Oscillator Oscillator: an oscillation is a type of feedback amplifier in which part of the output is feed back to the input via a feedback circuit. Tank circuit: a circuit which produces electrical oscillation of any desired frequency is known as an oscillatory circuit or tank circuit. Requirements for an oscillator:

1. Magnitude of the loop gain at least 1. 2. Total phase ship within the loop must be 00

Cause of using positive feedback: for an oscillation circuit, it is impossible to design without resistance. Because, of resistance oscillation of the circuit die out within short time. But we want a undamped oscillation. Here, positive feedback circuit provide power so that the oscillation cant die out. So, we get an undamped oscillation. This is the reason of using positive feedback.

Different types of transistor oscillator:

1. Tune collector oscillator 2. Colpitts oscillator 3. Hartley oscillator 4. Phase shift oscillator 5. Wein bridge oscillator 6. Crystal oscillator

Tune collector oscillator: It contains tuned circuit L1-C1 in the collector and hence the name. the frequency of oscillations depends upon the values of L1-C1 and is given by: f0=

Circuit operation: When switch s is closed , collector current starts increasing and charge the capacitor C1 .when this capacitor is fully charged, it discharges through coil L1. This oscillations induce some voltage in coil L2 by mutual induction. The frequency of voltage in coil L2 is the same as that of tank circuit but its magnetic depends upon the numbers of turns of L2 and coupling between L1 and L2. The voltage across L2 is applied between base and emitter and appears in the amplified fromin the collector circuit, thus overcoming the losses occurring in the tank circuit. The number of turns of L2 and coupling between L 1and L2 are so adjusted that oscillations across L2 are amplified to a level just sufficient to supply losses to the tank circuit.

Colpitts oscillator: It uses two capacitors and placed across a common inductor L and the center of the two capacitors is tapped. The tank circuit is made up of C1 , C2 and L . the frequency of oscillation is determined by the values of C1, C2 and L and is given by, f0=

Circuit operation: when the circuit is turned on, the capacitors C and C are charged. The capacitors discharged through L. the output voltage of the amplifier appears across C1 and feedback voltage is developed across C2. The voltage across it is 1800 out of phase with the voltage developed. Hartly oscillator: instead of using tapped capacitors, two inductors L1 ,L2 and C and the centre of the inductor is tapped as shown in fig, Where, f0=

Circuit operation: when the circuit is turned on, the capacitor is charged . when this capacitor is fully charged , it discharges through coil L1 and L2 setting up oscillations of frequency determined. The output voltage of the amplifier appears across L1 and feedback voltage across L2. the voltage across L2 is 180 out of phase with the voltage developed across L2. it is easy to see that voltage feedback to the transistor provides positive feedback. A phase shift of 180 is produced by the transistor and a further phase shift of 180 is produced by L1-L2 voltage divider. In this way , feedback is properly phased to produce continuous undamped oscillation. Frequency stability: The ability of oscillator circuit to oscillate at one exact frequency is known as frequency stabilities. The best frequency stable- crystal. Phase shift to oscillator:

1. It suffers from frequency instability and poor output wave form.

2. It cannot be used low frequencies because inductance and capacitance became bulky and expensive.

Phase shift oscillator

#Construction:

A phase shift oscillator consists of an op-amp as the amplifying stage and three RC cascaded networks as the feedback circuit.

#Drawbacks:

1. It suffers from frequency instability and poor output waveform. 2. It cannot be used at low frequencies because inductance and capacitance

became bulky and expensive.

#Circuit operation:

The feedback circuit provides feedback voltage from the output back to the input of the amplifier. The op-amp is used in the inverting mode; therefore, any signal that appears in the inverting terminal is shifted by 180 at the output. An additional

180 phase shift required for oscillation is provided by the cascaded RC networks. Thus the total phase shift around the loop is 360 (or 0 ).

The frequency of oscillation of a phase shift oscillator isgiven by

RCfo 62

1

#For a phase shift oscillator prove that 29,621

1

RR

RCf Fo

Proof:

First consider the feedback circuit consisting of RC combinations of the phase shift oscillator. For simplicity we use the Laplace transform again. Thus, the circuit is

represented in the S domain as shown in the figure. Let us determine )()(

SVSV

o

f for the

circuit.

Writing Kirchhoffs current law (KCL) at node )(1 SV

SCSVSV

RSV

SCSVSVo

/1)()()(

/1)()( 2111

Solving for )(1 SV

12))(()(

)( 21

RCS

RCSSVSVSV o

Writing KCL at node )(2 SV

SCSVSV

RSV

SCSVSV f

/1)()()(

/1)()( 2221

Solving for )(1 SV

)()12)(()( 21 SVRCSRCSSVSV f

If RR 1 in the circuit then 07 I A. This means that 65 II . Therefore using the voltage divider rule,

)()/1( 2

SVSCR

RV f

Or, )()/1(2 SVRSCRV f

Substituting the value of )(2 SV

12)()1(

12))((

)(1

RCS

SVRCSRCS

RCSSVSV fo

Also substituting the value of )(2 SV

)())((

)1)(12)(()(1 SVRCSRCS

RCSRCSSVSV f

f

Equating

BRCSSCRSCR

SCRSVSV

o

f

)156()(

)(222333

333

The op-amp part is redrawn below,

The voltage gain of op-amp is

1)()(

RR

SVSVA F

f

ov

For an oscillator 1. BAv

Or, 1)156( 222333

333

1

RCSSCRSCRSCR

RRF

Substituting jS

333

1

CjRRRF 1)5()6()( 222333 RCjCRCjR

Equating real parts

016 222 CR

RCfo 62

1

Equating imaginary parts

333

1

CjRRRF )5()( 333 RCjCjR

291

RRF

PROBLEM: Design the op-amp based phase shift oscillator so that Hzfo 200

SOLVE:

We know that oscillation frequency RC

fo 621

KKR 3.325.36101.02002

16

To prevent the loading effect of the amplifier because of RC network, it is necessary that RR 101 KRR 33101

MKRRF 1957332929 1

PROBLEM: It is desired to design a phase-shift oscillator using an FET having Krsg dm 40,5000 and feedback circuit value of KR 10 . Select the value of C

for oscillator operation at 1 KHz and DR for A>29 to ensure oscillation action.

SOLVE: Oscillation frequency, RC

fo 621

nFC 5.6

Lm RgA

KRL 8

Kr

RrRRd

LdDD 10

)(

Crystal Oscillator

# Limitation of LC and RC oscillator:

1. As the circuit operates, it will warm up. Consequently the values of resistors inductors, which are the frequency determining factors in LC and RC circuit, will change with temperature .this causes change in frequency of the oscillator.

2. If any component in the feedback network is changed, it will shift the operating frequency of the oscillator.

# in order to maintain constant frequency piezoelectric crystal is used in place of LC or RC circuits. Oscillators of this type are called crystal oscillator. The frequency of a crystal oscillator changes by less than 0.1% due to temperature and other changes.

#Piezoelectric crystal: Certain crystalline materials namely, Rochelle salt, quartz and tourmaline exhibit piezoelectric effect, i.e when an a.c voltage is applied across them, they vibrate at the frequency of the applied voltage. Conversely, when they are compressed or placed under mechanical strain, they produce an a.c voltage. Such crystals which exhibit piezoelectric effect are called piezoelectric crystal of the various piezoelectric crystals; quartz is most commonly used because it is inexpensive and readily available in nature.

Frequency of the crystal:

The natural frequency of the crystal is given by,

f =

Where k is a constant and t is the thickness of the crystal. It is clear that frequency is inversely proportional to crystal thickness. The thinner the crystal, the greater its natural frequency and vice versa. However extremely thin crystal may break because of vibration.

#working of quartz crystal:

Crystal

In order to use crystal in electronic circuit, it is placed between two metal plates. The arrangement then forms a capacitor with crystal as the dielectric. If an ac voltage is applied across the plates the crystal will start vibrating at the frequency of the applied

voltage. If the frequency of the applied voltage is made equal to the natural frequency of the crystal, response take place and crystal vibrations reach a maximum value. This natural frequency is almost constant.

#Equivalent circuit of crystal:

1. When the crystal is not vibrating, it is equivalent to capacitance CM because it has two metal plates separated by dielectric. This capacitance is known as mounting capacitance.

CM

2. When a crystal vibrates, it is equivalent to R-L-C series circuit. The equivalent circuit of vibrating crystal is R-L-C series circuit shunted by mounting capacitance CM.

Frequency response of the crystal:

1. At low frequencies the impedance of the crystal is controlled by extremely high values of cmX and cX . In other words, at low frequencies the impedance of the network and capacitive.

2. As the frequency is increased R-L-C branch approaches its resonant frequency. At some definite frequency, the reactance LX will be equal to cX . The crystal now acts as a series resonant circuit. For this condition impedance of the crystal is very low, being equal to R. The frequency at which the vibrating crystal

behaves as a series-resonant circuit is called series resonant frequency. For this condition impedance of the crystal is very low, being equal to R. the frequency at which the vibrating crystal behaves as a series resonant circuit is called series-resonant frequency.

3. At a slightly higher frequency, the net reactance of branch R-L-C becomes inductive and equal to cmX . The crystal act as a parallel-resonant circuit. For this condition, the crystal offers a very high frequency. The frequency at which vibrating crystal behaves as a parallel-resonant circuit is called parallel-resonant

frequency pf .

T

p LCf

21

M

MT CC

CCC

pf is always higher than sf .

4. At frequencies higher than pf , the value of CMX drops and crystal acts as a short circuit.

Transistor crystal oscillator:

The above circuit acts as a parallel tuned circuit. At parallel resonance, the impedance is maximum. This means maximum voltage is developed across the 1C . This in turn, will allow the maximum energy transfer through the feedback network

at pf . The feedback is positive. A phase shift of 180 is produced by the transistor.

A further phase shift is produced by the capacitor voltage divider. This oscillator

oscillates only at pf .

PROBLEM: calculate pf and sf of the previous circuit.

SOLVE:

KHzLCf s 15892

1

HL 1 ,

pFCCCCC

M

MT

31099.9

pFC 01.0 ,

KHzLCf

Tp 15902

1

pFCM 20

#Explain the operation of Wein bridge oscillator

The above circuit shows Wein bridge oscillator in which Wein bridge circuit is connected between the amplifier input and output terminal. The bridge has a series RC network in one arm and a parallel RC network in adjoining arm. In the remaining two arms of the bridge, resistors are connected.

The phase angle criterion for oscillation is that the total phase shift around the circuit must be 0 . This condition occurs when the bridge is balanced, that is at resonance. The frequency of oscillation,

RCfo 2

1

At this frequency the gain required for sustained oscillation,

31 v

A That is,

311

RRF

Or, 12RRF

#For a Wein bridge oscillator prove that the amplifier gain is, 3vA and 12RRF .

Proof:

First let us consider feedback circuit of the Wein bridge oscillator.

)()()()(

)(sZsZ

sVsZsV

sp

opf

11)(

RCSR

CSRsZ p

CSRCS

CSRsZ s

11)(

RCSRCSsVRCSsV of

2)1(

)()(

13222

RCSSCRRCS

VV

Bo

f

Now consider the op-amp part of the wein bridge oscillator

11)()(

RR

sVsVA F

f

ov

Requirement for oscillation is

1))(( BAv

1)13

)(1( 2221

RCSSCR

RCSRRF

Put jS

))(1(1

RCjRRF 13222 RCjCR

Comparing real part

01 222 CR

RC1

RCfo 2

1

Imaginary part

vF A

RR

311

21

RRF

12RRF

(Proved)

#Example the Wein bridge oscillator so that Hzfo 965

SOLVE:

Let

2105. CCFC

We know, KRC

fo 3.321

Let KR 121

KRRF 242 1

[Use a KRF 50 potentiometer]

KRR 3.32

Solution of Question 2007-2008

#For the following Hartley oscillator the component valuesare:C=250pF,L1=1.5mH,L2=1.5mH and M=0.5mH.Calculate the oscillator frequency fo.(2007)

LT=L1+L2+2M

=1.5+1.5+2*0.5

=4mH

fo =1/2(CLT)

=1/2*3.14*(250*10-12*4*10-3)

=159.236KHz

#Design BJT phase shift oscillator for 1KHz Oscillating frequency. Given Collector resistance RC=1k.Assume all other parameters required.(2008)

Given,RC=1K,f=1KHz, We assume that the value of R is 100K

For BJT,

f= 1/2RC(6+4(RC/R))

so,C=1/2*3.14*100*103*103(6+4(1/100))

=6.5*10-10

=650pF,which nears to 680pF available in market.

The circuit is given below: