VTU ANALOG ELECTRONICS CIRCUITS AMPLIFIERSModule 2 (2).pdf

AMPLIFIERS

1. Module-II
TransistorAmplifiers

2. Figure 1.1 a The basic configurations of transistor amplifiers. (a)–(c) For the MOSFET; (d)–(f) forthe
BJT.
Classification ofAmplifiers

3. •Grounded-source or common-source (CS) amplifier: The source
terminal is connected to ground, the input voltage signal vi is
applied between the gate and ground, and the output voltage signal
vo is taken between the drain and ground, across the resistanceR
D
•common-gate(CG)orgrounded-gateamplifier:Itisobtainedby
connectingthegatetoground,applyingtheinputvibetweenthe
sourceandground,andtakingtheoutputvoacrosstheresistance
R
Dconnectedbetweenthedrainandground
•common-drain (CD) or grounded-drain amplifier. It is obtained
by connecting the drain terminal to ground, applying the input
voltage signal vi between gate and ground, and taking the output
voltage signal between the source and ground, across a load
resistanceRL.
Classification ofAmplifiers

4. CharacterizingAmplifiers

5. •The input resistance R
in represents the loading effect of the
amplifier input on the signal source. It is foundfrom
Rin≡Vi/Ii (1.1)
•and together with the resistance R
sig forms a voltagedivider
that reduces v
sig to the value v
i that appears at theamplifier
input,
�??????=
????????????�
????????????�+??????�??????�
��??????
(1.2)
•The second parameter in characterizing amplifier
performance is the open-circuit voltage gain Avo , defined
as
Avo ≡ vo/viforRL=∞ (1.3)
CharacterizingAmplifiers

6. •The third and final parameter is the output resistance Ro. Observe from Fig.
6.35(b) that Ro is the resistance seen looking back into the amplifier output
terminal with set to zero. Thus Ro can be determined,as
Ro=V
X/I
X (1.4)
•The controlled source A
voV
i and the output resistance R
o represent theThevenin
equivalent of the amplifier output circuit, and the output voltage
vo canbe
foundfrom
??????�=
??????
????????????+????????????
�??????��?????? (1.5)
•Thus the voltage gain of the amplifier proper, A
v , can be foundas
�??????=
��
=�??????�
??????
(1.6)
�?????? ????????????+??????�
and the overall voltage gain G
v, Gv ≡Vo/Vsig
can be determined by combining Eqs. (1.4) and(1.6):
Gv= �??????�
????????????�????????????
????????????�+??????�??????�????????????+??????�
CharacterizingAmplifiers

7. Common Source Amplifier (Without Rs-
Bypassed resistanceRs)
Fig.6.35(a) AC equivalent circuit of Common SourceAmplifier, Biasing circuit is ommitted
(b) MOSFET replaced by its small signalmodel

8. Characteristic Parameters of the CSAmplifier
•Fig. 6.35(a) shows the small-signal model for the
common-source amplifier. Here, R
D is considered
part of the amplifier and is the resistance that one
measures between the drain and theground.
•The small-signal model can be replaced by its
hybrid-π model as shown in Fig. 6.35(b). Then the
current induced in the output port is i = −g
mv
gs as
indicated by the current source.Thus
v
o =−g
mv
gsR
D (1.8)
By inspection, one seesthat
Rin=∞,vi =vsig,vgs=vi(1.9)

9. •Thus the open-circuit voltage gainis
•To find the Norton equivalence resistance, one sets v
i = 0, which
will make the current source an open circuit with zero current.
And by the test-current method, the output resistanceis
R
o=R
D (2.1)
•If now, a load resistor, R
L is connected to the output acrossR
D,
then the voltagegainproperis
Av=-gm(R
D || R
L)(2.2)
•From the fact that R
in = ∞, then v
i = v
sig. The overall voltagegain,
G
v, is the same as the voltage gain proper, A
v,namely
•�??????=
�
��??????�
= −��(??????�∥????????????) (2.3)
Characteristic Parameters of the CSAmplifier

10. CommonSource Amplifier
(Without Rs)-Bypassed resistance
Rs)
Fig.6.36(a)ACequivalentcircuitofCommonSourceAmplifier,Biasingcircuitisomitted
(b) MOSFET replaced by its small signalmodel

11. Characteristic Parameters of the CSAmplifier
•Fig. 6.35(a) shows the small-signal model for the
common-source amplifier. Here, R
D is considered
part of the amplifier and is the resistance that one
measures between the drain and theground.
•The small-signal model can be replaced by its
hybrid-π model (λ ≠ 0) as shown in Fig. 6.35(b).
Then the current induced in the output port is i=
−g
mv
gs as indicated by the current source.Thus
v
o =−g
mv
gs(R
D||r
o ) (1.8)
By inspection, one seesthat
Rin=∞,vi =vsig,vgs=vi(1.9)

12. •Thus the open-circuit voltage gainis
•To find the Norton equivalence resistance, one sets v
i = 0, which
will make the current source an open circuit with zero current.
And by the test-current method, the output resistanceis
R
o = (R
D || r
o) (2.1)
•If now, a load resistor, R
L is connected to the output acrossR
D,
then the voltagegainproperis
Av=-gm (R
D || r
o || R
L)(2.2)
•From the fact that R
in = ∞, then v
i = v
sig. The overall voltagegain,
G
v, is the same as the voltage gain proper, A
v,namely
�
��??????�
D o
•�??????= =−��(R||r∥????????????) (2.3)
Characteristic Parameters of the CSAmplifier

13. Performing the Analysis Directly onthe
CircuitDiagram
In simple situations and after a lot of practice, one can perform thesmall-signal
analysis directly on the circuit schematic. Because in this way one remains
closer to the actual circuit, the direct analysis can yield greater insight into
circuitoperation.
Fi
.
gure shows the direct analysis of the CS amplifier. Observe that wehave
―pulled out‖ the resistance from the transistor, thus making the transistor drain
conduct g
mv
gs while still accounting for the effect ofr
o

14. Problem
•ACSamplifierutilizesaMOSFETbiasedatI
D=0.25mA
withV
OV=0.25VandR
D=20kΩ.ThedevicehasV
A=50V.
TheamplifierisfedwithasourcehavingRsig=100kΩ,
anda20kΩloadisconnectedtotheoutput.
•(a)FindRin,Ro,Avo,AvandGv.
•(b)Iftomaintainreasonablelinearity,thepeakofthe
inputsine-wavesignalislimitedto10%of(2V
OV)what
isthepeakofthesinewavevoltageattheoutput?

15. Solution

16. Solution
•The input resistance is Rin is obviouslyinfinite,
•Rin=∞
|��
|
•r
o=
??????
=200kΩ
m
2??????
�
??????
=2mA/V•g=
•
• =-36.4
R
o = (R
D || r
o )=18.18k
Av=-gm (R
D || r
o || R
L)=-19.05
�??????=
�
��??????�
= −��(R
D || r
o ∥????????????)=-19.05

17. Solution
•(b)
•Vgs=(10%)2Vov=0.05V
•V
O=Av*Vgs=0.9525V

18. Problem2-Homework
•A CS amplifier utilizes a MOSFET biased at I
D =0.25
mA with V
OV=0.25VandR
D=20kΩ. The amplifier is
fed with a source having Rsig=100 kΩ, and a 20 kΩ
load is connected to theoutput.

19. •The CS amplifiers has infinite input impedance (draws
no current at DC), and a moderately high output
resistance (easier to match for maximum power
transfer), and a high voltage gain (a desirable feature of
anamplifier).
•Reducing R
D reduces the output resistance of a CS
amplifier, but unfortunately, the voltage gain is also
reduced. Alternate design can be employed to reduce
the output resistance (to be discussedlater).
•A CS amplifier suffers from poor highfrequency
performance, as most transistor amplifiersdo.
Characteristic Parameters of the CSAmplifier

20. BridgeCourse
T model ofMOSFET

21. Bridge Course for Tmodel
•The development of T model, is illustrated in Fig. Figure (a) shows the
equivalent circuit studied. we have added a second g
mv
gs current source in series
with the original controlled source. This addition obviously does not change the
terminal currents and is thusallowed.
•The newly created circuit node, labeled X, is joined to the gate terminal G inFig.
(c). Observe that the gate current does not change—that is, it remains equal to
zero—and thus this connection does not alter the terminalcharacteristics.
•Wenownotethatwehaveacontrolledcurrentsourceg
mv
gsconnectedacrossits
controlvoltagevgs.Wecanreplacethiscontrolledsourcebyaresistanceaslong
asthisresistancedrawsanequalcurrentasthesource.
•Thus the value of the resistance is = vgs/gmvgs=1/gm . This replacement is
shown in Fig.(d), which depicts the alternative model. Observe that ig is still
zero, and Id=g
mv
gs, Ig=0, all the same as in the original model inFig.(a).
•The model of Fig.(d) shows that the resistance between gate and source looking
into the source is 1/gm. This observation and the T model prove useful in many
applications. Note that the resistance between gate and source, looking into the
gate, isinfinite.

22. BridgeCourse
T model with drain to source resistancer
0
•In developing the T modelwe
did not includero.
•If desired, this can be doneby
incorporating in the circuit of
Fig. (d) a resistance ro
between drain and source, as
shown in Fig.(e).
•An alternativerepresentation
of the T model, in which the
voltage-controlled current
source is replaced with a
current-controlled current
source, is shown in Fig.(f).
(e) (f)

23. Common-Source Amplifier with aSource
Resistance(unbypassedRs)
Fig. 1.4: ACS
amplifier with a
source resistance:
(top) detail circuit,
and (bottom)
equivalent circuitT
model

24. Common-Source Amplifier witha
SourceResistance
•As a consequence, vi = vsig. However, because of
the existence of the source resistance, less of the
input voltage is divided to vgs, by the voltage
divider formula.Thus
• v
o =−iR
D
Thus the open-circuit voltage gain (assume that R
D is part of the amplifier)is

25. Common-SourceAmplifier
with a SourceResistance
•When a load resistor R
L is added, then the voltage
gain proper (also called terminal voltage gain)is
•
•Because the input resistance is infinite, hence v
i=
v
sig and the overall voltage gain G
v =A
v.

26. Summary of the CSAmplifier
with SourceResistance
•The input resistance R
in isinfinite.
•The open-circuit voltage gain, A
vo, is reduced by a
factor of (1 + g
mR
s )as seen in(2.4).
•In general, the addition of the source resistance R
s
gives rise to a “negative” feedback factor 1+g
mR
s
that reduces voltage gain, but improves linearity,
and high-frequencyresponse.
•Because of the negative-feedback action of R
s, it is
also called the source-degenerationresistance.

27. Source Follower(Common
DrainAmplifier)

28. Source Follower
(Common DrainAmplifier)
•Since the gate current is zero for thiscircuit,
R
in=∞ (4.1)
•Using the voltage divider formula, it is seen that
voltage gain proper or terminal voltage gainis
•Open Circuit VoltageGain,
Avo=1

29. Source Follower
(Common DrainAmplifier)
•The output resistance is obtained by replacing the
proper part of the amplifier with a Thevenin’s
equivalence. To this end, with the use of the test-
current method, one sets the value of v
i = 0, and
thus
R
o =1/g
m
•overall voltage gain G
v (also called the total voltage
gain) is the same as the voltage gain properA
v

30. Figure 6.41 Illustrating the need for a unity-gain voltage bufferamplifier.

31. Comparisons of MOSFETand
BJTamplifiers.
•MOS amplifiers have high input impedance (except for CG amplifiers). Thisis
an advantage over BJTamplifiers.
•BJT’s have higher transconductance g
m than MOSFET’s giving BJTamplifiers
highergains.
•Discrete-circuit amplifiers, e.g., circuits assembled on printed-circuit
board(PCB), BJT’s are prevalent because of their longer history and wider
commercial availability.
•Because of easier fabrication, integrated circuit (IC) amplifiers are dominatedby
MOSFET’s.
•The CS and CE configurations are best suited for gain amplifiers becauseoftheir
larger than unity gain. A cascade of them can be used to increase thegain.
•The addition of R
s in a CS amplifier and R
e in a CE amplifier improvesthe
linearity of the circuit and better high frequencyperformance.
•TheCEandCSdesignshavebothhighvoltageandcurrentgains.TheCBand
CGdesignshavelowcurrentgain,butstillhighvoltagegain.TheCCandCD
designs(emitterandsourcefollowers)havelowvoltagegain,buthighcurrent
gain.

32. Comparison ofcharacteristics

33. Internal Capacitive Effects ofMOSFET
•InternalcapacitorsinaMOSFET.
•Anytwopiecesofconductivematerialsseparatedbyaninsulator(ora
regionoflowconductivity)willhaveacapacitancebetweenthem.This
Fig.showstheinternalcapacitancesasC
gs,C
gd,C
sb,andC
db

34. Internal CapacitiveEffects
•MOSFET has internal capacitances. In fact, we used
one of these, the gate-to-channel capacitance, to
understand the operation of the device and its
characteristics.
•There are basically two types of internal
capacitance in theMOSFET
•The gate capacitive effect:
•The source-body and drain-body depletion-layer
capacitances

35. Internal CapacitiveEffects
•Thegatecapacitiveeffect:
Thegateelectrode(polysilicon)formsaparallel-plate
capacitorwiththechannel,withtheoxidelayerserving
asthecapacitordielectric.Thiscapacitanceisexpressed
asgate(oroxide)capacitanceinperunitareaas(Cox).
•Thesource-bodyanddrain-bodydepletion-layer
capacitances:
Thesearethecapacitancesofthereverse-biasedpn
junctionsformedbythen+sourceregion(alsocalledthe
sourcediffusion)andthep-typesubstrateandbythen+
drainregion(thedraindiffusion)andthesubstrate.

36. Internal CapacitiveEffects
•These two capacitive effects can be modeled by
including capacitances in the MOSFET model
between its four terminals, G, D, S, and B. There
will be five capacitances intotal:
Cgs, Cgd, Cgb, Csb, and Cdb, where the subscripts
indicate the location of the capacitances in the
model. In the following, we show how the values of
the five model capacitances can bedetermined.

37. The Gate CapacitiveEffect
•The gate capacitive effect can be modeled by the
three capacitances Cgs, Cgd, and Cgb. The values of
these capacitances can be determined asfollows
•When the MOSFET is operating in the triode region
at small V
DS, the channel will be of uniform depth.
The gate-channel capacitance will be WLCox and
can be modeled by dividing it equally between the
source and drain ends; thus,:

38. TheGateCapacitiveEffect
•WhentheMOSFEToperatesinsaturation,the
channelhasataperedshapeandispinchedoffat
ornearthedrainend.Itcanbeshownthatthe
gate-to-channelcapacitanceinthiscaseis
approximatelyandcanbemodeledbyassigning
thisentireamounttoCgs,andazeroamountto
Cgd(becausethechannelispinchedoffatthe
drain);thus,

39. The Gate CapacitiveEffect
•When the MOSFET is cut off, the channel
disappears, and thus Cgs = Cgd = 0. However, we
can (after some rather complex reasoning) model
the gate capacitive effect by assigning a capacitance
WLCox to the gate-body model capacitance;thus,

40. Overlapcapacitance
•There is an additional small capacitive component
that should be added to Cgs and Cgd in all the
preceding formulas. This is the capacitance that
results from the fact that the source and drain
diffusions extend slightly under the gateoxide.
•If the overlap length is denoted Lov, we see that
the overlap capacitance componentis

41. TheJunctionCapacitances
•For the source diffusion, we have the source-body capacitance,Csb,
where C
sb0 is the value of Csb at zero body-source bias, V
SB is the
magnitude of the reverse bias voltage, and V
0 is the junction built-in
voltage (0.6 V to 0.8V).
•Similarly, for the drain diffusion, we have the drain-body capacitance
C
db, where C
db0 is the capacitance value at zero reverse-bias voltage, and
V
DB is the magnitude of this reverse-biasvoltage.

42. PROBLEM
•Forann-channelMOSFETwithtox=10nm,L=1.0μm,
W=10μm,Lov=0.05μm,Csb0=Cdb0=10fF,V
0=0.6
V,V
SB=1V,V
DS=2V,calculatethefollowingcapacitances
whenthetransistorisoperatinginsaturation:
•Cox,Cov,Cgs,Cgd,Csb,andCdb.

43. Solution
•��??????= (∈�??????)/��??????= (3.45 ∗10
−11
)/(10 ∗10^ − 9)
= 3.45 ∗10
−3
�
2
=3.45
���
??????�
2
•C
ov = �??????
�??????�
�??????= 10 ∗0.05 ∗3.45 = 172fF
gs
3 3
•C=
2
�??????��??????+ ��??????=
2
∗10 ∗1 ∗3.45 + 1.72 = 24.72��
•C
gd= C
ov = 1.72fF
sb
•C=
�
�??????0
1+
�??????0
�
0
=
10
1
= 6.1fF
•Cdb=
�????????????0
1+
���
�0
=
1+
0.6
10
1+
2+1
0.6
= 4.1fF

44. Frequency Response ofCommon
sourceamplifiers
•The frequency response of a discrete circuit is
affected by the coupling capacitors and bypass
capacitors at the low frequency end. At the high
frequency end, it is affected by the internal
capacitors (or parasitic capacitances) of thecircuit

45. Frequency Response of
Common sourceamplifiers
•Atlowerfrequencies,themagnitudeoftheamplifier
gainfallsoff.Thisoccursbecausethecouplingand
bypasscapacitorsno longer have low impedances.At
midbandfrequenciestheirimpedancesaresmall
enough to act as shortcircuits.
•Atlowfrequencies,asthefrequencyoftheinputsignal
islowered,thereactance1/jωCofeachofthese
capacitorsbecomessignificantandthisresultsina
decrease in the overall voltage gain of theamplifier.
•Lowe cut off frequency,frequencyf
L,defines thelower
endofthemidband.Itisusuallydefinedasthe
frequencyatwhichthegaindropsby3dBbelowits
valueinmidband.

46. Frequency Response of
Common sourceamplifiers
•BW = f
H − f
L, (discrete-circuitamplifiers),
•BW = f
H, (integrated-circuitamplifiers)
•A Figure of merit for amplifier design is the gain-
bandwidth product, definedas
GB =|A
M|BW

47. High-Frequency Model ofthe
MOSFET
•Gainoftheamplifierfallsoffatthehigh-frequency
end.Thisisduetointernalcapacitiveeffectsinthe
BJTandintheMOSFET.
•TheHighercutofffrequencyf
H,definestheupper
endofthemidband.Itisdefinedasthefrequency
atwhichthegaindropsby3dBbelowitsmidband
value.
•Thus, the amplifier bandwidth is defined by f
Land
f
H (0 and f
H for ICamplifiers).

48. High-Frequency Model of the
MOSFET
•The simplified High-frequency equivalent circuit of aMOSFET.

49. High-Frequency Model ofthe
MOSFET
Model can be further simplified by ignoring Cdb,
which is usuallysmall.

50. High-FrequencyResponse
oftheCSAmplifiers
•Frequency response of a direct-coupled (dc) amplifier.
Observe that the gain does not fall off at low frequencies,
and the midband gain A
M extends down to zerofrequency.
•Mid Band Gain is expressedas

51. Determining the high-frequency response of the CSamplifier:
(a)equivalentcircuit;
(b)the circuit of (a) simplified at the input and theoutput;
•(a)
•(b)

52. High-FrequencyResponse
oftheCSAmplifiers
•The equivalent output impedance can be expressed
as,
•Output voltage can be expressedas,
•Using Miller’s Theorem, circuit can be representedas

53. High-FrequencyResponse
oftheCSAmplifiers
•By Millers theorem, equivalent capacitance is givenby,
C
eq = (1 + A
v)C = (1 +A
v)C
gd
•Since input Voltage is V
gs, wehave
•Av =
�
0
′
= −
�
�
�
��
??????
??????
�
??????
�
=−��??????
′
??????
•Ceq=(1 + g
mR’
L)C
gd
•Total input capacitance Cin can be givenby,
•C
in = C
gs +C
eq
• = C
gs + (1 + g
mR’
L)C
gd

54. High-FrequencyResponse
oftheCSAmplifiers
•The total resistance is givenby,
•Rsig’= R
si ||R
G
•By considering input circuit as a simple time
constant circuit wehave,
•??????= RC =R’
sigC
in
H
??????
•??????=??????=
1
=
1
??????
′
�
�??????�??????�
H
•f=
0
1
2????????????
′�
�??????�??????�

55. Problem
•Find the midband gain A
M and the upper 3-dB
frequency f
H of a CS amplifier fed with a signal
source having an internal resistance Rsig = 100 kΩ.
The amplifier has R
G = 4.7 MΩ, R
D = R
L = 15 kΩ, g
m =
1 mA/V, r
o = 150 kΩ Cgs = 1 pF Cgd = 0.4pF

56. Solution

57. Solution

58. Oscillators
•Oscillators are the electronics circuits which generate
voltages of desired frequency amplitude andshape.
•Oscillators can be classified:
•Basedon outputwaveform
Example: Sinusoidal, Square wave or saw-toothetc
•Based on circuitcomponents
Example: RC Oscillator, LC Oscillator and Crystal Oscillatoretc
•Based on range offrequency
Example: Audio frequency Oscillator, Radio Frequency Oscillatoretc
•Based on presence offeedback
Example\; Oscillators with feedback circuit and withoutfeedback.

59. OscillatorsWorking
Principle
•Oscillators Employ PositiveFeedback
•Amplifiers Employ NegativeFeedback

60. Barkhausen’sPrinciple

61. Barkhausen’sPrinciple
•The output of inverting amplifier is givenby
V
o = AV
i..............(32)
•The feedback network decides the amount of
feedback to be given to the inputi.e.,
• V
f =βV
o................(33)
V
f= βAV
i .................(34)

62. LOOP GAINAβ
•If | Aβ | >1 circuit generates oscillations of growing
type
•If | Aβ | =1 the circuit generates oscillations of fixed
amplitude

63. LOOP GAINAβ
•If | Aβ | < 1 the circuit generates oscillations of
decayingnature

64. Phase-ShiftOscillator
-
K

65. FET Phase-ShiftOscillator
•The amplifier stage is
self biased with a
capacitor bypassed
source resistor R
S and a
drain bias resistor R
D.

66. FET Phase-ShiftOscillator
•Apracticalversionofaphase-shiftoscillatorcircuit
isdrawntoshowclearlytheamplifierandfeedback
network.
•Here FET amplifier introduces 180ᵒ phase shift. The
3 RC sections further introduce 180ᵒ phase shift.
Total phase shift around the loop is 360ᵒ ensuring
Positive feedback. The amplifier stage is self biased
with a capacitor bypassed source resistor R
S and a
drain bias resistor R
D.

67. FET Phase ShiftOscillators
•From FET amplifier theory,the
amplifier gain magnitude is
calculatedfrom
•Gain |A|> 29 tosustain
oscillations
•where R
L in this case is the parallel
resistance of R
D andr
d
•Frequency of oscillationsis

68. Features of RC Phase shiftOscillator
1.Circuit is simple todesign.
2.Can produce output over audio frequencyrange.
3.Produces sinusoidalwaveform.
4.It is a fixed frequency oscillator. Cannot achieve a
variablefrequency.
5.Frequency stability is poor due to temperature,
agingetc...

69. Problem

70. Problem

71. Tuned OscillatorCircuits
•Tuned oscillators use a parallel LC resonant circuit
(LC tank) to provide the oscillations. Frequency
range from KHZ to severalGHz.
•Frequency of generated oscillations is givenby,
•&#3627408467;=
1
.......(41)
2????????????&#3627408438;

72. Tuned OscillatorCircuits
•Basic form of LC oscillator circuit consists of an
Amplifier and a feedback network consisting of LC
resonant circuit. In the feedback network, either L
or C is broken down into two impedances Z
1 and Z
3.
Depending on the arrangement we have two types
of tuned oscillatorcircuits.
•Colpitts Oscillator—The feedback network consists
of two inductive and one capacitiveimpedances.
•Hartley Oscillator—The feedback network consists
oftwo inductive and one capacitiveimpedances

73. Colpitt’s OscillatorCircuit
•The oscillator
frequency canbe
found tobe
•Where,

74. Colpitt’soscillator
•Colpitt’s oscillator consists of tank circuit which is made up of two
capacitors connected in series and an inductor is connected parallel to
this series combination. The frequency of oscillator is determined bythe
value of capacitor andinductor.
•RFC (Radio Frequency Choke) is used to prevent highfrequency
oscillations passing through powersupply.
•In this circuit resistors R1 and R2 provide voltage divider biasing to the
transistor FET. Cc1 is the input DC decoupling capacitor while Cc2 isthe
output decoupling capacitor. RS is the Source resistance which is
bypassed by capacitorCs.
•If source bypass capacitor is not used then signal will dropacross
R
Eand gain will be reduced. Feedback is taken from junction of
capacitorC
2and inductor L & fed back to the Gate ofFET.

75. Colpitt’s OscillatorCircuit
•HereFETamplifierintroduces180ᵒphaseshift.The
LCfeedbacknetworkfurtherintroduces180ᵒphase
shift.Thus,Totalphaseshiftaroundtheloopis360ᵒ
ensuringPositivefeedback.

76. Transistorized Colpitt’soscillator
•Colpitt’s oscillator
consists of tankcircuit
which is made up of
two capacitors
connected in series
and an inductor is
connected parallel to
this series
combination.
•The frequency of
oscillator is
determined bytank
circuit.

77. Transistorized Colpitt’soscillator
•RFC (Radio Frequency Choke) is used to prevent high
frequency oscillations passing through powersupply.
•In this circuit resistor R1 and R2 provides voltage
divider biasing to the transistor Q1.Cin is the input DC
decoupling capacitor while Cout is the output
decouplingcapacitor.
•Re is the emitter resistor which is bypassed bycapacitor
Ce. If bypass capacitor is not used then AC signal will
drop across RE , hence it will alter DC bias condition of
transistor and reducesgain.
•Feedbackis takenfromjunctionofcapacitorC2and
inductor L & fed back to the base of transistorQ1.

78. Transistorized Colpitt’soscillator
•The frequency of oscillations of the colpitt’s
oscillator is givenby,
Where,

79. Hartley Oscillator usingFET
•The oscillator
frequency canbe
found tobe
•Where,

80. Hartley Oscillator usingFET
•Hartley oscillator consists of tank circuit which is made up of two
inductors connected in series and an capacitor is connected
parallel to this series combination. The frequency of oscillator is
determined by the value of capacitor andinductor.
•RFC (Radio Frequency Choke) is used to prevent high frequency
oscillations passing through power supply.
•In this circuit resistors R1 and R2 provide voltage divider biasing
to the transistor FET. Cc
2 is the input DC decoupling capacitor
while Cc
1 is the output decoupling capacitor. R
S is the Source
resistance which is bypassed by capacitor Cs.
•If source bypass capacitor is not used then signal will drop across
R
sand gain will be reduced. Feedback is taken from junction of
inductorL
2and capacitor C & fed back to the Gate ofFET.

81. HartleyOscillator
•HereFETamplifierintroduces180ᵒphaseshift.The
LCfeedbacknetworkfurtherintroduces180ᵒphase
shift.Thus,Totalphaseshiftaroundtheloopis360ᵒ
ensuringPositivefeedback.
•Note,however,thatinductorsL1andL2havea
mutualcouplingM,whichmustbetakeninto
accountindeterminingtheequivalentinductance
fortheresonanttankcircuit.

82. Hartley Oscillator usingBJT
•The oscillator frequency
can be found tobe
•Where,

83. Hartley Oscillator usingBJT
•Hartley oscillator consists of tank circuit which is made up of two
inductors connected in series and an capacitor is connected
parallel to this series combination. The frequency of oscillator is
determined by the value of capacitor andinductor.
•RFC (Radio Frequency Choke) is used to prevent high frequency
oscillations passing through power supply.
•In this circuit resistors R1 and R2 provide voltage divider biasing
to the transistor. Cc
2 is the input DC decoupling capacitor while
Cc
1 is the output decoupling capacitor. R
E is the Source resistance
which is bypassed by capacitorCs.
•If source bypass capacitor is not used then signal will drop across
R
Eand gain will be reduced. Feedback is taken from junction of
inductorL
2and capacitor C & fed back to the Gate of BJT
Amplifier.

84. CrystalOscillators
•The Quartz, Tourmaline and Rochelle salt exhibit piezo-electriceffect.
•Thismeansundertheinfluenceofmechanicalstress,Thevoltagegetsgenerated
acrossoppositefacesofcrystal.Conversely,ifelectricpotentialisappliedacross
thecrystal,itvibratescausingmechanicaldistortioninthecrystalsheet.This
mechanicalvibrationgenerateselectricalsignalsofconstantfrequency.
•OutofQuartz,TourmalineandRochellesalt,Rochellesaltexhibitshighpiezo-
electricactivityandRochellesaltaremechanicallyweakest.Theybreakeasily.
•Tourmaline is the strongest of three but is veryexpensive.
•Quartzcrystalisthecompromisebetweenthetwo,itisnaturallyavailablein
abundantquantityandexhibitsreasonablygoodpiezo-electricactivity.Quartz
crystalsareusedinallratiofrequencyoscillatorsandfilters.

85. AC Equivalent Circuit ofcrystal
•For the practical views, Crystal is cut as a rectangular slab and is held between
two mechanical plates. When the crystal is not vibrating it is equivalent to
parallel plate capacitor, i.e., a dielectric medium (Crystal slab) presentbetween
two metallic plates of the capacitor. This is represented as capacitance C
M.

86. CrystalOscillators
•When the crystal is vibrating there are internal
frictional loss which is denoted by resistance“R”.
The mass of crystal corresponds to its inertia
represented by “L”. The stiffness of crystalis
denoted by “C”. This forms a series RLC circuitand
corresponding series resonant frequencyis

87. CrystalOscillators
•Actually frequency of generated oscillations is
inversely proportional to thickness. There exists a
parallel resonant circuit also. This is formed due to
the presence of inductor L and capacitance Ceq. At
parallel resonant frequency, impedance is
maximum. It is givenby,
•Where

88. CrystalOscillators
•Actually fr and fp are very close to each other. The
graph of impedance vs Frequency is as shown in
Figure

89. Series Resonant Crystal Oscillator usingBJT
•Here Crystal is connected in feedbackpath
of theamplifier.
•The voltage feedback from collector to
base is a maximum when the crystal
impedance is minimum (inseries-resonant
mode). The resulting circuit frequency of
oscillation is set by the series-resonant
frequency of thecrystal.
•The resistance R
1, R
2 and R
E provide DC
bias while Capacitor C
E is emitter bypass
capacitor. RFC coil provides for dc bias
while decoupling any ac signal on the
power lines from affecting the output
signal.

90. Series Resonant Crystal Oscillator usingFET
•Here Crystal is connectedin
feedback path of the FET
amplifier.
•The voltage feedback from Drain
to Gate is a maximum when the
crystal impedance is minimum(in
series-resonantmode).
•The resulting circuit frequencyof
oscillation is set by the series-
resonant frequency of thecrystal.
•RFC coil provides for dc bias
while decoupling any ac signalon
the power lines from affecting the
outputsignal.
•

91. Millers crystaloscillator
•The Hartley oscillator can be
modified using crystal at one ofthe
inductors of the tankcircuit.
•The crystal behaves as inductor ―L‖
connected to base and ground. The
internal capacitance of transistoracts
as capacitors C
1 and C
2 of the tank
circuit.
•The crystal decides operating
frequency of oscillator. A tuned LC
circuit in the drain section isadjusted
to the crystal parallel-resonant
frequency.

92. Parallel Resonant CrystalOscillators
•Crystal is connected as a inductor in
the modified Colpitt’s oscillator
circuit.
•Attheparallel-resonantoperating
frequency,acrystalappearsasan
inductivereactanceoflargevalue.
•The resulting circuit frequency of
oscillation is set by the parallel-
resonant frequency of the crystal.
The resistance R
1, R
2 and RE
provide DC bias while Capacitor C
E
is emitter bypasscapacitor.
•RFC coil provides for dc bias while
decoupling any ac signal on the
power lines from affecting the
output signal.