69647_op amp basics
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OP-AMPS in SIGNAL PROCESSING
APPLICATIONS
Filters, Integrators, Differentiators,
and Instrumentation Amplifier
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Sensors :
definition and principles
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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
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What is electronics engineering all
about?
Process
Sensors/ transducers
Filters
Amplifiers Feedback
Actuators/transducers
New process
A/D converters Computers
D/A converters
Physicalvariables
ElectricalvariablesV, I, t, f
Physicalvariables
F, P, d, v, t, f
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What is a Signal Processor?
It selects useful partsof a signal
It cleans signal fromimpurities
It compares signals
It converts signals into
forms that can be
easily recognized
The ECG wave in its raw form
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Examples of
biomedical
signals
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Examples of Signal Processors
Basic amplifiers already discussed
Instrumentation amplifier an improveddifferential amplifier
Active filters
Integrators and differentiators
Extra Reading:
Precision rectifiers
Logarithmic amplifiers
Negative capacitance amplifiers
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Signal Conditioning
Ideal operational amplifier
Inverting, non-inverting and instrumentamplifier
Integrator, differentiator
Filters
Examples of signal conditioning module
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Ideal Operational Amplifier
Operational (OP) amplifier is a high-gain dc differentialamplifier. It is made of an integrated circuit chip. It has twoinputs, negative terminal v1 and positive terminal v2 and oneoutput v0. The effective input to the amplifier is the voltagedifference of the two inputs v1-v2.
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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.
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Basic Rules for Ideal OP Amplifier
When the OP amplifier output is in its linearrange, the two input terminals are at the same
voltage.
No current flows into either input terminal ofthe OP amp.
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Inverting Amplifier
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Non-inverting Amplifier
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Instrument Amplifier
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Active Filters
Low-pass
High-pass
Band-pass Band-stop (notch)
All-pass (phase-shift)
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Low-pass filter: circuit
i
f
Z
ZG =
V0ViA
-Ri Rf
Cf
GL = Rf/Ri ; f= RfCf; fc = 1/2f
Derive the equation for gain G
What are the important parameters?
)(
)(0
jV
jVG
i
=
i
ff
f
f
R
RCj
CjR
+
=
1
( ) fL
fi
f
iff
f
j
G
jR
R
RCRj
R
+=
+=
+=
11
1
1
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Low-pass filter: characteristics
Gain/GL
f/fc
0.1 1 10 100
0.1
0.01 Slope = -1
1
-3/2
Phase f/fc
-5/4
-
GL = Rf/Ri ;
f= RfCf; fc = 1/2
f
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V0Vi A
-Ri Cf
Integrator ideal: circuit
i
f
i
f
i RCj
ZZ
jVjVG
1
)()(0 ===
Virtual
Ground
i
i
+=
t
iCifi
VdtVCR
V00
1
dt
dVC
R
Vi f
i
i 0==
Express V0 in terms of Vi Write expression for gain G
ffi jCRj 11 ==
Where = RiCf
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Integrator ideal:
characteristics
G/GL
f/fc
0.1 1 10 100
0.1
0.01
0
10
100
Slope = -1
1
-3/2
Phase f/fc
-5/4
-
Draw G/GL versus f/fc
Draw phase versus f/fc
f
fG cc ====
1
== 2702/3
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The integrator with bias currents
Through which component the bias current flows?
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Gain/GL
f/fc
0.1 1 10 100
0.1
0.01 Slope = -1
Integrator - practical
V0Vi A
-Ri
Rf
Cf
f
L
fi
f
i j
G
jR
R
jV
jVG
+=
+==
11
1
)(
)(0
Without Rf
With Rf
Rfprovides route for DC bias currents
1Phase f/fc
-3/2
-5/4
-
Without Rf
Low-pass filter Integ.
fcfff fCR 2
1
; ==
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Charge amplifier
vo
Electrode
The piezoelectric sensor generates charge, which is transferred tothe capacitor,C, by the charge amplifier. Feedback resistorR
causes the capacitor voltage to decay to zero
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Charge amplifier: circuit
V0IsC
FET
-
Rf
Cf
IsRdtKdxidtdq ss // ==
Piezoelectricsensor
Is
VirtualGround
IsC = IsR = 0 ===t
ff C
KXdt
dt
Kdx
CVV
00
1
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Step response of a charge amplifierxi
Deflection
vo
vo max
vo max /e
Time
Time
The charge amplifierresponds to a step inputwith an output thatdecays to zero with atime constant=RfCf
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High-pass filter: circuit
V0ViA
-Ri RfCi
i
f
i Z
Z
jV
jVG ==
)(
)(0
GH = -Rf/Ri ; i = RiCi ; fc = 1/2i
Derive the equation for gain G
What are the important parameters?
i
Hi
i
i
i
f
j
Gj
j
j
R
R
+=
+=
11ii
if
ii
f
CRj
CRj
CjR
RG
+
=+
=11
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High-pass filter: characteristics
Gain/GH
f/fc
0.1 1 10 100
0.1
0.01Slope = +1
1
-
Phase f/fc
-3/4
-/2
Draw G/GH
versus f/fc
Draw phase versus f/fc
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Differentiator - ideal
V0ViA
-RfCi
II
dt
dVCRV
dt
dVCi
iif
ii
=
=
0
jCRjG
Cj
R
Z
Z
jV
jVG
if
i
f
i
f
i
==
===1)(
)(0
Slope = +1
fc = 1/2
Phase shift = - 90o
Gain
f/fc
0.1 1 10 100
0.1
0.01
1
10
100
Express V0 in terms of Vi
Write expression for gain G
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Differentiator - practical
V0ViA
-RfCi
Cf
f
f
f
i
ff
if
i
ff
ff
i
f
i j
j
C
C
CRj
CRj
Cj
CjR
CjR
Z
Z
jV
jVG
+
=
+
=
+
===
111
1
1
)(
)(0
In an ideal differentiator, the high
frequency response is limited by theopen-loop gain of the op-amp yieldinga very noisy output voltage.
f= RfCf; fc = 1/2f
Cfis used to limit the hf gain,
hence the hf noise.
The gain at hf: GH = - Ci/Cf
ff
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Differentiator characteristics
Slope = +1
1
-
Phase f/fc
-3/4
-/2
Gain/GH
f/fc
0.1 1 10 100
0.1
0.01
10 Without Cf
With Cf
With Cf
DifferentiatorHigh-pass filter
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Measurement of integrator and
differentiator characteristics
V0V i A-
1k5
15 k
10 nF
Integrator
Form your lab team and assignduties.
Go to the lab bench.
Built the circuit given:
integrator for team 1 anddifferentiator for team 2
Connect the power supply.
Connect the CRO
Do the experiment according tothe procedures in the sheet - 6.
Switch off the PS and signalgenerator
Return your seat
V0ViA
-15 k0.1F
10 nFDifferentiator
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Band-pass filter: circuit
V0ViA-
Ri Rf
Cf
Ci
( )( )
)1)(1()1)(1(
111
1
)(
)(0
fi
iMB
fi
i
i
f
iiff
fi
ii
ff
f
f
i
f
i
jj
jG
jj
j
R
R
G
CRjCRj
RCj
CjR
RCj
CjR
Z
Z
jV
jVG
++=++=
++=
+
+
===
Mid-band gain =GMB = - Rf/Ri
Time-constants: I= RiCi ; f= RfCf
Critical frequencies: fL = 1/2I ; fH = 1/2f
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Band-pass filter: characteristics
Gain/GH
f/fc
0.1 1 10 100
0.1
0.01
Slope = -1Slope = +1
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Band-pass filter (non-inverting)
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Frequency response of bpf
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BPF with bias compensation
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Comparators: Basic Rules
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 op-amp.
V0
V1
V2A
+
-
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Simple comparator (inverting)
V0Vref A
+
-
R1
R2
Vi
V-
Output stays at +VSAT if V-0
+10V
-10V
V0
Time
-Vref*R1/(R1+R2)
Vi*R2/(R1+R2)
V0
Vi
10V
10V-10V
-10V-Vref*R1/R2
-
VSAT
VSAT
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Comparator with hysteresis
V0Vref A
+
-
R1
R2
Vi
V-
R3R4
V
V- = (Vi*R2+Vref*R1)/(R1+R2)
V+= V0*R3 /(R3+R4)
Output stays at +VSAT if V-V+
V
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CharacteristicsV0
Vref A
+
-
R1
R2
Vi
V-
R3R4
V+
V0
Vi
10V
10V-10V
-10VVL
+VSAT
-VSAT
VH
Vhys
VH = -Vref*R1/R2 VSAT*R3/(R3+R4)
VL = -Vref*R1/R2 + VSAT*R3/(R3+R4)
+10V
-10V
V0
Time
-Vref*R1/(R1+R2)
Vi*R2/(R1+R2)
VH
VLVhys
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Op-Amps in Reality
Fabricated using integrated-circuit 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
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Instrumentation Amplifier
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Overall Circuit Structure
Overall differential gain is given by:
R4
R3
vo
R3
R4
R2
va
R2
R1
vb
VS+
VS+
VS+
1
221R
RG +=
3
4
R
RG =
+==
3
4
1
221R
R
R
RGGGD
DifferenceAmplifier
InputConditioner
VS
VS
VS
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In-Amp: Input Conditioner
Note that iR1 = iR2a = iR2b. Then the diff. outputvoltage is given by:
The diff. input voltage is equal to:
The differential gain is thus equal to:
R2
va
R2
R1
vb
VS+
VS+
va'
vb'
iR1
iR2a( )211 2'' RRivv Rab +=
iR2b
1
221''
R
R
vv
vvG
ab
abD +=
=
11Rivv Rab =VS
VS
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In-Amp: Input Conditioner
This input conditioner does not amplifycommon-mode signals!
Brief proof of principle:
Since vb v
a= iR1R1:
The output voltages are thusequal to:
01 === Rcmba ivvv
cmba vvv == ''
R2
va
R2
R1
vb
VS+
VS+
va'
vb'
iR1
iR2a
iR2b
VS
VS
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In-Amp: Difference Amplifier
1) From voltage divider principles:
2) Note that iR1a = iR2a. Thus:
R2
R1
vbvo
va
R1
R2
VS+
+=+
21
2
RR
Rvv b iR1a
iR2a
12 R
vv
R
vv ao =
1122
11
R
vv
RRR
v ao
+=
VS
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In-Amp: Difference Amplifier
3) Note that v
= v+. So from the voltage divider
expression:
4) Substituting the above into the
output voltage expression, we get:
R2
R1
vbvo
va
R1
R2
VS+iR1a
iR2a
22 )(Rv
GvvR
v o ===
+= 212
RR
R
vv b
112
2
12
12
2 R
vvRR
R
RR
RR
R
v ab
o
+
+=
VS
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