6/9/2015. amplifiers and oscillators 6/9/2015
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AMPLIFIERS AND OSCILLATORS
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Differential amplifier
An amplifier which amplifies the difference
between the two input signals is called
differential amplifier Fig 9.1.
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CLASSIFICATION
Differential amplifier is classified as
1 Voltage amplifier
2. Current amplifier
3. Tran conductance amplifier
4. Tran resistance amplifier
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CHARACTERISTICS
The characteristics of differential amplifier
1.Infinite input resistance
2.Zero output resistance
3.Infinite gain
4.Infinite Bandwidth
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BASIC DIFFERENTIAL AMPLIFIER
Basic differential amplifier has two input voltage signals.
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Cont
VO=Ad(V1-V2)
Where Ad is called differential gain, (V1-V2)
Is called differential voltage Vd= (V1-V2) .
Ad =VO / Vd if V1=V2 then Vd, and VO would
become zero. In practical differential
Amplifier depends on both difference
and average common level of two inputs.
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Cont….
Common mode signal is VC = (V1+V2) /2
VO=AC(V1+V2) /2 or VO=AC VC.
Where VC is called as common mode gain.
The Total output voltage is
VO= Ad Vd +AC VC.
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COMMON MODE REJECTION RATIO
The ability of a differential amplifier to reject a common mode signal is expressed by
CMMR =ρ=20 log Ad/ AC
Differential amplifier has two modes
Common mode and differential mode.
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DIFFERENTIAL MODE
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Cont…
The input signals are same in magnitude
but 180 degree phase shift Q1 is positive going and Q2 is negative
going. RE has same current but 180 degree
phase shift ,so AC current in RE is zero.
The output of collector one and two are equal but opposite in phase.
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Cont…..
Therefore the total output voltage of
differential mode is Vd= (V1-V2)
V1 = Voltage across collector one
V2 = Voltage across collector two
Therefore total output voltage is twice as
large as signal voltage from either collector
to ground.
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COMMON MODE
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Cont..
The input signals of Q1 and Q2 are equal
in magnitude and phase.
RE carries current (2IE) and provides negative feedback.
The output voltage at collector one and two are same in phase and magnitude
Therefore Vo= 0 in common mode .
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Feedback amplifiers and oscillators
Feedback is a process of injecting some energy from the output and then return it to the input .The amplifier which uses this is called feedback amplifier.
Classification of feedback amplifiers
1. Positive Feedback amplifiers
2. Negative Feedback amplifiers
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Feedback amplifiers and oscillators
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Feedback amplifiers and oscillators
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Feedback amplifiers and oscillators
Types of feedback connectionsTypes of feedback connections
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Feedback amplifiers and oscillators
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Feedback amplifiers and oscillators
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Feedback amplifiers and oscillators
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Feedback amplifiers and oscillators
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Feedback amplifiers and oscillators
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Feedback amplifiers and oscillators
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Feedback amplifiers and oscillators
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Feedback amplifiers and oscillators
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Feedback amplifiers and oscillators
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Feedback amplifiers and oscillators
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Feedback amplifiers and oscillators
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Oscillators
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Oscillator
• Introduction of Oscillator
• Linear Oscillator– Wien Bridge Oscillator– RC Phase-Shift Oscillator– LC Oscillator
• Stability
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Oscillators
Oscillation: an effect that repeatedly and regularly fluctuates about the mean value
Oscillator: circuit that produces oscillation
Characteristics: wave-shape, frequency, amplitude, distortion, stability
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Application of Oscillators
• Oscillators are used to generate signals, e.g.– Used as a local oscillator to transform the RF
signals to IF signals in a receiver;– Used to generate RF carrier in a transmitter– Used to generate clocks in digital systems;– Used as sweep circuits in TV sets and CRO.
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Linear Oscillators
1. Wien Bridge Oscillators
2. RC Phase-Shift Oscillators
3. LC Oscillators
4. Stability
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Integrant of Linear Oscillators
For sinusoidal input is connected “Linear” because the output is approximately sinusoidal
A linear oscillator contains:- a frequency selection feedback network- an amplifier to maintain the loop gain at unity
+
+Amplifier (A)
Frequency-SelectiveFeedback Network ()
Vf
Vs VoV
PositiveFeedback
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Basic Linear Oscillator
+
+
SelectiveNetwork(f)
Vf
Vs VoV
A(f)
)( fso VVAAVV and of VV
A
A
V
V
s
o
1
If Vs = 0, the only way that Vo can be nonzero is that loop gain A=1 which implies that
0
1||
A
A (Barkhausen Criterion)
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Wien Bridge OscillatorFrequency Selection NetworkLet
11
1
CX C
and
111 CjXRZ 2
2
1
CX C
22
22
1
222
11
C
C
C jXR
XjR
jXRZ
Therefore, the feedback factor,
)/()(
)/(
222211
2222
21
2
CCC
CC
i
o
jXRXjRjXR
jXRXjR
ZZ
Z
V
V
222211
22
))(( CCC
C
XjRjXRjXR
XjR
Vi Vo
R1 C1
R2C2
Z1
Z2
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can be rewritten as:
)( 2121221221
22
CCCCC
C
XXRRjXRXRXR
XR
For Barkhausen Criterion, imaginary part = 0, i.e.,
02121 CC XXRR
Supposing,
R1=R2=R and XC1= XC2=XC,
2121
2121
/1
11or
CCRR
CCRR
)(3 22CC
C
XRjRX
RX
0.2
0.22
0.24
0.26
0.28
0.3
0.32
0.34
Fee
dbac
k fa
ctor
-1
-0.5
0
0.5
1
Phas
e
Frequency
=1/3
Phase=0
f(R=Xc)
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Example
Rf
+
R
R
C
CZ1
Z2
R1
Vo
By setting , we get
Imaginary part = 0 and
RC
1
3
1
Due to Barkhausen Criterion,
Loop gain Av=1
where
Av : Gain of the amplifier
1
131R
RAA f
vv
21
R
R fTherefore, Wien Bridge Oscillator
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RC Phase-Shift Oscillator
+
Rf
R1
R R R
C C C
Using an inverting amplifier The additional 180o phase shift is provided by an RC
phase-shift network
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Applying KVL to the phase-shift network, we have
R R R
C C CV1 Vo
I1 I2 I3Solve for I3, we get
)2(0
)2( 0
)(
32
321
211
C
C
C
jXRIRI
RIjXRIRI
RIjXRIV
C
C
C
C
C
jXRR
RjXRR
RjXRR
jXRR
VRjXR
I
20
2
000
02
3
1
)2(])2)[(( 222
21
3CCC jXRRRjXRjXR
RVI
Or
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The output voltage,
)2(])2)[(( 222
31
3CCC
o jXRRRjXRjXR
RVRIV
Hence the transfer function of the phase-shift network is given by,
)6()5( 2323
3
1 CCC
o
XRXjRXR
R
V
V
For 180o phase shift, the imaginary part = 0, i.e.,
RC
R
XXRX CCC
6
1
6X
(Rejected) 0or 0622
C
23
and,
29
1
Note: The –ve sign mean the phase inversion from the voltage
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LC Oscillators
+
~
Av Ro
Z1 Z2
Z3
12
Zp
The frequency selection network (Z1, Z2 and Z3) provides a phase shift of 180o
The amplifier provides an addition shift of 180o
Two well-known Oscillators:• Colpitts Oscillator• Harley Oscillator
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+
~Av Ro
Z1 Z2
Z3
Zp
Vf Vo
321
312
312
)(
)//(
ZZZ
ZZZ
ZZZZ p
For the equivalent circuit from the output
po
pv
i
o
p
o
po
iv
ZR
ZA
V
V
Z
V
ZR
VA
or
Therefore, the amplifier gain is obtained,
)()(
)(
312321
312
ZZZZZZR
ZZZA
V
VA
o
v
i
o
oof VZZ
ZVV
31
1
Zp AvVi
Ro
+
+
Vo
Io
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The loop gain,
)()( 312321
21
ZZZZZZR
ZZAA
o
v
If the impedance are all pure reactances, i.e.,
332211 and , jXZjXZjXZ
The loop gain becomes,)()( 312321
21
XXXXXXjR
XXAA
o
v
The imaginary part = 0 only when X1+ X2+ X3=0 It indicates that at least one reactance must be –ve (capacitor) X1 and X2 must be of same type and X3 must be of opposite type
2
1
31
1
X
XA
XX
XAA vv
With imaginary part = 0,
For Unit Gain & 180o Phase-shift,1
2 1X
XAA v
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R L1
L2
C
RC1
C2
L
Hartley Oscillator Colpitts Oscillator
T
oLC
1
1
2
RC
Cgm
21
21
CC
CCCT
CLLo
)(
1
21
2
1
RL
Lgm
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Colpitts Oscillator
RC1
C2
L
Equivalent circuit
In the equivalent circuit, it is assumed that: Linear small signal model of transistor is used The transistor capacitances are neglected Input resistance of the transistor is large enough
+
V
gmVR C1C2
L
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)(11 LjiVV At node 1,
where,
VCji 21
)1( 22
1 LCVV
Apply KCL at node 1, we have
0111
2 VCjR
VVgVCj m
01
)1( 122
2
Cj
RLCVVgVCj m
0)(1
213
212
2
CLCCCj
R
LC
Rgm
For Oscillator V must not be zero, therefore it enforces,
+
V
gmVR C1C2
Lnode 1
I1
I2I3
I4
V
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Imaginary part = 0, we have
0)(1
213
212
2
CLCCCj
R
LC
Rgm
T
oLC
1
1
2
RC
Cgm
21
21
CC
CCCT
Real part = 0, yields
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THANK U
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