Oscillators: Basics and Operations
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RAMA ARORA,
PHYSICS DEPARTMENT
PGGCG-11, CHANDIGARH
OSCILLATORS
Oscillators
An oscillator is an electronic device which converts DC
power from the supply into AC power in the load without
the application of an external input signal. The essential
components of the oscillator are:
Tank circuit, Transistor
amplifier, and Feedback circuit
Tank circuit
Amplifier and Feedback diagram
Classification of Oscillators
Depending upon the method of producing oscillations.
(a) Feedback oscillators
(b) Negative resistance oscillators
Depending upon nature of generated waveform
(a) Sinusoidal or harmonic oscillators
(b) Non-sinusoidal or relaxation oscillators
Both sinusoidal and relaxation oscillators may be negative
resistance and feedback type.
Depending upon the frequency of generated voltage.
(a) Audio frequency (AF) oscillator
(b) Radio frequency (RF) oscillator
(c) very high frequency (VHF) oscillators
(d) ultrahigh frequency (UHF) oscillators
(e)Microwave oscillators
Fundamental Principle of Oscillators
In oscillator, a negative resistance is provided to compensate for
the losses in the circuit.
In a feedback oscillator, external positive feedback sufficient to
make the overall gain infinity, provides the negative resistance
required to overcome the natural damping of the oscillations.
In a negative resistance oscillator internal positive feedback is
present and serves to provide the required negative resistance.
In an oscillator no external signal is applied. The initial signal to
trigger the oscillations is ordinarily supplied by the
noise
voltage
. This noise voltage originates when the power supply is
switched on. Since the frequency spectrum of noise is very wide,
it always possesses a component voltage at a frequency that is
correct for the oscillator operation.
Feedback oscillators
The basic requirements of a feedback oscillator are:
An amplifier with positive feedback to provide
negative resistance in the circuit.
Some circuit non-linearity to define amplitude of
oscillators.
A frequency determining network to produce
oscillations at a desired frequency.
Dc power supply to act as energy source.
Tuned collector oscillator
The basic circuit of a tuned
collector oscillator is shown in
figure. It is called the tuned-
collector oscillator, because the
tuned circuit is connected to
the collector.
The tuned circuit, constituted
by the capacitor C and
transformer primary coiL,
forms the load impedance and
determines the frequency of
oscillation.
Hartley oscillator
Hartley oscillator is an electronic oscillator circuit that uses an inductor and a
capacitor in parallel to determine the frequency.
It is used in radio receiver as a local oscillator because
(i) It is easy to tune
(ii) It’s adaptability to a wide range of frequencies
Hartley Oscillator is generally of two types
:
1.
Series fed oscillator
2. Parallel or shunt fed Hartley oscillator
Series Fed
Hartley
oscillator
In series fed
Hartley
oscillator, the junction of two inductors of the tuned circuit is
directly connected to V
cc
and one end of the LC circuit is connected to the collector
of the transistor. The lower portion of the tank coil is inductively coupled to the
upper portion.
Shunt fed Hartley oscillator
Shunt fed Hartley oscillator uses a transistor in CE configuration, in which
the collector current is divided into two parallel paths.
One branch connects the collector
to the V
cc
through RFC and
provides the path for DC keeping
the AC out. The other branch
connects the collector to LC tank
through a capacitor and provides
the path for AC keeping the DC
out.
AC equivalent circuit of Hartley oscillator
The frequency of oscillation is given by,
(a)
(b)
Let the currents I
1
, I
2
and I
3
be non-zero. Applying Kirchhoff’s voltage law to loop (1), we get
Similarly, applying Kirchhoff’s voltage law to loop (2) and (3), we get
and
From fig. (b), we have
Rearranging the above eqns. We get
For non-zero I
1
, I
2
, I
3
, the determinant of above three eqns. must be zero.
At frequency of oscillation,
Taking real part of the equation equal to zero, we get
Since
h
re
< < 1and putting
h
ie
h
oe
–
h
fe
h
re
= ∆
h
e
, the above equation becomes,
In general, > > 4 ∆
h
e
Therefore,
This is the equation for sustained oscillations.
Taking imaginary part of the equation equal to zero, we get
This is the frequency of oscillations of Hartley oscillator.
15
Feedback amplifier with inductor L and
capacitors C
1
and C
2
in feedback
network.
Feedback is frequency dependent.
Aim to adjust components to get
positive feedback and oscillation.
Output taken at collector V
o
.
No input needed, noise at oscillation
frequency
o
is picked up and
amplified.
R
B1
and R
B2
are biasing resistors.
RFC is RF Choke (inductor) to allow dc
current flow for transistor biasing, but
to block ac current flow to ac ground.
Simplified circuit shown at
midband
frequencies
where large emitter bypass
capacitor C
E
and base capacitor C
B
are
shorts and transistor capacitances (C
and C
) are opens.
C
B
C
E
V
0
V
i
V
0
V
i
Voltage across C
2
is just V
Neglecting input current to transistor
(I
0),
Then, output voltage V
o
is
KCL at output node (C)
Setting s = j
AC equivalent circuit
I
π
≈ 0
sC
2
V
sC
2
V
V
0
Assuming oscillations have started, then V
≠
0 and V
o
≠ 0, so
To get oscillations, both the real and imaginary parts
of this equation must be set equal to zero.
From the imaginary part we get the expression for the
oscillation frequency
From the real part, we get the condition on the ratio
of C
2
/C
1
Given:
Design oscillator at
150 MHz
Transistor
g
m
= 100 mA/V, R = 0.5 K
Design:
Select L= 50 nH, then calculate C
2
, and
then C
1
Example
Based on op amp using inverting
input
Combination of R’s and C’s in
feedback loop so get additional
phase shift. Target 180
o
to get
oscillation.
Analysis assumes op amp is ideal.
V
0
V
X
R
I
C1
R
I
C2
I
C3
I
R1
I
R2
I
f
R
f
V
1
V
2
C
C
C
V
0
V
X
R
I
C1
R
I
C2
I
C3
I
R1
I
R2
I
f
R
f
V
1
V
2
Example
Oscillator specifications:
o
=1x10
6
rad/s
Note: We get 180
o
phase shift from
op amp since input is to inverting
terminal and another 180
o
from the
RC ladder.
C
C
C
WIEN BRIDGE OSCILLATOR
Wien bridge oscillator is a two stage amplifier. The first stage is CE amplifier and the second
stage is CC amplifier. The output of the second stage is fed back to the first stage through feed
back network consisting of R
1
C
1
in series and R
2
C
2
in parallel. It is advantageous over phase
shift oscillator as its frequency can be varied over a frequency range of 10:1.
The ratio of output voltage of the network to the input voltage is given by
If the imaginary term vanishes, the phase shift will be zero i.e.
Therefore, frequency of oscillation is,
Also, we have
Hence the oscillations will be sustained if the amplifier has a gain just exceeding 3.
The End
Oscillators are vital electronic devices that convert DC power to AC power without external input. They consist of components like the tank circuit, transistor amplifier, and feedback circuit. Learn about the classification, fundamental principles, feedback oscillators, and tuned collector oscillators in this insightful guide.
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Presentation Transcript
OSCILLATORS RAMA ARORA, PHYSICS DEPARTMENT PGGCG-11, CHANDIGARH
Oscillators An oscillator is an electronic device which converts DC power from the supply into AC power in the load without the application of an external input signal. The essential components of the oscillator are: Tank circuit, Transistor amplifier, and Feedback circuit Tank circuit Amplifier and Feedback diagram
Classification of Oscillators Depending upon the method of producing oscillations. (a) Feedback oscillators (b) Negative resistance oscillators Depending upon nature of generated waveform (a) Sinusoidal or harmonic oscillators (b) Non-sinusoidal or relaxation oscillators Both sinusoidal and relaxation oscillators may be negative resistance and feedback type. Depending upon the frequency of generated voltage. (a) Audio frequency (AF) oscillator (b) Radio frequency (RF) oscillator (c) very high frequency (VHF) oscillators (d) ultrahigh frequency (UHF) oscillators (e)Microwave oscillators
Fundamental Principle of Oscillators In oscillator, a negative resistance is provided to compensate for the losses in the circuit. In a feedback oscillator, external positive feedback sufficient to make the overall gain infinity, provides the negative resistance required to overcome the natural damping of the oscillations. In a negative resistance oscillator internal positive feedback is present and serves to provide the required negative resistance. In an oscillator no external signal is applied. The initial signal to trigger the oscillations is ordinarily supplied by the noise voltage. This noise voltage originates when the power supply is switched on. Since the frequency spectrum of noise is very wide, it always possesses a component voltage at a frequency that is correct for the oscillator operation.
Feedback oscillators The basic requirements of a feedback oscillator are: An amplifier with positive feedback to provide negative resistance in the circuit. Some circuit non-linearity to define amplitude of oscillators. A frequency determining network to produce oscillations at a desired frequency. Dc power supply to act as energy source.
Tuned collector oscillator The basic circuit of a tuned collector oscillator is shown in figure. It is called the tuned- collector oscillator, because the tuned circuit is connected to the collector. The tuned circuit, constituted by the capacitor C and transformer primary coiL, forms the load impedance and determines the frequency of oscillation.
Hartley oscillator Hartley oscillator is an electronic oscillator circuit that uses an inductor and a capacitor in parallel to determine the frequency. It is used in radio receiver as a local oscillator because (i) It is easy to tune (ii) It s adaptability to a wide range of frequencies Hartley Oscillator is generally of two types: 1. Series fed oscillator 2. Parallel or shunt fed Hartley oscillator
Series Fed Hartley oscillator In series fed Hartley oscillator, the junction of two inductors of the tuned circuit is directly connected to Vcc and one end of the LC circuit is connected to the collector of the transistor. The lower portion of the tank coil is inductively coupled to the upper portion.
Shunt fed Hartley oscillator Shunt fed Hartley oscillator uses a transistor in CE configuration, in which the collector current is divided into two parallel paths. One branch connects the collector to the Vcc through RFC and provides the path for DC keeping the AC out. The other branch connects the collector to LC tank through a capacitor and provides the path for AC keeping the DC out.
AC equivalent circuit of Hartley oscillator (a) (b) The frequency of oscillation is given by,
From fig. (b), we have h I I 1 fe = 2 V 2 h h oe oe Let the currents I1, I2 and I3 be non-zero. Applying Kirchhoff s voltage law to loop (1), we get + + = ( ) 0 h I h V jX I I 1 2 1 3 ie re L 1 Similarly, applying Kirchhoff s voltage law to loop (2) and (3), we get h I I 1 fe + + = 2 ( ) 0 jX I I 2 3 L h h 2 oe oe and + + = ( ) ( ) 0 jX I I jX I I jX I 3 1 3 2 3 L L c 1 2
Rearranging the above eqns. We get h h h fe h re + + = re 0 h jX I I jI X 1 2 3 ie L L h 1 1 oe oe h 1 fe + + + = 0 I jX I jI X 1 2 2 L L h h 2 1 oe oe + + + = ( ) 0 jX I jX I jX jX jX I 1 2 3 L L L L c 1 2 1 2 For non-zero I1, I2, I3, the determinant of above three eqns. must be zero. 1 = 2 At frequency of oscillation, + ( C ) L L 1 2 + = 0 jX jX jX L L c 1 2
Taking real part of the equation equal to zero, we get + + = 2 L 2 L ( ) ( ) 0 h h h h X h h X X X ie oe fe re re fe L L 2 1 2 1 Since hre < < 1and putting hie hoe hfe hre= he, the above equation becomes, + = 2 L 2 L 0 h X h X X X e fe L L 2 1 2 1 2 4 h h h fe h fe e = X X L L 2 2 1 e In general, > > 4 he h2 fe h fe Therefore, L L 2 1 h e This is the equation for sustained oscillations.
Taking imaginary part of the equation equal to zero, we get 1 = 1 h 2 + + oe ( ) L L C L L 1 2 1 2 h fe 1 = 1+ ( C ) L L 2 1 1+ = = f 2 2 ( C ) L L 2 1 = f 2 LC This is the frequency of oscillations of Hartley oscillator.
Colpitts LC-Tuned Oscillator Feedback amplifier with inductor L and capacitors C1 and C2 in feedback network. Feedback is frequency dependent. Aim to adjust components to get positive feedback and oscillation. Output taken at collector Vo. No input needed, noise at oscillation frequency o is picked up and amplified. RB1 and RB2 are biasing resistors. RFC is RF Choke (inductor) to allow dc current flow for transistor biasing, but to block ac current flow to ac ground. Simplified circuit shown at midband frequencies where large emitter bypass capacitor CE and base capacitor CB are shorts and transistor capacitances (C and C ) are opens. CB V0 CE Vi V0 Vi 15
Colpitts LC-Tuned Oscillator Voltage across C2 is just V V = = I sC V 2 2 C Z 2 C Neglecting input current to transistor (I 0), Z C 2 V = = = I I sC V 2 2 L C Then, output voltage Vo is Z I V V L L o = + = ( 1 ) 2 + = + ( )( ) V sC V sL V s LC 2 2 AC equivalent circuit KCL at output node (C) Assuming oscillations have started, then V 0 and Vo 0, so sC2V 1 R + + + = 0 sC V g V sC V 2 1 m o V0 ( 1 ) 1 R I 0 + + + + = 2 0 sC V g V sC V s LC 2 1 2 m sC2V 1 R LC ( ) + + + + + = 3 2 0 2 s LC C s s C C g 1 2 1 2 m R Setting s = j 0 = 2 1 R LC ( ) 3 + + + 2 gm j C C LC C 1 2 1 2 R
Colpitts LC-Tuned Oscillator To get oscillations, both the real and imaginary parts of this equation must be set equal to zero. + R R 0 = 2 1 LC ( ) 3 + + 2 gm j C C LC C 1 2 1 2 From the imaginary part we get the expression for the oscillation frequency ( ) + = 2 1 L + = 3 o 0 C C LC C 1 2 1 2 o C C 1 = 1 2 o LC C C C 1 2 + C C 1 2 From the real part, we get the condition on the ratio of C2/C1 LC R 2 1 + = 2 o 0 g m R + C C C 2 + = = = + 1 2 2 1 1 g R LC LC 2 2 m o LC C 1 C 2 1 C = 2 g R m C 1
Colpitts LC-Tuned Oscillator Given: Design oscillator at 150 MHz Example ( 150 ) 6= 8 = = 2 2 10 4 . 9 10 / f x x rad s o Transistor gm = 100 mA/V, R = 0.5 K Design: C m = = = 2 100 ( / 5 . 0 )( ) 50 g R mA V K C 1 Select L= 50 nH, then calculate C2, andthen C1 + 1 C C C = = + 1 2 2 1 o LC C 1 LC C 2 2 1 1 1 C 9 = + = + = = 2 1 1 ( 50 ) . 1 13 10 , 1 130 C x F pF 2 2 8 2 C 50 4 . 9 ( 10 ) L nH x 1 o , 1 130 C 50 pF = = = 2 23 C pF 1 50
Phase Shift Oscillator Rf If + 1 1 V V V = + = + = IC3 IC2 IC1 o o o I I I 2 1 1 C R C sCRR R R sCR V2 V1 f f f + 1 1 1 V V = = VX o o V V I Z C C C 2 1 2 C C sCR R sCR sC V0 IR1 f f R R IR2 1 V = + o 2 sCR sCR Based on op amp using inverting input Combination of R s and C s in feedback loop so get additional phase shift. Target 180o to get oscillation. Analysis assumes op amp is ideal. f 1 V V = = + 2 o 2 I 2 R R sCRR sCR f + 1 1 1 V V = + = + + o o 2 I I I 3 2 2 C R C sCRR sCR R sCR f f + 1 1 1 1 3 1 V V = + + = + + o o 2 1 V 2 R sCR sCR sCR R sCR ( ) sCR = = = o 0 V V so I I f f + 1 f C R Finally f V 1 3 1 I V V = = o V V I Z = = + + + 3 C o o 2 1 V V 1 1 C C sCR 2 X 2 sC sCR sCR sCR sCR ( ) sCR f f f 1 V V V 4 1 V = = = 1 o o I = + + o 3 1 R R R sCR sCRR 2 sCR sCR ( ) sCR f f f
Phase Shift Oscillator Rearrangin g 4 1 V If Rf = + + V 3 o IC3 IC2 IC1 X 2 ( ) sCR sCR sCR f V2 V1 we get for the loop gain VX C sCR V C C f = = = = 0 ( ) ( ) ( ) 1 L A V0 R R IR1 V 4 1 IR2 + + X 3 2 ( ) sCR sCR 2 2 j CR C RR f f = = 1 4 1 Example Oscillator specifications: o=1x106 rad/s + 4 3 j CR 3 j 2 CR term ( ) CR CR ns, oscillatio get To imaginary the need we to go to zero. = Selecting convenienc for e 10 , C nF frequency one at this achieve can We so o 1 1 1 = then from = = = 3 CR so o 3 RC 0 CR 3 RC 1 1 = = = = oscillatio get To ns, we also need 1 L( ) so 58 R o 6 3 3 10 1 ( 10 / ) C nF x rad s o 2 2 C RR Then 0 f = = 1 substituti and for ng we get L( ) o o 4 = = 12 58 ( ) . 0 67 R K f 2 2 2 C RR C RR 4 R 1 2 0 f f f = = = 1 so Note: We get 180o phase shift from op amp since input is to inverting terminal and another 180o from the RC ladder. 2 4 3 12 R C R = 12 R R f
WIEN BRIDGE OSCILLATOR Wien bridge oscillator is a two stage amplifier. The first stage is CE amplifier and the second stage is CC amplifier. The output of the second stage is fed back to the first stage through feed back network consisting of R1C1 in series and R2C2 in parallel. It is advantageous over phase shift oscillator as its frequency can be varied over a frequency range of 10:1.
The ratio of output voltage of the network to the input voltage is given by V impedance of parallel combinatio n o= V total impedance i . R jX c R jX = c RjX c R jX c R jX c V jRX 2 = o c 2 V 3 R X jRX i c c
If the imaginary term vanishes, the phase shift will be zero i.e. = 2 2 c 0 R X Xc= R 1 C= R 1 = RC 1 = = f Therefore, frequency of oscillation is, 2 LC 2 V jRX 1 = = o c Also, we have 3 3 V jRX i c Hence the oscillations will be sustained if the amplifier has a gain just exceeding 3.
