Transcript Chapter 5 - Oscillators (Part 2)

Oscillators
Oscillators with LC
Feedback Circuits
Oscillators
Oscillators With LC Feedback Circuits
•For frequencies above 1 MHz, LC feedback
oscillators are used.
•LC feedback oscillators use resonant
circuits in the feedback path.
•We will discuss the Colpitts, Hartley and
crystal-controlled oscillators.
•Transistors are used as the active device in
these types.
LC Oscillators – Colpitts
The Colpitts
oscillator utilizes a
tank circuit (LC) in
the feedback loop
to provide the
necessary phase
shift and to act as
a resonant filter
that passes only
the desired
frequency of
oscillation.
V CC
R1
R3
C5
V out
C3
R2
R4
C1 L C2
C4
Amplifier
Feedback
circuit
LC Oscillators – Colpitts
A popular LC oscillator is the Colpitts oscillator. It uses two
series capacitors in the resonant circuit. The feedback voltage is
developed across C1.
Vf
The effect is that the tank
circuit is “tapped”.
Usually C1 is the larger
capacitor because it
develops the smaller
voltage.
Av
Vout
L
Out
In
I
C1
C2
Vf
Vout
LC Oscillators – Colpitts
The resonant frequency can be determined by the formula below.
1
fr 
2 LCT
LC Oscillators – Colpitts
• Figure below shows the input impedance of the amplifier
acts as a load on the resonant feedback circuit and
reduces the Q of the circuit.
• The resonant frequency of a parallel resonant circuit
depends on the Q, according to the formula below:
1
fr 
2π LCT
Q2
Q2  1
Zin
Vout
L
C1
C2
LC Oscillators – Hartley
V CC
The Hartley
oscillator is similar
to the Colpitts
except that the
feedback circuit
consists of two
series inductors
and a parallel
capacitor.
R1
R3 C
2
C1
C4
V out
R2
R4
L1 C5 L2
C3
Amplifier
Feedback
circuit
LC Oscillators – Hartley
The frequency of oscillation for Q > 10 is:
1
fr 

2π LTC 2π
1
 L1  L2  C
where LT = L1 + L2
Vf
One advantage of a
Hartley oscillator is
that it can be tuned
by using a variable
capacitor
in
the
resonant circuit.
Vout
Av
C
Out
L1
L2
In
LC Oscillators – Crystal-Controlled
The crystal-controlled oscillator is the most stable and
accurate of all oscillators. A crystal has a natural
frequency of resonance. Quartz material can be cut or
shaped to have a certain frequency.
Quartz
wafer
XTAL
Cm
Ls
Cs
Rs
(a) Typical
packaged
crystal
(b) Basic
(b) Symbol
construction
(without case)
(b) Electrical
equivalent
LC Oscillators – Crystal-Controlled
V CC
Since crystal has
natural resonant
frequencies of 20 MHz
or less, generation of
higher frequencies is
attained by operating
the crystal in what is
called the overtone
mode
R1
R3
C2
V out
R2
R4
XTAL C C
C1
Oscillators
Relaxation Oscillators
Oscillators – Relaxation
Relaxation oscillators make use of an RC timing
and a device that changes states to generate a
periodic waveform (non-sinusoidal).
1. Triangular-wave
2. Square-wave
3. Sawtooth
Oscillators – Relaxation
Triangular-wave oscillator
Triangular-wave oscillator circuit is a combination
of a comparator and integrator circuit.
Comparator
A square wave
can be taken
as an output
here.
C
R1
V out
R2
Integrator
R3
Oscillators – Relaxation
Triangular-wave oscillator
• Assume that the output voltage of the comparator is at its
maximum negative level.
• This output is connected to the inverting input of the
integrator through R1, producing a positive-going ramp on
the output of the integrator.
• When the ramp voltage reaches the UTP, the comparator
switches to its maximum positive level.
•This positive level causes the integrator ramp to change
to a negative-going direction.
•The ramp continues in this direction until the LTP of the
comparator is reached and the cycle repeats.
Oscillators – Relaxation
Triangular-wave oscillator
+V max
Comparator
output
-V max
V UTP
V out
V LTP
Oscillators – Relaxation
Triangular-wave oscillator
Amplitude of the triangular output is set by establishing the UTP and LTP
voltages according to the following formulas:
VUTP
 R3 
 Vmax  
 R2 
VLTP
 R3 
 -Vmax  
 R2 
The frequency of both waveforms depends on the R1C time constant as well
as the amplitude-setting resistors, R2 and R3. By varying R1, the fr can be
adjusted without changing the output amplitude.
1  R2 
 
fr 
4 R1C  R3 
Oscillators – Relaxation - EXAMPLE
Determine the frequency of oscillation of the
circuit in figure below. To what value must R1 be
changed to make the frequency 20 kHz?
Answer: fr = 8.25 kHz, R1 = 4.13 kOhm
Oscillators – Square-wave
A square wave relaxation oscillator is like the
Schmitt trigger or Comparator circuit.
The charging and discharging of the capacitor
cause the op-amp to switch states rapidly and
produce a square wave.
The RC time constant determines the frequency.
Oscillators – Square-wave
R1
VC
C
V out
Vf
R2
R3
Oscillators – Square-wave
Oscillators – Sawtooth voltage controlled
oscillator (VCO)
Sawtooth VCO circuit
is a combination of a
Programmable
Unijunction Transistor
(PUT) and integrator
circuit.
Ri
VG
PUT
Vp
-
I
0V
V IN
+
V out
Oscillators – Sawtooth VCO
OPERATION
Initially, dc input = -VIN
• Vout = 0V, Vanode < VG
• The circuit is like an integrator.
• Capacitor is charging.
• Output is increasing positive going ramp.
Oscillators – Sawtooth VCO
OPERATION
VG
PUT
Vp
-
Ri
I
0V
V IN
+
V out
0
Oscillators – Sawtooth VCO
OPERATION
When Vout = VP
•
•
Vanode > VG , PUT turns ‘ON’
The capacitor rapidly
discharges.
•
Vout drop until Vout = VF.
•
Vanode < VG , PUT turns ‘OFF’
VP – maximum peak value
VF – minimum peak value
Oscillators – Sawtooth VCO
OPERATION
Oscillation frequency
VIN  1 


f 
Ri C  VP - VF 
Oscillators – Sawtooth VCO
EXAMPLE
In the following circuit, let VF = 1V.
a) Find;
(i) amplitude;
(ii) frequency;
b) Sketch the output waveform
Oscillators – Sawtooth VCO
EXAMPLE (cont’d)
Oscillators – Sawtooth VCO
EXAMPLE – Solution
a) (i) Amplitude
R4
10
 V  
15  7.5 V
VG 
R3  R4
10  10
VP  VG  7.5 V
and
So, the peak-to-peak amplitude is;
VP - VF  7.5 -1  6.5 V
VF  1 V
Oscillators – Sawtooth VCO
EXAMPLE – Solution
a) (ii) Frequency
VIN  1

f 
Ri C  VP - VF



R2
- V   -1.92 V
VIN 
R1  R2
Oscillators – Sawtooth VCO
EXAMPLE – Solution
a) (ii) Frequency
1.92
1


f 


100k 0.0047μ   7.5V - 1V 
 628 Hz
Oscillators – Sawtooth VCO
EXAMPLE – Solution
b) Output waveform
1
1
T 
 2 ms
f 628
7.5 V
V out
1V
2 ms
t
Oscillators
The 555 timer
as an oscillator
Oscillators
The 555 Timer As An Oscillator
The 555 timer is an integrated circuit that can be
used in many applications. The frequency of
output is determined by the external components
R1, R2, and C. The formula below shows the
relationship.
144
fr 
R1  2 R2 Cext
Oscillators
The 555 Timer As An Oscillator
Duty cycles can be adjusted by values of R1 and
R2. The duty cycle is limited to 50% with this
arrangement. To have duty cycles less than 50%,
a diode is placed across R2. The two formulas
show the relationship;
Duty Cycle > 50 %
 R1  R2 
100%
Duty cycle  
 R1  2 R2 
Oscillators
The 555 Timer As An Oscillator
Duty Cycle < 50 %
 R1 
100%
Duty cycle  
 R1  R2 
Oscillators
The 555 Timer As An Oscillator
Oscillators
The 555 Timer As An Oscillator
The 555 timer
may be
operated as a
VCO with a
control voltage
applied to the
CONT input
(pin 5).
END CHAPTER 5