Course Number
ELN - 131
Course Title
Analog circuits 1
REPORT TITLE Phase Shift Oscillator
Section
2
NUMBER Final Project
Semester Spring Year 2014 Date Performed
May 1
Instructor
Dennis Leak
Student Name
Ramon Martinez
Grade________
Date Submitted
May 13
Lab Instructor Dennis Leak
English and Spelling:____/25
Format:____/25
Content:____/50
Number of Misspelled words________
May 13, 2014 ELN131-02 Ramon Martinez Lab Report Final
Phase Shift Oscillator
1.- Purpose: To investigate the properties of a Bipolar Junction Transistor (BJT) Phase Shift Oscillator. A second BJT will be used to provide power amplification. To provide 12 volt power to the circuit a wall plug transformer and a three terminal voltage regulator will be used. 2.- Parts List: Parts
Designator
Quantity
Value
Cost each $
Resistor
R1
1
33kΩ
0.09
Resistor
R2, R3, R4
3
6.2kΩ
0.09
Resistor
R5, RC
2
3kΩ
0.09
Resistor
RE, R6
2
600Ω
0.09
Resistor
R7
1
22kΩ
0.09
Transistor
Q1, Q2
2
2N2222A
0.06
Capacitor
C1
1
0.1µF
0.09
Capacitor
C2
1
1µF
0.09
Capacitor
C3
2
220µF
0.09
Capacitor
C4
1
100µF
0.09
Capacitor
C5, C6
2
0.01µF
0.09
Capacitor
C7
1
47µF
0.09
DC Plug
J1
1
Jack, DC Power
0.45
LED
LED1
1
Green LED
0.15
Voltage Regulator
U2
1
LM7812-CT
0.45
Total Cost
2.10
Table 1: Parts List
All prices provided by Jameco, Inc.
3.- Equipment List: Equipment
Manufacturer
Model
Description
Hp 34401A Multimeter
Hewlett Packard
34401A
Digital Multimeter (DMM)
Breadboard
Jameco
WBU-208-R
7.3 x 7.5 3220 PNTS Breadboard
Oscilloscope
Sun Equipment Corp.
OS-221005
100 Mhz Oscilloscope
DC Power Supply
BK Precision
1673
Triple Output Power Supply
Table 2: Equipment List 4.- Procedure: 4.1. Study the attached schematic. • Capacitor values are given and are included in the kit provided. • Assume that the BJTs have a beta of 100. 4.2. Design the network with a Q1 Ic of 2 mA, and a VCE of 5 volts. Show all equations in your final report. Choose 5% values and show them in your final schematic and parts list. 4.3. Measure and record the voltages of Q1’s base, emitter and collector. Compare them to the calculated values. (Note: You may want to adjust R1 to get a better sine wave). 4.4. Calculate the output impedance of the network of Q2. Without R7, measure the output impedance of Q2 and compare it to the calculated value. Connect the speaker to the output. 4.5. Measure and record the voltages of Q2’s base, emitter and collector. Compare them to the calculated values. 4.6. Chose frequency selective resistors R3 and R4 to give (as close as possible) an output frequency of 1 kHz. (see research section). 5.- Research: 5.1. Do an online study of Phase Shift Oscillators. 5.2. What does the gain have to be, before oscillation is achieved? 5.3. What does the phase shift have to be, to obtain oscillation? 5.4. Can you find a formula to predict the frequency of oscillation? Does the frequency of your circuit change when you put your hand near some of the components? Why?
4.- Procedure: Given Schematic
4.2. Design the network with a Q1 Ic of 2 mA, and a VCE of 5 volts. Show all equations in your final report. IC = 2mA, VCE = 5V, VCC = 12V a.- Calculate RE: RE = VE/IE
assuming
VE =12V/10= 1.2V
then
VE =1/10 VCC RE =1.2V/2mA = 600立
b.- Use KVL to determine value of RC: VCC - ICRC - VCE - VE = 0 then RC = VCC - VCE - VE IC so 12V - 5V - 1.2V 5.8V = = RC = 2mA 2mA 2.9k立
c.- Use “Rule of Thumb” to determine R2: BRE≥10R2
100 x 600Ω then R2 = BRE = = 600Ω 10 10 d.- Determine VB: VE= VB - VBE, then VB = VE + VBE, so VB = 1.2V + 0.7V = 1.9V e.- Determine R1: VB = R2 x VCC then R1 = R2 x VCC - R2 R1 + R2 VB so R1 = 6kΩ x 12V - 6kΩ = 31.894kΩ 1.9V f.- Determine VC: VC= VCC - VCE - VE, then VC = 12V - 5V - 1.2V = 5.8V
4.3. Measure and record the voltages of Q1’s base, emitter and collector. Compare them to the calculated values. (Note: You may want to adjust R1 to get a better sine wave). Calculated Values:
Measured Values:
VB = 1.9V
VB = 1.85V
VC = 5.8V
VC = 6.15V
VCE = 5V
VE = 1.72V
VE = 1.9V RE = 600Ω RC = 2.9kΩ R1 = 31.894kΩ R2 = 600Ω
4.4. Calculate the output impedance of the network of Q2. Without R7, measure the output impedance of Q2 and compare it to the calculated value. Connect the speaker to the output. Q2 is part of an emitter follower circuit so: Zo = re re = 26mV IE IE = IC re = 26mV = 13Ω 2µA Measured Impedance: 9.5Ω 4.5. Measure and record the voltages of Q2’s base, emitter and collector. Compare them to the calculated values. Calculated Values:
Measured Values:
VB = 5.8V
VB = 5.95V
VC = 12V
VC = 11.49V
VE = 5.1V
VE = 5.65V
4.6. Choose frequency selective resistor R3 and R4 to give (as close as possible) an output frequency of 1kHz. From the formula: ƒ= 1 2πRC√6
so:
R can be obtained: R= 1 2πƒC√6
Same values for R3 and R4
R=
1 = 6.497kΩ 2π x 0.01 µF x 1kHz x √6
CIRCUIT ON THE PROTOBOARD
SINEWAVE PRODUCED BY THE OSCILLATOR CIRCUIT
6.- Conclusion A phase-shift oscillator is a linear electronic oscillator circuit that produces a sine wave output. It consists of an amplifying element such as a transistor with its output fed back to its input through a phase-shift network consisting of resistors and capacitors. The feedback network ‘shifts’ the phase of the amplifier output by 180 degrees at the oscillation frequency to give positive feedback. Phase-shift oscillators are often used at audio frequency as audio oscillators. The most common phase-shift network cascades three identical resistor-capacitor stages that produce a phase shift of 60° per stage. Phase shift oscillators have distortion, but they achieve low-distortion output voltages because cascaded RC sections act as distortion filters. In this experiment Q1 is configured as a divider amplifier, and its output (collector) signal is fed back to its input (base) through a three-stage RC network, which includes R2 and C3, R3 and C1, and R4 and C2. Each of the three RC stages in this ladder introduces a 60° phase shift between its input and output terminals so the sum of those three phase shifts provides the overall 180° required for oscillation. The phase shift per stage depends on both the frequency of the input signal and the values of the resistors and capacitors in the network. The values of the three RC ladder network capacitors C1, C2, and C3 are equal (0.01µF) as are the values of the the three resistors R2, R3, and R4 (6.5k)(6.2k resistors were used since 6.5k are not available on the market and they fall into the 5% tolerance required for the experiment). With this component values, the 180° phase shift occurs at about 1.05 Hz. Oscillators use a fraction of the output signal created by the feedback network as the input signal. It is the noise voltage that provides the inital boost to the circuit to acheive oscillation. The gain has to be 1, in this case the signal reverses to the opposite direction creating the sine wave. If the gain exceeds 1 the circuit becomes stable and oscillation stops. Q2 is part of an emitter follower circuit which is used to match the high output impedance produced by the voltage divider’s Q1 and the high input impedance at Q2, this results in an almost non existant voltage gain but a high power gain with a very low output impedance. In the experiment It was observed that a minor change in frequency took place when a hand was apprached to some of the components, this effect is due to what its called “stray capacitance” , which occurs when two conducting elemts are close to each other, one being the circuit and the other the hand. The behavioir is the same as a capacitor itself where the two conducting plates are separated by a semi conductor, in this case air.
There were differences between the measured and calculated values for Q1’s and Q2’s base, emitter and collector voltages in this experiment. Even though there were not significant, this has been the case in almost all experiment involving BJT’s throughout the semester, not only in this one. It is due largely to the manufacturing process since not all transistors can be matched exactly to one another while keeping the price low enough to make them affordable.
Q1 Calculated Values:
Measured Values:
VB = 1.9V
VB = 1.85V
VC = 5.8V
VC = 6.15V
VCE = 5V
VE = 1.72V
VE = 1.9V RE = 600Ω RC = 2.9kΩ R1 = 31.894kΩ R2 = 600Ω
Q2 Calculated Values:
Measured Values:
VB = 5.8V
VB = 5.95V
VC = 12V
VC = 11.49V
VE = 5.1V
VE = 5.65V
When building the voltage regulator section of the circuit, the calculations required for a 1,020kΩ resistor to be used but it had to be replaced by a 1kΩ, since the 1,020k is not available and also the 1k falls within the 5% tolerance range.