6. n 6 -tyar Sigeyaeyanartaynal Generatorsyaneny.pdf
23021784
29 views
92 slides
Aug 28, 2025
Slide 1 of 92
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
About This Presentation
Qenrtwetaeymaze5kyjae
Slide Content
1
Lecture #6
Electronic Oscillation Generators
Mai Linh, PhD
Email: [email protected] ; [email protected]
VNU-University of Engineering and Technology
Faculty of Electronics and Telecommunications
Lecture #6
Electronic Oscillation Generators
Mai Linh, PhD
Email: [email protected] ; [email protected]
VNU-University of Engineering and Technology
Faculty of Electronics and Telecommunications
6.0 Introduction
❑ Signal-generator circuits (or oscillators) generate signals in standard waveforms
such as sinusoidal, square, triangular, or pulse.
❑ There are 2 sorts of the generation of sinusoids:
✓1
st
approach employs a positive-feedback loop consisting an amplifier and an
RC or LC frequency-selective network. It generates sine waves utilizing
resonance phenomena, are known as linear oscillators.
✓2
nd
approach: a sine wave is obtained by appropriately shaping a triangular
waveform.
❖Circuits that generate square, triangular, pulse waveforms are called non-linear
oscillators or function generators. These circuits employ circuit building blocks
known as multivibrators (bistable, astable, and monostable).
2
Mai Linh, PhD
❑ Signal-generator circuits (or oscillators) generate signals in standard waveforms
such as sinusoidal, square, triangular, or pulse.
❑ There are 2 sorts of the generation of sinusoids:
✓1
st
approach employs a positive-feedback loop consisting an amplifier and an
RC or LC frequency-selective network. It generates sine waves utilizing
resonance phenomena, are known as linear oscillators.
✓2
nd
approach: a sine wave is obtained by appropriately shaping a triangular
waveform.
❖Circuits that generate square, triangular, pulse waveforms are called non-linear
oscillators or function generators. These circuits employ circuit building blocks
known as multivibrators (bistable, astable, and monostable).
2
Mai Linh, PhD
6.1 Basic principles of sinusoidal oscillators
❑Positive Feedback
Fig.: The basic structure of sinusoidal oscillators
➢Designed to produce an output even thought the input is zero
➢Use positive feedback through frequency-selective feedback network to
ensure sustained oscillation at one frequency component: ω
0
3
A basic structure of a sinusoidal
oscillator consists of an
amplifier and a frequency
selective network connected in
a positive-feedback loop.
Mai Linh, PhD
❑Positive Feedback
Fig.: The basic structure of sinusoidal oscillators
➢Designed to produce an output even thought the input is zero
➢Use positive feedback through frequency-selective feedback network to
ensure sustained oscillation at one frequency component: ω
0
3
A basic structure of a sinusoidal
oscillator consists of an
amplifier and a frequency
selective network connected in
a positive-feedback loop.
Mai Linh, PhD
6.1 Basic principles of sinusoidal oscillators
An amplifier with positive feedback results in oscillations
if the following conditions are satisfied:
✓The loop gain (product of the gain of the amplifier - A &
the gain of the feedback network - ) is unity (= 1)
✓The total phase shift in the loop is 0
❖Switch is open → no oscillations.
❖If V
i is fed to the circuit & the switch is closed → V
o = A
VV
i and V
f = V
o are fed back to the circuit.
❖If V
f = V
i : even if remove V
i away, the output continues to exist!
V
o = A
VV
i and V
f = V
o ➔ V
f = A
VV
i
➢If let V
f = V
i, (above equation) ➔ A
V =1. (unity loop gain)
Thus, by closing the switch and removing the input, we are able to get the oscillations at the output if
A
V =1, where A
V is called the Loop gain.
Positive feedback refers to the fact that the fed back signal is in phase with the input signal.
4Amplifier A
Freq.-Selective
Network
x0
xf
+
Switch
Mai Linh, PhD
An amplifier with positive feedback results in oscillations
if the following conditions are satisfied:
✓The loop gain (product of the gain of the amplifier - A &
the gain of the feedback network - ) is unity (= 1)
✓The total phase shift in the loop is 0
❖Switch is open → no oscillations.
❖If V
i is fed to the circuit & the switch is closed → V
o = A
VV
i and V
f = V
o are fed back to the circuit.
❖If V
f = V
i : even if remove V
i away, the output continues to exist!
V
o = A
VV
i and V
f = V
o ➔ V
f = A
VV
i
➢If let V
f = V
i, (above equation) ➔ A
V =1. (unity loop gain)
Thus, by closing the switch and removing the input, we are able to get the oscillations at the output if
A
V =1, where A
V is called the Loop gain.
Positive feedback refers to the fact that the fed back signal is in phase with the input signal.
4Amplifier A
Freq.-Selective
Network
x0
xf
+
Switch
Mai Linh, PhD
6.1 Basic principles of sinusoidal oscillators
❑Oscillator Criterion
Feedback signal x
f is summed with a positive sign
The gain-with-feedback or closed-loop gain:
➔ At a specific of ω
0 the phase of the loop gain should be zero and the magnitude of the loop
gain should be unity. (named as Barkhausen criterion)
An alternative approach to the study of oscillator circuits consists of examining the circuit poles,
which are the roots of the characteristic equation. For the circuit to produce sustained
oscillations at a frequency ω
0, the characteristic equation has to have roots at s = ± ω
0.()
()
1()()
f
As
As
Ass
=
−
(Note: s = j)
Open-loop gain
Feedback gain
5Amplifier A
Freq.-Selective
Network
x0
xf
+
S
+
xS
Loop gain: L(s) A(s)(s)
Characteristic equation: 1 – L(s) = 0
The oscillation criterion: L(j) A(j)(j) = 1
Mai Linh, PhD
❑Oscillator Criterion
Feedback signal x
f is summed with a positive sign
The gain-with-feedback or closed-loop gain:
➔ At a specific of ω
0 the phase of the loop gain should be zero and the magnitude of the loop
gain should be unity. (named as Barkhausen criterion)
An alternative approach to the study of oscillator circuits consists of examining the circuit poles,
which are the roots of the characteristic equation. For the circuit to produce sustained
oscillations at a frequency ω
0, the characteristic equation has to have roots at s = ± ω
0.()
()
1()()
f
As
As
Ass
=
−
(Note: s = j)
Open-loop gain
Feedback gain
5Amplifier A
Freq.-Selective
Network
x0
xf
+
S
+
xS
Loop gain: L(s) A(s)(s)
Characteristic equation: 1 – L(s) = 0
The oscillation criterion: L(j) A(j)(j) = 1
Mai Linh, PhD
6.1 Basic principles of sinusoidal oscillators
❖Barkhausen criterion:
1) The phase of loop gain should be zero at
0 .
2) The magnitude of the loop gain A(s)(s) = 1 at
0 .
The characteristic equation has roots at s = ± j
0 .
The stability of ω
o will be determined by the manner in
which the phase of the feedback loop (ω) varies with
frequency.
Fig.: Dependence of the oscillator-frequency
stability on the slope of the phase response.
If d/d is large →
o is small (for a given change
in phase , resulting from a change – due to
temperature - in a circuit component)
Applications of Oscillator Circuits
➢Clock input for CPU, DSP chips …
➢Local oscillator for radio receivers, mobile receivers, etc.
➢As a signal generators in the lab
➢Clock input for analog-digital and digital-analog converters
6
Mai Linh, PhD
❖Barkhausen criterion:
1) The phase of loop gain should be zero at
0 .
2) The magnitude of the loop gain A(s)(s) = 1 at
0 .
The characteristic equation has roots at s = ± j
0 .
The stability of ω
o will be determined by the manner in
which the phase of the feedback loop (ω) varies with
frequency.
Fig.: Dependence of the oscillator-frequency
stability on the slope of the phase response.
If d/d is large →
o is small (for a given change
in phase , resulting from a change – due to
temperature - in a circuit component)
Applications of Oscillator Circuits
➢Clock input for CPU, DSP chips …
➢Local oscillator for radio receivers, mobile receivers, etc.
➢As a signal generators in the lab
➢Clock input for analog-digital and digital-analog converters
6
Mai Linh, PhD
6.1 Basic principles of sinusoidal oscillators
Analysis of Oscillator Circuits
Analysis of a given oscillator circuit to determine the frequency of oscillation and
the condition for the oscillations:
1. Break the feedback loop to determine the loop gain L(s) = A(s)β(s).
2. The oscillation frequency ω
0 is found as the frequency for which the phase
angle of A(jω)β(jω) = 0 or, equivalently, 360.
3. The condition for the oscillations to start is found from
|A(jω
0)β(jω
0)|≥ 1
7
Mai Linh, PhD
Analysis of Oscillator Circuits
Analysis of a given oscillator circuit to determine the frequency of oscillation and
the condition for the oscillations:
1. Break the feedback loop to determine the loop gain L(s) = A(s)β(s).
2. The oscillation frequency ω
0 is found as the frequency for which the phase
angle of A(jω)β(jω) = 0 or, equivalently, 360.
3. The condition for the oscillations to start is found from
|A(jω
0)β(jω
0)|≥ 1
7
Mai Linh, PhD
8
Element by corresponding transform impedance (in s-domain)
Component Z(s) Y(s)
R R
1
�
=??????
L sL
1
????????????
C
1
????????????
sC
6.1 Basic principles of sinusoidal oscillators
Mai Linh, PhD
Element by corresponding transform impedance (in s-domain)
Component Z(s) Y(s)
R R
1
�
=??????
L sL
1
????????????
C
1
????????????
sC
6.1 Basic principles of sinusoidal oscillators
Mai Linh, PhD
6.1 Basic principles of sinusoidal oscillators
Read the Example 18.1 in the
S&S textbook @ home!
9
Mai Linh, PhD
Read the Example 18.1 in the
S&S textbook @ home!
9
Mai Linh, PhD
Oscillator Circuits
Op Amp-RC Oscillator Circuits
➢ The Wien-Bridge Oscillator
➢ The phase-Shift Oscillator
LC-Tuned Oscillator
➢ Colpitts oscillator
➢ Hartley oscillator
Crystal Oscillator
10
Mai Linh, PhD
Op Amp-RC Oscillator Circuits
➢ The Wien-Bridge Oscillator
➢ The phase-Shift Oscillator
LC-Tuned Oscillator
➢ Colpitts oscillator
➢ Hartley oscillator
Crystal Oscillator
10
Mai Linh, PhD
6.1 Basic principles of sinusoidal oscillators
❑Nonlinear Amplitude Control
✓ Suppose we make loop gain L(s) = A(s)(s) = 1 at =
0,
If the temperature changes: Thus
➢Loop gain A < 1 → oscillations die out
➢Loop gain A > 1 → oscillations grow and clip at supply rails
We therefore need a nonlinear circuit for gain control to force Aβ to remain equal 1 at the
desired value of output amplitude.
✓ The function of gain-control mechanism is as follows:
oTo ensure that oscillation will start, Aβ is designed greater than unity, lightly.
oAs the oscillation grow in amplitude and reaches the desired level, the nonlinear network
reduces the loop gain to exactly unity (poles will be pulled back to j axis).
oIf the loop gain is reduced below unity, the amplitude will diminish and the nonlinear
network increases the loop gain to exactly unity.
11+VCC -VEE
Supply rails
Mai Linh, PhD
❑Nonlinear Amplitude Control
✓ Suppose we make loop gain L(s) = A(s)(s) = 1 at =
0,
If the temperature changes: Thus
➢Loop gain A < 1 → oscillations die out
➢Loop gain A > 1 → oscillations grow and clip at supply rails
We therefore need a nonlinear circuit for gain control to force Aβ to remain equal 1 at the
desired value of output amplitude.
✓ The function of gain-control mechanism is as follows:
oTo ensure that oscillation will start, Aβ is designed greater than unity, lightly.
oAs the oscillation grow in amplitude and reaches the desired level, the nonlinear network
reduces the loop gain to exactly unity (poles will be pulled back to j axis).
oIf the loop gain is reduced below unity, the amplitude will diminish and the nonlinear
network increases the loop gain to exactly unity.
11+VCC -VEE
Supply rails
Mai Linh, PhD
❑Nonlinear Amplitude Control
Two basic approaches to the implementation of the nonlinear amplitude-
stabilization mechanism.
1.Use of a limiter circuit. Oscillations are allowed to grow until the amplitude
reaches the level to which the limiter is set. When the limiter comes into
operation, the amplitude remains constant.
2.Utilizes an element whose resistance can be controlled by the amplitude of the
output sinusoid. In fact, diodes, or JFETs operated in the triode region, are
commonly employed to implement the controlled-resistance element.
6.1 Basic principles of sinusoidal oscillators
12
Mai Linh, PhD
Two basic approaches to the implementation of the nonlinear amplitude-
stabilization mechanism.
1.Use of a limiter circuit. Oscillations are allowed to grow until the amplitude
reaches the level to which the limiter is set. When the limiter comes into
operation, the amplitude remains constant.
2.Utilizes an element whose resistance can be controlled by the amplitude of the
output sinusoid. In fact, diodes, or JFETs operated in the triode region, are
commonly employed to implement the controlled-resistance element.
6.1 Basic principles of sinusoidal oscillators
12
Mai Linh, PhD
1
(1)
f
OI
R
vv
R
=− Limiter circuit
❑Nonlinear Amplitude Control
✓Limiter Circuit for Amplitude Control
o For small v
I, → D
1 & D
2 off:
➔ slope is linear = -R
f /R
1
Limiter gain
13
6.1 Basic principles of sinusoidal oscillators3 2
2 3 2 3
54
4 5 4 5
(2)
(3)
AO
BO
R R
v Vv
RRRR
RR
v Vv
RRRR
=+
++
=−+
++
Mai Linh, PhD
(1)
f
OI
R
vv
R
=− Limiter circuit
❑Nonlinear Amplitude Control
✓Limiter Circuit for Amplitude Control
o For small v
I, → D
1 & D
2 off:
➔ slope is linear = -R
f /R
1
Limiter gain
13
6.1 Basic principles of sinusoidal oscillators3 2
2 3 2 3
54
4 5 4 5
(2)
(3)
AO
BO
R R
v Vv
RRRR
RR
v Vv
RRRR
=+
++
=−+
++
Mai Linh, PhD
2 3 3
2 2 3
33
22
1
D
D
RRR
L VV
RRR
RR
VV
RR
−
+
=−−
+
=−−+
Fig.: Transfer characteristic
of the limiter circuit
❑Nonlinear Amplitude Control
✓Limiter Circuit for Amplitude Control
o As v
I continues to go positive, v
O goes negative, equ. (1) → D
1 on &
D
2 off, until v
A = – V
D; the value of v
o at which D
1 conducts named as
the negative limiting level (let) v
o = L
–
If v
I > the value of L
– /(-R
f /R
1) → R
3||R
f
➔ Incremental Gain (slop of the transfer characteristic):
equ. (2)
14
6.1 Basic principles of sinusoidal oscillators31
()/
f
RRR− vI
L_
L+
0
Slope=
Rf
R1
Slope=
Rf ||R3
R1
Slope=
Rf ||R4
R1
vO
Mai Linh, PhD
2 2 3
33
22
1
D
D
RRR
L VV
RRR
RR
VV
RR
−
+
=−−
+
=−−+
Fig.: Transfer characteristic
of the limiter circuit
❑Nonlinear Amplitude Control
✓Limiter Circuit for Amplitude Control
o As v
I continues to go positive, v
O goes negative, equ. (1) → D
1 on &
D
2 off, until v
A = – V
D; the value of v
o at which D
1 conducts named as
the negative limiting level (let) v
o = L
–
If v
I > the value of L
– /(-R
f /R
1) → R
3||R
f
➔ Incremental Gain (slop of the transfer characteristic):
equ. (2)
14
6.1 Basic principles of sinusoidal oscillators31
()/
f
RRR− vI
L_
L+
0
Slope=
Rf
R1
Slope=
Rf ||R3
R1
Slope=
Rf ||R4
R1
vO
Mai Linh, PhD
vI
L_
L+
0
Slope=
Rf
R1
Slope=
Rf ||R3
R1
Slope=
Rf ||R4
R1
vO ❑Nonlinear Amplitude Control
✓Limiter Circuit for Amplitude Control
oSimilarly, as v
I continues to go negative, v
O goes
positive. D
1 off & D
2 on, from equ. (3) to find the
positive limiting level L
+
➔ Slope of the transfer characteristic:
Fig.: Transfer characteristic of the
limiter circuit
15
6.1 Basic principles of sinusoidal oscillators44
55
1
D
RR
LVV
RR
+
=++
44
55
1
D
RR
LVV
RR
+
=++
41
()/
f
RRR−
Mai Linh, PhD
L_
L+
0
Slope=
Rf
R1
Slope=
Rf ||R3
R1
Slope=
Rf ||R4
R1
vO ❑Nonlinear Amplitude Control
✓Limiter Circuit for Amplitude Control
oSimilarly, as v
I continues to go negative, v
O goes
positive. D
1 off & D
2 on, from equ. (3) to find the
positive limiting level L
+
➔ Slope of the transfer characteristic:
Fig.: Transfer characteristic of the
limiter circuit
15
6.1 Basic principles of sinusoidal oscillators44
55
1
D
RR
LVV
RR
+
=++
44
55
1
D
RR
LVV
RR
+
=++
41
()/
f
RRR−
Mai Linh, PhD
❑Nonlinear Amplitude Control
✓Limiter Circuit for Amplitude Control
Popular Limiter circuit Transfer characteristic When �
?????? is removed
16
6.1 Basic principles of sinusoidal oscillatorsvI
L_
L+
0
Slope=
Rf
R1
Slope=
Rf ||R3
R1
Slope=
Rf ||R4
R1
vO vI
L_
L+
0
vO
Slope=
R4
R1
Slope=
R3
R1
Mai Linh, PhD
✓Limiter Circuit for Amplitude Control
Popular Limiter circuit Transfer characteristic When �
?????? is removed
16
6.1 Basic principles of sinusoidal oscillatorsvI
L_
L+
0
Slope=
Rf
R1
Slope=
Rf ||R3
R1
Slope=
Rf ||R4
R1
vO vI
L_
L+
0
vO
Slope=
R4
R1
Slope=
R3
R1
Mai Linh, PhD
Exercise 6.1
For the circuit as shown in figure with V = 15 V,
R
1 = 30 kΩ, R
f = 60 kΩ, R
2 = R
5 = 9 kΩ, and
R
3 = R
4 = 3 kΩ, find the limiting levels and the
value of ??????
?????? at which the limiting levels are
reached. Also determine the limiter gain and the
slope of the transfer characteristic in the positive
and negative limiting regions.
Assume that ??????
�=0.7 ??????.
❑Nonlinear Amplitude Control
17
6.1 Basic principles of sinusoidal oscillators
Mai Linh, PhD
For the circuit as shown in figure with V = 15 V,
R
1 = 30 kΩ, R
f = 60 kΩ, R
2 = R
5 = 9 kΩ, and
R
3 = R
4 = 3 kΩ, find the limiting levels and the
value of ??????
?????? at which the limiting levels are
reached. Also determine the limiter gain and the
slope of the transfer characteristic in the positive
and negative limiting regions.
Assume that ??????
�=0.7 ??????.
❑Nonlinear Amplitude Control
17
6.1 Basic principles of sinusoidal oscillators
Mai Linh, PhD
Exercise 6.1 – Sol.
❑Nonlinear Amplitude Control
Slope in the limiting regions
18
6.1 Basic principles of sinusoidal oscillators44
55
33
22
33
1150.7150.935.93()
99
33
1150.715.93()
99
D
D
RR
LVV V
RR
RR
LVV V
RR
+
−
=++=++=+=+
=−++=−−+=−
1
60
Limiter gain 2/
30
5.93
Thus limiting occurs at 2.97()
2
f
R
VV
R
V
− −
===−
= 4
1
|| 60||3
0.095/
30
f
RR
VV
R
−
==−=−
Mai Linh, PhD
❑Nonlinear Amplitude Control
Slope in the limiting regions
18
6.1 Basic principles of sinusoidal oscillators44
55
33
22
33
1150.7150.935.93()
99
33
1150.715.93()
99
D
D
RR
LVV V
RR
RR
LVV V
RR
+
−
=++=++=+=+
=−++=−−+=−
1
60
Limiter gain 2/
30
5.93
Thus limiting occurs at 2.97()
2
f
R
VV
R
V
− −
===−
= 4
1
|| 60||3
0.095/
30
f
RR
VV
R
−
==−=−
Mai Linh, PhD
6.2 Op Amp – RC Oscillator Circuits
❑Wien-Bridge
The Wien’s bridge use in AC circuits for determining the value
of unknown frequency. It can also be used for the measurement
of an unknown capacitor with great accuracy.
@ balancing condition: Z
1Z
4 = Z
2Z
3
Z
1 = R
1 – j/C
1 ; Y
3 = 1/Z
3 = 1/R
3 + jC
1; Z
2 = R
2; Z
4 = R
4
Thus Z
2 = Z
1Z
4Y
3 or2 4 1 3
13
1j
R R R j C
CR
= − +
341 4 4
2 3 1 4
3 1 1 3
CRR R R
R j C R R
R C C R
= + − −
Equating the real & imaginary terms:3414
2
31
CRRR
R
RC
=+ 4
3 1 4
13
0
R
C R R
CR
−= 321
4 3 1
CRR
R R C
=+ 31
13 1 1 3 3 1 1 3 3
1 1 1
2
C R f
CR C RC R C RC R
= → = → =
Max Wien (1866–1938) was a
Prussian physicist. He theoretically
developed the concept of the
Wien-bridge oscillator in 1891. At
that time, Wien did not have a
means of developing electronic
gain, so a workable oscillator could
not be achieved. Based on Wien’s
work, William Hewlett, co-founder
of Hewlett-Packard, was successful
in building a practical Wien-bridge
oscillator in 1939.
19
Mai Linh, PhD
❑Wien-Bridge
The Wien’s bridge use in AC circuits for determining the value
of unknown frequency. It can also be used for the measurement
of an unknown capacitor with great accuracy.
@ balancing condition: Z
1Z
4 = Z
2Z
3
Z
1 = R
1 – j/C
1 ; Y
3 = 1/Z
3 = 1/R
3 + jC
1; Z
2 = R
2; Z
4 = R
4
Thus Z
2 = Z
1Z
4Y
3 or2 4 1 3
13
1j
R R R j C
CR
= − +
341 4 4
2 3 1 4
3 1 1 3
CRR R R
R j C R R
R C C R
= + − −
Equating the real & imaginary terms:3414
2
31
CRRR
R
RC
=+ 4
3 1 4
13
0
R
C R R
CR
−= 321
4 3 1
CRR
R R C
=+ 31
13 1 1 3 3 1 1 3 3
1 1 1
2
C R f
CR C RC R C RC R
= → = → =
Max Wien (1866–1938) was a
Prussian physicist. He theoretically
developed the concept of the
Wien-bridge oscillator in 1891. At
that time, Wien did not have a
means of developing electronic
gain, so a workable oscillator could
not be achieved. Based on Wien’s
work, William Hewlett, co-founder
of Hewlett-Packard, was successful
in building a practical Wien-bridge
oscillator in 1939.
19
Mai Linh, PhD
❑Wien-Bridge Oscillator
Wien – Bridge oscillator without amplitude stabilization
A circuit in which output voltage
leads or lags the input voltage is
called a lead-lag network.
@ high frequencies, C
2 acts as short
→ no output V
0.
@ low frequencies, C
1 acts as open →
no output V
0.
At low frequencies, phase angle is
positive and the circuit acts as a lead
network. At high frequencies, phase
angle is negative and circuit acts as a
lag network.
Z
p & Z
s provide frequency selection.1
p
R
Z
jRC
=
+
20
6.2 Op Amp – RC Oscillator Circuits
??????
??????=
1+????????????�??????
??????????????????Va
R1
-
R2
+
CR
R
Vo
C Zs
Zp
(a)
(b)
(c)
(d)
Mai Linh, PhD
Wien – Bridge oscillator without amplitude stabilization
A circuit in which output voltage
leads or lags the input voltage is
called a lead-lag network.
@ high frequencies, C
2 acts as short
→ no output V
0.
@ low frequencies, C
1 acts as open →
no output V
0.
At low frequencies, phase angle is
positive and the circuit acts as a lead
network. At high frequencies, phase
angle is negative and circuit acts as a
lag network.
Z
p & Z
s provide frequency selection.1
p
R
Z
jRC
=
+
20
6.2 Op Amp – RC Oscillator Circuits
??????
??????=
1+????????????�??????
??????????????????Va
R1
-
R2
+
CR
R
Vo
C Zs
Zp
(a)
(b)
(c)
(d)
Mai Linh, PhD
Break the loop @ P to find loop gain L(s):
❑Wien-Bridge Oscillator22
1
11
(s)1(s)(s),1
II
RR
V VAVA
RR
=+= =+
1
(s)
(s)(s) ,
(s)(s)
p
O
sp
Z
VV sj
ZZ
==
+ (j)A
1
3
O
I
p
ps
V
L
V
Z A
A
ZZ
jRC
RC
==
==
+
+−
Imaginary part
must be zero(j) 1
1
3
o
o
o
A
L
jRC
RC
==
+−
Oscillations start @
o if:11
0
oo
o
RC
RC RC
−=→=
Put
o = 1/RC into L(j
o) = 1, hence: A = 3. ( let R
2/R
1 = 2)
To initiate oscillation, choose
�
2
�
1
=2+??????δ is a small number
21
6.2 Op Amp – RC Oscillator CircuitsR1
-
R
+
C
R2
R
A
C
P 2
1
1
R
A
R
=+ Zs
vO
R1
-
R
+
v1
R2
C
R
C
vI
Zp
Mai Linh, PhD
❑Wien-Bridge Oscillator22
1
11
(s)1(s)(s),1
II
RR
V VAVA
RR
=+= =+
1
(s)
(s)(s) ,
(s)(s)
p
O
sp
Z
VV sj
ZZ
==
+ (j)A
1
3
O
I
p
ps
V
L
V
Z A
A
ZZ
jRC
RC
==
==
+
+−
Imaginary part
must be zero(j) 1
1
3
o
o
o
A
L
jRC
RC
==
+−
Oscillations start @
o if:11
0
oo
o
RC
RC RC
−=→=
Put
o = 1/RC into L(j
o) = 1, hence: A = 3. ( let R
2/R
1 = 2)
To initiate oscillation, choose
�
2
�
1
=2+??????δ is a small number
21
6.2 Op Amp – RC Oscillator CircuitsR1
-
R
+
C
R2
R
A
C
P 2
1
1
R
A
R
=+ Zs
vO
R1
-
R
+
v1
R2
C
R
C
vI
Zp
Mai Linh, PhD
❑Wien-Bridge Oscillator
The amplitude of oscillation can be determined and stabilized by using a
nonlinear control network.
Wien – Bridge oscillator with a
limiter used for amplitude control
A Wien-bridge oscillator with an alternative method
for amplitude stabilization.
22
6.2 Op Amp – RC Oscillator Circuits
Mai Linh, PhD
The amplitude of oscillation can be determined and stabilized by using a
nonlinear control network.
Wien – Bridge oscillator with a
limiter used for amplitude control
A Wien-bridge oscillator with an alternative method
for amplitude stabilization.
22
6.2 Op Amp – RC Oscillator Circuits
Mai Linh, PhD
Exercise 6.2: For the circuit in figure.
(a)Find the location of the closed-loop
poles (calculate the value of s).
(b)Find the frequency of oscillation.
(c)Find the amplitude of the output sine
wave (assume that the diode drop is 0.7
V).
❑Wien-Bridge Oscillator
23
6.2 Op Amp – RC Oscillator Circuits
Mai Linh, PhD
(a)Find the location of the closed-loop
poles (calculate the value of s).
(b)Find the frequency of oscillation.
(c)Find the amplitude of the output sine
wave (assume that the diode drop is 0.7
V).
❑Wien-Bridge Oscillator
23
6.2 Op Amp – RC Oscillator Circuits
Mai Linh, PhD
Exercise 6.2 – Sol.
where R = 10 k and C = 16 nF. Thus,
The closed-loop poles are found by setting L(s) = 1; that is, they are the values of s, satisfying
24
❑Wien-Bridge Oscillator
6.2 Op Amp – RC Oscillator Circuits2
1
()()1
P
PS
RZ
La s
RZZ
=+
+
20.3 1 3.03
1
11110
31 sCRR sC
sCRsCR
+=
+++++
5
5
3.03
()
1
31610
1610
Ls
s
s
−
−
=
++
( )
5
5
5
1 10
31610 3.03 0.015
1610 16
s s j
s
−
−
++ ==
Mai Linh, PhD
where R = 10 k and C = 16 nF. Thus,
The closed-loop poles are found by setting L(s) = 1; that is, they are the values of s, satisfying
24
❑Wien-Bridge Oscillator
6.2 Op Amp – RC Oscillator Circuits2
1
()()1
P
PS
RZ
La s
RZZ
=+
+
20.3 1 3.03
1
11110
31 sCRR sC
sCRsCR
+=
+++++
5
5
3.03
()
1
31610
1610
Ls
s
s
−
−
=
++
( )
5
5
5
1 10
31610 3.03 0.015
1610 16
s s j
s
−
−
++ ==
Mai Linh, PhD
❑Wien-Bridge Oscillator
Exercise 6.2 – Sol.
(b) The frequency of oscillation is (10
5
/16) rad/s or approximately 1 kHz.
(c) From the circuit, at the positive peak ??????
�, the voltage at node b will be one diode
drop (0.7 V) above the voltage V
1 which is about 1/3 of V
o; thus V
b = 0.7 + ??????
??????
/3.
Now if we neglect the current through D
2 in comparison with the currents through R
5
and R
6, we find that
From symmetry, we see that the negative peak is equal to the positive peak. Thus,
the output peak-to-peak voltage is 21.36 V.
25
6.2 Op Amp – RC Oscillator Circuits56
ˆ ˆ ˆ
(15) 15 44
ˆ
50.7510.68()
13 333
o b b o b b o
ob
VVV VVV V
VV V
RR
−−− −+
→==+=++=
Mai Linh, PhD
Exercise 6.2 – Sol.
(b) The frequency of oscillation is (10
5
/16) rad/s or approximately 1 kHz.
(c) From the circuit, at the positive peak ??????
�, the voltage at node b will be one diode
drop (0.7 V) above the voltage V
1 which is about 1/3 of V
o; thus V
b = 0.7 + ??????
??????
/3.
Now if we neglect the current through D
2 in comparison with the currents through R
5
and R
6, we find that
From symmetry, we see that the negative peak is equal to the positive peak. Thus,
the output peak-to-peak voltage is 21.36 V.
25
6.2 Op Amp – RC Oscillator Circuits56
ˆ ˆ ˆ
(15) 15 44
ˆ
50.7510.68()
13 333
o b b o b b o
ob
VVV VVV V
VV V
RR
−−− −+
→==+=++=
Mai Linh, PhD
❑Phase-shift Oscillator
✓Phase-shift Oscillator (aka. RC Oscillator) produces a sine wave output signal
using regenerative feedback obtained from the resistor-capacitor (RC) ladder
network.
✓In the figure, the phase-shifter consists of a negative gain amplifier (-K) with a
third order RC ladder network in the feedback.
✓The circuit will oscillate at the frequency for which the phase shift of the RC
network is 180
o
. Only at the frequency will the total phase shift around the loop
be 0
o
or 360
o
.
✓The minimum number of RC sections is 3 because it is capable of producing a
180
o
phase shift at a finite frequency.
26
6.2 Op Amp – RC Oscillator Circuits
Mai Linh, PhD
✓Phase-shift Oscillator (aka. RC Oscillator) produces a sine wave output signal
using regenerative feedback obtained from the resistor-capacitor (RC) ladder
network.
✓In the figure, the phase-shifter consists of a negative gain amplifier (-K) with a
third order RC ladder network in the feedback.
✓The circuit will oscillate at the frequency for which the phase shift of the RC
network is 180
o
. Only at the frequency will the total phase shift around the loop
be 0
o
or 360
o
.
✓The minimum number of RC sections is 3 because it is capable of producing a
180
o
phase shift at a finite frequency.
26
6.2 Op Amp – RC Oscillator Circuits
Mai Linh, PhD
❑Phase-shift OscillatorRC Phase-Shift Network
➢A single R-C network (single-
pole network) has output
voltage “leads” the input
voltage by some angle ≤ 90
o
.
➢By cascading together 3 such
RC networks in series →
produce a phase shift in the
circuit of 180
o
at the chosen
frequency and this forms the
bases of a “RC Oscillator”
otherwise known as aPhase
Shift Oscillator
27
6.2 Op Amp – RC Oscillator Circuits
3 Cascaded High Pass Filters
Mai Linh, PhD
➢A single R-C network (single-
pole network) has output
voltage “leads” the input
voltage by some angle ≤ 90
o
.
➢By cascading together 3 such
RC networks in series →
produce a phase shift in the
circuit of 180
o
at the chosen
frequency and this forms the
bases of a “RC Oscillator”
otherwise known as aPhase
Shift Oscillator
27
6.2 Op Amp – RC Oscillator Circuits
3 Cascaded High Pass Filters
Mai Linh, PhD
❑Phase-shift OscillatorRC Phase-Shift Network
Consider the figure. V
b = (I
2Z
C) + V
o, V
o = I
2R → I
2 = V
o/R
??????
??????=
??????�
??????
??????
�+??????
� = ??????
�1+
1
??????????????????�
Apply KCL12
11
12
b o o o
V V V V
II
R R j RC R R j RC
= + = + + = +
1
1 2
1 11 31
1 2 1
()
o
a b C b o o
VI
VVIZVV V
jCjRCRjRCjC jRCRC
==+=+=+++ =+−
Apply KCL at V
a1 2
41
3
()
ao
oo
VV
I I I
R R j RC RC
= + = + −
Now: 23
6 5 1
1
( ) ( )
o
i a o C a o
I
V V I Z V V
j C j RC RC j RC
= + = + = + − −
23
1
561
1
() ()
o
i
V
V
RCjRCjRC
→=
−+−
equating the imaginary parts to
find the specific
o
(*)1
C
Z
jC
=
28
6.2 Op Amp – RC Oscillator Circuits
Mai Linh, PhD
Consider the figure. V
b = (I
2Z
C) + V
o, V
o = I
2R → I
2 = V
o/R
??????
??????=
??????�
??????
??????
�+??????
� = ??????
�1+
1
??????????????????�
Apply KCL12
11
12
b o o o
V V V V
II
R R j RC R R j RC
= + = + + = +
1
1 2
1 11 31
1 2 1
()
o
a b C b o o
VI
VVIZVV V
jCjRCRjRCjC jRCRC
==+=+=+++ =+−
Apply KCL at V
a1 2
41
3
()
ao
oo
VV
I I I
R R j RC RC
= + = + −
Now: 23
6 5 1
1
( ) ( )
o
i a o C a o
I
V V I Z V V
j C j RC RC j RC
= + = + = + − −
23
1
561
1
() ()
o
i
V
V
RCjRCjRC
→=
−+−
equating the imaginary parts to
find the specific
o
(*)1
C
Z
jC
=
28
6.2 Op Amp – RC Oscillator Circuits
Mai Linh, PhD
❑Phase-shift OscillatorRC Phase-Shift Network
And on equating the real part of equ. (*) we get at
o:2
2
2
55
1 1 (1 30)
1()
()
6( )
i o o o
o
V V V V
RC
RC
RC
= − = − = −
Hence:1
29
o
i
V
V
−
= 3
61 1 1
0 or
() 6 26
oo
oo
f
RCRC RC RC
−=→= =
❖This network is providing attenuation -1/29. Here, the negative sign indicates that
there exists a phase shift of 180⁰.
❖Clearly, to produce a total loop gain of -1, this 3-stage RC network above, the
amplifier gain must be ≥ 29 to compensate for the attenuation of the RC network.
29
6.2 Op Amp – RC Oscillator Circuits
(frequency of oscillation)
Mai Linh, PhD
And on equating the real part of equ. (*) we get at
o:2
2
2
55
1 1 (1 30)
1()
()
6( )
i o o o
o
V V V V
RC
RC
RC
= − = − = −
Hence:1
29
o
i
V
V
−
= 3
61 1 1
0 or
() 6 26
oo
oo
f
RCRC RC RC
−=→= =
❖This network is providing attenuation -1/29. Here, the negative sign indicates that
there exists a phase shift of 180⁰.
❖Clearly, to produce a total loop gain of -1, this 3-stage RC network above, the
amplifier gain must be ≥ 29 to compensate for the attenuation of the RC network.
29
6.2 Op Amp – RC Oscillator Circuits
(frequency of oscillation)
Mai Linh, PhD
30
❑Phase-shift Oscillator
6.2 Op Amp – RC Oscillator Circuits
RC Phase Shift Oscillator Using a Bipolar Transistor
This circuit uses multiple filters to cause a exactly 180° phase shift at
feedback circuit can be produced. The CE Amp., which also has a phase
shift of 180° between base and collector, the filters produce positive
feedback to cause oscillation to take place.
A sine wave of approximately 3V
pp with minimum distortion is produced across
the load resistor R5. The frequency of oscillation is given by:
�
�=
1
2??????6�??????
BJT phase shift oscillator is a useful as a source of low frequency sine waves but does have several drawbacks:
✓There can be a wide difference between the calculated frequency value and the actual frequency produced.
✓The waveform amplitude is generally not well stabilized, so a good wave shape is not guaranteed without
additional circuitry.
✓They are difficult to design in variable frequency form as this would involvegangingtogether either 3 variable
capacitors or 3 large variable resistors, and such components are not readily available.
Fig.: RC Phase Shift Oscillator Using
a Bipolar Transistor
Mai Linh, PhD
❑Phase-shift Oscillator
6.2 Op Amp – RC Oscillator Circuits
RC Phase Shift Oscillator Using a Bipolar Transistor
This circuit uses multiple filters to cause a exactly 180° phase shift at
feedback circuit can be produced. The CE Amp., which also has a phase
shift of 180° between base and collector, the filters produce positive
feedback to cause oscillation to take place.
A sine wave of approximately 3V
pp with minimum distortion is produced across
the load resistor R5. The frequency of oscillation is given by:
�
�=
1
2??????6�??????
BJT phase shift oscillator is a useful as a source of low frequency sine waves but does have several drawbacks:
✓There can be a wide difference between the calculated frequency value and the actual frequency produced.
✓The waveform amplitude is generally not well stabilized, so a good wave shape is not guaranteed without
additional circuitry.
✓They are difficult to design in variable frequency form as this would involvegangingtogether either 3 variable
capacitors or 3 large variable resistors, and such components are not readily available.
Fig.: RC Phase Shift Oscillator Using
a Bipolar Transistor
Mai Linh, PhD
Fig.: A practical phase-shift oscillator with a limiter for
amplitude stabilization
A feedback limiter, consisting of
diodes D
1 & D
2 and resistors R
1, R
2,
R
3, and R
4 for amplitude stabilization.
❑Phase-shift Oscillator
To start oscillations, R
f has to be made
slightly greater than the minimum
required value. It stabilizes more
rapidly and provides sine waves with
more stable amplitude
But, if R
f is made much larger than this
minimum ➔ an increased output distortion.
31
6.2 Op Amp – RC Oscillator Circuits
Mai Linh, PhD
amplitude stabilization
A feedback limiter, consisting of
diodes D
1 & D
2 and resistors R
1, R
2,
R
3, and R
4 for amplitude stabilization.
❑Phase-shift Oscillator
To start oscillations, R
f has to be made
slightly greater than the minimum
required value. It stabilizes more
rapidly and provides sine waves with
more stable amplitude
But, if R
f is made much larger than this
minimum ➔ an increased output distortion.
31
6.2 Op Amp – RC Oscillator Circuits
Mai Linh, PhD
vO1
vO2
X
v
V
+2
2
O
v
v=
V a) A quadrature-oscillator circuit
b) Equivalent circuit at the input of
op amp 2
❑Quadrature Oscillator (Bộ dao động pha vuông góc)
The quadrature oscillator is another type of phase-
shift oscillator. The difference is that the quadrature
oscillator uses an op amp integrator to obtain a full 90°
phase shift from a single RC segment. Because of that
90° phase shift, it needs 2 op-amps to produce both
sine & cosine wave outputs. Note: waveform outputs
are both sine & cosine (quadrature or a 90° phase shift
between op amp outputs).
But, it still produces a usable output voltage.
Amplifier 1 is connected as an inverting Miller
integrator with a limiter in the feedback for
amplitude control. Amplifier 2 is connected as
a noninverting integrator.
32
6.2 Op Amp – RC Oscillator Circuits
Mai Linh, PhD
vO2
X
v
V
+2
2
O
v
v=
V a) A quadrature-oscillator circuit
b) Equivalent circuit at the input of
op amp 2
❑Quadrature Oscillator (Bộ dao động pha vuông góc)
The quadrature oscillator is another type of phase-
shift oscillator. The difference is that the quadrature
oscillator uses an op amp integrator to obtain a full 90°
phase shift from a single RC segment. Because of that
90° phase shift, it needs 2 op-amps to produce both
sine & cosine wave outputs. Note: waveform outputs
are both sine & cosine (quadrature or a 90° phase shift
between op amp outputs).
But, it still produces a usable output voltage.
Amplifier 1 is connected as an inverting Miller
integrator with a limiter in the feedback for
amplitude control. Amplifier 2 is connected as
a noninverting integrator.
32
6.2 Op Amp – RC Oscillator Circuits
Mai Linh, PhD
??????
0=2??????�=
1
�??????
The loop will oscillate at frequency:
33
6.2 Op Amp – RC Oscillator Circuits vO1
vO2
X
v
V
+2
2
O
v
v=
V
a) A quadrature-oscillator circuit
b) Equivalent circuit at the input of
op amp 2
❑Quadrature Oscillator
R
f is made equal to 2R, and
thus –R
f cancels 2R.12
2
10
1
1
0
2
2
(Loop gain)
21
2
2
11
1
()
()
t
Oo
O
o
t
oX
O
x
o
x
vV
vvdt
CRVsRC
Vv
v dt
CRVsRC
V
Ls
VsRC
==→=
=→=
=−
Mai Linh, PhD
0=2??????�=
1
�??????
The loop will oscillate at frequency:
33
6.2 Op Amp – RC Oscillator Circuits vO1
vO2
X
v
V
+2
2
O
v
v=
V
a) A quadrature-oscillator circuit
b) Equivalent circuit at the input of
op amp 2
❑Quadrature Oscillator
R
f is made equal to 2R, and
thus –R
f cancels 2R.12
2
10
1
1
0
2
2
(Loop gain)
21
2
2
11
1
()
()
t
Oo
O
o
t
oX
O
x
o
x
vV
vvdt
CRVsRC
Vv
v dt
CRVsRC
V
Ls
VsRC
==→=
=→=
=−
Mai Linh, PhD
❑Active – Filter – Tuned Oscillator
Assume the oscillations have already started. The output of the bandpass filter will be a sine
wave whose frequency is equal to the center frequency of the filter.
The sine-wave signal is fed to the limiter and then produces a square wave.
Block diagram of Active-Filter-Tuned Oscillator
Hard limiter
(op-amp)
34
6.2 Op Amp – RC Oscillator Circuits
Mai Linh, PhD
Assume the oscillations have already started. The output of the bandpass filter will be a sine
wave whose frequency is equal to the center frequency of the filter.
The sine-wave signal is fed to the limiter and then produces a square wave.
Block diagram of Active-Filter-Tuned Oscillator
Hard limiter
(op-amp)
34
6.2 Op Amp – RC Oscillator Circuits
Mai Linh, PhD
A practical implementation of the active-filter-tuned oscillator
35
Mai Linh, PhD
35
Mai Linh, PhD
6.3 LC and Crystal Oscillators
✓Oscillators utilizing transistors (FETs or BJTs), with LC circuits (called
LC oscillators) or crystal oscillators are used in the frequency range of
100 kHz to hundreds of megahertz (greater than RC types)
✓LC oscillators are difficult to tune over wide ranges.
✓Crystal oscillators operate at a single frequency.
36
Mai Linh, PhD
✓Oscillators utilizing transistors (FETs or BJTs), with LC circuits (called
LC oscillators) or crystal oscillators are used in the frequency range of
100 kHz to hundreds of megahertz (greater than RC types)
✓LC oscillators are difficult to tune over wide ranges.
✓Crystal oscillators operate at a single frequency.
36
Mai Linh, PhD
37
6.3 LC and Crystal Oscillators
Because most of the current flowing in the oscillator is flowing just around the resonant tank circuit
rather than though the amplifier section of the oscillator, LC oscillators generally produce a sine wave
with very little amplifier sourced distortion.
Resonance occurs when an LC circuit is driven from an external
source at an angular frequency ω
0 (called resonant frequency), the
reactances of L and C cancel each other out, allowing a maximum
transfer of energy between the two components.
??????
�=
1
????????????
or �
�=
1
2?????????????????? Fig.: LC Tank Circuit
The Tuned LC (Tank) Circuit: consists of a C and L in parallel, that use magnetic resonance to
store electrical energy oscillating at a certain resonating frequency. It is used to produce electric
oscillations of any desired frequency. This tank circuit is applied in circuits such as oscillators,
filters, tuners and frequency mixers.
A tank circuit acts as a resonant circuit because of the movement of charges between C and L. The
natural frequency or the frequency of the oscillations in the circuit is determined by the C and L
values.
Mai Linh, PhD
6.3 LC and Crystal Oscillators
Because most of the current flowing in the oscillator is flowing just around the resonant tank circuit
rather than though the amplifier section of the oscillator, LC oscillators generally produce a sine wave
with very little amplifier sourced distortion.
Resonance occurs when an LC circuit is driven from an external
source at an angular frequency ω
0 (called resonant frequency), the
reactances of L and C cancel each other out, allowing a maximum
transfer of energy between the two components.
??????
�=
1
????????????
or �
�=
1
2?????????????????? Fig.: LC Tank Circuit
The Tuned LC (Tank) Circuit: consists of a C and L in parallel, that use magnetic resonance to
store electrical energy oscillating at a certain resonating frequency. It is used to produce electric
oscillations of any desired frequency. This tank circuit is applied in circuits such as oscillators,
filters, tuners and frequency mixers.
A tank circuit acts as a resonant circuit because of the movement of charges between C and L. The
natural frequency or the frequency of the oscillations in the circuit is determined by the C and L
values.
Mai Linh, PhD
❑LC-Tuned Oscillators
LC-tuned oscillators: Colpitts
LC-tuned oscillators: Hartley
A parallel LC circuit connected between C & B (or between D & G if a FET is used) with a fraction of the
tuned-circuit voltage fed to the E terminal.
Frequency of operation:
38
6.3 LC and Crystal Oscillators12
12
1
o
CC
L
CC
=
+
12
1()
o
LLC=+
Each Cap causes
90
o
phase shift →
total 180
o
phase
shift.
Feedback fraction: 2
1
C
C
= 1
2
L
L
=
Invented in 1915
Invented in 1918 Edwin H. Colpitts
1872 – 1949
Ralph Hartley
1888 – 1970
Each L causes
90
o
phase shift
→ total 180
o
phase shift.C2
L
C1
R
Vcb
+
_Veb
+
-
IC
Tank circuit L2R
C
L1
Vcb
+
_
IC
Veb
+
-
Tank circuit
Mai Linh, PhD
LC-tuned oscillators: Colpitts
LC-tuned oscillators: Hartley
A parallel LC circuit connected between C & B (or between D & G if a FET is used) with a fraction of the
tuned-circuit voltage fed to the E terminal.
Frequency of operation:
38
6.3 LC and Crystal Oscillators12
12
1
o
CC
L
CC
=
+
12
1()
o
LLC=+
Each Cap causes
90
o
phase shift →
total 180
o
phase
shift.
Feedback fraction: 2
1
C
C
= 1
2
L
L
=
Invented in 1915
Invented in 1918 Edwin H. Colpitts
1872 – 1949
Ralph Hartley
1888 – 1970
Each L causes
90
o
phase shift
→ total 180
o
phase shift.C2
L
C1
R
Vcb
+
_Veb
+
-
IC
Tank circuit L2R
C
L1
Vcb
+
_
IC
Veb
+
-
Tank circuit
Mai Linh, PhD
39
❑Complete circuit for a Hartley oscillator
6.3 LC and Crystal Oscillators
The top of L1 is connecting to +V
CC means
it is connected to ground via the very low
impedance of C5 for ac signals. Therefore,
waveform X across L1, and waveform Y
across the whole circuit tank are in phase.
As a common base amplifier is being
used, the collector and emitter signals are
also in phase, hence the tank circuit is
providing positive feedback.
Fig.: Hartley Oscillator using CB amplifier
Tapped inductorC3
Collector
Emitter
To 0 V via
+VCC & C5
L2
L1
X
Y OUT
C3L2
L1
X
Y
C2C1
C4
R1
R2
R3
TR1
+VCC
+
C5
Feedback
Fig.: Waves in LC Tank Circuit
Mai Linh, PhD
❑Complete circuit for a Hartley oscillator
6.3 LC and Crystal Oscillators
The top of L1 is connecting to +V
CC means
it is connected to ground via the very low
impedance of C5 for ac signals. Therefore,
waveform X across L1, and waveform Y
across the whole circuit tank are in phase.
As a common base amplifier is being
used, the collector and emitter signals are
also in phase, hence the tank circuit is
providing positive feedback.
Fig.: Hartley Oscillator using CB amplifier
Tapped inductorC3
Collector
Emitter
To 0 V via
+VCC & C5
L2
L1
X
Y OUT
C3L2
L1
X
Y
C2C1
C4
R1
R2
R3
TR1
+VCC
+
C5
Feedback
Fig.: Waves in LC Tank Circuit
Mai Linh, PhD
40
❑Alternative Hartley Designs
6.3 LC and Crystal Oscillators
Hartley Oscillator using CE
amplifier
Hartley Oscillator with Tuned Feedback
The top and bottom ends of the
tapped inductor L1/L2 are in anti-
phase as in this design.
The base in a CE amplifier is also
in anti-phase with the collector
waveform, resulting in positive
feedback via C2.
Automatic class C bias is again
used but in this circuit the value of
the emitter decoupling capacitor C
E
will be critical, and smaller than in
a normal class A amplifier.
It will only partially decouple R
E, the time constant of R
E/C
E
controlling the amount of class C bias applied.
The tapping on the inductor is connected to
ground rather than +V
CC, and two DC blocking
capacitors are used to eliminate any DC from
the tuned circuit. The collector load is now
provided by a RF choke which simply provides
a high impedance at the oscillator frequency.CE
C4
RFC
R2
RE
TR1
+VCC
OUT
C1
L2
L1
X
Y
C3
R3
C2
R1
Feedback C2
CE
C3
R1
R2
RE
TR1
+VCC
+
C5
OUT
C1
L2
L1
X
Y
Feedback
Mai Linh, PhD
❑Alternative Hartley Designs
6.3 LC and Crystal Oscillators
Hartley Oscillator using CE
amplifier
Hartley Oscillator with Tuned Feedback
The top and bottom ends of the
tapped inductor L1/L2 are in anti-
phase as in this design.
The base in a CE amplifier is also
in anti-phase with the collector
waveform, resulting in positive
feedback via C2.
Automatic class C bias is again
used but in this circuit the value of
the emitter decoupling capacitor C
E
will be critical, and smaller than in
a normal class A amplifier.
It will only partially decouple R
E, the time constant of R
E/C
E
controlling the amount of class C bias applied.
The tapping on the inductor is connected to
ground rather than +V
CC, and two DC blocking
capacitors are used to eliminate any DC from
the tuned circuit. The collector load is now
provided by a RF choke which simply provides
a high impedance at the oscillator frequency.CE
C4
RFC
R2
RE
TR1
+VCC
OUT
C1
L2
L1
X
Y
C3
R3
C2
R1
Feedback C2
CE
C3
R1
R2
RE
TR1
+VCC
+
C5
OUT
C1
L2
L1
X
Y
Feedback
Mai Linh, PhD
❑Equivalent circuit of the Colpitts oscillator
41
6.3 LC and Crystal Oscillators
Frequency of oscillation:
For oscillators to start, the loop gain must be
made greater than unity: �
??????�>Τ??????
2??????
1
➢KCL @ node C
s = jω
For oscillations= 0 = 0( )
2
2 1 2
1
10
m
sCVgVsCsLCV
R
++++ =
( )( )
32
12 2 1 2
1
/0
m
sLCCsLCRsCCg
R
+ ++++=
2
32
1 2 12
1
() 0
m
LC
g jCCLCC
RR
+−++− =
12
12
1
o
CC
L
CC
=
+
r
_
+
g
m
V
V
R
L
C2
C1
CVO
VC = V(1+s
2
LC2)
sC2V
sC2V V
C1
L
C2
R
+
VO
+
_
_
Fig: Simplistic Small signal model. Assume r
π =
Note: biasing resistors are not shown
Mai Linh, PhD
41
6.3 LC and Crystal Oscillators
Frequency of oscillation:
For oscillators to start, the loop gain must be
made greater than unity: �
??????�>Τ??????
2??????
1
➢KCL @ node C
s = jω
For oscillations= 0 = 0( )
2
2 1 2
1
10
m
sCVgVsCsLCV
R
++++ =
( )( )
32
12 2 1 2
1
/0
m
sLCCsLCRsCCg
R
+ ++++=
2
32
1 2 12
1
() 0
m
LC
g jCCLCC
RR
+−++− =
12
12
1
o
CC
L
CC
=
+
r
_
+
g
m
V
V
R
L
C2
C1
CVO
VC = V(1+s
2
LC2)
sC2V
sC2V V
C1
L
C2
R
+
VO
+
_
_
Fig: Simplistic Small signal model. Assume r
π =
Note: biasing resistors are not shown
Mai Linh, PhD
❑Complete circuit for a Colpitts oscillator
✓Oscillator amplitude:
oLC tuned oscillators are known as self-limiting oscillators. (As oscillation grown
in amplitude, transistor gain is reduced below its small-signal value).
oOutput voltage signal will be a sinusoid of high purity because of the filtering
action of the LC tuned circuit.
42
6.3 LC and Crystal Oscillators
Fig. A Colpitts Oscillator using
a CE Amp.
Fig.: The Colpitts Oscillator using a
CB Amp.C4
RE
TR1
RFC
CE
C2
C3
C1
L
R1
R2
OUTPUT
+VCC C4
C1
TR1
C2
C3
L
R1
R2
OUTPUT
+ C5
RE
+VCC
Values of C2 & C3
are chosen so that
the ratio of the
values produces the
necessary proportion
of feedback signal.
Mai Linh, PhD
✓Oscillator amplitude:
oLC tuned oscillators are known as self-limiting oscillators. (As oscillation grown
in amplitude, transistor gain is reduced below its small-signal value).
oOutput voltage signal will be a sinusoid of high purity because of the filtering
action of the LC tuned circuit.
42
6.3 LC and Crystal Oscillators
Fig. A Colpitts Oscillator using
a CE Amp.
Fig.: The Colpitts Oscillator using a
CB Amp.C4
RE
TR1
RFC
CE
C2
C3
C1
L
R1
R2
OUTPUT
+VCC C4
C1
TR1
C2
C3
L
R1
R2
OUTPUT
+ C5
RE
+VCC
Values of C2 & C3
are chosen so that
the ratio of the
values produces the
necessary proportion
of feedback signal.
Mai Linh, PhD
43
❑Application example:
6.3 LC and Crystal Oscillators2k4k
Microphone
Q1
5 V
60
nH
X
2mF
Antenna
30k
2k
2mF
200
80pF
80pF
Oscillator
Q2
FM output
The figure shows an FM transmitter operating in ~ 100 MHz. A self-biased
CE stage amplifies the microphone signal, applying the result to an
oscillator. The audio signal “modulates” (varies) the frequency of the
oscillator, creating an FM output.
Fig.: Simple FM transmitter
Mai Linh, PhD
❑Application example:
6.3 LC and Crystal Oscillators2k4k
Microphone
Q1
5 V
60
nH
X
2mF
Antenna
30k
2k
2mF
200
80pF
80pF
Oscillator
Q2
FM output
The figure shows an FM transmitter operating in ~ 100 MHz. A self-biased
CE stage amplifies the microphone signal, applying the result to an
oscillator. The audio signal “modulates” (varies) the frequency of the
oscillator, creating an FM output.
Fig.: Simple FM transmitter
Mai Linh, PhD
44
❑Crystal Oscillators
6.3 LC and Crystal Oscillators
The most stable and accurate type of feedback oscillator uses a
piezoelectric crystal in the feedback loop to control the
frequency.
Quartz is one type of crystalline substance found in nature that
exhibits a property called the piezoelectric effect.
When a changing mechanical stress is applied across the crystal
to cause it to vibrate, a voltage develops at the frequency of
mechanical vibration.
Conversely, when an ac voltage is applied across the crystal, it
vibrates at the frequency of the applied voltage.
The greatest vibration occurs at the crystal’s natural resonant
frequency, which is determined by the physical dimensions and
by the way the crystal is cut.
Mai Linh, PhD
❑Crystal Oscillators
6.3 LC and Crystal Oscillators
The most stable and accurate type of feedback oscillator uses a
piezoelectric crystal in the feedback loop to control the
frequency.
Quartz is one type of crystalline substance found in nature that
exhibits a property called the piezoelectric effect.
When a changing mechanical stress is applied across the crystal
to cause it to vibrate, a voltage develops at the frequency of
mechanical vibration.
Conversely, when an ac voltage is applied across the crystal, it
vibrates at the frequency of the applied voltage.
The greatest vibration occurs at the crystal’s natural resonant
frequency, which is determined by the physical dimensions and
by the way the crystal is cut.
Mai Linh, PhD
❑Crystal Oscillators
✓Piezoelectric crystal (is high-Q device)
45
6.3 LC and Crystal Oscillators
frequency stability is determined by the ability of the
circuit to select a particular frequency
in tuned circuits this is described by the quality factor, Q
piezoelectric crystals act like resonant circuits with a
very high Q – as high as 100,000
Crystal reactance versus frequency( )
2
2
,2
11
,
22
,
LS S
CS CP
SP
SCP
S S LS CS P
S CP
RRZL
ZZ
fC fC
ZZ
ZRZZZ
ZZ
==
==
=+− =
+
Impedances:
Mai Linh, PhD
✓Piezoelectric crystal (is high-Q device)
45
6.3 LC and Crystal Oscillators
frequency stability is determined by the ability of the
circuit to select a particular frequency
in tuned circuits this is described by the quality factor, Q
piezoelectric crystals act like resonant circuits with a
very high Q – as high as 100,000
Crystal reactance versus frequency( )
2
2
,2
11
,
22
,
LS S
CS CP
SP
SCP
S S LS CS P
S CP
RRZL
ZZ
fC fC
ZZ
ZRZZZ
ZZ
==
==
=+− =
+
Impedances:
Mai Linh, PhD
❑Crystal Oscillators
There are two resonance frequencies:
✓A series resonance at ??????
??????=Τ1????????????
??????
✓A parallel resonance at ??????
�=ൗ1??????
�??????��
�??????+��
Let ??????=????????????, we have:
????????????=
??????
2
????????????
??????+1
�??????
??????
1
????????????
�
??????
2
????????????
??????+1
�??????
??????
+
1
????????????
�
=
??????
2
????????????
??????+1
????????????
�??????
2
????????????
??????+1+????????????
??????
=
1
????????????
�
??????
2
+Τ1????????????
??????
??????
2
+ Τ(??????
�+??????
??????)????????????
????????????
�
(assume R
s = 0 because the Q factor is very high)
46
6.3 LC and Crystal Oscillators22
22
1
()
s
pp
Zjj
C
−
=−
−
Crystal impedance:
Mai Linh, PhD
There are two resonance frequencies:
✓A series resonance at ??????
??????=Τ1????????????
??????
✓A parallel resonance at ??????
�=ൗ1??????
�??????��
�??????+��
Let ??????=????????????, we have:
????????????=
??????
2
????????????
??????+1
�??????
??????
1
????????????
�
??????
2
????????????
??????+1
�??????
??????
+
1
????????????
�
=
??????
2
????????????
??????+1
????????????
�??????
2
????????????
??????+1+????????????
??????
=
1
????????????
�
??????
2
+Τ1????????????
??????
??????
2
+ Τ(??????
�+??????
??????)????????????
????????????
�
(assume R
s = 0 because the Q factor is very high)
46
6.3 LC and Crystal Oscillators22
22
1
()
s
pp
Zjj
C
−
=−
−
Crystal impedance:
Mai Linh, PhD
✓The crystal reactance is inductive over the narrow frequency
band between ??????
?????? and ??????
�. Thus, may use the crystal to replace
the inductor of the Colpitts oscillator.
✓→ Colpitts crystal oscillator
47
6.3 LC and Crystal Oscillators
Note:
➢??????
�>??????
??????
➢Since ??????
�≫??????
?????? then ??????
�≈??????
??????
➢Resonance frequency: ??????
0≈Τ1????????????
??????=??????
??????
C
s << C
p , C
1, C
2 1
os
s
LC
= V
Cs
Cp
L
R
C
gmV
+
_ C1C2
Mai Linh, PhD
band between ??????
?????? and ??????
�. Thus, may use the crystal to replace
the inductor of the Colpitts oscillator.
✓→ Colpitts crystal oscillator
47
6.3 LC and Crystal Oscillators
Note:
➢??????
�>??????
??????
➢Since ??????
�≫??????
?????? then ??????
�≈??????
??????
➢Resonance frequency: ??????
0≈Τ1????????????
??????=??????
??????
C
s << C
p , C
1, C
2 1
os
s
LC
= V
Cs
Cp
L
R
C
gmV
+
_ C1C2
Mai Linh, PhD
6.4 Bistable Multivibrators
❑ Multivibrators (3 types)
✓Bistable: two stable states
✓Monostable: one stable state
✓Astable: no stable state
❑Bistable
✓Has two stable states
✓Can be obtained by using an amplifier with
positive-feedback loop having a loop
gain greater than unity. i.e. ????????????>1 where
??????=Τ�
1(�
1+�
2)
48
(Các bộ đa hài ổn định kép)
Fig.: A physical analogy for the operation
of the bistable circuit. The ball cannot remain at the
top of the hill for any length of time
Mai Linh, PhD
❑ Multivibrators (3 types)
✓Bistable: two stable states
✓Monostable: one stable state
✓Astable: no stable state
❑Bistable
✓Has two stable states
✓Can be obtained by using an amplifier with
positive-feedback loop having a loop
gain greater than unity. i.e. ????????????>1 where
??????=Τ�
1(�
1+�
2)
48
(Các bộ đa hài ổn định kép)
Fig.: A physical analogy for the operation
of the bistable circuit. The ball cannot remain at the
top of the hill for any length of time
Mai Linh, PhD
❑ Transfer characteristics of the Bistable circuit
❑ Triggering the Bistable circuit
❑The bistable circuit as a memory element
❑Schmitt trigger
How to make the bistable circuit change state?
49
6.4 Bistable Multivibrators
(a) The bistable circuit(b) The transfer characteristic
for increasing v
I
(c) The transfer characteristic
for decreasing v
I
(d) complete transfer
characteristics
Mai Linh, PhD
❑ Triggering the Bistable circuit
❑The bistable circuit as a memory element
❑Schmitt trigger
How to make the bistable circuit change state?
49
6.4 Bistable Multivibrators
(a) The bistable circuit(b) The transfer characteristic
for increasing v
I
(c) The transfer characteristic
for decreasing v
I
(d) complete transfer
characteristics
Mai Linh, PhD
❑A Bistable circuit with Noninverting Transfer characteristics
50
6.4 Bistable Multivibrators
A bistable circuit: applying v
I through R
1.The transfer characteristic of the circuit
Apply the input signal v
I (the
trigger signal) to the terminal of R
1
v
O = L
+, v
+ = 0, and v
I = V
TL.
➢If the circuit is in the positive
stable state with v
O = L
+,
positive values for v
I will have
no effect. To trigger the circuit
into the L
− state, v
Imust be
made negative and of such a
value as to make v
+ decrease
below zero.
v
O = L
− and v
+ = 0
Positive triggering signal v
I
(> V
TH) causes the circuit to
switch to the positive state (v
O
goes from L
− to L
+). → the
transfer characteristic of this
circuit is noninverting.21
1 2 1 2
IO
RR
vv v
RRRR
+
=+
++ ( )
12
/
TL
V LRR
+
=− ( )
12
/
TL
V LRR
−
=−
Mai Linh, PhD
50
6.4 Bistable Multivibrators
A bistable circuit: applying v
I through R
1.The transfer characteristic of the circuit
Apply the input signal v
I (the
trigger signal) to the terminal of R
1
v
O = L
+, v
+ = 0, and v
I = V
TL.
➢If the circuit is in the positive
stable state with v
O = L
+,
positive values for v
I will have
no effect. To trigger the circuit
into the L
− state, v
Imust be
made negative and of such a
value as to make v
+ decrease
below zero.
v
O = L
− and v
+ = 0
Positive triggering signal v
I
(> V
TH) causes the circuit to
switch to the positive state (v
O
goes from L
− to L
+). → the
transfer characteristic of this
circuit is noninverting.21
1 2 1 2
IO
RR
vv v
RRRR
+
=+
++ ( )
12
/
TL
V LRR
+
=− ( )
12
/
TL
V LRR
−
=−
Mai Linh, PhD
51
Exercise 6.3: The op amp in the bistable circuit shown in Fig. has output saturation
voltages of 13V.
Design the circuit to obtain threshold voltages of 5 V. For R
1 = 10 k, find the
value required for R
2.
6.4 Bistable Multivibrators
The bistable circuit with the negative input terminal
Mai Linh, PhD
Exercise 6.3: The op amp in the bistable circuit shown in Fig. has output saturation
voltages of 13V.
Design the circuit to obtain threshold voltages of 5 V. For R
1 = 10 k, find the
value required for R
2.
6.4 Bistable Multivibrators
The bistable circuit with the negative input terminal
Mai Linh, PhD
52
Exercise 6.3: Sol.
V
TH = |V
TL| = β|L|
6.4 Bistable Multivibrators
R
1/R
2 = 1.6
➔ R
2 = 16 k1
12
5 13
R
RR
=
+
Mai Linh, PhD
Exercise 6.3: Sol.
V
TH = |V
TL| = β|L|
6.4 Bistable Multivibrators
R
1/R
2 = 1.6
➔ R
2 = 16 k1
12
5 13
R
RR
=
+
Mai Linh, PhD
53
Exercise 6.4: If the op-amp in the circuit of Fig. has 10 V output saturation
levels.
Design the circuit to obtain threshold voltages of 5V. Give suitable component
values.
6.4 Bistable Multivibrators
Bistable circuit derived from the positive-feedback loop:
apply v
I through R
1.
Mai Linh, PhD
Exercise 6.4: If the op-amp in the circuit of Fig. has 10 V output saturation
levels.
Design the circuit to obtain threshold voltages of 5V. Give suitable component
values.
6.4 Bistable Multivibrators
Bistable circuit derived from the positive-feedback loop:
apply v
I through R
1.
Mai Linh, PhD
54
Exercise 6.4: Sol.
V
TH = V
TL =
??????
1
??????2
|L| → 5=
??????
1
??????2
×10 → �
2=2�
1
Possible choice: R
1 = 10 k; R
2 = 20 k
6.4 Bistable Multivibrators
Mai Linh, PhD
Exercise 6.4: Sol.
V
TH = V
TL =
??????
1
??????2
|L| → 5=
??????
1
??????2
×10 → �
2=2�
1
Possible choice: R
1 = 10 k; R
2 = 20 k
6.4 Bistable Multivibrators
Mai Linh, PhD
55
Exercise 6.5: Consider a bistable circuit with a noninverting transfer characteristic
and let L
+ = −L
− = 10 V and V
TH = −V
TL =5 V. If v
I is a triangular wave with a 0-V
average, a 10-V peak amplitude, and a 1 ms period, sketch the waveform of v
O. Find
the time interval between the zero crossings of v
I and v
O.
6.4 Bistable Multivibrators
Mai Linh, PhD
Exercise 6.5: Consider a bistable circuit with a noninverting transfer characteristic
and let L
+ = −L
− = 10 V and V
TH = −V
TL =5 V. If v
I is a triangular wave with a 0-V
average, a 10-V peak amplitude, and a 1 ms period, sketch the waveform of v
O. Find
the time interval between the zero crossings of v
I and v
O.
6.4 Bistable Multivibrators
Mai Linh, PhD
56
6.4 Bistable Multivibrators
Exercise 6.5: Sol.
Mai Linh, PhD
6.4 Bistable Multivibrators
Exercise 6.5: Sol.
Mai Linh, PhD
57
Exercise 6.6: In the circuit of below Fig. and let L
+ = −L
− = 10 V and R
1 = 1 k.
Find a value for R
2 that gives a hysteresis of 100-mV width.
6.4 Bistable Multivibrators
Mai Linh, PhD
Exercise 6.6: In the circuit of below Fig. and let L
+ = −L
− = 10 V and R
1 = 1 k.
Find a value for R
2 that gives a hysteresis of 100-mV width.
6.4 Bistable Multivibrators
Mai Linh, PhD
58
6.4 Bistable Multivibrators
Exercise 6.6: Sol.
|V
TH| = |V
TL| = 100 mV /2 = 50 mV
50×10
-3
= 10×R
1/R
2 → R
2/R
1 = 10/0.05 → R
2 = 200R
1
Case R
1 = 1 k → R
2 = 200 k
Mai Linh, PhD
6.4 Bistable Multivibrators
Exercise 6.6: Sol.
|V
TH| = |V
TL| = 100 mV /2 = 50 mV
50×10
-3
= 10×R
1/R
2 → R
2/R
1 = 10/0.05 → R
2 = 200R
1
Case R
1 = 1 k → R
2 = 200 k
Mai Linh, PhD
❑Applications of the Bistable circuit as a comparator
59
6.4 Bistable Multivibrators
Block diagram representation for
a comparator Transfer characteristic of
comparator
Comparator characteristic with
hysteresis
The comparator is an analog-circuit building block that is used in a variety of
applications ranging from detecting the level of an input signal relative to a preset
threshold value to the design of analog-to-digital (A/D) converters.
Mai Linh, PhD
59
6.4 Bistable Multivibrators
Block diagram representation for
a comparator Transfer characteristic of
comparator
Comparator characteristic with
hysteresis
The comparator is an analog-circuit building block that is used in a variety of
applications ranging from detecting the level of an input signal relative to a preset
threshold value to the design of analog-to-digital (A/D) converters.
Mai Linh, PhD
Illustrating the using hysteresis in the comparator characteristics as a
means of rejecting interference
60
6.4 Bistable Multivibrators
Design a circuit that detects and counts the zero
crossings of an arbitrary waveform.
❖If the signal has interference superimposed on
it (say of a frequency much higher than that
of the signal.) → counting would obviously
be in error.
❖This problem can be solved by introducing
hysteresis of appropriate width in the
comparator characteristics.
Fig.: Illustrating the use of hysteresis in the comparator
characteristics as a means of rejecting interference.
Mai Linh, PhD
means of rejecting interference
60
6.4 Bistable Multivibrators
Design a circuit that detects and counts the zero
crossings of an arbitrary waveform.
❖If the signal has interference superimposed on
it (say of a frequency much higher than that
of the signal.) → counting would obviously
be in error.
❖This problem can be solved by introducing
hysteresis of appropriate width in the
comparator characteristics.
Fig.: Illustrating the use of hysteresis in the comparator
characteristics as a means of rejecting interference.
Mai Linh, PhD
❑Applications of the Bistable circuit as a comparator
3bit ADC
61
6.4 Bistable Multivibrators
Mai Linh, PhD
3bit ADC
61
6.4 Bistable Multivibrators
Mai Linh, PhD
62
6.4 Bistable Multivibrators
Exercise 6.7: Consider an op amp having saturation levels of 12 V used without
feedback, with the inverting input terminal connected to +3 V and the
noninverting input terminal connected to v
I . Characterize its operation as a
comparator. What are L
+, L
−, and V
R, as defined in Fig.
Comparator having a reference, or threshold, voltage V
R.
Mai Linh, PhD
6.4 Bistable Multivibrators
Exercise 6.7: Consider an op amp having saturation levels of 12 V used without
feedback, with the inverting input terminal connected to +3 V and the
noninverting input terminal connected to v
I . Characterize its operation as a
comparator. What are L
+, L
−, and V
R, as defined in Fig.
Comparator having a reference, or threshold, voltage V
R.
Mai Linh, PhD
63
6.4 Bistable Multivibrators
Exercise 6.7: Sol.
A comparator with a threshold of 3 V and output levels of ±12 V.
Mai Linh, PhD
6.4 Bistable Multivibrators
Exercise 6.7: Sol.
A comparator with a threshold of 3 V and output levels of ±12 V.
Mai Linh, PhD
❑ Making the output levels more precise
The output levels of the bistable circuit can be made more precise than the saturation voltages of
the op amp are by cascading the op amp with a limiter circuit.
64
6.4 Bistable Multivibrators
Fig.: Limiter circuits are used to obtain more precise output levels for the bistable circuit. In both
circuits the value of R should be chosen to yield the current required for the proper operation of the
zener diodes. (a) For this circuit L
+ = V
Z
1
+ V
D and L
– = –(V
Z
2
+ V
D), where V
D is the forward diode
drop. (b) For this circuit L
+ = V
Z + V
D
1
+ V
D
2
and L
– = –(V
Z + V
D
3
+ V
D
4
).
Mai Linh, PhD
The output levels of the bistable circuit can be made more precise than the saturation voltages of
the op amp are by cascading the op amp with a limiter circuit.
64
6.4 Bistable Multivibrators
Fig.: Limiter circuits are used to obtain more precise output levels for the bistable circuit. In both
circuits the value of R should be chosen to yield the current required for the proper operation of the
zener diodes. (a) For this circuit L
+ = V
Z
1
+ V
D and L
– = –(V
Z
2
+ V
D), where V
D is the forward diode
drop. (b) For this circuit L
+ = V
Z + V
D
1
+ V
D
2
and L
– = –(V
Z + V
D
3
+ V
D
4
).
Mai Linh, PhD
6.5 Generation of square and Triangular wave forms using astable
multivibrators
❑ Astable multivibrators (Đa hài không bền/Đa hài tự dao động)
Connecting a bistable multivibrator with inverting transfer characteristics in a
feedback loop with a circuit results in a square-wave generator.
65
Fig.: (a) Connecting a bistable multivibrator with
inverting transfer characteristics in a feedback loop with
an RC circuit results in a square-wave generator.
Mai Linh, PhD
multivibrators
❑ Astable multivibrators (Đa hài không bền/Đa hài tự dao động)
Connecting a bistable multivibrator with inverting transfer characteristics in a
feedback loop with a circuit results in a square-wave generator.
65
Fig.: (a) Connecting a bistable multivibrator with
inverting transfer characteristics in a feedback loop with
an RC circuit results in a square-wave generator.
Mai Linh, PhD
❑ Square-wave generator
66( )
/
,
t
vLLLeCR
−
− + + −
=−− = 1
atvLtT
−−
== 12
1(/) 1(/)
ln ,ln
11
LL LL
TT
−+ +−
−−
==
−− 12
1
2ln
1
TTT
+
=+=
−
Fig.: (Continued) (b) The circuit obtained when the bistable multivibrator is
implemented with the circuit of Fig. (a) – slide#43. (c) Waveforms at various
nodes of the circuit in (b). This circuit is called an astable multivibrator.
Mai Linh, PhD
66( )
/
,
t
vLLLeCR
−
− + + −
=−− = 1
atvLtT
−−
== 12
1(/) 1(/)
ln ,ln
11
LL LL
TT
−+ +−
−−
==
−− 12
1
2ln
1
TTT
+
=+=
−
Fig.: (Continued) (b) The circuit obtained when the bistable multivibrator is
implemented with the circuit of Fig. (a) – slide#43. (c) Waveforms at various
nodes of the circuit in (b). This circuit is called an astable multivibrator.
Mai Linh, PhD
67
6.5 Astable multivibrators
Exercise 6.8: For the circuit in Fig., let the op-amp saturation voltages be 10 V,
R
1 =100 k, R
2 = R = 1 M , and C = 0.01 μF. Find the frequency of oscillation.
Mai Linh, PhD
6.5 Astable multivibrators
Exercise 6.8: For the circuit in Fig., let the op-amp saturation voltages be 10 V,
R
1 =100 k, R
2 = R = 1 M , and C = 0.01 μF. Find the frequency of oscillation.
Mai Linh, PhD
68
Exercise 6.8: Sol.:
Period T:
= 0.00365 (sec) → f
o = 1/T = 1/0.00365 274 (Hz)1
12
100
0.091/
1001000
R
VV
RR
== =
++ 1
2ln
1
T
+
=
− 66 1.091
20.011010ln
10.091
−
=
−
Mai Linh, PhD
6.5 Astable multivibrators
Exercise 6.8: Sol.:
Period T:
= 0.00365 (sec) → f
o = 1/T = 1/0.00365 274 (Hz)1
12
100
0.091/
1001000
R
VV
RR
== =
++ 1
2ln
1
T
+
=
− 66 1.091
20.011010ln
10.091
−
=
−
Mai Linh, PhD
6.5 Astable multivibrators
v+
v– 69
Exercise 6.9: Consider the circuit shown in below Fig. Given L
+ = −L
− = 12 V, and the
diode voltage as a constant denoted V
D.
Find an expression for frequency as a function of V
D.
If V
D = 0.70 V at 25°C with a TC of –2 mV/°C, find the frequency at 0°C, 25°C, 50°C, and
100°C.
Note that the output of this circuit can be sent to a
remotely connected frequency meter to provide a
digital readout of temperature.
Mai Linh, PhD
6.5 Astable multivibrators
v– 69
Exercise 6.9: Consider the circuit shown in below Fig. Given L
+ = −L
− = 12 V, and the
diode voltage as a constant denoted V
D.
Find an expression for frequency as a function of V
D.
If V
D = 0.70 V at 25°C with a TC of –2 mV/°C, find the frequency at 0°C, 25°C, 50°C, and
100°C.
Note that the output of this circuit can be sent to a
remotely connected frequency meter to provide a
digital readout of temperature.
Mai Linh, PhD
6.5 Astable multivibrators
70
Exercise 6.9: Sol.:v+
v–
This is a bistable multivibrator which generates
square waveform.
So the charging & discharge intervals:
T
1 = T
2 = T/26312 12
2 ln 2 0.1 10 10 10 ln
12 12
DD
DD
VV
T
VV
− ++
= =
−−
1 500
12
ln
12
D
D
f
T V
V
==
+
−
So the frequency:
During T
1, the voltage v
− across the capacitor: v
− = L
+ − (L
+ − L
−)e
−t/
at t = T/2 : v
− = V
D so
@ 25ºC:25
500
4281()
12.7
ln
11.3
C
f Hz
==
Mai Linh, PhD
6.5 Astable multivibrators
Exercise 6.9: Sol.:v+
v–
This is a bistable multivibrator which generates
square waveform.
So the charging & discharge intervals:
T
1 = T
2 = T/26312 12
2 ln 2 0.1 10 10 10 ln
12 12
DD
DD
VV
T
VV
− ++
= =
−−
1 500
12
ln
12
D
D
f
T V
V
==
+
−
So the frequency:
During T
1, the voltage v
− across the capacitor: v
− = L
+ − (L
+ − L
−)e
−t/
at t = T/2 : v
− = V
D so
@ 25ºC:25
500
4281()
12.7
ln
11.3
C
f Hz
==
Mai Linh, PhD
6.5 Astable multivibrators
71
Exercise 6.9: Sol.:v+
v–
At 0ºC, V
D = 0.7 + 0.05 = 0.75 (V)
At 50ºC, V
D = 0.7 − 0.05 = 0.65 (V)
At 100ºC, V
D = 0.7 − 0.15 = 0.55 (V)0
500
3995()
12.75
ln
11.25
C
f Hz
==
50
500
4611()
12.65
ln
11.35
C
f Hz
==
100
500
5451()
12.55
ln
11.45
C
f Hz
==
Mai Linh, PhD
6.5 Astable multivibrators
Exercise 6.9: Sol.:v+
v–
At 0ºC, V
D = 0.7 + 0.05 = 0.75 (V)
At 50ºC, V
D = 0.7 − 0.05 = 0.65 (V)
At 100ºC, V
D = 0.7 − 0.15 = 0.55 (V)0
500
3995()
12.75
ln
11.25
C
f Hz
==
50
500
4611()
12.65
ln
11.35
C
f Hz
==
100
500
5451()
12.55
ln
11.45
C
f Hz
==
Mai Linh, PhD
6.5 Astable multivibrators
❑ Generation of triangular waveforms
721
TH TL
VVL
TCR
+
−
= 1
TH TL
VV
TCR
L
+
−
= 2
TH TL
VVL
TCR
−
−−
= 2
TH TL
VV
TCR
L
−
−
=
−
Mai Linh, PhD
721
TH TL
VVL
TCR
+
−
= 1
TH TL
VV
TCR
L
+
−
= 2
TH TL
VVL
TCR
−
−−
= 2
TH TL
VV
TCR
L
−
−
=
−
Mai Linh, PhD
73
❑ Generation of triangular waveforms
Exercise 6.10: Consider the circuit of Fig. (a) with the bistable circuit realized by
the circuit in Fig. (b). If the op amps have saturation voltages of 10 V, and if a
capacitor C = 0.01 μF and a resistor R
1 = 10 k are used, find the values of R and
R
2 (note that R
1 & R
2 are associated with the bistable circuit of Fig. (b) such that
the frequency of oscillation is 1 kHz and the triangular waveform has a 10 V
peak-to-peak amplitude.
Fig. (a) Fig. (b) Mai Linh, PhD
❑ Generation of triangular waveforms
Exercise 6.10: Consider the circuit of Fig. (a) with the bistable circuit realized by
the circuit in Fig. (b). If the op amps have saturation voltages of 10 V, and if a
capacitor C = 0.01 μF and a resistor R
1 = 10 k are used, find the values of R and
R
2 (note that R
1 & R
2 are associated with the bistable circuit of Fig. (b) such that
the frequency of oscillation is 1 kHz and the triangular waveform has a 10 V
peak-to-peak amplitude.
Fig. (a) Fig. (b) Mai Linh, PhD
74
Exercise 6.10: Sol.
❑ Generation of triangular waveforms
To obtain a triangular waveform with 10 V peak-to-peak amplitude, we should have
V
TH = −V
TL = 5 V
But 1
2
22
10
5 10 20 (k )
TL
R
V L R
RR
+
= − → − = − =
For 1 kHz frequency, T = 1 ms.
Thus,36
/ 2 0.5 10 0.01 10 10/10
TH TL
VV
T CR R
L
−−
+
−
= = =
Hence R = 50 k
Mai Linh, PhD
Exercise 6.10: Sol.
❑ Generation of triangular waveforms
To obtain a triangular waveform with 10 V peak-to-peak amplitude, we should have
V
TH = −V
TL = 5 V
But 1
2
22
10
5 10 20 (k )
TL
R
V L R
RR
+
= − → − = − =
For 1 kHz frequency, T = 1 ms.
Thus,36
/ 2 0.5 10 0.01 10 10/10
TH TL
VV
T CR R
L
−−
+
−
= = =
Hence R = 50 k
Mai Linh, PhD
6.6 Generation of a standardized pulse – The monostable Multivibrator
❑ Monostable Multivibrator – One shot
(Đa hài đơn bền)
75( )
13
/
1
()
()
tCR
BD
B
vtLLVe
vTL
−
−−
−
=−−
= 1
13
ln
D
VL
TCR
LL
−
−−
−
=
−
1
For||,
D
VL
−
13
1
ln
1
TCR
−
Mai Linh, PhD
❑ Monostable Multivibrator – One shot
(Đa hài đơn bền)
75( )
13
/
1
()
()
tCR
BD
B
vtLLVe
vTL
−
−−
−
=−−
= 1
13
ln
D
VL
TCR
LL
−
−−
−
=
−
1
For||,
D
VL
−
13
1
ln
1
TCR
−
Mai Linh, PhD
76
6.6 Generation of a standardized pulse – The monostable Multivibrator
❑ Monostable Multivibrator – One shot
Exercise 6.11: For the monostable circuit of Fig., find the value of R
3 that will result in
a 100 μs output pulse for C
1 = 0.1 μF, β = 0.1, V
D = 0.7 V, and L
+ = −L
− = 12 V.
Mai Linh, PhD
6.6 Generation of a standardized pulse – The monostable Multivibrator
❑ Monostable Multivibrator – One shot
Exercise 6.11: For the monostable circuit of Fig., find the value of R
3 that will result in
a 100 μs output pulse for C
1 = 0.1 μF, β = 0.1, V
D = 0.7 V, and L
+ = −L
− = 12 V.
Mai Linh, PhD
77
6.6 Generation of a standardized pulse – The monostable Multivibrator
❑ Monostable Multivibrator – One shot
Exercise 6.11: Sol.
The duration T of the output pulse is determined from
C
1 = 0.1μF, β = 0.1, V
D = 0.7 V, and L
+ = −L
− = 12 V.
→ R
3 = 6171 1
13
ln
D
VL
TCR
LL
−
−−
−
=
−
66
3
12.7
100100.110ln
10.8
R
−−
=
Mai Linh, PhD
6.6 Generation of a standardized pulse – The monostable Multivibrator
❑ Monostable Multivibrator – One shot
Exercise 6.11: Sol.
The duration T of the output pulse is determined from
C
1 = 0.1μF, β = 0.1, V
D = 0.7 V, and L
+ = −L
− = 12 V.
→ R
3 = 6171 1
13
ln
D
VL
TCR
LL
−
−−
−
=
−
66
3
12.7
100100.110ln
10.8
R
−−
=
Mai Linh, PhD
78
6.7 Integrated-Circuit Timers
Fig.: 555 Timer IC As A-Stable Multivibrator
A
B
Mai Linh, PhD
❑ 555 IC Timer
✓invented by Hans Camenzind of
Signetics in 1968. The 555 timer is
very stable and easy to use which is
probably why it’s so popular.
✓Main uses are as monostable, bistable,
and astable oscillators.
6.7 Integrated-Circuit Timers
Fig.: 555 Timer IC As A-Stable Multivibrator
A
B
Mai Linh, PhD
❑ 555 IC Timer
✓invented by Hans Camenzind of
Signetics in 1968. The 555 timer is
very stable and easy to use which is
probably why it’s so popular.
✓Main uses are as monostable, bistable,
and astable oscillators.
79
PinNamePurpose
1GNDGround reference voltage, low level (0 V)
2TRIG
The OUT pin goes high and a timing interval starts when this input falls below 1/2 of CTRL
voltage (which is typically 1/3 Vcc, CTRL being 2/3 Vcc by default if CTRL is left open). In
other words, OUT is high as long as the trigger is low. The output of the timer totally depends
upon the amplitude of the external trigger voltage applied to this pin.
3OUTThis output is driven to approximately 1.7 V below +Vcc, or to GND.
4RESET
A timing interval may be reset by driving this input to GND, but the timing does not begin again
until RESET rises above approximately 0.7 volts. Overrides TRIG which overrides threshold.
5CTRLProvides “control” access to the internal voltage divider (by default, 2/3 Vcc).
6THR
The timing (OUT high) interval ends when the voltage at the threshold is greater than that at
CTRL (2/3 Vcc if CTRL is open).
7DISOpen collector output which may discharge a capacitor between intervals. In phase with output.
8VccPositive supply voltage, which is usually between 3 and 15 V depending on the variation.
6.7 Integrated-Circuit Timers
❑ 555 IC Timer
Mai Linh, PhD
PinNamePurpose
1GNDGround reference voltage, low level (0 V)
2TRIG
The OUT pin goes high and a timing interval starts when this input falls below 1/2 of CTRL
voltage (which is typically 1/3 Vcc, CTRL being 2/3 Vcc by default if CTRL is left open). In
other words, OUT is high as long as the trigger is low. The output of the timer totally depends
upon the amplitude of the external trigger voltage applied to this pin.
3OUTThis output is driven to approximately 1.7 V below +Vcc, or to GND.
4RESET
A timing interval may be reset by driving this input to GND, but the timing does not begin again
until RESET rises above approximately 0.7 volts. Overrides TRIG which overrides threshold.
5CTRLProvides “control” access to the internal voltage divider (by default, 2/3 Vcc).
6THR
The timing (OUT high) interval ends when the voltage at the threshold is greater than that at
CTRL (2/3 Vcc if CTRL is open).
7DISOpen collector output which may discharge a capacitor between intervals. In phase with output.
8VccPositive supply voltage, which is usually between 3 and 15 V depending on the variation.
6.7 Integrated-Circuit Timers
❑ 555 IC Timer
Mai Linh, PhD
❑ 555 IC Timer connected to implement a
monostable multivibrator
80
6.7 Integrated-Circuit Timers
( )
/
/
()
(0)for0
1 (0) 0
2
at. ln31.1
3
tCR
C CC CC
tCR
CC CEsat
C TH CC
vVVVe tT
Ve VV
vVVtTTCRCR
−
−
=−−
=−
====
(a)
Fig.: (a) The 555 timer
connected to implement a
monostable multivibrator. (b)
Waveforms of the circuit in (a).
Substituting
Mai Linh, PhD
monostable multivibrator
80
6.7 Integrated-Circuit Timers
( )
/
/
()
(0)for0
1 (0) 0
2
at. ln31.1
3
tCR
C CC CC
tCR
CC CEsat
C TH CC
vVVVe tT
Ve VV
vVVtTTCRCR
−
−
=−−
=−
====
(a)
Fig.: (a) The 555 timer
connected to implement a
monostable multivibrator. (b)
Waveforms of the circuit in (a).
Substituting
Mai Linh, PhD
❑ 555 IC Timer connected to implement an astable multivibrator
81
6.7 Integrated-Circuit Timers( )
( ) ( )
/( )
:
21
at.
33
ln20.69
AB
CTL TH
tCRR
C CC CC TL
C TH CC H TL CC
H A B A B
VVV
vVVVe
vVVtTVV
TCRR CRR
−+
→
=−−
====
=+ +
❶
Fig.: (a) The 555 timer connected to implement an astable
multivibrator. (b) Waveforms of the circuit in (a)./
:
2
3
1
=at.
3
ln20.69
B
CTH TL
tCR
C TH TH CC
C TL CC L
L B B
VVV
vVeVV
vVVtT
TCR CR
−
→
==
==
=
❷
❸ Period T of the output square wave: ( )0.692.The duty cycle
2
H A B
H L A B
H L A B
TRR
TTTRRC
TTRR
+
=+=+ =
++
Mai Linh, PhD
81
6.7 Integrated-Circuit Timers( )
( ) ( )
/( )
:
21
at.
33
ln20.69
AB
CTL TH
tCRR
C CC CC TL
C TH CC H TL CC
H A B A B
VVV
vVVVe
vVVtTVV
TCRR CRR
−+
→
=−−
====
=+ +
❶
Fig.: (a) The 555 timer connected to implement an astable
multivibrator. (b) Waveforms of the circuit in (a)./
:
2
3
1
=at.
3
ln20.69
B
CTH TL
tCR
C TH TH CC
C TL CC L
L B B
VVV
vVeVV
vVVtT
TCR CR
−
→
==
==
=
❷
❸ Period T of the output square wave: ( )0.692.The duty cycle
2
H A B
H L A B
H L A B
TRR
TTTRRC
TTRR
+
=+=+ =
++
Mai Linh, PhD
82
❑ 555 IC Timer connected to implement an astable multivibrator
6.7 Integrated-Circuit Timers
Exercise 6.12: Using a 10-nF
capacitor C, find the value of R that
yields an output pulse of 100 μs in
the monostable circuit of Fig.
Fig.: The 555 timer connected to implement a
monostable multivibrator
Mai Linh, PhD
❑ 555 IC Timer connected to implement an astable multivibrator
6.7 Integrated-Circuit Timers
Exercise 6.12: Using a 10-nF
capacitor C, find the value of R that
yields an output pulse of 100 μs in
the monostable circuit of Fig.
Fig.: The 555 timer connected to implement a
monostable multivibrator
Mai Linh, PhD
83
❑ 555 IC Timer connected to implement an astable multivibrator
6.7 Integrated-Circuit Timers
Exercise 6.12: Sol.
T= 1.1CR ⇒R = T/1.1C = 9.1 (k)
Mai Linh, PhD
❑ 555 IC Timer connected to implement an astable multivibrator
6.7 Integrated-Circuit Timers
Exercise 6.12: Sol.
T= 1.1CR ⇒R = T/1.1C = 9.1 (k)
Mai Linh, PhD
84
❑ 555 IC Timer connected to implement an astable multivibrator
6.7 Integrated-Circuit Timers
Exercise 6.13: For the circuit in
Fig., with a 1 nF capacitor, find the
values of R
A & R
B that result in an
oscillation frequency of 100 kHz
and a duty cycle of 75%.
Fig.: The 555 timer connected to implement
an astable multivibrator
Mai Linh, PhD
❑ 555 IC Timer connected to implement an astable multivibrator
6.7 Integrated-Circuit Timers
Exercise 6.13: For the circuit in
Fig., with a 1 nF capacitor, find the
values of R
A & R
B that result in an
oscillation frequency of 100 kHz
and a duty cycle of 75%.
Fig.: The 555 timer connected to implement
an astable multivibrator
Mai Linh, PhD
85
❑ 555 IC Timer connected to implement an astable multivibrator
6.7 Integrated-Circuit Timers
Exercise 6.13: Sol.:
The period T of the output square wave T = T
H + T
L = 0.69C(R
A + 2R
B)
The duty cycle of the output square wave:
→0.75 0.75 14.49 10.88(k ) (2)
2
AB
AB
AB
RR
RR
RR
+
= → + = =
+
(1) − (2) ⇒ R
B = 3.61 k, then R
A = 7.27 k( )
3 13
34
11
0.6910102 2 14.49()(1)
10010 0.6910
A B A B
RRRR k
−
−
=+→+= =
duty cycle
2
H A B
H L A B
TRR
TTRR
+
=
++
Mai Linh, PhD
❑ 555 IC Timer connected to implement an astable multivibrator
6.7 Integrated-Circuit Timers
Exercise 6.13: Sol.:
The period T of the output square wave T = T
H + T
L = 0.69C(R
A + 2R
B)
The duty cycle of the output square wave:
→0.75 0.75 14.49 10.88(k ) (2)
2
AB
AB
AB
RR
RR
RR
+
= → + = =
+
(1) − (2) ⇒ R
B = 3.61 k, then R
A = 7.27 k( )
3 13
34
11
0.6910102 2 14.49()(1)
10010 0.6910
A B A B
RRRR k
−
−
=+→+= =
duty cycle
2
H A B
H L A B
TRR
TTRR
+
=
++
Mai Linh, PhD
6.8 Oscillators with transistors
86
This is a classicoscillatorcircuit.
✓If T
1 turns ON, it changes the voltage of C
1 → T
2 turns off.
✓After a short while, the voltage of C
1 rises back up and turns on the T
2.
✓When T
2 turns on, it changes the voltage of C
2 → T
1 turns off.T2
T1
R1 R2 R3 R4
C1 C2
+EC
Vo2Vo1
UB2UB1
➢The voltage on the left side of C
2 controls T
1.
➢The voltage on the right side of C
1 controls T
2.
Superficial
explanation!
❖Sort of astable (free-running)multivibrator,
oroscillator.
❖Circuit has two outputs but no inputs.
❖ R
1 = R
4, R
2 = R
3, C
1 = C
2, T
1 & T
2 are as close
as is possible in their operating characteristics.
T
1 & T
2 work as switches
Mai Linh, PhD
86
This is a classicoscillatorcircuit.
✓If T
1 turns ON, it changes the voltage of C
1 → T
2 turns off.
✓After a short while, the voltage of C
1 rises back up and turns on the T
2.
✓When T
2 turns on, it changes the voltage of C
2 → T
1 turns off.T2
T1
R1 R2 R3 R4
C1 C2
+EC
Vo2Vo1
UB2UB1
➢The voltage on the left side of C
2 controls T
1.
➢The voltage on the right side of C
1 controls T
2.
Superficial
explanation!
❖Sort of astable (free-running)multivibrator,
oroscillator.
❖Circuit has two outputs but no inputs.
❖ R
1 = R
4, R
2 = R
3, C
1 = C
2, T
1 & T
2 are as close
as is possible in their operating characteristics.
T
1 & T
2 work as switches
Mai Linh, PhD
87
Detailed explanation the way it works:
T
1 is ON (~ +0.7 V on its base) → left side of C
2 is at ~ +0.7V.
The right side of C
2 connects to +E
C through R
4 & collector of
T
2, so C
2 is charging, and the voltage is rising.
A capacitor charges exponentially, which means the voltage
rises quickly in the beginning, then slows down more and more.
The voltage reaches nearly +E
C quickly, but from there the
voltage rises slowly.
Since T
2 is off (its base must be < 0.7 V) → The right side of C
1 is also < 0.7 V.
But the right side of C
1 is also connected to +E
C through R
2, → it is being charged (voltage is below
0.7 V but rising), then it reaches 0.7V ➔ T
2 turns ON
But when T
2 turns on, something interesting happens with the voltages we had over C
2…T2
T1
R1 R2 R3 R4
+EC
Vo2Vo1
UB2UB1
C1 C2
Right side of C
1 < +0.7 V → T
2 is OFF
6.8 Oscillators with transistors
Mai Linh, PhD
Detailed explanation the way it works:
T
1 is ON (~ +0.7 V on its base) → left side of C
2 is at ~ +0.7V.
The right side of C
2 connects to +E
C through R
4 & collector of
T
2, so C
2 is charging, and the voltage is rising.
A capacitor charges exponentially, which means the voltage
rises quickly in the beginning, then slows down more and more.
The voltage reaches nearly +E
C quickly, but from there the
voltage rises slowly.
Since T
2 is off (its base must be < 0.7 V) → The right side of C
1 is also < 0.7 V.
But the right side of C
1 is also connected to +E
C through R
2, → it is being charged (voltage is below
0.7 V but rising), then it reaches 0.7V ➔ T
2 turns ON
But when T
2 turns on, something interesting happens with the voltages we had over C
2…T2
T1
R1 R2 R3 R4
+EC
Vo2Vo1
UB2UB1
C1 C2
Right side of C
1 < +0.7 V → T
2 is OFF
6.8 Oscillators with transistors
Mai Linh, PhD
88
6.8 Oscillators with transistors
C
2 had +0.7 V on its left side and ~ +E
C V on its right side (or the
left side was ~(E
C – 0.7) V lower than the right side).
But now that T
2 turns on, the voltage on the right side of C
2 is
suddenly pulled down to 0 V through T
2.
The internal charge of the capacitor does not change though, so the
left side of C
2 keeps being ~(E
C – 0.7) V lower than the right side.
But now that the right side is 0 V, that means the left side becomes
~ –(E
C – 0.7) V → T
1 turns OFF
Left side of C
2 < +0.7 V → T
1 is OFFT2
T1
R1 R2 R3 R4
+EC
Vo2Vo1
UB2UB1
C1 C2
So now, T
2 is ON
The left side of C
2 starts at –(E
C – 0.7) V and is charged through resistor R
3 and therefore rising.
Since it connects to the base of transistor T
1, when it reaches +0.7 V, T
1 turns ON again.
And so it continues.
T
1 & T
2 keep alternating between ON and OFF.
Mai Linh, PhD
6.8 Oscillators with transistors
C
2 had +0.7 V on its left side and ~ +E
C V on its right side (or the
left side was ~(E
C – 0.7) V lower than the right side).
But now that T
2 turns on, the voltage on the right side of C
2 is
suddenly pulled down to 0 V through T
2.
The internal charge of the capacitor does not change though, so the
left side of C
2 keeps being ~(E
C – 0.7) V lower than the right side.
But now that the right side is 0 V, that means the left side becomes
~ –(E
C – 0.7) V → T
1 turns OFF
Left side of C
2 < +0.7 V → T
1 is OFFT2
T1
R1 R2 R3 R4
+EC
Vo2Vo1
UB2UB1
C1 C2
So now, T
2 is ON
The left side of C
2 starts at –(E
C – 0.7) V and is charged through resistor R
3 and therefore rising.
Since it connects to the base of transistor T
1, when it reaches +0.7 V, T
1 turns ON again.
And so it continues.
T
1 & T
2 keep alternating between ON and OFF.
Mai Linh, PhD
89
DEMO circuit video
https://www.falstad.com/circuit/e-multivib-a.html : simulation circuit
Mai Linh, PhD
DEMO circuit video
https://www.falstad.com/circuit/e-multivib-a.html : simulation circuit
Mai Linh, PhD
90
6.8 Oscillators with transistors
Monostable (One Shot) Multivibrator
When +V
CC is firstly applied, B of TR2is
connected to +V
CC via the biasing
resistor,R
T → TR2 is “ON” and
turningTR1“OFF”. This then represents the
circuits “Stable State” with zero output. The
current flowing into the B ofTR2will be
equal toI
b = (V
CC – 0.7)/R
T. C
T is fully
charged maintaining ~ +V
CC on the B of TR2
If a negative trigger pulse is now applied at
the input, it will pass C
1to the B ofTR1via
the blocking diode, turning TR1 “ON”.
The C ofTR1which was previously at+V
CCdrops quickly to below zero volts effectively giving C
Ta reverse charge of
-0.7V across its plates. →TR2now having a minus base voltage @ pointXholding TR2 fully “OFF”. This then
represents the circuits second state, the “Unstable State” with an output voltage equal to +V
CC .
Timing capacitor,C
Tbegins to discharge this -0.7v through the timing resistorR
T, attempting to charge up to the supply
voltage +V
CC. This negative voltage at the base of transistorTR2begins to decrease gradually at a rate determined by the
time constant = R
TC
T.
Mai Linh, PhD
6.8 Oscillators with transistors
Monostable (One Shot) Multivibrator
When +V
CC is firstly applied, B of TR2is
connected to +V
CC via the biasing
resistor,R
T → TR2 is “ON” and
turningTR1“OFF”. This then represents the
circuits “Stable State” with zero output. The
current flowing into the B ofTR2will be
equal toI
b = (V
CC – 0.7)/R
T. C
T is fully
charged maintaining ~ +V
CC on the B of TR2
If a negative trigger pulse is now applied at
the input, it will pass C
1to the B ofTR1via
the blocking diode, turning TR1 “ON”.
The C ofTR1which was previously at+V
CCdrops quickly to below zero volts effectively giving C
Ta reverse charge of
-0.7V across its plates. →TR2now having a minus base voltage @ pointXholding TR2 fully “OFF”. This then
represents the circuits second state, the “Unstable State” with an output voltage equal to +V
CC .
Timing capacitor,C
Tbegins to discharge this -0.7v through the timing resistorR
T, attempting to charge up to the supply
voltage +V
CC. This negative voltage at the base of transistorTR2begins to decrease gradually at a rate determined by the
time constant = R
TC
T.
Mai Linh, PhD
91
6.8 Oscillators with transistors
Monostable (One Shot) Multivibrator
▪Timers: to create timers that produce a delay or a timeout after a trigger
signal. Ex., to turn off a device after a certain time period or to activate an
alarm after a preset interval.
▪Delay circuits: to create delay circuits that introduce a lag or a phase shift
between an input signal and an output signal. Ex., they can be used to
synchronize signals in digital systems or to delay signals in analog
systems.
▪Gated circuits: create gated circuits that enable or disable an output signal based on a trigger signal. Ex., they
can be used to gate an oscillator or a counter or to switch between two signals.
▪Frequency dividers: to create frequency dividers that reduce the frequency of an input signal by a factor of
two or more. Ex., they can be used to divide a clock signal or a carrier signal by using multiple stages of
monostable multivibrators.
▪Pulse generators: to create pulse generators that produce single or multiple pulses with precise widths and
intervals. Ex., they can be used to generate pulses for sampling signals or for testing circuits.
▪Pulse shapers: to create pulse shapers that modify or improve the shape or quality of an input signal. Ex.,
they can be used to sharpen or smooth pulses or to eliminate noise or glitches.
Applications:Negative
trigger pulse
Time
Time
Time
Short time
constant
Longer time
constant
-Vin
longer
short
Vout
Vout
Mai Linh, PhD
6.8 Oscillators with transistors
Monostable (One Shot) Multivibrator
▪Timers: to create timers that produce a delay or a timeout after a trigger
signal. Ex., to turn off a device after a certain time period or to activate an
alarm after a preset interval.
▪Delay circuits: to create delay circuits that introduce a lag or a phase shift
between an input signal and an output signal. Ex., they can be used to
synchronize signals in digital systems or to delay signals in analog
systems.
▪Gated circuits: create gated circuits that enable or disable an output signal based on a trigger signal. Ex., they
can be used to gate an oscillator or a counter or to switch between two signals.
▪Frequency dividers: to create frequency dividers that reduce the frequency of an input signal by a factor of
two or more. Ex., they can be used to divide a clock signal or a carrier signal by using multiple stages of
monostable multivibrators.
▪Pulse generators: to create pulse generators that produce single or multiple pulses with precise widths and
intervals. Ex., they can be used to generate pulses for sampling signals or for testing circuits.
▪Pulse shapers: to create pulse shapers that modify or improve the shape or quality of an input signal. Ex.,
they can be used to sharpen or smooth pulses or to eliminate noise or glitches.
Applications:Negative
trigger pulse
Time
Time
Time
Short time
constant
Longer time
constant
-Vin
longer
short
Vout
Vout
Mai Linh, PhD
92
End of Lecture #6
Mai Linh, PhD
End of Lecture #6
Mai Linh, PhD
Tags
Categories
EducationDownload
Download Slideshow Get the original presentation fileQuick Actions
Statistics
- Views 29
- Slides 92
- Age 107 days
Related Slideshows
11
TLE-9-Prepare-Salad-and-Dressing.pptxkkk
MaAngelicaCanceran
45 views
12
LESSON 1 ABOUT MEDIA AND INFORMATION.pptx
JojitGueta
35 views
60
GRADE-8-AQUACULTURE-WEEKQ1.pdfdfawgwyrsewru
MaAngelicaCanceran
59 views
26
Feelings PP Game FOR CHILDREN IN ELEMENTARY SCHOOL.pptx
KaistaGlow
53 views
![Jeopardy_Figures_of_Speech_Template.pptx [Autosaved].pptx](https://slidepub.com/thumb/5828/284015828.jpg)
54
Jeopardy_Figures_of_Speech_Template.pptx [Autosaved].pptx
acecamero20
31 views
7
Jeopardy_Figures_of_Speech.pptxvdsvdsvsdvsd
acecamero20
32 views