AC Voltmeter: PMMC Based
Example: A PMMC instrument has FSD of 100 µA and a coil resistance of 1700 Ω is used in the
half-wave rectifier voltmeter. The silicon diode (D1) must have a minimum (peak) forward current
of 100 µA. When the measured voltage is 20% of FSD. The voltmeter is to indicate 50 Vrms at full
scale Calculate the values of RS and RSH.
Known: FSD of Im Rm
Solution:
RS = 139.5 kΩ RSH = 778 Ω
Rs
D1
Vp
Rm
D2
RSH
Vrms
Vav
AC Voltmeter: PMMC Based
Example The symmetrical square-wave voltage is applied to an average-responding ac
voltmeter with a scale calibrated in terms of the rms value of a sine wave. If the
voltmeter is the full-wave rectified configuration. Calculate the error in the meter
indication. Neglect all voltage drop in all diodes.
E
Solution 11%
Em
t
T
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
The standard resistor R3 can be adjusted to null or balance the circuit.
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
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
B
Thévenin Resistance (RTH)
(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
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)
Deflection Method
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
V
Thévenin Voltage (VTH)
C
D
V
VTH = VCD = V
R
R+∆R
B
∆R / R
4 + 2∆R / R
Thévenin Resistance (RTH)
Small unbalance
occur by the external
RTH ≈ R
environment
In an unbalanced condition, the magnitude of the current or voltage drop for the
meter or galvanometer portion of a bridge circuit is a direct indication of the
change in resistance in one arm.
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.
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.
5 kΩ
R v (kΩ )
5 kΩ
6V
Rv Output
signal
5 kΩ
6
5
4
3
2
1
0
4.5 kΩ
0
20
40
80
Temp (oC)
(b)
(a)
60
100 120
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.
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
V
I1
Z1 = R + jω L = 200 + j100 Ω
I2
C
D
A
Z3
SOLUTION
Z2
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
Z1 Z 4 = Z 2 Z 3
Operational Amplifier: Op Amp
VCC(+)
Inverting
Input
_
Non-inverting
Input
+
I1
_
I2
Output
V-
+
V+
VEE(-)
(a) Electrical Symbol for the op amp
(b) Minimum connections to an op amp
Ideal Op Amp Rules:
1. No current flows in to either input terminal
2. There is no voltage difference between the two input terminals
Rule 1: I1 = I2 = 0; R+/- = ∞
Rule 2: V+ = V-; Virtually shorted
Vout
Inverting Amplifier
Rf
KCL
Use KCL at point A and apply Rule 1:
(no current flows into the inverting input)
R1
_
A
vin
+
+
vout
-
v A − vin v A − vout
+
=0
R1
Rf
Rearrange
 1
1   vin vout 
vA 
+
+
 − 
 = 0
R
 1 R f   R1 R f 
Apply Rule 2: (no voltage difference between inverting and non-inverting inputs)
Since V+ at zero volts, therefore V- is also at zero volts too.
vin vout
+
=0
R1 R f
Rf
vout
=−
vin
R1
vA = 0
Inverting Amplifier: another approach
Rf
No current flows
into op amp
vin
R1
_
i
+
i
From Rule 2: we know that V- = V+ = 0,
and therefore
0
vin
i=
R1
−
−vin + iR1 − V = 0
+
Since there is no current into op amp
vout (Rule 1)
-
vout = −iR f
−V − + iR f + vout = 0
Combine the results, we get
mV
Rf
vout
=−
vin
R1
vout
60
40
20
Given vin = 5sin3t, R1=4.7 kΩ and Rf =47 kΩ
vout = -10vin = -50 sin 3t
0
mV
-20
-40
-60
vin
1
2
3
4
5
6
time
Non-inverting Amplifier
Rf
KCL
R1
Use KCL at point A and apply Rule 1:
v A v A − vout
+
=0
R1
Rf
_
A
+
vin
+
vin = v A
Apply Rule 2:
vout
-
Rf
vout
= 1+
vin
R1
mV
vout
60
Given vin = 5sin3t, R1=4.7 kΩ and Rf =47 kΩ
40
20
vout = 11vin = 55 sin 3t
mV
-20
-40
-60
vin
1
2
3
4
5
6
time
Summing Amplifier: Mathematic Operation
i = i1 + i2 + i3
Use KCL and apply Rule 1:
i
i1
R
i2
R
i3
v1
v2
R
vA
v A − v1 v A − v2 v A − v3 v A − vout
+
+
+
=0
R
R
R
Rf
Rf
Since vA = 0 (Rule 2)
_
vB
+
+
vout
-
vout = −
Rf
R
( v1 + v2 + v3 )
Sum of v1, v2 and v3
v3
Difference Amplifier: Mathematic Operation
Use KCL and apply Rule 1:
R4
R1
R2
v1
v2
vA
v A − v1 v A − vout
+
=0
R1
R4
_
vB
+
R3
Substitute eq. (2) into eq. (1), we get
If R1 = R2 = R and R3 = R4 = Rf
(1)
Since vA = vB (Rule 2) and
+
vout
-
 R3 
v A = vB = 
 v2
 R2 + R3 
vout  R1 + R4   R3 
v1
=
v
−
 2

R4  R1 R4   R2 + R3 
R1
vout =
Rf
R
( v2 − v1 )
Difference of v1and v2
(2)
Differentiator and Integrator: Mathematic Operation
R
i
vout = − iR
C
i
_
+
vin
But
i=C
dvC
and
dt
vout
dvin
= − RC
dt
+
vout
-
Differentiator
i
C
R
vout = − vC
+ vc -
t
1
But vC (t ) = ∫ idt + vC (0) and
C 0
_
i
+
vin
Integrator
vin = vC
+
vout
-
t
vout
vin = iR
1
=−
vin dt + vC (0)
∫
RC 0