by Earl Boysen
B. Write the formula for the impedance of the circuit at the resonance frequency. _____
C. What is the AC voltage gain at this frequency? _____
Answers
A. The resonance frequency
B.
where r is the DC resistance of the coil
C.
13 Because of the low DC resistance of the coil, the DC voltage at the collector is usually close to the supply voltage (VS). In addition, the AC output voltage positive peaks can exceed the DC level of the supply voltage. With large AC output, the positive peaks can actually reach 2VS, as shown in Figure 9.10.
Figure 9.10
Question
Indicate which of the following is an accurate description of the circuit in Figure 9.10:
A. Oscillator
B. Tuned amplifier
C. Common base circuit
D. Common emitter circuit
Answer
B. Tuned amplifier
Feedback
14 To convert an amplifier into an oscillator, you must connect a portion of the output signal to the input. This feedback signal must be in phase with the input signal to induce oscillations.
Figure 9.11 shows three methods you can use to provide a feedback signal from a parallel LC circuit. Each is named for its inventor.
Figure 9.11
In the Colpitts method, the feedback signal is taken from a connection between two capacitors that form a voltage divider. In the Hartley method, the feedback signal is taken from a tap partway down the coil, or from a connection between two inductors. Therefore, an inductive voltage divider determines the feedback voltage. The Armstrong method uses a step-down transformer (an inductor with an extra coil with fewer turns than the main coil). In all three of these methods, between one-tenth and one-half of the output must be used as feedback.
Questions
A. Where is the feedback taken from in a Colpitts oscillator? _____
B. What type of oscillator uses a tap on the coil for the feedback voltage? _____
C. What type does not use a voltage divider? _____
Answers
A. A capacitive voltage divider
B. Hartley
C. Armstrong
15 The output voltage appears at one end of the parallel LC circuit shown in Figure 9.12, and the other end is effectively at ground. The feedback voltage Vf is taken between the junction of the two capacitors.
Figure 9.12
Question
Using the voltage divider formula, what is Vf? _____
Answer
which becomes
16 To find the resonance frequency in this circuit, first find the equivalent total capacitance CT of the two series capacitors. You then use CT in the resonance frequency formula.
Questions
A. What is the formula for CT? _____
B. What is the resonance frequency formula for the Colpitts oscillator?_____
Answers
A.
B.
if Q is equal to or greater than 10
Note If Q is less than 10, you can use one of the following two formulas to calculate the resonance frequency for a parallel LC circuit:
17 Figure 9.13 shows a parallel LC circuit in which the feedback voltage is taken from a tap N1 turns from one end of a coil, and N2 turns from the other end.
Figure 9.13
You can calculate the feedback voltage with a voltage divider formula that uses the number of turns in each part of the coil.
The manufacturer should specify N1 and N2.
Questions
A. Who invented this feedback method? _____
B. When you divide Vf by Vout, what is the result?
Answers
A. Hartley
B. Between one-tenth and one-half
18 Figure 9.14 shows a parallel LC circuit in which the feedback voltage is taken from the secondary coil of a transformer. The formula used to calculate the output voltage of a secondary coil is covered in problem 6 in Chapter 10, “The Transformer.”
Figure 9.14
Question
Who invented this type of oscillator? _____
Answer
Armstrong
19 For each of the feedback methods described in the last few problems, the voltage fed back from the output to the input is a fraction of the total output voltage ranging between one-tenth and one-half of Vout.
To ensure oscillations, the product of the feedback voltage and the amplifier voltage gain must be greater than 1.
It is usually easy to achieve this because Av is much greater than 1.
No external input is applied to the oscillator. Its input is the small part of the output signal that is fed back. If this feedback is of the correct phase and amplitude, the oscillations start spontaneously and continue as long as power is supplied to the circuit.
The transistor amplifier amplifies the feedback signal to sustain the oscillations and converts the DC power from the battery or power supply into the AC power of the oscillations.
Questions
A. What makes an amplifier into an oscillator? _____
B. What input does an amplifier need to become an oscillator?_____
Answer
A. A resonant LC circuit with feedback of the correct phase and amount.
B. None. Oscillations happen spontaneously if the feedback is correct.
Inside the Inductor
When you use inductors, you should know how to deal with the different ways manufacturers label them.
Inductors are simply a coil of wire wrapped around a core, and some manufacturers leave them just like that. These inductors come with no markings,so you must keep them with the label from the packaging to identify them.
You can also find inductors that have a plastic coating around the wire coil. That coating is often marked with a numerical code that identifies the value of the inductor. The first two numbers are the first and second significant digits of the inductance value; the third number is the multiplier. (The units are μH, so an inductor marked with 101 has a value of 100 μH.)
Another method to mark inductors involves the same color code used for resistors. With this method,an inductor is marked with four color bands to show its value and tolerance. Some inductors have a wide silver band (about twice the width of the other bands) at the front of the color code bands. This wider band indicates that the component was built to a U.S. military specification and is not used to determine the inductance value.
The value of each color used in the bands is shown in the following table (with units in μH):
The first two colored bands are the first and second significant digits of the inductance value. The third band is the multiplier and the fourth band represent the tolerance. For example, if an inductor is marked with blue, gray, red, and silver bands, its nominal inductance value is 6800 μH (6.8 mH) with a tolerance of 610 percent .
Finally, you see inductors that have the value simply printed on them. This generally occurs on higher-value inductors, which are also physically larger.
Your best bet is to save the label on the packaging that the inductor comes in until you can check out the markings on the component.
The Colpitts Oscillator
20 Figure 9.15 shows a Colpitts oscillator circuit, the simplest of the LC oscillators to build.
Figure 9.15
The feedback signal is taken from the capacitive voltage divider and fed to the emitter. This connection provides a feedback signal to the emitter in the phase required to provide positive feedback.
In this circuit, the reactance of capacitor CB is low enough for the AC signal to pass through it, rather than passing through R2. Capacitor CB should have a reactance, XCB, of less than 160 ohms at the oscillation frequency. If R2 happens to be smaller than 1.6 kΩ, choose a value of XCB that is less than one-tenth of R2.
Question
For the circuit shown in Figure 9.15, what is your first estimate for CB? Assume that fr is equal to 1
kHz, and that Xc equals 160 ohms._____
Answer
Therefore, CB = 1 μF; larger values of CB also work.
21 Use the Colpitts oscillator component values shown in Figure 9.15 to answer the following questions.
Questions
A. What is the effective total capacitance of the two series capacitors in the tuned circuit?
CT = _____
B. What is the oscillator frequency?
fr = _____
C. What is the impedance of the tuned circuit at this frequency?
Z = _____
D. What fraction of the output voltage is fed back?
Vf = _____
E. What is the reactance of CB at the frequency of oscillation?
XCB = _____
Answers
A. CT = 0.067 μF.
B. Because Q is not known, use the formula that includes the resistance of the coil (see problem 16):
If you use the calculated value of fr to calculate Q, as in problem 20 of Chapter 7, you find that Q = 4.2. Therefore, it is appropriate to use the formula that includes the resistance of the coil to calculate fr.
C. Use the following:
D. Use a voltage divider with the capacitor values:
E. XCB = about 6 ohms, which is a good value (much less than the 8200 ohm value of R2).
22 Figure 9.16 shows a Colpitts oscillator circuit that uses a different method for making feedback connections between the parallel LC circuit and the transistor.
Figure 9.16
Question
List the differences between this circuit and the one shown in Figure 9.15. _____
Answer
The feedback is connected to the base instead of the emitter, and the ground is connected to the center of the capacitive voltage divider. The capacitor CE has been added. (This connection provides a feedback signal to the base in the correct phase to provide positive feedback.)
23 In the circuit shown in Figure 9.16, capacitor CE should have a reactance of less than 160 ohms at the oscillation frequency. If the emitter resistor RE is smaller in value than 1.6 kΩ, then CE should have a reactance that is less than RE/10 at the oscillation frequency.
Question
If you use an emitter resistor of 510 ohms in a 1 kHz oscillator, what value of capacitor should you use for CE?_____
Answers
So, CE = 3.2 μF. Thus, you should use a capacitor larger than 3 μF.
Project 9.1: The Colpitts Oscillator
Objective
The objective of this project is to demonstrate that an oscillator generates a sine wave when feedback is applied to either the emitter or base.
General Instructions
When the Colpitts oscillator circuit with feedback to the emitter is set up, you use your oscilloscope to measure the period of the waveform. Then you change the circuit to provide feedback to the base, and again use your oscilloscope to measure the period of the waveform. This data enables you to calculate the frequency of the sine wave generated in each case.
Parts List
You need the following equipment and supplies:
One 10 k, 0.25-watt resistor.
One 510 , 0.25-watt resistor.
One 82 k, 0.25-watt resistor.
One 8.2 k, 0.25-watt resistor.
Two 1 μF capacitors (This value of capacitor is available in either polarized or unpolarized versions.You should get unpolarized capacitors for this application.)
One 0.1 μF capacitor.
One 0.22 μF capacitor.
One 4.7 μF capacitor. (This value of capacitor is usually polarized, which is fine for this position in the circuit.)
One 0.5 mH inductor. (Suppliers may also refer to this value as 500 μH.)
One 9-volt battery pack.
One breadboard.
One oscilloscope.
One PN2222 transistor. Figure 9.17 shows the pinout diagram for PN2222 transistors.
Figure 9.17
Step-by-Step Instructions
Set up Circuit #1, the Colpitts oscillator circuit with feedback to the emitter, as shown in Figure 9.18. If you have some experience in building circuits, this schematic (along with the previous parts list) should provide all the information you need to build the circuit. If you need a bit more help building the circuit, look at the photos of the completed circuit in the “Expected Results” section.
Figure 9.18
Carefully check your circuit against the diagram.
After you have checked your circuit, follow these steps and record your measurements in the blank table following the steps.
1. Connect the oscilloscope probe for Channel 1 to a jumper wire connected to Vout. Connect the ground clip to a jumper wire attached to the ground bus.
2. Measure and record the period of the sine wave.
3. Disconnect the battery from the circuit, and make the changes required to set up Circuit #2, the Colpitts oscillator circuit with feedback to the base, as shown in Figure 9.19.
Figure 9.19
4. Repeat steps 1 and 2.
Circuit # Period (μsec) Frequency (kHz)
1
2
Expected Results
Figure 9.20 shows the breadboarded Colpitts oscillator with feedback to the emitter (Circuit# 1).
Figure 9.20
Figure 9.21 shows an oscilloscope attached to the circuit.
Figure 9.21
Figure 9.22 shows the sine wave generated by the Colpitts oscillator with feedback to the emitter. You can determine the period of this waveform by counting the number of horizontal divisions the waveform takes to complete one cycle, and then multiplying the number of divisions by the TIME/DIV setting.
Figure 9.22
As you measure the period, you may need to adjust the TIME/DIV, the horizontal POSITION, and the vertical POSITION controls on the oscilloscope. The controls shown in Figure 9.23 are adjusted to measure the period for the Colpitts oscillator.
Figure 9.23
Figure 9.24 shows the breadboarded Colpitts oscillator with feedback to the base (Circuit# 2).
Figure 9.24
Figure 9.25 shows the sine wave generated by the Colpitts oscillator with feedback to the base. You can determine the period of this waveform by counting the number of horizontal divisions the waveform takes to complete one cycle, and then multiplying the number of divisions by the TIME/DIV setting.
The oscilloscope connections and oscilloscope control panel settings for the Colpitts oscillator with feedback to the base are not shown. They are the same as the oscilloscope connections and oscilloscope control panel for the Colpitts oscillator with feedback to the emitter.
Figure 9.25
Your values should be close to those shown in the following table:
Circuit # Period (μsec) Frequency (kHz)
1 34 29.4
2 34 29.4
Notice that the frequency of the sine wave generated by both circuits is the same. This demonstrates that an oscillator can function with feedback to either the emitter or base of the transistor.
24 Figure 9.26 shows a Colpitts oscillator with the parallel LC circuit connected between the collector and the supply voltage. As with the circuits shown in Figures 9.15 and 9.16, this circuit provides a feedback signal to the transistor (in this case, the emitter) in the correct phase to provide positive feedback.
Figure 9.26
The following table shows possible values you might use for C1 and C2 in the circuit shown in Figure 9.26:
Questions
A. Calculate CT and fr for each row of the preceding table._____
B. Does increasing C2, while holding C1 constant, increase or decrease the resonance frequency?_____
C. What effect does increasing C1 have on the resonance frequency?_____
D. What is the condition that results in the highest possible resonance frequency?_____
E. What would be the highest resonance frequency if C1 is fixed at 0.01 μF, and C2 can vary from 0.005 μF to 0.5 μF?_____
>
Answers
A. The following table shows the values of CT and fr:
B. Increasing C2 decreases the resonance frequency, and, therefore, decreases the output frequency of the oscillator.
C. Increasing C1 also decreases the resonance frequency and the output frequency of the oscillator.
D. When CT is at its lowest possible value.
E. When C2 is 0.005 μF, CT will be 0.0033 μF, which is its lowest possible value. Therefore, the frequency is at the highest possible value, or approximately 6.9 kHz. The lowest frequency occurs when C2 is at its highest setting of 0.5 μF.
The Hartley Oscillator
25 Figure 9.27 shows a Hartley oscillator circuit. In this type of circuit, the feedback is taken from a tap on the coil, or from a connection between two inductors.
Figure 9.27
Capacitor CL stops the emitter DC voltage from being pulled down to 0 volts through the coil. CL should have a reactance of less than RE/10, or less than 160 ohms at the oscillator frequency.
Questions
Work through the following calculations:
A. What is the resonance frequency?
fr = _____
B. What is the approximate impedance of the load? Z = _____
C. What missing information prevents you from calculating the fraction of the voltage drop across the coil that is fed back to the emitter?
Answers
A. 80 Hz (approximately).