Download - Biomedical Instrumentation
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Prof. Dr. Nizamettin AYDIN
[email protected]@ieee.org
http://www.yildiz.edu.tr/~naydin
Biomedical Instrumentation
Amplifiers and Signal Processing
Applications of Operational AmplifierIn 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 amplificationsOp-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 transistors11 resistors1 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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Inverting Amplifier
(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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Rii
o
i
Rf
i
+
(a)
10 V
10 V
(b)
i
o
Slope = -Rf / Ri
-10 V
-10 V
i
f
i
oi
i
fo R
RvvGv
RR
v
Summing Amplifier
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2
2
1
1
Rv
RvRv fo
1
o
+
R2
R1Rf
2
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.
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i
vb
i
o
o
+
+15V
+10
0Time
i + b /2
-10
(a) (b)
5 k
-15 V
Rb
20 k
Ri
10 kRf
100 k
Vol
tage
, V
44
1025
)10(100i
bib
b
b
i
o
vvRR
Rv
Rv
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.
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oi +
1 Gvv io
Noninverting Amplifier
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o
10 V
10 V
i
Slope = (Rf + Ri )/ Ri
-10 V
-10 V
Rf
oi
i
+
-
iRi
i
f
i
ifi
i
ifo R
RR
RRGv
RRR
v 1
Differential Amplifiers
• Differential Gain Gd
• 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
• Disadvantage of one-op-amp differential amplifier is its low input resistance
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3
4
34 RR
vvvG o
d
v3
v4
)( 343
4 vvRRvo
c
d
GGCMRR
Instrumentation Amplifiers
Differential Mode Gain
Advantages: High input impedance, High CMRR, Variable gain
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1
12
21
43
121
21243
2
)(
RRR
vvvvG
iRvvRRRivv
d
Comparator – No Hysteresis
o
iref
10 V
-10 V
-10 V
v2
+15
-15
i
o
+
R1
R1
R2
ref
If (vi+vref) > 0 then vo = -13 V else vo = +13 VR1 will prevent overdriving the op-amp
v1 > v2, vo = -13 Vv1 < v2, vo = +13 V
Comparator – With Hysteresis
• Reduces multiple transitions due to mV noise levels by moving the threshold value after each transition.
Width of the Hysteresis = 4VR3
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i
o
+
R1
R1
R2
R3
ref
o
i- ref
10 V
-10 V
With hysteresis
-10 V 10 V
Rectifier
• 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
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10 V
(b)
-10 V
o
i
-10 V 10 V
(a)
D2
vo
i
+
xR (1-x)R
+
(a)
D3
R
R
i
+
D2D1
D4
xR (1-x)R
xvv i
o
Rectifier
• 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
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10 V
(b)
-10 V
o
i
-10 V 10 V
+
(a)
D3
R
R
i
+
D2D1
D4
xR (1-x)R
xvv i
o
(b)
D4
voi
+
xRi R
One-Op-Amp Full Wave Rectifier
• 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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(c)
D
vo
i
+
Ri = 2 k Rf = 1 k
RL = 3 k
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
(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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(a)
Rf
Ic
Rf /9
o
Rii
+
1310log06.0
i
io R
vv
S
CBE I
IV log06.0
Logarithmic Amplifiers
(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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(b)
10 V
-10 V
vo
i
-10 V
1
10
10 V
(a)
Rf
Ic
Rf /9
o
Rii
+
VBE
VBE
9VBE
Integrators
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ffc
i
f
i
o
CRf
RR
vv
21
for f < fc
CRj
RRjV
jV
CRRjRR
jVjV
ZZ
jVjV
vdtvCR
v
if
ii
o
ifi
f
i
o
i
f
i
o
t
icifi
o
1
)()(
1 1
0
A large resistor Rf is used to prevent saturation
• 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.
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RCjjVjV
ZZ
jVjV
dtdvRCv
i
o
i
f
i
o
io
)()()()(
Active Filters- Low-Pass Filter
• A low-pass filter attenuates high frequencies
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ffi
f
i
o
CRjRR
jVjV
1
1 G Gain
|G|
freq fc = 1/2RiCf
Rf/Ri
0.707 Rf/Ri
+
Ri
Rf
(a)
i
o
Active Filters (High-Pass Filter)
• A high-pass filter attenuates low frequencies and blocks dc.
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Ci
+
Rii
o
(b)
Rf ii
ii
i
f
i
o
CRjCRj
RR
jVjV
1
G Gain
|G|
freq fc = 1/2RiCf
Rf/Ri
0.707 Rf/Ri
Active Filters (Band-Pass Filter)
• A bandpass filter attenuates both low and high frequencies.
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iiff
if
i
o
CRjCRjCRj
jVjV
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|G|
freq fcL = 1/2RiCi
Rf/Ri0.707 Rf/Ri
fcH = 1/2RfCf
+
i o
(c)
RfCi Ri
Cf
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
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di
iai RA
ivR )1(
+
Rdii
Ro
RL CL
ioAd
d
o
i
+
1
AR
ivR o
o
oao
Typical value of Rd = 2 to 20 M Typical value of Ro = 40
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
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Used in many medical instruments for signal detection, averaging, and Noise rejection
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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