Positive-Feedback Oscillators

1
Positive-Feedback Oscillators: Illustrations
© Eugene PAPERNO, 2006
I. PHASE-SHIFT OSCILLATOR
β ( s) =
1
1 ⎞
⎛
⎜1 +
⎟
sRC
⎝
⎠
3
Σ
Sin
Sε
So
AOL
RC = 1
β(s)
A f ( s) =
AOL
1 − AOL β ( s )
Im[β (s)]=0
Fig. 3. Abs[Af(s)] for AOL=8. [AOLβ(ω1) =1.]
Fig. 1. Abs[β(s)].
AOL=2
1
3
AOL=3.3
1
0.3
1
2
0.2
0.75
0.75
0.125
0.1
0.08
0.5
0.06
Im[β(s)]=0
0.25
jω
AOL=8
AOL=12.5
AOL=25
0.25
0.04
0.02
0
jω
− 0.25
− 0.5
− 0.5
− 0.75
− 0.75
− 1.5
−1
− 0.5
σ
Fig. 2. Abs[β(s)].
0
0.5
Im[β(s)]=0
0
− 0.25
−2
AOL=8
0.5
− 1.5
−1
− 0.5
0
σ
Fig. 4. Abs[Af(s)] for AOL=8. [AOLβ(ω1) =1.]
0.5
2
Note that AOLβ(ω1) >1 shifts the poles to the right of the jω axis.
1
1
0.75
0.75
AOL=8
0.5
0.5
A OL=25
0.25
jω
Im[β(s)]=0
0.25
jω
0
0
− 0.25
− 0.25
− 0.5
− 0.5
− 0.75
− 0.75
− 1.5
−1
− 0.5
0
Im[β(s)]=0
0.5
−1.5
−1
σ
1
0.75
0.75
A OL=12.5
0.25
jω
0
− 0.25
− 0.5
− 0.5
− 0.75
− 0.75
− 0.5
0
σ
Fig. 6. Abs[Af(s)] for AOL=12.5. [AOLβ(ω1) =1.56.]
0.5
Im[β(s)]=0
0
− 0.25
−1
A OL=3.3
0.5
Im[β(s)]=0
− 1.5
0.5
Fig. 7. Abs[Af(s)] for AOL=25. [AOLβ(ω1) =3.125.]
1
0.5
jω
0
σ
Fig. 5. Abs[Af(s)] for AOL=8. [AOLβ(ω1) =1.]
0.25
− 0.5
− 1.5
−1
− 0.5
0
σ
Fig. 8. Abs[Af(s)] for AOL=3.3. [AOLβ(ω1) =0.41.]
0.5
3
II. WIEN-BRIDGE (HEWLETT) OSCILLATOR
1
sC
1
R+
⎞
⎛1
sC
− ⎜ − 0.1⎟
β ( s) =
1
⎠
⎝3
R
sC + R + 1
1
sC
R+
sC
R
Σ
Sin
Sε
So
AOL
β(s)
R =1
C =1
A f ( s) =
AOL
1 − AOL β ( s )
Fig. 1. Abs[β(s)].
Fig. 3. Abs[Af(s)] for AOL=8. [AOLβ(ω1) =1.]
1.5
1.5
1
1
0.5
0.5
0
0
− 0.5
− 0.5
−1
−1
− 1.5
−1
− 0.5
Fig. 2. Abs[β(s)].
0
0.5
1
− 1.5
−1
− 0.5
0
Fig. 4. Abs[Af(s)] for AOL=10. [AOLβ(ω1) =1.]
0.5
1
4
Note that AOLβ(ω1) >1 shifts the poles to the right of the jω axis.
1.5
1.5
1
1
0.5
0.5
0
0
− 0.5
− 0.5
−1
−1
− 1.5
−1
− 0.5
0
0.5
− 1.5
1
1.5
1.5
1
1
0.5
0.5
0
0
− 0.5
− 0.5
−1
−1
−1
− 0.5
0
Fig. 6. Abs[Af(s)] for AOL=20. [AOLβ(ω1) =2.]
− 0.5
0
0.5
1
0.5
1
Fig. 7. Abs[Af(s)] for AOL=106. [AOLβ(ω1) =105.]
Fig. 5. Abs[Af(s)] for AOL=10. [AOLβ(ω1) =1.]
− 1.5
−1
0.5
1
− 1.5
−1
− 0.5
0
Fig. 8. Abs[Af(s)] for AOL=5. [AOLβ(ω1) =0.5.]
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III. HARTLEY-COLPITTS OSCILLATORS
1
sC sL
1
R+
sC
β ( s) =
1
R
sC + sL
1
R+
sC
R
Σ
Sin
Sε
R =1
L =1
C =1
A f ( s) =
β(s)
AOL
1 − AOL β ( s )
Fig. 3. Abs[Af(s)] for AOL=8. [AOLβ(ω1) =1.]
Fig. 1. Abs[β(s)].
1.5
1.5
1
1
0.5
0.5
0
0
− 0.5
− 0.5
−1
−1
− 1.5
−1
− 0.5
Fig. 2. Abs[β(s)].
So
AOL
0
0.5
1
− 1.5
−1
− 0.5
0
Fig. 4. Abs[Af(s)] for AOL=1. [AOLβ(ω1) =1.]
0.5
1
6
Note that AOLβ(ω1) >1 shifts the poles to the right of the jω axis.
1.5
1.5
1
1
0.5
0.5
0
0
− 0.5
− 0.5
−1
−1
− 1.5
−1
− 0.5
0
0.5
− 1.5
1
1.5
1.5
1
1
0.5
0.5
0
0
− 0.5
− 0.5
−1
−1
−1
− 0.5
0
Fig. 6. Abs[Af(s)] for AOL=1.4. [AOLβ(ω1) =1.4.]
− 0.5
0
0.5
1
0.5
1
Fig. 7. Abs[Af(s)] for AOL=2. [AOLβ(ω1) =2.]
Fig. 5. Abs[Af(s)] for AOL=1. [AOLβ(ω1) =1.]
− 1.5
−1
0.5
1
− 1.5
−1
− 0.5
0
Fig. 8. Abs[Af(s)] for AOL=0.8. [AOLβ(ω1) =0.8.]
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IV. OSCILLATOR WITH AN UNSTABLE FEEDBACK NETWORK
β (s) =
−25
1 − ( −25)
Sin
1
1 ⎞
⎛
⎜1 +
⎟
sRC ⎠
⎝
Σ
Sε
So
AOL
3
−25
Σ
RC = 1
b(s)
AOL
A f ( s) =
1 − AOL β ( s )
Phase-shift oscillator
Fig. 1. Abs[β(s)].
Fig. 3. Abs[Af(s)] for AOL=0.85. [AOLβ(ω1) =1.]
1
1
0.75
0.75
0.5
0.5
0.25
0.25
0
0
− 0.25
− 0.25
− 0.5
− 0.5
− 0.75
− 0.75
− 0.1
0
Fig. 2. Abs[β(s)].
0.1
0.2
− 0.1
0
0.1
Fig. 4. Abs[Af(s)] for AOL=0.085. [AOLβ(ω1) =1.]
0.2
8
Note that AOLβ(ω1) >1 shifts the poles to the left of the jω axis.
Fig. 5. Abs[Af(s)] for AOL=0.085. [AOLβ(ω1) =1.]
Fig. 7. Abs[Af(s)] for AOL=0.01. [AOLβ(ω1) =0.012.]
Fig. 6. Abs[Af(s)] for AOL=0.05. [AOLβ(ω1) =0.59.]
Fig. 8. Abs[Af(s)] for AOL=0.2. [AOLβ(ω1) =2.35.]
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SUMMARY
Hartley-Colpitts Oscillators
AOL1 < AOL2 < AOL3
AOL1 β(s) =1
AOL2 β( jω1) =1
AOL3 β(s) =1
β(s)
AOL
Im[β (s)]=0
Oscillator with an unstable feedback network
Im[β (s)]=0
β(s)
AOL1 β(s) =1
AOL
AOL2 β( jω1) =1
AOL1 > AOL2 > AOL3
AOL3 β(s) =1
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Title:
Positive-Feedback Oscillators
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Creation Date:
12/28/2006 1:06:00 PM
Change Number:
147
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