OP-AMPS in SIGNAL PROCESSING
APPLICATIONS
Filters, Integrators, Differentiators,
and Instrumentation Amplifier
Sensors :
definition and principles
Sensors : taxonomies
• Measurand
– physical sensor
– chemical sensor
– biological sensor(cf : biosensor)
• Invasiveness
– invasive(contact) sensor
– noninvasive(noncontact) sensor
• Usage type
– multiple-use(continuous monitoring) sensor
– disposable sensor
• Power requirement
– passive sensor
– active sensor
What is electronics engineering all
about?
Physical
variables
Process
Sensors/ transducers
Electrical
variables
V, I, t, f
Filters
Amplifiers
Feedback
A/D converters
Physical
variables
F, P, d, v, t, f
Actuators/transducers
New process
Computers
D/A converters
What is a Signal Processor?
The ECG wave in it’s raw form
• It selects useful parts
of a signal
• It cleans signal from
impurities
• It compares signals
• It converts signals into
forms that can be
easily recognized
Examples of
biomedical
signals
Examples of Signal Processors
• Basic amplifiers – already discussed
• Instrumentation amplifier – an improved
differential amplifier
• Active filters
• Integrators and differentiators
• Extra Reading:
– Precision rectifiers
– Logarithmic amplifiers
– Negative capacitance amplifiers
Signal Conditioning
• Ideal operational amplifier
• Inverting, non-inverting and instrument
amplifier
• Integrator, differentiator
• Filters
• Examples of signal conditioning module
Ideal Operational Amplifier
Operational (OP) amplifier is a high-gain dc differential
amplifier. It is made of an integrated circuit chip. It has two
inputs, negative terminal v1 and positive terminal v2 and one
output v0. The effective input to the amplifier is the voltage
difference of the two inputs v1-v2.
Properties of Ideal OP Amplifier
•
•
•
•
•
Gain if infinity (A=∞).
V0 = 0, when v1 = v2
Input impedance is infinity.
Output impedance is zero.
Bandwidth is infinitely wide and has no phase
shift.
Basic Rules for Ideal OP Amplifier
• When the OP amplifier output is in its linear
range, the two input terminals are at the same
voltage.
• No current flows into either input terminal of
the OP amp.
Inverting Amplifier
Non-inverting Amplifier
Instrument Amplifier
Active Filters
•
•
•
•
•
Low-pass
High-pass
Band-pass
Band-stop (notch)
All-pass (phase-shift)
Low-pass filter: circuit
Cf
Derive the equation for gain G
What are the important parameters?
Ri
Vi
GL = Rf/Ri ; τf = RfCf ; fc = 1/2πτf



=− 
+
Rf
A
V0
G=
V0 ( jω )
Vi ( jω )
 Rf

Zf


G=−
j ωC f 

Zi


1
+ Rf 
jωC f 
Rf
Rf
GL
1
 =−
=−
=
(1 + jωR f C f )Ri Ri 1 + jωτ f 1 + jωτ f
Ri
Low-pass filter: characteristics
Gain/GL
0.1
1
10
100
f/fc
0.1
0.01
-π
Phase
Slope = -1
1
f/fc
-5π/4
-3π/2
GL = Rf/Ri ; τf = RfCf ; fc = 1/2πτf
Integrator – ideal: circuit
1
V0 = −
Ri C f
Virtual
Ground
Vi
dV0
= Cf
i=
Ri
dt
Ri
-
Vi
i
+
Express V0 in terms of Vi
Zf
V0 ( jω )
=−
=−
G=
Vi ( jω )
Zi
Where τ = RiCf
Cf
A
i
t
∫ V dt + V
0
i
iC
V0
Write expression for gain G
1
j ωC f
Ri
1
1
=−
=−
jωRi C f
jωτ f
G/GL
Integrator – ideal:
characteristics
100
10
0
0.1
1
10
100
f/fc
0.1
ωc fc
=
=
G ==
ωτ ω
f
1
Slope = -1
0.01
-π
Phase
f/fc
Draw phase versus f/fc
-5π/4
-3π/2
1
Draw G/GL versus f/fc
θ = −3π / 2 = −270°
The integrator with bias currents
Through which component the bias current flows?
Integrator - practical
Gain/GL
Without Rf
Rf
Ri
Vi
-
0.1
1
10
f/fc
Cf
V0
A
0.1
With Rf
0.01
+
Slope = -1
Low-pass filter
Rf provides route for DC bias currents
-π
Rf
V0 ( jω )
GL
1
G=
=−
=
Vi ( jω )
Ri 1 + jωτ f 1 + jωτ f
τ f = R f C f ; fc =
100
1
2πτ f
-3π/2
Phase
1
-5π/4
Without Rf
Integ.
f/fc
Charge amplifier
Electrode
vo
The piezoelectric sensor generates charge, which is transferred to
the capacitor, C, by the charge amplifier. Feedback resistor R
causes the capacitor voltage to decay to zero
Charge amplifier: circuit
Rf
Virtual
Ground
Cf
dqs / dt = is = Kdx / dt
IsC
Piezoelectric
sensor
IsC = IsR = 0
IsR
Is
+
V0
FET
1
V0 = −V = −
Cf
∫
t
0
Kdx
KX
dt = −
dt
Cf
Step response of a charge amplifier
xi
Deflection
vo
Time
vo max
vo max /e
Time
The charge amplifier
responds to a step input
with an output that
decays to zero with a
time constant τ = RfCf
High-pass filter: circuit
Derive the equation for gain G
Ci Ri
-
Vi
What are the important parameters?
=−
j ωC i
j ωR f C i
A
V0
+
Zf
V0 ( jω )
G=
=−
Vi ( jω )
Zi
GH = -Rf/Ri ; τi = RiCi ; fc = 1/2πτi
Rf
G=−
Ri + 1
Rf
jωτ i
jωτ i GH
=−
=
1 + jωRi Ci
Ri 1 + jωτ i 1 + jωτ i
Rf
High-pass filter: characteristics
Gain/GH
0.1
1
10
100
f/fc
0.1
Draw G/GH versus f/fc
0.01
-π/2
Phase
-3π/4
-π
Slope = +1
1
f/fc
Draw phase versus f/fc
Differentiator - ideal
Ci
Vi
I
Rf
A
I
Express V0 in terms of Vi
V0
100
+
dVi
i = Ci
dt
Gain
10
fc = 1/2πτ
dVi
V0 = − R f Ci
dt
1
0.1
1
10
100
f/fc
0.1
Slope = +1
0.01
Zf
Rf
V0 ( jω )
=−
=−
G=
1
Vi ( jω )
Zi
j ωC i
Phase shift = - 90o
G = − jωR f Ci = − jωτ
Write expression for gain G
Vi
Differentiator - practical
Cf
Ci
-
Rf
A
V0
+
In an ideal differentiator, the high
frequency response is limited by the
open-loop gain of the op-amp yielding
a very noisy output voltage.
Cf is used to limit the hf gain,
hence the hf noise.
The gain at hf: GH = - Ci/Cf
Rf 1
τf = RfCf ; fc = 1/2πτf
j ωC f
Rf + 1
Zf
j ωC f
jωR f Ci
V0 ( jω )
Ci jωτ f
G=
=−
=−
=−
=−
1
Vi ( jω )
Zi
1 + j ωR f C f
C f 1 + jωτ f
j ωC i
Differentiator characteristics
10
Gain/GH
0.1
1
10
Without Cf
100
f/fc
With Cf
0.1
Differentiator
0.01
-π/2
Phase
-3π/4
-π
High-pass filter
Slope = +1
1
f/fc
With Cf
Measurement of integrator and
differentiator characteristics
15 k
10 nF
1k5
-
Vi
A
+
Differentiator
0.1µF
Vi
V0
Integrator
10 nF
15 k
+
A
V0
• Form your lab team and assign
duties.
• Go to the lab bench.
• Built the circuit given:
integrator for team 1 and
differentiator for team 2
• Connect the power supply.
• Connect the CRO
• Do the experiment according to
the procedures in the sheet - 6.
• Switch off the PS and signal
generator
• Return your seat
Band-pass filter: circuit
Cf
Ci Ri
Mid-band gain =GMB = - Rf/Ri
Vi
Time-constants: τI= RiCi ; τf = RfCf
+
Rf
A
V0
Critical frequencies: fL = 1/2πτI ; fH = 1/2πτf
 Rf



j
C
ω
f 

 1


 jωC  + R f 

Zf
j ωCi R f
f 
V0 ( jω )



=−
=−
G=
=−
(
1 + jωR f C f )(1 + jωRi Ci )
Vi ( jω )
Zi

Ri +  1

 j ωC i 
Rf
jωτ i
GMB jωτ i
G=−
=
Ri (1 + jωτ i )(1 + jωτ f ) (1 + jωτ i )(1 + jωτ f )
Band-pass filter: characteristics
Gain/GH
0.1
1
10
100
f/fc
0.1
Slope = -1
Slope = +1
0.01
Band-pass filter (non-inverting)
Frequency response of bpf
BPF with bias compensation
Comparators: Basic Rules
V1
-
V0
A
V2
+
•
•
•
•
•
V0=A(V2-V1)
V0 = 0 if V1 = V2
V0 = +VSAT if V1 < V2
V0 = -VSAT if V1 > V2
No current flows into either
input terminals of the opamp.
Simple comparator (inverting)
Vi
Vref
V-
+10V
-
V0
V0
-Vref*R1 /(R1 +R2 )
R1
R2
Time
A
+
Vi*R2 /(R1 +R2 )
-10V
V0
10V
V- = (Vi*R2+Vref*R1)/(R1+R2)
VSA
T -10V
10V Vi
Output stays at +VSAT if V-<0
Output goes to -VSAT if V->0
-Vref*R1 /R2
-10V
VSAT
Comparator with hysteresis
V- = (Vi*R2+Vref*R1)/(R1+R2)
VVi
V+ = V0*R3 /(R3+R4)
R1
-
Vref
R2
V0
A
+
V+
R3
R4
Output stays at +VSAT if V-<V+
Output goes to -VSAT if V->V+
Characteristics
VVi
R1
-
Vref
V0
A
R2
VH = -Vref*R1/R2 – VSAT*R3/(R3+R4)
+
V+
VL = -Vref*R1/R2 + VSAT*R3/(R3+R4)
R4
R3
+VSAT
+10V
V0
VH
10V
Vhys
-10V
10V Vi
-10V
-VSAT
-Vref*R1 /(R1 +R2 )
Vhys
VL
Time
VH
VL
V0
Vi*R2 /(R1 +R2 )
-10V
Op-Amps in Reality
• Fabricated using integratedcircuit technologies
– Inside an op-amp:
1) Transistors
2) Parasitic capacitors
3) Internal resistors
• Op-amp chip configuration that we will use:
Quad-channel (i.e. 4-in-1)
– 14 pins in the op-amp chip
Instrumentation Amplifier
Overall Circuit Structure
• Overall differential gain
is
given
by:
 2 R  R 
GD = Gα Gβ = 1 + 2  4 
R1  R3 

Gα = 1 +
R2
2 R2
R1
VS+
va
R4
VS–
R2
VS+
vb
VS–
R4
R3
VS+
R3
R1
Input
Conditioner
Gβ =
vo
R3
R4
VS–
Difference
Amplifier
In-Amp: Input Conditioner
• Note that iR1 = iR2a = iR2b. Then the diff. output
voltage is given by:
R2
vb '−va ' = iR1 (R1 + 2 R2 )
iR2a
• The diff. input voltage is equal to:v
a
vb − va = iR1 R1
va'
iR1
R1
VS–
R2
• The differential gain is thus equal to:
vb '−va '
2 R2
GD =
= 1+
vb − va
R1
VS+
iR2b
VS+
vb'
vb
VS–
In-Amp: Input Conditioner
• This input conditioner does not amplify
common-mode signals!
R
• Brief proof of principle:
2
iR2a
– Since vb – va = iR1R1:
va = vb = vcm
⇒ iR1 = 0
va
va'
iR1
R1
VS–
R2
– The output voltages are thus
equal to:
va ' = vb ' = vcm
VS+
iR2b
VS+
vb'
vb
VS–
In-Amp: Difference Amplifier
1) From voltage divider principles:
 R2 

v+ = vb 
 R1 + R2 
2) Note that iR1a = iR2a. Thus:
vo − v− v− − va
=
R2
R1
vo  1
va
1
⇒
=  + v− −
R2  R2 R1 
R1
iR2a
iR1a
va
vb
R2
VS+
R1
vo
R1
R2
VS–
In-Amp: Difference Amplifier
3) Note that v– = v+. So from the voltage divider
i
R
expression:
R2a
 R2 

v− = vb 
 R1 + R2 
iR1a
va
vb
2
VS+
R1
vo
R1
4) Substituting the above into the
output voltage expression, we get:
VS–
R2
vo  R2 + R1  R2 
va
vb −

= 
R2  R2 R1  R2 + R1 
R1
vo
R2
R2
⇒ vo =
(vb − va ) ⇒ Gd =
=
R1
vb − va R1