Download EXPERIMENT 3. SINGLE TUNED AMPLIFIERS Introduction

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EXPERIMENT 3. SINGLE TUNED AMPLIFIERS
Introduction
In this laboratory, characteristics of single tuned amplifiers will be examined. In this type of
amplifiers, one parallel tuned circuit is used as a load. They have several limitations such as
small bandwidth, small gain bandwidth product, and do not provide flatten response. On the
other hand, their alignment is more straightforward than double tuned amplifiers.
Theory
Many communications system requirements exceed the high-frequency limits of op-amps. In
cases such as these, tuned amplifiers are often used. The basic principle underlying the design
of tuned amplifiers is the use of a parallel RLC (tank) circuit as the load, or at the input of a BJT
or FET amplifier. Single tuned amplifier investigated in this experiment is a small signal voltage
amplifier. The basic principle of single tuned amplifiers is illustrated using a BJT with tunedcircuit loads in Figure 1.
Figure 1: A Basic BJT single tuned amplifier
A single tuned amplifier is usually consists of an amplifier and a tuning stages. In the first stage,
the input is amplified. The amplified signal may contain different frequency components. The
tuning stage is transparent to the components which has frequencies falling in the passband
of the circuit. For example, if the tuning circuit is tuned to ω0, only the frequency component
at ω0 is reflected to the output while the others are rejected.
Consider an ideal tuning circuit shown in Figure 2.
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Figure 2: Ideal tuning stage
If the input is
𝑥(𝑡) = 7 cos(2𝜔0 𝑡) + 5𝑐𝑜𝑠(𝜔0 𝑡) + 13𝑐𝑜𝑠(7𝜔0 𝑡)
Then the output is 𝑦(𝑡) = 5𝐴 cos(𝜔0 𝑡), where A is a constant depending on the component
values of the tuning stage.
In a single tuned amplifier, tuning stage is a simple narrowband resonant circuit, such as the
one given in Figure 3.
Figure 3: Parallel RLC (Tank) circuit
Let Vo(s) and I(s) denote the Laplace Transforms of Vo(t) and i(t), respectively. The transfer
function H(s) can be written as
𝟏
𝒔
𝑽𝒐 (𝒔)
𝟏
𝑪
𝑯(𝒔) =
=
=
𝟏 𝟏
𝟏
𝟏
𝑰(𝒔)
+
+ 𝒔𝑪 𝒔𝟐 +
𝒔+
𝑹 𝒔𝑳
𝑹𝑪
𝑳𝑪
𝟏
𝒔
𝑪
=
(𝒔 − 𝒔𝟏 )(𝒔 − 𝒔𝟐 )
where
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𝒔𝟏,𝟐 =
−𝟏
𝟐𝑹𝑪
∓ √(
𝟏
𝟐
𝟏
) −
𝟐𝑹𝑪
𝑳𝑪
s1 and s2 are the poles of the transfer function. Zeros of the transfer function are at the
origin and at infinity. Following conventional notations are used for H(s):
𝛼=
𝜔0 =
1
2𝑅𝐶
1
√𝐿𝐶
𝛽 = √𝜔02 − 𝛼 2
𝑠1,2 = −𝛼 ∓ √𝛼 2 − 𝜔02
Depending on the values of 𝛼 and 𝜔0 the two poles may be either real or a complex conjugate
pair. If 𝜔0 > 𝛼 or 𝑅 > 𝜔0 𝐿/2 then the poles will be complex conjugate and can be written
as:
𝑠1,2 = −𝛼 ∓ 𝑗√𝜔02 − 𝛼 2
The distance of the poles to the origin is 𝜔0 . If 𝛼 is increased (R is decreased) the poles move
into the left half plane along the semicircular trajectory of radius w0 until they meet on the
real axis for 𝛼 = 𝜔0 .
Figure 4: The position of the poles in complex plane
Sinusoidal steady state response for the tank circuit is obtained by inserting s=jw in H(s):
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−1
𝑗𝜔
𝑗𝜔
𝐶
𝐻(𝑗𝜔) =
2 =
2
2
(𝑗𝜔) + 𝑗𝜔 + 𝜔0 𝐶(𝜔0 − 𝜔 2 ) + 𝑗2𝛼𝜔
|𝐻(𝑗𝜔)| =
𝜔
𝐶√4𝛼 2 𝜔 2 + (𝜔02 − 𝜔 2 )
∠𝐻(𝑗𝜔) =
𝜋
2𝛼𝜔
− 𝑡𝑎𝑛−1 2
2
𝜔0 − 𝜔 2
𝜔0 is the frequency at which the imaginary part of H(jw) is zero and hence it is real. At
frequency 𝜔0 , H(jw)=R as seen in Figure 5.
Figure 5: Amplitude of the impedance of capacitor and inductor, and the amplitude of the transfer
function
Figure 6: Phase of the transfer function
The frequencies 𝜔1 and 𝜔2 are the Half Power or 3 dB cut-off frequencies. The real power
dissipated by the tank circuit at 𝜔1 and at 𝜔2 is the half of that for 𝜔 = 𝜔0 . Solving Eq. 1 for
𝜔1 and 𝜔2
𝜔1 − 𝜔2 = 2𝛼
𝜔1 𝜔2 = 𝜔02
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The tank circuit is frequency selective and this behavior can be seen in Figure 5. An ideal tank
circuit passes only the frequency components in the interval [𝜔1, 𝜔2 ]. The sharper the
characteristics, the better the attenuation of unwanted frequency components. The quality
factor (Q) determines how good a tnak circuit is.
𝑄=
𝑐𝑒𝑛𝑡𝑒𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
𝜔
=
3 𝑑𝐵 𝑏𝑎𝑛𝑑𝑤𝑖𝑑𝑡ℎ
2𝛼
In practice, inductors and capacitors used in tank circuits have some losses. These losses
are represented as a series resistance for inductors and a parallel resistance for the capacitors.
Capacitor leakage may be neglected but the series loss resistance of inductors has an
equivalent parallel resistance around the resonant frequency 𝜔0 and it should be taken into
account, as given in Figure 7.
Figure 7: LC circuit with loss resistance
The unloaded Q of the inductor is defined as:
𝑄=
𝜔0 𝜔0 𝐿
𝐿
=
=
2𝛼
𝑟
𝜔0 𝑟𝐶
If QL >10 then around 𝜔0 , the circuit in Figure 7a can be replaced by the equivalent model in
Figure 7b.
𝑅𝑝 = 𝑄𝐿2 𝑟 =
1
𝐿
=
(𝜔0 𝑟𝐶)2 𝑟𝐶
Overall quality factor of the tank circuit when loaded with a parallel resistor RL is:
𝑄𝑟 =
𝜔0
= 𝜔0 𝑅𝑒𝑞 𝐶
2𝛼
where 𝑅𝑒𝑞 = 𝑅𝑝 //𝑅𝐿
Consider the following single tuned amplifier configuration given in Figure 8. We want to
obtain AC and DC load line equations of this configuration. To obtain DC load line equation
short circuit the inductor and open circuit the capacitors. Then 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼𝐶 (𝑅𝐸 + 𝑟)
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Figure 8: Tuned amplifier
To find AC load line, find the equivalent parallel resistance RP of the inductor, short circuit
CS, CL, CE and assume L and C are in resonance. Put the small signal model of the
transistor into the amplifier configuration (Figure 9).
Figure 9: Small signal model
AC load line:
𝐼𝐶 = (𝐼𝐶𝑄 +
𝑉𝐶𝐸𝑄
𝑉𝐶𝐸
)−
𝑅𝑎𝑐
𝑅𝑎𝑐
where 𝑅𝑎𝑐 = 𝑅𝑃 //𝑅𝐿
Obviously, slope of DC load line is -1/(RE+r) while that of AC load line is -1/Rac. In general,
Rac>>RE+r, which means AC load line intersects the VCE axis at a voltage larger than VCC
(Figure 10).
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Figure 10: DC and AC load lines
Preliminary work
1. Each of the students is tested before starting experiment by research assistant. This
test may be written or oral examination. Theoretical knowledge of the experiment
that should be gotten from experimental sheet and other sources (lecture notes, books
etc.). The theoretical information about experiment is not limited to study only
experimental sheet, students have to research other sources to get enough
knowledge.
2. Students should know the purpose of the experiment. They should know how the
experiment can be done and which measuring elements can be used. They should
also get measuring elements catalog information.
3. In the following, there are two video recordings about the use of oscilloscope with
Agilent Vee.
https://drive.google.com/open?id=0B24exSh3SWAqNm9XS2NnSE92bFU
https://drive.google.com/open?id=0B24exSh3SWAqQm9wMng5eFdIZVE
Each student have to watch these videos (If necessary several times) and take some
necessary notes for working properly with agilent vee oscilloscope.
PLEASE NOTE THAT: During the lab, there WON’T be an introductory section about
how to work with agilent vee and oscilloscope.
4. The response y(t) of a linear time invariant system with a transfer function h(t) to a
sinusoidal input x(t) is given as
𝑥(𝑡) = 𝐴𝑐𝑜𝑠(𝜔0 𝑡)
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𝑦(𝑡) = 𝐴|𝐻(𝑗𝜔0 )|𝑐𝑜𝑠(𝜔0 𝑡 + ∠𝐻(𝑗𝜔0 ))
Find the response of the same system to the following inputs
a. 𝑥(𝑡) = 𝐴𝑐𝑜𝑠(𝜔1 𝑡) + 𝐵𝑐𝑜𝑠(𝜔2 𝑡)
b. 𝑥(𝑡) = ∑∞
𝑛=0 𝑎𝑛 𝑐𝑜𝑠(𝑛𝜔0 𝑡)
c. Discuss what would be observed at the output if the above system were a tank
circuit with center frequency 𝜔0 and high Q, and the input is a square wave of
frequency 𝜔0
5. Find the Norton equivalent of the voltage source and 100 kΩ resistor and find the
equivalent parallel resistance of the circuit given below (Figure 11). Simulate the circuit
given in Figure 11, plot the magnitude-frequency response and phase-frequency
response of the tank circuit on the scale for requencies from f 0-5 kHz up to f0+5 kHz in
100 Hz steps by assuming that the impedance of the loaded tank circuit is much smaller
than the parallel resistor of the Norton current source. Find f 0, Q, and the bandwidth.
Comment on the results.
Figure 11: Tuned circuit
6. For the circuit given in Figure 12
a. Calculate ICQ, VCQ, and VCEQ. Draw the DC load line.
b. Draw the small signal model. Draw the AC load line.
c. Calculate 𝜔0 , f0, QT, and bandwidth.
d. Find the gain of the circuit at the resonant frequency, and calculate V o(t) for
Vin(t)=0.001cos(𝜔0 t).
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Figure 12: Tuned amplifier
Experiment
1. Construct the circuit of Figure 11. Set the input to 200 mV peak-to-peak.
a. Find the center frequency f0
b. Find and plot magnitude- and phase-frequency responses in steps of 100 Hz
beginning from f0-5 kHz up to f0+5 kHz by using VEE program you designed in
the preliminary work.
2. Construct the circuit given in Figure 12 with a BD 135 transistor. Measure ICQ, VCQ, and
VCEQ. Apply a sine wave of 200 mV peak-to-peak and then change the input level to
have a maximum output peak-to-peak swing at the center frequency.
a. Calculate the magnitude-frequency response.
b. Calculate half-power frequencies and the values of the magnitude-frequency
response at these frequencies.
c. Plot the magnitude- and phase-frequency responses using the VEE program
you designed. To plot ∠H(jw) you should use the oscillioscope. Both of the
signals should be seen clearly at the screen of the scope (there should be no
clipping). The program will measure the phase difference between the two
signals. Compare the values calculated in a and b with the values you observed
from the plot.
3. Using the same circuit as in the previous step,
a. Apply a square wave at resonance frequency to the input. You should observe
a signal very close to a sine wave. Why? Adjust the input level such that the
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output has a maximum peak to peak swing. Observe the emitter and collector
voltages.
a. Decrease the square wave at resonance frequency towards f0/2 and see how
the output voltage waveform changes. Adjust the input frequency around f 0/2
to obtain the maximum undistorted swing at the output. Observe the emitter
and collector voltages. Comment on what you have observed.
SINGLE TUNED AMPLIFIERS
NAMES:
1.
2.
SECTION:
1.
f0=
Comment:
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2.
ICQ=
VCQ=
a) |H(jw0)|=
b) |H(jw1)|=
ang H(jw1)
c)
Comment:
|H(jw2)|=
ang H(jw2)
VCEQ=
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3.
a)
b)
Comment:
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