Lecture 2-3.

Basic Concepts of Medical

Electronic System
3rd year
MDE Dept.
Al-Qalam
LECTURE two-three
Dr.Wassan Adnan Hashim
 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.
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 Inside the Op-Amp (IC-chip)

20 transistors
11 resistors
1 capacitor
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 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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 i
Inverting Amplifier o

Rf 10 V

i
Ri
i - -10 V 10 V
o i
+
Slope = -Rf / Ri
(a)
-10 V

Rf vo Rf (b)

vo  - vi G -
Ri vi Ri

(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
R1 Rf
1
-
R2 o
2 +

 v1 v2 
vo  - R f   
 R1 R2 

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 Example 2.1
• 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 2.1
• 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 - 104 (-10)
  0  Rb    2 104 W
Ri Rb vi 5
Ri Rf +10
10 kW 100 kW i
i

- Voltage, V i + b /2
+15V Rb o 0
20 kW Time
5 kW +
vb

-15 V

-10 o
(a) (b) 10
 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.
-

o
i +

vo  vi G 1

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 Noninverting Amplifier
o
i i 10 V
Ri Rf Slope = (Rf + Ri )/ Ri

-10 V 10 V

i
-

o
i
-10 V
+

R f  Ri R f  Ri  Rf 
vo  vi G  1  
Ri Ri  Ri 

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 Differential Amplifiers
• Differential Gain Gd v3
vo R4
Gd  
v4 - v3 R3 v4

• Common Mode Gain Gc

– For ideal op amp if the inputs are equal
then the output = 0, and the Gc = 0. R4
– No differential amplifier perfectly rejects vo  (v4 - v3 )
the common-mode voltage. R3
• Common-mode rejection ratio CMMR
Gd
– Typical values range from 100 to 10,000 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
+15
– No Hysteresis
v2
v1 > v2, vo = -13 V
v1 < v2, vo = +13 V
-15
o
10 V
R1
i
- -10 V
o ref
ref
R1 i
+
R2
-10 V

If (vi+vref) > 0 then vo = -13 V else vo = +13 V

R1 will prevent overdriving the op-amp
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 Comparator – With Hysteresis
• Reduces multiple transitions due to mV noise levels
by moving the threshold value after each transition.
o
R1
i 10 V With hysteresis
-

R1 o -10 V 10 V
ref + - ref
R2 i

R3
-10 V

Width of the Hysteresis = 4VR3

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 R
Rectifier
D1 D2
xR (1-x)R o
10 V
-
-10 V 10 V
i +
R i
D4 vi
D3
vo  -10 V
-
x
(b)

+
(a)
xR (1-x)R
vo
• Full-wave precision rectifier: - D2

– For i > 0, D2 and D3 conduct, whereas D1 i +

and D4 are reverse-biased. (a)

Noninverting amplifier at the top is active

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 R
D1 D2
Rectifier o
xR (1-x)R
10 V
-
-10 V 10 V
i +
R
i

D4 vi
vo  -10 V
D3 x
- (b)

+
(a)
xRi R
i vo
• Full-wave precision rectifier: - D4

– For i < 0, +

D1 and D4 conduct, whereas D2 and D3 are (b)

reverse-biased.
Inverting amplifier at the bottom is active

BME 311 LECTURE NOTE 2 - ALİ IŞIN, 2014 18

 One-Op-Amp Full Wave Rectifier
Ri = 2 kW Rf = 1 kW
i
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
 IC 
Ic
Rf /9
VBE  0.06 log 
 IS 

 vi 
Rf

i
Ri
- vo  0.06 log 
-13 
+
o
 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 Rf /9 vo
Ic
10 V

VBE Rf -10 V 10 V
9VBE
Ri
i - 1 i
o
+
-10 V 10
(b)
(a)

(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
t1
1
vo  -
Ri C f  v dt  v
0
i ic

Vo ( j ) Zf
-
Vi ( j ) Zi
Vo  j  - Rf

Vi  j  Ri  jR f Ri C
Vo  j  -1 vo - R f
 
Vi  j  Ri  jR C vi Ri for f < fc
i
Rf 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
Vo ( j ) Zf
-
Vi ( j ) Zi
Vo ( j )
 - jRC
Vi ( j )
BME 311 LECURE NOTE 2 - ALİ IŞIN, 2014 24
 Active Filters- Low-Pass Filter
• A low-pass filter attenuates high frequencies
Vo  j  - R f 1
Gain  G   Rf
Vi  j 
Ri
Ri 1  jR f C f i -
o
+
|G| (a)

Rf/Ri
0.707 Rf/Ri

freq
fc = 1/2RiCf
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 Active Filters (High-Pass Filter)
• A high-pass filter attenuates low frequencies
and blocks dc.
Vo  j  - R f jRi Ci
C R R f
 - i i

Gain  G  
i
o
Vi  j  Ri 1  jRi Ci +

(b)
|G|

Rf/Ri
0.707 Rf/Ri

fc = 1/2RiCf freq
BME 311 LECTURE NOTE 2 - ALİ IŞIN, 2014 26
 Active Filters (Band-Pass Filter)
• A bandpass filter attenuates both low and
high frequencies. C f

Vo  j  - jR f Ci

Vi  j  1  jR f C f 1  jRi Ci  Ci R
i
-
Rf
i
o
+
|G|
(c)

Rf/Ri
0.707 Rf/Ri

fcL = 1/2RiCi fcH = 1/2RfCf freq

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Phase Modulator for Linear variable differential
transformer LVDT

+
-

+
-

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Phase Modulator for Linear variable differential
transformer LVDT

+
-

+
-

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Phase-Sensitive Demodulator

Used in many medical instruments

for signal detection, averaging, and
Noise rejection

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 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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