Transcript Document

EIE 211 Electronic Devices and Circuit Design II
EIE 211 : Electronic Devices and
Circuit Design II
Lecture 11:Stability and
Oscillator Circuits
King Mongkut’s University of Technology Thonburi
7/17/2015 1
EIE 211 Electronic Devices and Circuit Design II
4. Amplifier with a two-pole response
Consider an amplifier whose open-loop transfer function is characterized by 2 real-axis
poles:
… (1)
In this case, the closed-loop poles are obtained from 1 + A(s)β = 0, which lead to
… (2)
Thus, the closed-loop poles are given by
… (3)
From eq (3):
 When the loop gain A0β = 0, we’ll get s1, s2 = - ωP1, - ωP2
 When the loop gain A0β increases to the point where (ωP1 + ωP2)2 = 4(1+ A0β)ωP1ωP2,
then s1, s2 = -0.5(ωP1 + ωP2). This means the two poles are at the same location.
 When the loop gain A0β increases further, the term (ωP1 + ωP2)2 - 4(1+ A0β)ωP1ωP2
would become negative, making the poles be complex conjugate pair.
 Therefore, as the loop gain A0β increases from 0, the poles are brought together, then
coincide and become complex conjugate and move along a vertical line.
King Mongkut’s University of Technology Thonburi
7/17/2015 2
EIE 211 Electronic Devices and Circuit Design II
The plot that shows locus of the poles for
increasing loop gain is called a root-locus
diagram. From this diagram, we also see
that the feedback amp is unconditionally
stable.
As is the case with 2nd order responses
generally, the closed-loop response can
show a peak. The characteristic eqn of 2nd
order network can be written in standard
form as

s2  s
o
Q
 o2  0
where ωo is called the pole frequency and Q is called pole Q factor. The poles are
complex if Q is greater than 0.5.
(1  A0  ) P1 P 2
Q
 P1   P 2
7/17/2015 3
EIE 211 Electronic Devices and Circuit Design II
ωo is the radial distance of the poles from the origin and Q
indicates the distance of the poles from the jω axis. Poles on
the jω axis have Q = ∞.
The response of the feedback amplifier shows no peaking for
Q ≤ 0.707 or Q  1 / 2 . The boundary case (Q = 0.707)
(poles at 45o angles) results in the maximally flat response.
King Mongkut’s University of Technology Thonburi
7/17/2015 4
EIE 211 Electronic Devices and Circuit Design II
5. Amplifier with 3 or more poles
From the root-locus diagram, as the loop gain
increases from zero, the highest-freq pole
moves outward while the 2 other poles are
brought closer together. As Aoβ increases
further, the two poles become coincident and
then become complex and conjugate.
A value of Aoβ exists at which this pair of
complex conjugate poles enters the right half of
the s plane, thus causing the amp to become
unstable.
Thus, one can always maintain amp stability by keeping the loop gain Aoβ smaller than
the value corresponding to the poles entering the right half plane.
King Mongkut’s University of Technology Thonburi
7/17/2015 5
EIE 211 Electronic Devices and Circuit Design II
Gain and Phase Margins
The difference between the value of |Aβ|
at ω180 (the freq of 180o phase shift) and
unity, called the gain margin, is usually
expressed in decibels.
The gain margin represents the amount by
which the loop gain can be increased while
stability is maintained.
The difference between the phase angle at
|Aβ| = 1 and 180o is termed the phase
margin. If at the freq where |Aβ| = 1, the
phase angle is less than 180o, then the
amplifier is stable.
King Mongkut’s University of Technology Thonburi
7/17/2015 6
EIE 211 Electronic Devices and Circuit Design II
An alternative approach for investigating stability
We can investigate the network stability by constructing a Bode plot for the open-loop
gain 20log|A(jω)| and 20log|1/β| which is a straight line, assuming that β is independent
of frequency. The difference between the two curves will be
which is the loop gain (in dB). We may therefore study stability by examining the difference
between the two plots.
For example, consider an amplifier whose open-loop transfer function is characterized by
three poles at 0.1 MHz, 1 MHz and 10MHz, as shown on the next page. Note that because
the poles are widely spaced, the phase is approximately -45o at the first pole freq, -135o at
the 2nd pole freq and -225o at the 3rd. The freq at which the phase of A(jω) is -180o lies on
the -40-dB/decade segment. The amplitude and phase of the open-loop gain of this amp
can be expressed as
105
A( f ) 
(1  jf / 105 )(1  jf / 106 )(1  jf / 107 )
 ( f )  [tan 1 ( f / 105 )  tan 1 ( f / 106 )  tan 1 ( f / 107 )]
King Mongkut’s University of Technology Thonburi
7/17/2015 7
EIE 211 Electronic Devices and Circuit Design II
We can either use the plot or the equations to help estimate where the frequency f180 at
which the phase angle is 180o. The f180 is found to be at 3.34x106 Hz and at this freq, the
magnitude is 58.2 dB.
Consider next the straight line labeled (a) which represents a feedback factor having
20log(1/β) = 85 dB, which corresponds to β = 5.623x10-5. Since the loop gain is the
difference between the |A| curve and the 1/β line, the point of intersection X1
corresponds to the freq at which |Aβ| = 1. From the inspection of the plot, this freq is
5.6x105 Hz. At this freq, the phase angle is approximately -108o. Thus, the closed-loop
amplifier for which 20log(1/β) = 85 dB will be stable with a phase margin of 72o. The gain
margin can be obtained from the plot; it is 25 dB.
King Mongkut’s University of Technology Thonburi
7/17/2015 9
EIE 211 Electronic Devices and Circuit Design II
Oscillators
King Mongkut’s University of Technology Thonburi
7/17/2015 10
EIE 211 Electronic Devices and Circuit Design II
Oscillators
The basic structure of a sinusoidal oscillator. A positive-feedback loop is formed by an amplifier and a frequencyselective network. In an actual oscillator circuit, no input signal will be present; here an input signal xs is employed
to help explain the principle of operation.
*
Feedback amplifier but frequency dependent feedback
As 
A f s  
1   f ( s) A( s)
*
*
Positive feedback, i.e. βf(s)A(s) < 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 )
King Mongkut’s University of Technology Thonburi
7/17/2015 11
EIE 211 Electronic Devices and Circuit Design II
Nonlinear Amplitude Control
The oscillation condition, the Barkhausen criterion, guarantees sustained oscillations in
a mathematical sense. However, the parameters of any physical system cannot be
maintained constant for any length of time. In other words, suppose Aβ = 1 at ω = ωo
and then the temperature changes and Aβ becomes slightly less than unity. Oscillations
will cease in this case. Conversely, if Aβ exceeds unity, oscillations will grow in
amplitude. We therefore need a mechanism for forcing Aβ to remain equal to unity at
the desired value of output amplitude. This task is accomplished by providing a
nonlinear circuit for gain control. A popular limiter circuit is shown on the next page.
King Mongkut’s University of Technology Thonburi
7/17/2015 12
EIE 211 Electronic Devices and Circuit Design II
(a) A popular limiter circuit. (b) Transfer characteristic of the limiter circuit. (c) When Rf is
removed, the limiter turns into a comparator with the characteristic shown.
King Mongkut’s University of Technology Thonburi
7/17/2015 13
EIE 211 Electronic Devices and Circuit Design II
Consider in the case small vi (≈ 0) and a small vo so that vA is
positive and vB is negative. Both diodes D1 and D2 will be off.
Thus,
…. (1)
v  ( R / R )v
O
f
1
I
This is the linear portion of the limiter transfer characteristic. We
now can use superposition to find the voltages at nodes A and B
in terms of +/- V and vo as
vA  V
R3
R2
 vO
R3  R2
R3  R2
vB  V
R5
R4
 vO
R4  R5
R4  R5
….(2)
….(3)
As vi goes positive, vo goes negative (from (1)) and vB becomes more negative (from (3)),
thus keeping D2 off. From (2), vA also becomes less positive and as vi increases to the
point where vA becomes -0.7 V (called VD), diode D1 conducts. To find the value of vo at
which D1 conducts (called this vo as “L- “), we’ll get the negative limiting level given by
King Mongkut’s University of Technology Thonburi
7/17/2015 14
EIE 211 Electronic Devices and Circuit Design II
The corresponding value of vi can be found by dividing L- by the limiter gain –Rf/R1. As vi
is increased, more current flows through the diode D1 and vA remains at approximately
constant. R3 appears in effect in parallel with Rf and the incremental gain is
-(Rf||R3)/R1.
The transfer characteristic for the negative vi can be found in a similar manner. The
positive limiting level L+ can be found to be
and the slope of the transfer characteristic in the positive
limiting region is -(Rf||R4)/R1.
Finally, increasing Rf results in higher gain the linear region.
Removing Rf altogether results in transfer characteristic of a
comparator. That is, the ckt compares vi with the comparator
reference value of 0 V: vi > 0 results in vo = L-, vi < 0 results in
vo = L+ .
7/17/2015 15
EIE 211 Electronic Devices and Circuit Design II
Wien Bridge Oscillator
*
*
*
R2
R1
V0
Vi
*
ZS
If
ZP
*
King Mongkut’s University of Technology Thonburi
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.
7/17/2015 16
EIE 211 Electronic Devices and Circuit Design II
Define
R1
R2
Z S  R  ZC  R 
V0
Vi
1
1 1  1
   sC 
Z P  R Z C   

 R ZC   R
ZS
If
1 1  sRC

sC
sC
ZP

1
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

sC 
R1  sCR 
1  sCR
 R  1  sCR   sCR  (1  sCR ) 2
1
King Mongkut’s University of Technology Thonburi
Z2  Z S  R  ZC 
Z2
1  sRC
sC
7/17/2015 17
EIE 211 Electronic Devices and Circuit Design II
Wien Bridge Oscillator
I1
I2
Amplifier gain including loading effects
R2
R1
Vi
V0
IS
IS
IS
Z2
Z1
If
Xo

To get
V0
Vo
, we use I1  I 2 
and
Vi
R1  R2
ZS
Since I   0,
V0
If
Vo
Vo
R1 so
R1  R2
V0 R1  R2
R

 1 2
Vi
R1
R1
ZP
Xf
V0 V0 Vi

I S Vi I S
Vi  V  V  I1 R1 
Feedback factor
f 
Ar 

1
ZS
sC

1  sRC
King Mongkut’s University of Technology Thonburi
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 )
7/17/2015 18
EIE 211 Electronic Devices and Circuit Design II
Wien Bridge Oscillator
Oscillation condition
P hase of  f Ar equal t o 180o. It already is since  f Ar  0.

R 
sCR
T hen need only  f Ar  1  2 
1
2
R
sCR

(
1

sCR
)
1 

Rewrit ing

R 
sCR
 f Ar  1  2 
R1  sCR  (1  sCR ) 2


 1 


 1 

R2 
sCR

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


R2 
sCR
R2 
1




1

2 2 2

R1  1  3sCR  s C R
R1  3  1  sCR

sCR

R 
1
 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

R 1
R
1  2   1 or 2  2
R1  3
R1

King Mongkut’s University of Technology Thonburi
7/17/2015 19
EIE 211 Electronic Devices and Circuit Design II
Wien Bridge Oscillator - Example
Oscillator specifications: ωo = 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
King Mongkut’s University of Technology Thonburi
7/17/2015 20
EIE 211 Electronic Devices and Circuit Design II
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.
King Mongkut’s University of Technology Thonburi
7/17/2015 21
EIE 211 Electronic Devices and Circuit Design II
*
Once oscillations start, a limiting circuit is needed to prevent
them from growing too large in amplitude
EIE 211 Electronic Devices and Circuit Design II
Phase Shift Oscillator
IC3
VX
V2
IC2
C
C
R
IR2
V1
IC1
If
Rf
I C 2  I R1  I C1 
C
I
R R1
V0 V  V  I Z   Vo  Vo 1  1  1


2
1
C2 C
sCR f

*
*
*
Vo
V
V 
1 
 o  o 1 

sCRR f R f R f  sCR 
Rf 
sCR  sC
Vo 
1 
2 

sCR f  sCR 
I R2 
Vo 
 V2
1 

2 

R
sCRR f  sCR 
Based on op amp using inverting input
Combination of R’s and C’s in feedback
Vo 
1  Vo
IC3  I R2  IC 2 
2


loop so get additional phase shift. Target
sCRR f  sCR  R f
180o to get oscillation.
V 
1  1 
1  Vo
 o 1 

2 
 
Analysis assumes op amp is ideal.
V
V  V  0 so I f  o  I C1
Rf
V1  V  I C1Z C  
I R1 
Vo
sCR f
 V1  1  Vo


R
R  sCR f
R f 
Finally
V X  V2 

  Vo
 sCRR f


sCR  sCR 
sCR 
1 

1 

 sCR 

3
1 
1




R f  sCR ( sCR ) 2 
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 
King Mongkut’s University of Technology Thonburi
7/17/2015 23
EIE 211 Electronic Devices and Circuit Design II
Phase Shift Oscillator
IC3
VX
IC2
V2
V1
C
C
R
IC1
If
Rf
IR1
V0
Example
Oscillator specifications: ωo = 1x106 rad/s
Selecting for convenience C  10 nF,
then from  o 
R
1
3 oC


4
1 
3  sCR  ( sCR ) 2 


we get for the loop gain
VX  
C
IR2 R
Rearranging
1
3RC
1
 58 
3 10nF (1x106 rad / s)
Vo
sCR f
L( )   ( ) A( ) 

 jCR f
 sCR f
V0

1
VX 
4
1 
3  sCR  ( sCR ) 2 


 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
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
o
Note: We get 180 phase shift from op amp
4
4 3R 2C 2 12R
since input is to inverting terminal and
R f  12R
T hen
another 180o from the RC ladder.
http://www.electronics-tutorials.ws/oscillator/rc_oscillator.html
24
EIE 211 Electronic Devices and Circuit Design II
Colpitts LC-Tuned Oscillator
CB
V0
CE
Vi
V0
Vi
* 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.
King Mongkut’s University of Technology Thonburi
7/17/2015 25
EIE 211 Electronic Devices and Circuit Design II
*
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
sC2Vπ
Iπ ≈ 0
sC2Vπ
V0

KCL at output node (C)
Assuming oscillations have started, then Vπ ≠ 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 

King Mongkut’s University of Technology Thonburi
7/17/2015 26
EIE 211 Electronic Devices and Circuit Design II
* 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, we2 get the condition on the ratio
1  LC
of C2/C1
gm   o 2  0
R
R
 C  C2 
C2
1  g m R   o2 LC2  LC2  1

1


C1
 LC1C2 
C2
 gm R
C1
King Mongkut’s University of Technology Thonburi
7/17/2015 27
EIE 211 Electronic Devices and Circuit Design II
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 
King Mongkut’s University of Technology Thonburi
7/17/2015 28
EIE 211 Electronic Devices and Circuit Design II
King Mongkut’s University of Technology Thonburi
7/17/2015 29
EIE 211 Electronic Devices and Circuit Design II
King Mongkut’s University of Technology Thonburi
7/17/2015 30