3
D
IB
Z4
Y
and  1   4   3  3
The null or balanced condition occurs when detector
current becomes Zero (no voltage difference occurs
from Y to X). This means that the impedance (Z) of
the detector circuit appears infinite (∞), or as an
apparent open circuit, so that each leg of the bridge is
isolated from the other leg.
For this balance condition,
VZ1 VZ3 VZ 2 VZ 4
IAZ1 = IBZ3 IAZ2 = IBZ4
I A Z1 I B Z 3
Z1 Z 3


so
IA Z2 IBZ4
Z2 Z4
and
Z1Z4 = Z2Z3
and
Z1Z4 = Z2Z3
Z indicates both magnitude and phase
Page 1 of 8
This is the general bridge equation and applies to any four arm bridge circuit whether
branches are pure resistances or combinations of R, C, and L. Most of the time, the balance
equation is not dependent upon frequency.
When the bridge is not balanced, these equations are not correct.
The circuit is complex and Z must be written in complex form Z  Z  .
AC Bridge Calculations
200Ω
150Ω
Z3
 60°
Z1
D
Z2
100 + j 300Ω
Z4
UNKNOWN
For example, the AC Bridge at right is balanced.
In order to use the bridge equation, complex forms must be
multiplied. When multiplying or dividing, the polar
complex form is easiest to use. When adding or
subtracting, the rectangular complex form is easiest to use.
Since Z1and Z3 are already in polar form, change Z2 to polar
form and solve for Z4.
Conversion of Z2, 100 + j 300Ω, to polar form: tan  
jXL = 300
X L 300

3.0
R
100
Sin  
R = 100
XL
Z
Z=
 Arc tan  3.0  71.6o
XL
300

316
sin 71.6
sin 71.6
Z 2 31671.6
Z1Z4 = Z2Z3 The bridge equation
Z4 
Z 2 Z3
Z1
Z4 
(316 1.6 )(150) 47400 1.6

2371.6
200 0
200 0
Z4 = 237 11 .6
Sin  
Conversion of 237 11.6 to rectangular form
j
XL
Z
Cos  
237
 11.6
R
Z
X L Sin 237(Sin 11.6 )
R = Z Cos
 237(Cos 11.6 )
Z4 = 232 + j47.7 Ω
Page 2 of 8
Review of Basic AC Concepts
Resistance: Resistors limit current and dissipate power. There is no phase shift across pure
resistance. Changes in frequency cause no change in the value of the resistance.
R
V
I
Z = R 0
R
no change with frequency
freq.
Vg
R=
Vg
Page 3 of 8
Inductance: Inductance opposes any change in current. X is the symbol for Inductive
Reactance which is the opposition to the flow of AC current measured in ohms. An inductor
causes the current through the inductor to lag the voltage across the inductor by 90 degrees.
L
j
N 2 A

XL 
VL
IL
Z X L  90
XL =2πƒL
frequency
Capacitance: Capacitance opposes any change in voltage. XC is the symbol for Capacitive
Reactance which is the opposition to the flow of AC current measured in ohms.
Capacitance
causes the voltage across the capacitor to lag the current through the
capacitor by 90 degrees.
Q
V
V=
XC =
1
jC
C
XC
Q
C
Z X C   90
Q = charge
Vg
0
-j
LR Circuits:
I is the same everywhere in a series circuit.
IT 
L
Vg
Z
R
IT 
∞
VR
R
X L  jL
∞
XL
Vg
Z R  jX L
R
|Z|
Z = R + jXL
0
ƒ
ƒ
ƒ
at DC, XL = 0 so Z = R  0  0
+j90°
at mid freq. Z = R + jXL, when XL = R,  45
at high freq. XL = ∞ Z = XL  90
If Vg is constant as ƒ increases, R remains constant, both XL and 
 increase, and IT decreases.
Z = R + jXL
Page 4 of 8
RC Circuits:
Z = R jXC
1
2fC
XC =
0
C
∞
XC
R
-j∞
Vg
|Z|
0
0
ƒ
ƒ
ƒ
R
at DC, XC =∞, Z = ∞   90
at mid freq. XC = R,   45 Z = R - jXC   45
at high freq. XC = 0Ω Z = R 0
RLC Circuits
Z = R + j (XL – XC)
at low freq. (dc) XL = 0, Z = R  jXC Z ∞   90
The circuit is open!
at ƒco min, XC > XL, Z = R – j(XC – XL) and Z = R –jX, when R = X,  = -45
at ƒ0 , Z = R  0 , XL = XC so they cancel each other leaving Z = R
at ƒco max, XC < XL, Z = R + j(XL – XC) = R + jX when R = X,  = 45
at high freq. XL = ∞, XC = 0, Z = R + jXL, Z = ∞   90
C
L
R
ƒo min
ƒo max
∞
ƒo
+
|Z:
-
I
ƒo
0
ƒ→
Impedance vs. freq.
ƒ→
Current vs. freq.
Practical problem:
Find the value of Z4 to balance this AC
bridge.
400
ƒo is the resonant frequency when XL = XC
XL = 2πƒL =
XL3 = 100
200 in series with
0.01592H
XC 
2 1000 15.92 10  3
1
1

2fC 2 1000 4 10  7
XC2 = 397.89
Z1 = 400
0
Z2 = 300 – j398 = 500-53.1
Z3 = 200 + j100 = 223.626.6
300 in series with
0.4F
UNKNOWN
C=
Z4 =
Z2 Z3  223.6 26.6  500  53.1

280  26.4
Z1
400 0
1 = 250  j124  6
Z4
1.28 10
21000  124
250
1.28f
Page 5 of 8
Capacitor Equivalent Circuits:
Except for electrolytic capacitors, capacitors have almost no intrinsic resistance. Electrolytic
leakage and dissipation factors are modeled using parallel/series resistances. RP is called leakage
resistance and is in parallel with the actual capacitance. The RP/CP parallel circuit shown below is
equivalent to the series RS/CS circuit. Electrolytic capacitors have a relatively low RP; other types of
capacitors have extremely high RP. The higher the RP the less the leakage current so the capacitor can
better maintain its charge and voltage. If RP decreases, the capacitor no longer acts as a capacitor; the
leakage increases to the point that the capacitor appears as a resistor.
RS = RP
D
2
1 D 2
CS = CP(1 + D 2 )
CP

RP
Bridges measure the ratio of reactance to resistance for the component under test. This ratio is called the
dissipation factor, D, which is directly proportional to the power loss per cycle. A capacitor with a low
leakage loss (keeps its charge when disconnected) is a high quality capacitor.
D
D
XP
1

R P 2f R P C P
1

2f R P C P
1
1  D2
C
2f
RS  S 2
2
D
1 D

D2
2f R SCS

D 2f R SCS
Inductor Equivalent Circuits:
Inductors are formed by winding a coil of wire. This wire has resistance. Each inductor then has
a series resistance along with its inductance. This ratio of Inductive Reactance to its coil wire resistance
is the storage factor or quality factor, Q. This Q factor is directly proportional to the energy stored per
cycle. The greater the dissipation factor, the lower the Q of the coil.
Q=
LS
RP
R
S
LP
X S 2fL S

RS
RS
R
X
1
Q= D
Q=
P
P

R P
2 fL
P
Page 6 of 8
Capacitance Bridges:
If the bridge is balanced, an apparent open circuit occurs from Y to X. At that balance point,
the ratio of each leg to the other is such that the voltage drops and the phase angles are the same, just as
in a balanced Impedance Bridge. The standard frequency of the voltage source for these AC bridges is
1000 Hz. The following formulae show that frequency is not necessary for a capacitance bridge to
operate at balance.
V1
R 1
R
 2
X C1
X C2
R1
1 V
1kHz
0Deg
R
R2
+
0.000
V
R
C1
Capacitance Bridge
C2
R
1
1
2 fC
U1
1
X
C

1
2 fC
2
 1
2 fC
2
1 2  fC 1   R 2 2  fC 2 
2 f R 1C
1
  2 f R
2
C
2

(Divide out 2πf)
R 1C1 R 2C 2
C1 C2

R 2 R1
R1 R 2

C2 C1
C2 
R1
C1
R2
Example: For C2 unknown, R2 = 50k, C1 = 0.01F,
and R1 = 300k for balance, then C2 = 0.06F
Since reactance is a function of frequency, it is possible with capacitors to obtain a very low reactance
for C2 (the unknown) for a frequency of 1 kHz. For example, if C2 = 10F then its reactance is 15.9.
For an R2 fixed at 10k, the source voltage divider ratio for R2-C2 would be only 1.5% or 15mV for a
1v RMS source. The sharpest nulls are obtained for divider ratios of 50%, so a 1.5% ratio would give a
very weak null. However, if the source frequency were lowered to 50 Hz and R2 lowered to 1k, the
C2 reactance would be 318.3 and the divider ratio would be 30%. This ratio gives a much better null.
So although frequency divides out of the balance equation, it can and does impact bridge accuracy.
Other Bridges:
This discussion has covered the basic concepts of bridges. Other bridges are Inductance, Similar
Angle, Opposite Angle, Wien, Radio Frequency, Schering, and Owen bridges. A Maxwell-Wien Bridge
is shown below (often called the Maxwell bridge), and is used to measure unknown inductances in terms
of calibrated resistance and capacitance. Calibration-grade inductors are more difficult to manufacture
than capacitors of similar precision, and so the use of a simple "symmetrical" inductance bridge is not
always practical. Because the phase shifts of inductors and capacitors are exactly opposite each other, a
capacitive impedance can balance out an inductive impedance if they are located in opposite legs of a
bridge, as they are here.
Page 7 of 8
R1
V1
RX
1V
1kHz
0Deg
+
-
0.000
V U1
R3
C1 50%
LX
R2
Maxwell-Wien Bridge
The balance of the Maxwell-Wien bridge is independent of source frequency, (with the understanding of
divider ratios already discussed) and in some cases this bridge can be made to balance in the presence of
mixed frequencies from the AC voltage source, the limiting factor being the inductor's stability over a
wide frequency range.
Bridge Equations:
LX = R1R2C1
RX = R1R2/R3
Page 8 of 8