2141-375
Measurement and Instrumentation
Analog Electrical Devices and
Measurements
Analog Devices: Current Measurements
Force on a conductor
I
A conductor is placed in a uniform magnetic field B T,
at an angle of θ. The current flow in the conductor is I
A. Force exerted on the conductor can be calculated
from
r
r r
F = IL × B
θ
F
F = ILB sinθ
B
Analog Devices: Current Measurements
Force on a conductor
With no current flowing through the conductor,
the spring will be at its unstretched length. As
current flows through the conductor, the spring
will stretch and developed force required to
balance the electromagnetic force.
kx = ILB
Fs = kx
I
I
k = spring constant, x = the total distance
moved by the spring and θ is 90o
B
F = IBL
k
I=
x
BL
D’Arsonval or PMMC Instrument
Important parts of PMMC Instrument
•Permanent magnet with two softiron poles
• Moving coil
• Controlling or restoring spring
PMMC = Permanent Magnet Moving Coil
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
F = IBL
F = IBL
F = IBL
D
L = the length of coil perpendicular to the
paper
Since the force acts on each side of the coil, the total force for a coil of
N turns is
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:
TC = Kθ
Where K = the spring constant. For a given deflection, the controlling
and deflecting torques are equal
BLIND = Kθ
Since all quantities except θ and I are constant for any given
instrument, the deflection angle is
θ = CI
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
• Galvanometers are often used to detect zero current or voltage in a
circuit rather than to measure the actual level of current or voltage. In
this situation, the instrument is referred as a null detector.
DC Ammeter
• An ammeter is always connected in
series with a circuit.
PMMC
instrument
• The internal resistance should be very
low
Ammeter
shunt
• 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
Rs
Rm
Coil
resistance
Im
I = Is + Im
Is
Vm
Rs
I
Shunt
resistance
Vm = Vs
I m Rm = I s Rs
I m Rm
Rs =
Is
I m Rm
Rs =
I − Im
DC Ammeter
Example: An ammeter has a PMMC instrument with a coil resistance of Rm = 99 Ω and FSD
current of 0.1 mA. Shunt resistance Rs = 1 Ω. 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:
Im (mA)
Is (mA)
I = Im + Is (mA)
FSD
0.5FSD
0.25FSD
0.1
0.05
0.025
9.9
4.95
2.475
10
5
2.5
DC Ammeter
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 mA
and (b) FSD = 1 A.
Known: FSD of Im Rm
Solution:
(a) FSD = 100 mA: Rs = 1.001 Ω
(b) FSD = 1 A: Rs = 0.10001 Ω
DC Ammeter: Multirange
Rm
C
Rs1
B
D
Rs2
E
Rs3
Rs4
C
B
A
D
E
A
Make-before break switch
Multirange ammeter using switch shunts
• 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.
DC Ammeter: Ayrton shunt
Rm
Im
VS
Im
R1
+
R2
I
R3
B
I
Is
C
Is
-
A
D
R1 + R2 + R3 in parallel with Rm
Rm
Im
VS
Im
R1
+
R2
R3
B
I
Is
Is
C
A
D
R1 + R2 in parallel with Rm + R3
I
An Ayrton shunt used with an
ammeter consists of several seriesconnected resistors all connected in
parallel with the PMMC instrument.
Range change is effected by switch
between resistor junctions
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:
Rm
Im
Im
R1
+
R2
R3
B
I
Position
B
C
D
IFSD (mA)
10.05
100.05
1000.05
VS
Is
Is
C
A
I
D
DC Voltmeter
• 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
PMMC
instrument
V = I m Rs + I m Rm
Series resistance
or “multiplier”
Given V = Range
V
Multiplier
resistance
Rs
V
Rs = − Rm
Im
Coil
resistance
Rm
Im
V
Range
Rs =
− Rm
Im
The reciprocal of full scale current is the
voltmeter sensitivity (kΩ
Ω/V)
The total voltmeter resistance =
Sensitivity X Range
DC Voltmeter: Multirange
Multiplier
resistors
• Multirange voltmeter using switched
multiplier resistors
R1
Meter
resistance
R2
V = I m (Rm + R)
R3
Rm
Where R can be R1, R2, or R3
V
Rm
R1
R2
R3
• Multirange voltmeter using seriesconnected multiplier resistor
V = I m (Rm + R)
Where R can be R1, R1 + R2, or R1 + R2 + R3
V
DC Ammeter
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:
Multiplier
resistors
R1
Meter
resistance
Rm
R2
R1
R2
R3
R3
Rm
V
V
(b)
(a)
R1
R2
R3
R1
R2
R3
198.3 kΩ
Ω
998.3 kΩ
Ω
1.9983 MΩ
Ω
198.3 kΩ
Ω
800 kΩ
Ω
1 MΩ
Ω
Ohmmeter: 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
= Rx + A
Measured Rx: Rmeas = = x
I
I
I
Rmeas ≈ Rx
if Vx>>VA
Therefore this circuit is suitable for measure
large resistance
Rx
Measured Rx: Rmeas =
if Ix>>IV
V
V
Rx
=
=
I I x + IV 1 + IV / I x
Rmeas ≈ Rx
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
Resistance to
be measured
Nonlinear scale
Standard
resistance
15k
Rx
k
45
Vs
− R1 − Rm
I
Basic series ohmmeter
75
A
0µ
10
Rx =
25
0
VS
Meter Infinity
resistance
0
Rm
∞
R1
Battery
5k
50
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.
Loading Effect: Voltage Measurement
Rth
a
Linear
Circuit
Vab
V
Vth
Rm
Vab
b
Undisturbed condition: Rm = ∞
Measured condition: Rm ≠ ∞
General equation:
Measurement error:
a
V
Rm
b
Vab = Vu = Vth
Vab = Vm =
Vm =
Rm
Vth
Rm + Rth
1
Vu
1 + Rth / Rm
Vm − Vu
× 100
Vu
Rth
1
=−
×100% = −
×100%
Rm + Rth
1 + Rm / Rth
error =
Therefore, in practice, to get the acceptable results, we must have Rm ≥ 10 Rth (error ~ 9%)
Loading Effect
R1
100kΩ
Ω
100kΩ
Ω
5V
6.7 V
10 V
10 V
R2
100kΩ
Ω
5V
3.3 V V
100kΩ
Ω
Vmeas =
100 // 100
10 V = 3.3 V
100 + 100 // 100
Circuit under measurement
Circuit before measurement
100kΩ
Ω
100kΩ
Ω
100kΩ
Ω
6V
5.2 V
10 V
10 V
100kΩ
Ω
Vmeas =
4V
200 // 100
10 V = 4.0 V
100 + 200 // 100
V
100kΩ
Ω
200kΩ
Ω
Vmeas =
4.8 V V
1000 // 100
10 V = 4.8 V
100 + 1000 // 100
1000kΩ
Ω
Loading Effect
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
SOLUTION The voltage drop across Rb without the voltmeter connection
Rb
5k
V=
× 50 = 5 V
Ra + Rb
45 k + 5 k
Vb =
Ra
45kΩ
Ω
On the 5 V range
45 V
50 V
Rb
5kΩ
Ω
5V
Rm = S × range = 20kΩ / V × 5V = 100 kΩ
Req =
Rm Rb
100 k × 5 k
=
= 4.76 kΩ
Rm + Rb 100 k + 5 k
The voltmeter reading is
Vb =
Req
Ra + Req
V=
4.76 k
50 = 4.782 V
45 k + 4.76 k
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 1 = ± 0.05 V
100
∴% of error on the 5 V range:
=
− 0.218 V ± 0.05 V
×100 = ±5.36%
5V
Range
(V)
Vb .
(V)
Loading
error (V)
Meter
error (V)
Total
error (V)
% error
5
4.78
-0.22
± 0.05
± 0.27
± 5.36
10
4.88
-0.12
± 0.1
± 0.22
± 4.40
30
4.95
-0.05
± 0.3
± 0.35
± 6.10
Loading Effect: Current Measurement
a
Linear
Circuit
Vab
Rth
I
A
Rm
Vth
b
Undisturbed condition: Rm = 0
Measured condition: Rm ≠ 0
General equation:
Measurement error:
I = I u = Vth / Rth
a
Vab
I
A
Rm
b
I = I m = Vth / (Rth + Rm )
I m = I u / (1 + Rm / Rth )
I m − Iu
×100
Iu
Rm
1
=−
×100% = −
× 100%
Rm + Rth
1 + Rth / Rm
error =
Therefore, in practice, to get the acceptable results, we must have Rm ≤ Rth /10 (error ~ 9%)
AC Voltmeter: PMMC Based
Waveform
Amplitude
Average
RMS
A
0
A
2
A
A
A
2
A
D
W
π
2A
π
A
2
A
0
A
3
A
0
A
A
D
A
D +W
D
A
D +W
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.
•Using 4 diodes
Full-wave Rectifier Voltmeter
Multiplier
resistors
Rs
D1
D3
Rm
Vrms
Vp
Vav
D2
•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
D4
•Actually voltage to be indicated in ac
measurements is normally the rms quantity
•The scale is calibrated for
pure sine with the scale factor
of 1.11 (A/√
√2 / 2A/π
π)
AC Voltmeter: PMMC Based
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 rms
input is 75 V and (c) the sensitivity of the voltmeter
Known: FSD of Im Rm
Solution:
Multiplier
resistors
Rs
D1
D3
Rm
Vrms
Vp
Vav
D2
D4
(a) Rs = 890.7 kΩ
(b) 0.75 of FSD
(c) 9 kΩ/V
AC Voltmeter: PMMC Based
Half-wave Rectifier Voltmeter
Rs
D1
Vp
Rm
D2
Vrms
RSH
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
π)
AC Voltmeter: PMMC Based
Example: A PMMC instrument has FSD of 50 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
RS = 139.5 kΩ RSH = 778 Ω
Solution:
Rs
D1
Vp
Rm
D2
Vrms
RSH
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.
Balance condition:
A
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)
A
I1
VS
VCD = VAC − VAD = I1 R1 − I 2 R2
I2
R1
C
R2
G
R3
where I1 =
D
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 = V AD − 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
The galvanometer current
C
VTH
2.77 mV
=
= 3.32 µ A
RTH + Rg 734 Ω + 100 Ω
Ig=3.34 µA
VTH
2.77 mV
G
D
(c)
Rg= 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
Please correct
A
R
R
V
Thévenin Voltage (VTH)
C
V
D
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.
5 kΩ
Ω
5 kΩ
Ω
6V
Rv Output
signal
5 kΩ
Ω
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
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
Z1, Z2, Z3 and Z4 are the impedance of bridge arms
Z3
Z4
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 I1Z1 = 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
I1
V
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
+
VV+
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
R1
Use KCL at point A and apply Rule 1:
(no current flows into the inverting input)
v A − vin v A − vout
+
=0
R1
Rf
_
A
+
+
vout
-
vin
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
No current flows
into op amp
vin
Rf
i
From Rule 2: we know that V- = V+ = 0,
and therefore
0
vin
i=
R1
−
R1
_
i
+
−vin + iR1 − V = 0
Since there is no current into op amp
(Rule 1)
+
vout
-
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 − v 2 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
+
+
vout
-
vin
Integrator
vin = vC
t
vout
vin = iR
1
=−
vin dt + vC (0)
∫
RC 0