Chapter 3Webster Amplifiers and Signal Processing. Applications of Operational Amplifier In Biological Signals and Systems. The three major operations done on biological signals using OpAmp: Amplifications and Attenuations DC offsetting: add or subtract a DC
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Applications of Operational AmplifierIn Biological Signals and Systems
Most bioelectric signals are small and require amplifications
Figure 3.1 Opamp 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 outputvoltage source. Any current flowing to the output terminal vo must pass through the output resistance Ro.
1 A = (gain is infinity)
2 Vo = 0, when v1 = v2 (no offset voltage)
3 Rd = (input impedance is infinity)
4 Ro = 0 (output impedance is zero)
5 Bandwidth = (no frequency response limitations) and no phase shift
Rule 1
When the opamp 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.
10 V
i
10 V
10 V
Rf
i
i
Slope = Rf / Ri
Ri

i
o
10 V
+
(b)
(a)
Inverting AmplifierFigure 3.3 (a) An inverting amplified. Current flowing through the input resistor Ri also flows through the feedback resistor Rf . (b) The inputoutput plot shows a slope of Rf / Ri in the central portion, but the output saturates at about ±13 V.
The output of a biopotential preamplifier that measures the electrooculogram is an undesired dc voltage of ±5 V due to electrode halfcell potentials, with a desired signal of ±1 Vsuperimposed. 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.
+10
Rf
Ri
i
100 kW
10 kW
i

i + b /2
+15V
Rb
o
0
Voltage, V
20 kW
Time
5 kW
+
v
b
15 V
10
o
(a)
(b)
Used as a buffer, to prevent a high source resistance from being loaded down by a lowresistance load. In another word it prevents drawing current from the source.

o
i
+
Differential Gain Gd
R4
R3
v3
R3
v4
vo
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 commonmode voltage.
R4
Commonmode rejection ratio CMMR
Typical values range from 100 to 10,000
Disadvantage of oneopamp differential amplifier is its low input resistance
Differential Mode Gain
Advantages: High input impedance, High CMRR, Variable gain
R1
ui

v2
uo
R1
uref
+
15
R2
uo
10 V
10 V
uref
ui
10 V
Comparator – No Hysteresisv1 > v2, vo = 13 V
v1 < v2, vo = +13 V
If (vi+vref) > 0 then vo = 13 V else vo = +13 V
R1 will prevent overdriving the opamp
R1
With hysteresis
10 V
ui

uo
10 V
R1
10 V
uref
+
 uref
ui
R2
R3
10 V
Comparator – With HysteresisReduces multiple transitions due to mV noise levels by moving the threshold value after each transition.
Width of the Hysteresis = 4VR3
Rf = 1 kW
i
v
o
D

RL = 3 kW
+
(c)
OneOpAmp Full Wave RectifierFor 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.
Ic
Rf
Ri

i
o
+
(a)
Logarithmic Amplifiers
VBE
Figure 3.8 (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 107 to 102 and the range of vo is .36 to 0.66 V.
o
10 V
10 V
10 V
i
1
10 V
10
(b)
Logarithmic Amplifiers
VBE
Rf/9
Ic
VBE
9VBE
Rf
Ri

i
o
+
(a)
Figure 3.8 (a) With the switch thrown in the alternate position, the circuit gain is increased by 10. (b) Inputoutput characteristics show that the logarithmic relation is obtained for only one polarity; 1 and 10 gains are indicated.
Figure 3.9 A threemode 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.
is
C

dqs/ dt = is = K dx/dt
uo
isC
isR
+
FET
Piezoelectric
sensor
Example 3.2
The output of the piezoelectric sensor may be fed directly into the negative input of the integrator as shown below. Analyze the circuit of this charge amplifier and discuss its advantages.
isC = isR = 0
vo = vc
Long cables may be used without changing sensor sensitivity or time constant.
Figure 3.11 A differentiator The dashed lines indicate that a small capacitor must usually be added across the feedback resistor to prevent oscillation.
Active Filters LowPass Filter
Cf
Rf
Ri
Gain = G =

ui
uo
+
(a)
G
Rf/Ri
0.707 Rf/Ri
freq
fc = 1/2RiCf
Active filters(a) A lowpass filter attenuates high frequencies
Active Filters (HighPass Filter)
Rf
Ci
Ri

ui
uo
Gain = G =
+
(b)
G
Rf/Ri
0.707 Rf/Ri
freq
fc = 1/2RiCf
Active filters(b) A highpass filter attenuates low frequencies and blocks dc.
Active Filters (BandPass Filter)
Cf
Ci
Rf
Ri

ui
uo
+
G
(c)
Rf/Ri
0.707 Rf/Ri
freq
fcL = 1/2RiCi
fcH = 1/2RfCf
Active filters(c) A bandpass filter attenuates both low and high frequencies.
Frequency Response of opamp and Amplifier
OpenLoop Gain
Compensation
ClosedLoop Gain
Loop Gain
Gain Bandwidth Product
Slew Rate
Offset Voltage and Bias Current
Read section 3.12
Nulling, Drift, Noise
Read section and 3.13
Differential bias current, Drift, Noise
Ro
uo
Rd
ud
ii
+
io
Aud

ui
RL
+
CL
Input and Output Resistance
Typical value of Ro = 40
Typical value of Rd = 2 to 20 M
Used in many medical instruments for signal detection, averaging, and Noise rejection
If vc is positive then D1 and D2 are forwardbiased and vA = vB. So vo = vDB
If vc is negative then D3 and D4 are forwardbiased and vA = vc. So vo = vDC
vc 2vi