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Biomedical Instrumentation
Prof. Dr. Nizamettin AYDIN
[email protected]
[email protected]
http://www.yildiz.edu.tr/~naydin
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Amplifiers and Signal Processing
Applications of Operational Amplifier
In Biological Signals and Systems
• The three major operations done on biological
signals using Op-Amp:
– Amplifications and Attenuations
– DC offsetting:
• add or subtract a DC
– Filtering:
• Shape signal’s frequency content
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Ideal Op-Amp
• Most bioelectric signals are small and require amplifications
Op-amp equivalent circuit:
The two inputs are 1 and  2. A differential voltage between them causes
current flow through the differential resistance Rd. The differential voltage
is multiplied by A, the gain of the op amp, to generate the output-voltage
source. Any current flowing to the output terminal vo must pass through the
output resistance Ro.
4
Inside the Op-Amp (IC-chip)
20 transistors
11 resistors
1 capacitor
5
Ideal Characteristics
•
•
•
•
•
A =  (gain is infinity)
Vo = 0, when v1 = v2 (no offset voltage)
Rd =  (input impedance is infinity)
Ro = 0 (output impedance is zero)
Bandwidth =  (no frequency response limitations) and no
phase shift
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Two Basic Rules
• Rule 1
– When the op-amp output is in its linear range, the two input terminals
are at the same voltage.
• Rule 2
– No current flows into or out of either input terminal of the op amp.
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Inverting Amplifier
o
i
Rf
10 V
i
i
Ri
-
-10 V
10 V
o
i
+
Slope = -Rf / Ri
(a)
-10 V
vo  -
Rf
Ri
vi
Rf
vo
G vi
Ri
(b)
(a) An inverting amplified. Current flowing through the input
resistor Ri also flows through the feedback resistor Rf .
(b) The input-output plot shows a slope of -Rf / Ri in the central
portion, but the output saturates at about ±13 V.
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Summing Amplifier
Rf
R1
1
-
o
R2
2
+
 v1 v2 
vo  - R f   
 R1 R2 
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Example 3.1
• The output of a biopotential preamplifier that
measures the electro-oculogram is an undesired
dc voltage of ±5 V due to electrode half-cell
potentials, with a desired signal of ±1 V
superimposed. Design a circuit that will
balance the dc voltage to zero and provide a
gain of -10 for the desired signal without
saturating the op amp.
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Answer 3.1
• We assume that vb, the balancing voltage at vi=5 V. For vo=0,
the current through Rf is zero. Therefore the sum of the currents
through Ri and Rb, is zero.
vo vb
- Ri vb - 10 4 (-10)

 0  Rb 

 2 10 4 W
Ri Rb
vi
5
+10
Rf
100 kW
Ri
10 kW
i
+15V
Rb
20 kW
5 kW
o
+
Voltage, V
i
i + b /2
0
Time
vb
-15 V
-10
(a)
o
(b)
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Follower ( buffer)
• Used as a buffer, to prevent a high source resistance
from being loaded down by a low-resistance load. In
another word it prevents drawing current from the
source.
-
i
vo  vi
o
+
G 1
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Noninverting Amplifier
o
i
Ri
i
Rf
10 V
Slope = (Rf + Ri )/ Ri
-10 V
10 V
i
-
i
o
+
vo 
R f  Ri
Ri
vi
-10 V
G
R f  Ri
Ri
 Rf
 1 
Ri




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Differential Amplifiers
• Differential Gain Gd
vo
R4
Gd 

v4 - v3 R3
v3
v4
• Common Mode Gain Gc
– For ideal op amp if the inputs are equal
then the output = 0, and the Gc = 0.
– No differential amplifier perfectly rejects
the common-mode voltage.
• Common-mode rejection ratio CMMR
– Typical values range from 100 to 10,000
R4
vo 
(v4 - v3 )
R3
Gd
CMRR 
Gc
• Disadvantage of one-op-amp differential amplifier is its low
input resistance
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Instrumentation Amplifiers
Differential Mode Gain
v3 - v4  i( R2  R1  R2 )
v1 - v2  iR1
v3 - v4 2 R2  R1
Gd 

v1 - v2
R1
Advantages: High input impedance, High CMRR, Variable gain
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Comparator – No Hysteresis
+15
v1 > v2, vo = -13 V
v1 < v2, vo = +13 V
v2
-15
o
i
ref
10 V
R1
-
-10 V
o
R1
ref
+
R2
-10 V
If (vi+vref) > 0 then vo = -13 V
else
R1 will prevent overdriving the op-amp
vo = +13 V
i
Comparator – With Hysteresis
• Reduces multiple transitions due to mV noise levels
by moving the threshold value after each transition.
o
i
ref
R1
With hysteresis
10 V
-
o
R1
-10 V
10 V
- ref
+
R2
R3
i
-10 V
Width of the Hysteresis = 4VR3
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Rectifier
R
xR
(1-x)R
D1
D2
o
10 V
-10 V
i
+
i
R
D4
-
10 V
D3
vo 
vi
x
-10 V
(b)
+
(a)
xR
• Full-wave precision rectifier:
– For i > 0, D2 and D3 conduct, whereas
D1 and D4 are reverse-biased.
Noninverting amplifier at the top is active
(1-x)R
-
i
vo
D2
+
(a)
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Rectifier
R
xR
(1-x)R
D1
D2
o
10 V
-10 V
i
+
i
R
D4
-
10 V
D3
vo 
vi
x
-10 V
(b)
+
(a)
xRi
• Full-wave precision rectifier:
– For i < 0,
D1 and D4 conduct, whereas D2 and D3 are
reverse-biased.
Inverting amplifier at the bottom is active
R
i
-
vo
D4
+
(b)
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One-Op-Amp Full Wave Rectifier
i
Ri = 2 kW
Rf = 1 kW
v
-
o
D
RL = 3 kW
+
(c)
• For i < 0, the circuit behaves like the inverting
amplifier rectifier with a gain of +0.5. For i > 0, the
op amp disconnects and the passive resistor chain
yields a gain of +0.5.
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Logarithmic Amplifiers
• Uses of Log Amplifier
–
–
–
–
Multiply and divide variables
Raise variable to a power
Compress large dynamic range into small ones
Linearize the output of devices
Rf /9
Ic
VBE
Rf
i
Ri
-
o
+
 IC
 0.06 log 
 IS



 vi


vo  0.06 log 
-13 
 Ri 10 
(a)
(a) A logarithmic amplifier makes use of the fact that a transistor's VBE is
related to the logarithm of its collector current.
For range of Ic equal 10-7 to 10-2 and the range of vo is -.36 to -0.66 V.
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Logarithmic Amplifiers
VBE
Ic
Rf /9
vo
10 V
VBE
i
Ri
9VBE
Rf
-10 V
10 V
-
1
i
o
+
(a)
(b)
-10 V
10
(a) With the switch thrown in the alternate position, the
circuit gain is increased by 10. (b) Input-output
characteristics show that the logarithmic relation is
obtained for only one polarity; 1 and 10 gains are
indicated.
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Integrators
1
vo  Ri C f
t1
 v dt  v
i
ic
0
Zf
Vo ( j )
Vi ( j )
Zi
- Rf
Vo  j 

Vi  j  Ri  jR f Ri C
Vo  j 
-1

Vi  j  Ri  jR C
i
Rf
vo - R f

vi
Ri
for f < fc
1
fc 
2R f C f
A large resistor Rf is used to prevent saturation
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• A three-mode integrator
With S1 open and S2 closed, the dc circuit behaves as an inverting amplifier.
Thus o = ic and o can be set to any desired initial conduction. With S1
closed and S2 open, the circuit integrates. With both switches open, the
circuit holds o constant, making possible a leisurely readout.
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Differentiators
• A differentiator
– The dashed lines indicate that a small capacitor must
usually be added across the feedback resistor to prevent
oscillation.
dvi
vo  - RC
dt
Zf
Vo ( j )
Vi ( j )
Zi
Vo ( j )
 - jRC
Vi ( j )
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Active Filters- Low-Pass Filter
• A low-pass filter attenuates high frequencies
Vo  j  - R f
1
Gain  G 

Vi  j 
Ri 1  jR f C f
|G|
i
Ri
-
Rf
o
+
(a)
Rf/Ri
0.707 Rf/Ri
fc = 1/2RiCf
freq
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Active Filters (High-Pass Filter)
• A high-pass filter attenuates low frequencies
and blocks dc.
R
C R
Vo  j  - R f jRi Ci

Gain  G 

Vi  j 
Ri 1  jRi Ci
+
i
i
f
i
o
(b)
|G|
Rf/Ri
0.707 Rf/Ri
fc = 1/2RiCf
freq
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Active Filters (Band-Pass Filter)
• A bandpass filter attenuates both low and high
frequencies.
C
- jR f Ci
Vo  j 

Vi  j  1  jR f C f 1  jRi Ci 
f
i
Ci R
i
-
Rf
o
+
|G|
(c)
Rf/Ri
0.707 Rf/Ri
fcL = 1/2RiCi
fcH = 1/2RfCf
freq
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Frequency Response of op-amp and Amplifier
•
•
•
•
•
•
Open-Loop Gain
Compensation
Closed-Loop Gain
Loop Gain
Gain Bandwidth Product
Slew Rate
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Input and Output Resistance
Rd
ii
d
+
i
+
o
Ro
-
Ad
io
RL
CL
vi
Rai 
 ( A  1) Rd
ii
vo
Ro
Rao 

io A  1
Typical value of Rd = 2 to 20 MW
Typical value of Ro = 40 W
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Phase Modulator for Linear variable differential
transformer LVDT
+
+
-
31
Phase Modulator for Linear variable differential
transformer LVDT
+
+
-
32
Phase-Sensitive Demodulator
Used in many medical
instruments for signal detection,
averaging, and Noise rejection
33
The Ring Demodulator
• If vc is positive then D1 and D2 are forward-biased and vA = vB. So vo = vDB
• If vc is negative then D3 and D4 are forward-biased and vA = vc. So vo = vDC
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