# Lecture Note 7

```MET 487-Lecture 7
Tuesday, July 13, 2010
7:18 PM
Lab Activities:
1. Low Pass filter modification: change from R= 10kOhms, C = 0.1 uF, cutoff
frequency (3dB frqeuency) f = 1/(2*pi*R*C) = 159 Hz
2. New RC values: R = 1kOhms, C = 0.01 uF so that fc - cutoff (3dB) frequency can be
increased to f = 1/(2*pi*R*C) = 15.9kHz
3. Revision of Lab 5
a. Set up the low pass filter with this new R and C parts
b. Connect function generator (Vi) to the input of the RC circuit
c. Connect digital scope CH1 to Vi, and CH2 to Vout as shown below.
d. Adjust amplitude adj volume of the function generator to output 10v Peakto-peak if possible
e. Increase frequency of Vi (function generator) to see Vout drop to 10v*0.707
≈ 7 v, record all the digital displayed info: fc, vpp, period, and others
f. Prepare to gather data for Table 1
1) 0.01 * fc = 159 Hz: Vin = 10, Vout (measured), A = Vout/Vin
(Computed) --- repeat this for step 2) through step 4)
2) 0.1 * fc = 1.59 kHz
3) 1 * fc = 15.9 kHz (copy the value from step 3e
4) 10 * fc = 159 kHz
g. Prepare the Bode-plot as shown in the lab
Chapter 5. Analog Signal Processing Using Operational Amplifiers
5.1 Introduction
• Figure 5.6 - 741 Internal Design and packaging
 Inverting input - Pin 2
 Non-inverting input - Pin 3
 Output - Pin 5
 V+ -- Pin 7
 V- -- pin 4
• LM 741 IC part - demo
5.2 Amplifiers
• Figure 5.1 Amplifier Model: Av = Vo/Vin, Zin = Vi/Ii, Zo = Vo/Io
5.3 Operational Amplifiers (building blocks)
• Amplifiers: example Vo = 5* Vi; using op-amp to produce a gain of 5
• Integrators:
• Differentiators
• Comparators: monitoring some predefined values
• A/D and D/A converters
• Active Filters
• Sample and Hold Circuits
5.4 Ideal Model for the Op-Amp
• Figure 5.2 Op-Amp terminology and schematic representation
• Figure 5.3 Op-Amp feedback
• Figure 5.4 Op-Amp equivalent circuit
• Assumption:
1. It has infinite impedance
2. It has infinite gain
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2. It has infinite gain
3. It has zero output impedance
5.5 Inverting Amplifier
Figure 5.7 Inverting amplifier
Figure 5.8 Equivalent circuit for an inverting amplifier
Use KCL at Node C: Sum of incoming current = Sum of out going currents
Ii = -Io
Ii = Vi/R, Io = Vo/RF substitute into the equation above
Vi/R = -Vo/RF
Av = Vout/Vin = - RF/Ri
Consider the implementation of the
equation: Vout = 5*Vi
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