Transcript Oscillators * Feedback amplifier but frequency dependent feedback A f s   As  1   f ( s) A( s) * Positive.

Oscillators
* Feedback amplifier but frequency dependent feedback
A f s  
As 
1   f ( s) A( s)
* Positive feedback, i.e. βf () A () < 0
* Oscillator gain defined by
A f s  
As 
1   f ( s) A( s)
* Oscillation condition at ω = ωo (Barkhausen’s criterion) Af (ωo) = 
L( o )   f ( o ) A( o )   f ( o ) A( o ) e j (o )  1
 ( o )  phase of  f ( o ) A( o )
ECES 352 Winter 2007
Ch 13 Oscillators
1
Wien Bridge Oscillator
*
*
*
R2
R1
V0
Vi
*
ZS
If
ZP
*
ECES 352 Winter 2007
Based on op amp
Combination of R’s and C’s in
feedback loop so feedback factor
βf has a frequency dependence.
Analysis assumes op amp is ideal.
 Gain A is very large
 Input currents are negligibly
small (I+  I_  0).
 Input terminals are virtually
shorted (V+  V_ ).
Analyze like a normal feedback
amplifier.
 Determine input and output
loading.
 Determine feedback factor.
 Determine gain with feedback.
Shunt-shunt configuration.
Ch 13 Oscillators
2
Wien Bridge Oscillator
Define
R1
Z S  R  ZC  R 
R2
V0
Vi
1
1 

Z P  R Z C   
R
Z
C 

ZS
If

ZP
1
1

  sC 
R

R
1  sCR
Output Loading
Input Loading
ZS
Z1
ZS
V0 = 0
ZP
 1
1 
Z1  Z P Z S  


 ZP ZS 
Vi = 0
ZP
1

Z2  Z S  R  ZC 
sC 
R1  sCR 
1  sCR
 R  1  sCR   sCR  (1  sCR ) 2
1
ECES 352 Winter 2007
1 1  sRC

sC
sC
Ch 13 Oscillators
Z2
1  sRC
sC
3
1
Wien Bridge Oscillator
I1
I2
Amplifier gain including loading effects
R2
R1
Vi
V0
IS
Z2
Z1
ZS
Since I   0,
V0
ZP
Xo

If
Vo
sC

1  sRC
ECES 352 Winter 2007
Vo
R1 so
R1  R2
V0 R1  R2
R

 1 2
Vi
R1
R1
If
Xf
V0
Vo
, we use I1  I 2 
and
Vi
R1  R2
Vi  V  V  I1 R1 
Feedback factor
f 
V0 V0 Vi

I S Vi I S
To get
IS
IS
Ar 

1
ZS
Vi
 Z1 and
IS
 R 
V0 Vi
 Z1 1  2 
Vi I S
 R1 
R1  sCR 
where Z1 
so
sCR  (1  sCR ) 2
Ar 
 R  R1  sCR 
Ar  1  2 
2
 R1  sCR  (1  sCR )
Ch 13 Oscillators
4
Wien Bridge Oscillator
Oscillation condition
P hase of  f Ar equal t o 180o. It already is since  f Ar  0.

sCR
R 
1
T hen need only  f Ar  1  2 
R1  sCR  (1  sCR ) 2

Rewrit ing

R 
sCR
 f Ar  1  2 
R1  sCR  (1  sCR ) 2


 1 


 1 

sCR
R2 

R1  sCR  1  2 sCR  s 2C 2 R 2


sC  R2 
R 1  sCR 


 
1 
R1  sCR  (1  sCR ) 2
 1  sCR 
 R 
sCR
 1  2 
R1  sCR  (1  sCR ) 2


1
R 
sCR
R2 

 1  2 
2 2 2
R1  3  1  sCR
R1  1  3sCR  s C R

sCR

1
R 
 1  2 
1 
R1 


3  j  CR 

CR 

T hen imaginary t erm 0 at t he oscillat ion frequency
1
  o 
RC
T hen,we can get  f Ar  1 by select ing t he resist ors R1 and R2
Gain with feedback is
appropriately using
Loop Gain
sC 

 Ar
1

sCR


 f Ar   
Arf 
Ar
1   f Ar
ECES 352 Winter 2007

R
R 1
1  2   1 or 2  2
R1
R1  3

Ch 13 Oscillators
5
Wien Bridge Oscillator - Example
Oscillator specifications: o=1x106 rad/s
Selecting for convenience C  10 nF, then from o 
R
1
RC
1
1

 100 
oC 10nF (1x106 rad / s )
Choosing R 1  10K , then since R 2  2 R 1 we get
R2  2(10K )  20K
ECES 352 Winter 2007
Ch 13 Oscillators
6
Wien Bridge Oscillator
Final note: No input signal is needed. Noise at the desired oscillation frequency
will likely be present at the input and when picked up by the oscillator when
the DC power is turned on, it will start the oscillator and the output will
quickly buildup to an acceptable level.
ECES 352 Winter 2007
Ch 13 Oscillators
7
Wien Bridge Oscillator
*
Once oscillations start, a limiting circuit is needed to prevent
them from growing too large in amplitude
ECES 352 Winter 2007
Ch 13 Oscillators
8
Phase Shift Oscillator
IC3
VX
V2
C
C
R
*
*
*
IC2
IR2
V1
If
IC1
C
I
R R1
Rf
I C 2  I R1  I C1 
V0
Vo
V
V 
1 
 o  o 1 

sCRR f R f R f  sCR 
V2  V1  I C 2 Z C  

Vo
V 
1  1
 o 1 

sCR f R f  sCR  sC
Vo 
1 
2 

sCR f  sCR 
Based on op amp using inverting input
Vo 
V
1 
I R2  2 
2 

Combination of R’s and C’s in
R
sCRR f  sCR 
feedback loop so get additional phase
Vo 
1  Vo 
1 
IC3  I R2  IC 2 
2 

1 

shift. Target 180o to get oscillation.
sCRR f  sCR  R f  sCR 
Analysis assumes op amp is ideal.
V  V  0 so I f 
V1  V  I C1Z C  
I R1 
Vo
 I C1
Rf
Vo
sCR f
 V1  1  Vo


R
R  sCR f
ECES 352 Winter 2007

1  1 
1  Vo 
3
1 
1


2


1









2
 sCR  sCR  sCR  R f  sCR ( sCR ) 
Finally

Vo
Rf
V X  V2 

  Vo
 sCRR f


IC3
V 
1  Vo
  o 2 

sC
sCR f  sCR  sCR f

3
1 

1 
2
 sCR ( sCR ) 
Vo 
4
1 

3 

sCR f  sCR ( sCR ) 2 
Ch 13 Oscillators
9
Phase Shift Oscillator
Rearranging
IC3
VX
IC2
V2
V1
C
C
R
IC1
If
C
IR2 R

4
1 
3  sCR  ( sCR ) 2 


we get for the loop gain
Rf
IR1
VX  
V0
Example
Oscillator specifications: o=1x106 rad/s
Selecting for convenience C  10 nF,
then from  o 
R
1
3 oC

1
3RC
1
 58 
3 10nF (1x106 rad / s)
Vo
sCR f
 sCR f
V0

1
VX 
4
1 
3  sCR  ( sCR ) 2 


 jCR f
 2C 2 RR f



1 
4
1  

3  j CR  (CR ) 2  4  j  3CR  CR 



 
T o get oscillations, we need the imaginary term to go to zero.
We can achieve this at one frequency o so
L( )   ( ) A( ) 
3CR 
1
1
so   0 
CR
3RC
T o get oscillations, we also need L(ωo )  1 so
0 2C 2 RR f
L(ωo ) 
 1 and substituting for ωo we get
4
R f  12(58 )  0.67K
0 2C 2 RR f C 2 RR f 1
Rf


 1 so
2
2
Note: We get 180o phase shift from op amp
4
4 3R C 12R
R f  12R
since input is to inverting terminal and
T hen
another 180o from the RC ladder.
ECES 352 Winter 2007
Ch 13 Oscillators
10
Colpitts LC-Tuned Oscillator
CB
V0
CE
Vi
V0
Vi
ECES 352 Winter 2007
* Feedback amplifier with inductor L and
capacitors C1 and C2 in feedback network.
 Feedback is frequency dependent.
 Aim to adjust components to get
positive feedback and oscillation.
 Output taken at collector Vo.
 No input needed, noise at oscillation
frequency o is picked up and
amplified.
* RB1 and RB2 are biasing resistors.
* RFC is RF Choke (inductor) to allow dc
current flow for transistor biasing, but to
block ac current flow to ac ground.
* Simplified circuit shown at midband
frequencies where large emitter bypass
capacitor CE and base capacitor CB are
shorts and transistor capacitances (C and
C) are opens.
Ch 13 Oscillators
11
Colpitts LC-Tuned Oscillator
*
Voltage across C2 is just V
*
V
 sC2V
ZC 2
Neglecting input current to transistor (I  0),
V
I L  IC 2    sC2V
ZC 2
*
Then, output voltage Vo is
IC 2 

Vo  V  I L ZL  V  (sC2V )(sL)  V 1  s2 LC2
AC equivalent circuit
*
sC2V
Iπ ≈ 0
sC2V
V0
KCL at output node (C)
Assuming oscillations have started, then V ≠ 0 and Vo ≠ 0, so
1

sC 2V  g mV    sC1 Vo  0
R

1

sC 2V  g mV    sC1 V 1  s 2 LC2   0
R

1
 LC 

s 3 LC1C2  s 2  2   s C1  C2    g m    0
R
 R 

*
Setting s = j



1  2 LC2 
 gm  
  j  C1  C2    3 LC1C2  0

R
R 

ECES 352 Winter 2007
Ch 13 Oscillators
12

Colpitts LC-Tuned Oscillator
* To get oscillations, both the real and imaginary parts
of this equation must be set equal to zero.



1  2 LC2 
 gm  
  j  C1  C2    3 LC1C2  0

R
R 

* From the imaginary part we get the expression for
the oscillation frequency
 o C1  C2    o3 LC1C2  0
o 
C1  C2

LC1C2
1
 CC 
L 1 2 
 C1  C2 
* From the real part, we get the condition on the ratio
1  o2 LC2
of C2/C1
g  
0
m
R
R
 C  C2 
C
1  g m R   o2 LC2  LC2  1
1 2

C1
 LC1C2 
C2
 gm R
C1
ECES 352 Winter 2007
Ch 13 Oscillators
13
Colpitts LC-Tuned Oscillator
Example
* Given:
 Design oscillator at 150 MHz


o  2f  2 150x106  9.4x108 rad / s
Transistor gm = 100 mA/V, R = 0.5 K
* Design:


o 
C2
 g m R  (100m A/ V )(0.5K )  50
C1
Select L= 50 nH, then calculate C2, and then C1
C1  C2

LC1C2
1  C2 
1 

LC2  C1 
1  C2 
1


1


(1  50)  1.13x109 F  1,130pF
2
8 2

L o  C1  50nH (9.4 x10 )
C
1,130pF
C1  2 
 23 pF
50
50
C2 
ECES 352 Winter 2007
Ch 13 Oscillators
14
Summary of Oscillator Design
Wien Bridge Oscillator
*
*
Phase Shift Oscillator
*
*
Colpitts LC-Tuned Oscillator
*
ECES 352 Winter 2007
Shown how feedback can be used with
reactive components (capacitors) in the
feedback path.
Can be used to achieve positive feedback.
 With appropriate choice of the resistor
sizes, can get feedback signal in phase
with the input signal.
 Resulting circuit can produce large
amplitude sinusoidal oscillations.
Demonstrated three oscillator circuits:
 Wien Bridge oscillator
 Phase Shift oscillator
 Colpitts LC-Tuned oscillator
Derived equations for calculating resistor and
capacitor sizes to produce oscillations at the
desired oscillator frequency.
Key result: Oscillator design depends
primarily on components in feedback
network, i.e. not on the amplifier’s
characteristics.
Ch 13 Oscillators
15