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Inleiding Meten en Modellen – 8C120. Prof.dr.ir. Bart ter Haar Romeny Dr. Andrea Fuster Faculteit Biomedische Technologie Biomedische Beeld Analyse www.bmia.bmt.tue.nl. Chapter 3 - Webster Operational Amplifiers and Signal Processing.
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Inleiding Metenen Modellen – 8C120 Prof.dr.ir. Bart ter Haar Romeny Dr. Andrea Fuster Faculteit Biomedische Technologie Biomedische Beeld Analyse www.bmia.bmt.tue.nl
Chapter 3 - WebsterOperational Amplifiers and Signal Processing
Applications of Operational AmplifierIn Biological Signals and Systems • The three major operations done on biological signals using Op-Amps: • Amplifications and Attenuations • DC offsetting: add or subtract a DC • Filtering: Shape signal’s frequency content
Ideal Op-Amp Most bioelectric signals are small and require amplifications Figure 3.1 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.
Inside the Op-Amp (IC-chip) 20 transistors 11 resistors 1 capacitor
Ideal Characteristics 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
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.
o 10 V i -10 V 10 V Rf i i Slope = -Rf / Ri Ri - i o -10 V + (b) (a) Inverting Amplifier Figure 3.3 (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.
Summing Amplifier Rf R1 1 - R2 o 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. +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)
Follower (buffer) Used as a buffer, to prevent a high source resistance from being loaded down by a low-resistance load. In other words: it prevents drawing current from the source. - o i +
Noninverting Amplifier o i i 10 V Rf Ri Slope = (Rf + Ri )/ Ri -10 V 10 V i - o -10 V i +
Differential Amplifiers R4 R3 v3 Differential Gain Gd R3 v4 vo R4 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
InstrumentationAmplifiers Differential Mode Gain Advantages: High input impedance, High CMRR, Variable gain
+15 R1 ui - v2 uo R1 uref + -15 R2 uo 10 V -10 V uref ui -10 V Comparator – No Hysteresis v1 > 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 op-amp
uo R1 With hysteresis 10 V ui - uo -10 V R1 10 V uref + - uref ui R2 R3 -10 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
o 10 V R D1 D2 -10 V 10 V (1-x)R xR i - -10 V i + R (b) i D4 o= x D3 - + (a) Rectifier xR (1-x)R v o D2 - + i (a) Full-wave precision rectifier:a) For i > 0, D2 and D3 conduct, whereas D1 and D4 are reverse-biased.Noninverting amplifier at the top is active
o 10 V R D1 D2 -10 V 10 V (1-x)R xR i - -10 V i + R (b) i D4 o= x D3 - + (a) Rectifier xRi R i v o D4 - + (a) Full-wave precision rectifier:b) For i < 0, D1 and D4 conduct, whereas D2 and D3 are reverse-biased. Inverting amplifier at the bottom is active
Ri = 2 kW Rf = 1 kW i v o D - RL = 3 kW + (c) Rectifier 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.
Rf/9 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 to 10-7 to 10-2 the range of vo is -.36 to -0.66 V.
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) Input-output characteristics show that the logarithmic relation is obtained for only one polarity; 1 and 10 gains are indicated.
Logarithmic Amplifiers • Uses of Log Amplifier: • Multiply variable • Divide variable • Raise variable to a power • Compress large dynamic range into small ones • Linearize the output of devices
Integrators for f < fc A large resistor Rf is used to prevent saturation
Integrators Figure 3.9 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.
R is C - dqs/ dt = is = K dx/dt uo isC isR + FET Piezo-electric 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.
Differentiators 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- Low-Pass Filter Cf |G| Gain = G = Rf Ri - ui uo Rf/Ri + (a) 0.707 Rf/Ri MMA freq fc = 1/2RiCf Active filters(a) A low-pass filter attenuates high frequencies
Active Filters (High-Pass Filter) Rf Ci Ri - ui uo + Gain = G = (b) |G| Rf/Ri 0.707 Rf/Ri MMA freq fc = 1/2RiCf Active filters(b) A high-pass filter attenuates low frequencies and blocks dc.
Cf Active Filters (Band-Pass Filter) Ci Rf Ri - ui uo + (c) |G| Rf/Ri 0.707 Rf/Ri MMA freq fcL = 1/2RiCi fcH = 1/2RfCf Active filters(c) A bandpass filter attenuates both low and high frequencies.
Frequency Response of op-amp and Amplifier Open-Loop Gain Compensation Closed-Loop Gain Loop Gain Gain Bandwidth Product Slew Rate Offset voltage Bias current 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
Chapter 3 - WebsterOperational Amplifiers and Signal Processing Wikipedia:OperationalAmplifier