Download DISP-2003: Introduction to Digital Signal Processing

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Dr. Hugh Blanton
ENTC 4307/ENTC 5307
RADIO FREQUENCY OSCILLATORS
• In the most general sense, an oscillator
is a non –linear circuit that converts DC
power to an AC waveform.
• Most oscillators used in wireless systems
provide sinusoidal outputs, thereby
minimizing undesired harmonics and
noise sidebands.
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• The basic conceptual operation of a
sinusoidal oscillator can be described
with the linear feedback circuit.
Vi (w)
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Vo(w)
3
• An amplifier with voltage gain A has an
output voltage Vo.
• This voltage passes through a feedback
network with a frequency dependent
transfer function H(w) and is added to the
input Vi of the circuit.
• Thus the output voltage can be expressed as
Vo (w )  AVi (w )  H (w ) AVo (w )
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Vo (w )  AVi (w )  H (w ) AVo (w )
Vo (w )  H (w ) AVo (w )  AVi (w )
Vo (w )1  H (w ) A)  AVi (w )
A
Vo (w ) 
Vi (w )
1  H (w ) A
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• If the denominator of the previous
equation becomes zero at a particular
frequency, it is possible to achieve a
nonzero output voltage for a zero input
voltage, thus forming an oscillator.
• This is known as the Barkhausen
criterion.
• In contrast to the design of an amplifier, where
we design to achieve maximum stability,
oscillator design depends on an unstable
circuit.
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General Analysis
• There are a large number of possible RF
oscillator circuits using bipolar or fieldeffect transistors in either common
emitter/source, base/gate, or
collector/drain configurations.
• Various types of feedback networks lead to the
well-known oscillator circuits:
• Hartley,
• Colpitts,
• Clapp, and
• Pierce
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• All of these variations can be
represented by a general oscillator
circuit.
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• The equivalent circuit on the right-hand
side of the figure is used to model either a
bipolar or a field-effect transistor.
• We can simplify the analysis by assuming real
input and output admittances of the transistor,
defined as Gi and Go, respectively, with a
transistor transconductance gm.
• The feedback network on the left side of the
circuit is formed from three admittances in a
bridged-T configuration.
• These components are usually reactive elements
(capacitors or inductors) in order to provide a
frequency selective transfer function with high Q.
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• A common emitter/source configuration
can be obtained by setting V2 = 0, while
common base/gate or common
collector/drain configurations can be
modeled by setting either V1 = 0 or V4 =
0, respectively.
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• The feedback path is achieved by
connecting node V3 to node V4.
• Writing Kirchoff’s current law for the four
voltage nodes of the circuit gives the
following matrix equation:
 Y1  Gi )
 Y3
0  V1 
 (Y1  Y3  Gi )
  (Y  G  g ) (Y  Y  G  G  g )
 V 

Y

G
1
i
m
1
2
i
o
m
2
o  2


 Y3
 Y2
(Y2  Y3 )
0  V3 

 
g

(
G

g
)
0
G
m
o
m
o  V4 

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• Recall from circuit analysis that if the ith
node of the circuit is grounded, so that
V = 0, the matrix will be modified by
eliminating the ith row and column,
reducing the order of the matrix by
one.
• Additionally, if two nodes are connected
together, the matrix is modified by adding
the corresponding rows and columns.
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Oscillators Using a Common Emitter BJT
• Consider an oscillator using a bipolar
junction transistor in a common emitter
configuration.
• V2 = 0, with feedback provided from the
collector, so that V3 = V4.
• In addition, the output admittance of the
transistor is negligible, so we set Go = 0.
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• These conditions serve to reduce the
matrix to the following:
 Y3 
(Y1  Y3  Gi )
 g  Y )

(
Y

Y
)
m
3
2
3 

V1 
V   0
 
• where V = V3 = V4
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• If the circuit is to operate as an oscillator,
then the new determinant must be satisfied
for nonzero values of V1 and V, so the
determinant of the matrix must be zero.
• If the feedback network consists only of lossless
capacitors and inductors, then Y1,Y2, and Y3 must
be imaginary, so we let Y1 = jB1, Y2 = jB2. and
Y3 = jB3.
• Also, recall that the transconductance, gm , and
transistor input conductance are Gi, are real.
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• Then the determinant simplifies to
 j B1  B3 )  Gi
 g  jB
m
3

 jB3 
0

j ( B2  B3 )
 B1  B3 )( B2  B3 )  B 32  jGi ( B2  B3 )  jgm B3  0
1
1
1


 0  X1  X 2  X 3  0
B3 B2 B1
gm
1
1
1
 0  X1 
X2
B3
B2
Gi
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• Since gm and Gi are positive, X1 and X2
must have the same sign, and
therefore are either both capacitors or
both inductors.
• Since X1 and X2 have the same sign, X3
must be opposite in sign from X1 and X2,
and therefore the opposite type of
component.
• This conclusion leads to two of the most
commonly used oscillator circuits.
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Colpitts Oscillator
• If X1 and X2 are capacitors and X3 is an
inductor, we have a Colpitts oscillator.
1
X1  
w oC1
1
X2  
w o C2
X 3  w o L3

1 1
1 
    wo L3  0  wo 
wo  C1 C2 
g
C2
 m
C1
Gi
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1  C1  C2 


L3  C1C2 
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Hartley Oscillator
• If we choose X1 and X2 to be inductors, and
X3 to be a capacitor, we have a Hartley
oscillator.
X 1  w o L1
X 2  w o L2
1
X3  
w o C3
1
1
wo L1  L2 ) 
 0  wo 
woC3
C3 L1  L2 )
g
L1
 m
L2
Gi
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Lab 5
• Implement the following Colpitts
oscillator using PSpice.
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• Determine the frequency of the tank circuit—
which sets the oscillation frequency.
• When we display the output waveform (from
0 to 10 ms), there is no signal!
• The problem is one of insufficient spark.
• One solution is to pre-charge one of the tank
capacitors.
• Using either CT1 or CT2, initialize either capacitor
with a small voltage (such as .1 v).
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• Again, display the output waveform
from 0 to 10 ms.
• This time the signal exists—but clearly, it
has not reached steady-state conditions
by 10 ms.
• Using the No-Print Delay option, display the
waveform from 200 to 210 ms.
• Measure the resonant frequency and
compare it to the calculated value.
• Are they similar?
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• Add a plot of Vf (the feedback signal
shown in the figure).
• Is Vf 180 out of phase with Vout?
• Generate a frequency spectrum for the
Colpitts oscillator.
• Is there a DC component?
• Does the fundamental frequency
approximately equal measured the timedomain frequency?
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