The Phase-Shift Oscillator: The following figure shows the circuit diagram of the phase-shift oscillator. Oscillation occurs at the frequency where the total phase shift through the three RC feedback circuits is 180. The inversion of the op-amp itself provides the another 180 phase shift to meet the requirement for oscillation of a 360 (or 0) phase shift around the feedback loop. The feedback circuit in the phase-shift oscillator is shown in the following figure. In the derivation we assume, R1 = R2 = R3 = R and C1 = C2 = C3 = C Using mesh analysis we have,

i21 VRII)Cj/1R( ... ... (1)

0RII)Cj/1R2(RI 321 ... ... (2)

32 I)Cj/1R2(RI ... ... (3)

In order to get V0, we must solve for I3 using determinants:

)Cj/1R2(R0

R)Cj/1R2(R

0R)Cj/1R(

0R0

0)Cj/1R2(R

VR)Cj/1R(

I

i

3

)Cj/1R2(RR)Cj/1R2()Cj/1R(

VR222

i2

Rf

+V

-V

0V Vo

C1

R1 R2 R3

C2 C3

C

R R R

C C

Vi Vo I3 I2 I1

Now, )Cj/1R2(RR)Cj/1R2()Cj/1R(

R

V

RI

V

V222

3

i

3

i

0

)Cj/1R2(R)Cj/1R(R)Cj/1R2)(Cj/1R(

R222

3

)RCj/12()RCj/11()RCj/12)(RCj/11(

12

)RCj/23()CR/1RCj/44)(RCj/11(

1222

)RCj/23)CRj/1CR/4RCj/4CR/1RCj/44(

1333222222

)CRj/1RCj/6CR/51(

1333222

)CR/1RC/6(j)CR/51(

1333222

... ... (4)

For oscillation in the phase-shift amplifier, the phase shift through the RC circuit must be equal to 180. For this condition to exist, the j term must be 0 at the frequency of oscillation 0.

0CR/1RC/6 33300

0CR

1CR6333

0

2220

01CR6 2220

22

20

CR6

1

6RC

10

6RC2

1f0

Now, from the equation (4) we have,

29

1

)6x51(

1

V

V

i

0

The negative sign results from the 180 inversion by the circuit. Thus, the value of voltage gain by the RC circuit is,

29

1

V

V

i

0

To meet the greater-than-unity loop gain requirement, the closed-loop voltage gain of the op-amp must be greater than 29. So, Rf 29 R3

Exercise:

The Colpitts Oscillator: The following figure shows the circuit diagram of the Colpitts oscillator. Oscillation occurs at the frequency where the L-C feedback circuits is at resonance. Assuming R1>>XC1 we have the impedance of the L-C circuit,

)jXjXjX(

)jXjX)(jX(Z

1CL2C

1CL2C

)XXX(j

)XX(XZ

1C2CL

1CL2C

At parallel resonance the impedance will be maximum and we can write,

0XXX 1C2CL

2C1CL XXX ... ... (1)

21 C/1C/1L

21

2121

CC

CC1C/1C/1L

)CC/(CC

1

L

1

2121

2

)CC/(CLC

1

2121

TLC

1 where

21

21T

CC

CCC

TLC2

1f

Again, the voltage gain of the LC circuit,

1CL

1C

1CL

1C

1

2

XX

X

jXjX

jX

V

V

Here negative sign is for 180 phase shift by the circuit. So magnitude of the voltage gain is,

V1 C1 C2

L

V2

L-C feedback circuit Z

Rf

+V

-V

Vo

R1

C1 C2

L

Colpitts oscillator

1CL

1C

XX

X

2C

1C

X

X (from equation (1))

1

2

C

C

For oscillation to sustain, the loop gain must be greater than unity. Therefore, the voltage gain of the amplifier should be,

1A V

2

1

C

C

1R

Rf

Design the Colpitts oscillator to produce a 40 kHz output frequency. Use a 100 mH inductor and an OP-AMP with a 10 V supply. SOLUTION We know,

pF 8.153mH 100kHz) 40(4

1

4

12222T

Lf

C

For ,10 21 CC value)standard pF 1500 (use pF 1538pF 8.1531010 T1 CC

value)standard pF 180 (use pF 177

)pF 15001()pF 8.1531(

1

)1()1(

1

1T

2

CCC

k 22pF 180kHz 402

1

2

1

2

C2 fCX

amplifier theof 0C2 ZX

k 65.2pF 1500kHz 402

1

2

1

1

C1 fCX

Since 11 CXR , we select

value)standard k 27 (use k 26.5k 65.21010 11 CXR

Now 33.8pF 180

pF 1500

2

1CL(min)

C

CA

value)standard k 270 (use k 225k 2733.81CL(min)2 RAR

value)standard k 27 (use k 5.24k 270k 27213 RRR

The OP-AMP full-power bandwidth (fp) must be a minimum of 40 kHz when

.33.8 and V 9 CL0 AV

Since , therefore,pCL2 fAf

kHz 333kHz 4033.82 f

and

Slew rate, sVfSR V/ 262.2V 9kHz 4022 pp

Example: Design of OP-AMP Colpitts Oscillator

The Hartley Oscillator: The following figure shows the circuit diagram of the Hartley oscillator. Oscillation occurs at the frequency where the C-L feedback circuits is at resonance. Assuming R1>>XL1 we have the impedance of the C-L circuit,

)jXjXjXjXjX(

)jXjXjX)(jXjX(Z

M1LCM2L

M1LCM2L

)X2XXX(j

)XXX)(XX(Z

M1LC2L

M1LCM2L

At parallel resonance the impedance will be maximum and we can write,

0X2XXX M1LC2L

)XX(XXX M2LCM1L ... ... (1)

21 LC/1M2L

C/1M2LL 21

)M2LL(

1

C

1

21

2

)M2LL(C

1

21

TCL

1 where M2LLL 21T

TCL2

1f

Again, the voltage gain of the C-L circuit,

CM1L

M1L

CM1L

M1L

1

2

XXX

XX

jXjXjX

jXjX

V

V

CM1L

M1L

XXX

XX

M2L

M1L

XX

XX

(from equation (1))

Rf

+V

-V

Vo

R1

L1

CL

Hartley oscillator

L2

M

V1

V2

C-L feedback circuit Z

L1

CL

L2

M

Here negative sign is for 180 phase shift by the circuit. So magnitude of the voltage gain is,

M2L

M1L

XX

XX

M2L

ML1

For oscillation to sustain, the loop gain must be greater than unity. Therefore, the voltage gain of the amplifier should be,

1A V

ML

ML

1R

Rf

1

2

If the inductors are wound on separate core, then mutual inductance M = 0 and we can write,

1

2

L

L

1R

Rf

Design the Hartley oscillator to produce a 100 kHz output frequency with an amplitude of 8 V. For simplicity, assume that there is no mutual inductance between L1 and L2. SOLUTION VCC = (V0 + 1 V) = (8 V + 1 V) = 9 V

amplifier theof 0L2 ZX

Select k 1L2X

value)standard mH 1.5 (use mH 59.1kHz 1002

k 1

2L2

L2

f

XX

Select value)(standard H 15010

mH 5.1

102

1 L

L

mH 1.65H 150mH 5.121T LLL

Now,

necessary) if e,capacitanc parallel additional with pF 1500 (use pF 1535

mH 65.1kHz) 100(4

1

4

122

T22

LfC

ecapacitancstray C

94.2H 150kHz 10022 1L1 fLX

L11 XR

Select value)(standard k 11 R

Therefore,

10H 150

mH 5.1

1

2CL(min)

L

LA

Example: Design of OP-AMP Hartley Oscillator

value)(standard k 10k 1101CL(min)2 RAR

value)standard k 1 (use 909k 10k 1213 RRR

The OP-AMP full-power bandwidth (fp) must be a minimum of 100 kHz when

.10 and V 8 CL0 AV

Since , therefore,pCL2 fAf

MHz 1kHz 100102 f

and Slew rate, sVfSR V/ 5V 8kHz 10022 pp

Wein Bridge Oscillator: The following figures show the circuit diagram of the Wein Bridge oscillator. Oscillation occurs at the particular frequency when ac balance is obtained in the Wein Bride. At the balanced condition of the bridge we can write,

43

04

21

02

ZZ

VZ

ZZ

VZ

43

4

21

2

ZZ

Z

ZZ

Z

43

4

222211

2222

RR

R

)Cj/1R/()Cj/1)(R()Cj/1R(

)Cj/1R/()Cj/1)(R(

43

4

222211

22

RR

R

)Cj/1)(R()Cj/1R)(Cj/1R(

)Cj/1)(R(

43

4

122211

12

RR

R

CRj)CRj1)(CRj1(

CRj

43

4

122211

12

RR

R

CRj)CRj1)(CRj1(

CRj

43

4

1222112

2211

12

RR

R

CRj)CRCRCRjCRj1(

CRj

43

4

12221122112

12

RR

R

)CRCRCR(j)CRCR1(

CRj

... ... (1)

Since, the right hand side of the above equation is a real term, the left hand side must also be a real term. So, we can write,

0CRCR1 22112

2211 CRCR

1 ... ... (2)

From equation (1) we have,

43

4

122211

12

RR

R

)CRCRCR(

CR

3

4

2211

12

R

R

CRCR

CR

1

2

2

1

12

2211

4

3

C

C

R

R

CR

CRCR

R

R

... ... (3)

The op-amp along with the two resistors R3 and R4 constitutes a non-inverting amplifier who’s voltage gain is,

4

3V

R

R1A

Using the of R3/R4 obtained in equation (3) we have,

1

2

2

1V

C

C

R

R1A

This corresponds that the attenuation of the feedback network is,

1

2

2

1

C

C

R

R1/1

Therefore, AV must be equal to or greater than

1

2

2

1

C

C

R

R1 to sustain oscillation.

Mathematically,

1

2

2

1V

C

C

R

R1A ... ... (4)

1

2

2

13

C

C

R

R1

4R

R1

1

2

2

1

4

3

C

C

R

R

R

R

For R1 = R2 = R and C1 = C2 = C we have,

2R

R

4

3

43 R2R

Also from equation (4) we have,

3A V

From equation (2) we have,

RC

1

RC2

1f

Design the Wein bridge oscillator to produce a 100 kHz output frequency with an amplitude of 9 V. Design the amplifier to have a closed-loop gain of 3. SOLUTION VCC = (V0 + 1 V) = (9 V + 1 V) = 10 V

For ACL = 3, R1 = R2 = R and C1 = C2 = C Also, R3 =2R4

Select, C1 = 1000 pF (standard value) C2 = C1 = 1000 pF

Therefore,

value)standard k 1.5 (use k 1.59

pF 1000kHz 1002

1

2

1

fCR

Select, value)(standard k 1.5 24 RR

value)standard k 3.3 (use k 3 k 1.52 43 RR

The OP-AMP full-power bandwidth (fp) must be a minimum of 100 kHz when

.3 and V 9 CL0 AV

Since , therefore,pCL2 fAf

kHz 300kHz 10032 f

and

Slew rate, sVfSR V/ 7.5V 9kHz 10022 pp

+V

-V

Wein Bridge

C1

R1

R2

C2

R3

R4

V0

+V

-V

C1

R1

R2 C2 R3

R4

V0 Fee

db

ack

net

wo

rk

Noninverting amplifier

Example: Design of Wein Bridge Oscillator

9.12 CRYSTAL OSCILLATORS If a piezoelectric crystal, usually quartz, has electrodes plated on opposite faces and if a potential is applied between these electrodes, forces will be exerted on the bound charges within the crystal. If this device is properly mounted, deformations take place within the crystal, and an electromechanical system is formed which will vibrate when properly excited. The resonant frequency and the Q depend upon the crystal dimensions, how the surfaces are oriented with respect to its axes, and how the device is mounted. Frequencies ranging from a few kilohertz to a few megahertz, and Q’s in the range from several thousand to several hundred thousand, are commercially available. These extraordinarily high values of Q and the fact that the characteristics of quartz are extremely stable with respect to time and temperature account for the exceptional frequency stability of oscillators incorporating crystals. Crystal oscillators are used whenever great stability is required, for example, in communication transmitters and receivers. The electrical equivalent circuit of a crystal is indicated in Figure 9.25. The inductor L, capacitor C, and resistor R are the analogs of the mass, the compliance (the reciprocal of the spring constant), and the viscous-damping factor of the mechanical system. The typical values for a 90-kHz crystal are L = 137 H, C = 0.0235 pF, and R = 15 k, corresponding to Q = 5,500. The

dimensions of such a crystal are 30 by 4 by 1.5 mm. Since /C represents the electrostatic capacitance between electrodes with the crystal as a dielectric, its magnitude (~3.5 pF) is very much larger than C.

If we neglect the resistance R, the impedance of the crystal (Z in Figure 9.25(b)) is given by,

21 ZZZ

21

21

ZZ

ZZ

/

/

11

)1)(1(

CjCjLj

CjLjCj

)11(

)1(// LCLCL

LCL

C

j

)11(1

1

/2

2

/

CCL

LC

C

j

(a)

C

L

R

/C

X

0

Reactance (inductive)

Reactance (capacitive)

p

s

(b) (c)

Figure 9.25 A piezoelectric crystal: (a) Symbol, (b) electrical model, and (c) the reactance function (if R = 0).

Z

2p

2

2s

2

/

C

j (9.64)

WhereLC

1 s = series resonant frequency and

/p

111

CCL = parallel resonant

frequency. Equation (9.64) can be written, in terms of reactance, as

2p

2

2s

2

/

C

jjXZ (9.65)

Therefore, reactance of the crystal is

1

2p

2

2s

2

/

CX (9.66)

The plot of Equation (9.66) is shown in Figure 9.25(c). Since /C >> C, .sp For the crystal

whose parameters are specified above, the parallel frequency is only three-tenths of 1 percent

higher than the series frequency. For ,ps the reactance is inductive, and outside this

range it is capacitive, as indicated in Figure 9.25(c). In order to use the crystal properly it must be connected in a circuit so that its low impedance in the series resonant operating mode or high impedance in the parallel resonant operating mode is selected. Series Resonant Circuits To excite a crystal for operation in the series resonant mode it may be connected as a series element in a feedback path. At the series resonant frequency of the crystal its impedance is smallest and the amount of (positive) feedback is largest. A typical transistor circuit is shown in Figure 9.26. Resistors R1, R2, and RE provide a voltage divider stabilized dc bias circuit. Capacitor CE provides ac bypass of the emitter resistor and the RFC coil provides for dc bias while decoupling any ac signal on the power lines from affecting the output signal. The voltage feedback from collector to base is a maximum when the crystal (XTAL) impedance is minimum (in series resonant mode). The coupling capacitor CC has negligible impedance at the circuit operating frequency but blocks ant dc between collector and base.

(a)

Q1

RE CE

R1 RFC

R2

CC

VCC

V0

XTAL

RFC

RG

CC

VCC

V0

XTAL

(b)

Figure 9.26 Crystal-controlled oscillator using crystal in series feedback path: (a) BJT circuit, and (b) FET circuit.

The resulting circuit frequency of oscillation is set by the series resonant frequency of the crystal. Changes in supply voltage, transistor device parameters, and so on, have no effect on the circuit operating frequency which is held stabilized by the crystal. The circuit frequency stability is set by the crystal frequency stability, which is good. Parallel Resonant Circuits Since the parallel resonant impedance of a crystal is a maximum value, it is connected in parallel/shunt. At the parallel resonant operating frequency a crystal appears as an inductive reactance of largest value. Figure 9.27(a) shows a crystal connected as the inductor connected in a modified Colpitts circuit. The basic dc bias circuit should be evident. Maximum voltage is developed across the crystal at its parallel resonant frequency. The voltage is coupled to the emitter by a capacitor voltage divider capacitors C1 and C2. A Miller crystal-controlled oscillator circuit is shown in Figure 9.27(b). A tuned LC circuit in the drain section is adjusted near the crystal parallel resonant frequency. The maximum gate-source signal occurs at the crystal parallel resonant frequency controlling the circuit operating frequency. OP-AMP Crystal Oscillator An OP-AMP can be used in a crystal oscillator as shown in Figure 9.28. The crystal is connected in the series resonant path and operates at the crystal series resonant frequency. The present circuit has a high gain so that an output square-wave signal results as shown in the figure. A pair of Zener diodes is shown at the output to provide output amplitude at exactly the Zener voltage (VZ).

(a)

Q1

RE C2

R1 RFC

R2 CB

VCC

V0

XTAL

L

RS

C

VCC

XTAL

(b)

Figure 9.27 Crystal-controlled oscillators in parallel resonant operating mode: (a) BJT circuit, and (b) FET circuit.

C1

CS

V0

RG

RFC

VZ

+

V0

+VCC

Rf

R1

100 k

VEE

XTAL 0.1 F

1 k

100 k V0

t

VZ

0

Figure 9.28 Crystal oscillator using OP-AMP.