at the end of the lesson, you should be able to: • analyse

________________________________________________________________________________________

INTRODUCTION________________________________________________________________________________________

In this lesson, we investigate some forms of wave-form generation using op

amps. Of course, we could use basic transistor circuits, but it makes sense to

simplify the analysis by considering the ideal op amp. We also see in this

lesson how the op amp can be used in switching applications.

________________________________________________________________________________________

YOUR AIMS________________________________________________________________________________________

At the end of the lesson, you should be able to:

• analyse and design a range of sine-wave oscillators

• use an op amp as a switch

• understand the design of multivibrators.

________________________________________________________________________________________

STUDY SKILLS________________________________________________________________________________________

There is a fair amount of mathematical manipulation in this unit. Make sure

you follow the steps. When you come to do the Self-Assessment Questions,

remember that there are many ways of using electrical theory to analyse a

circuit. Use that with which you are most comfortable.

1

Teesside University Open Learning(Engineering)

© Teesside University 2012

________________________________________________________________________________________

OSCILLATORS________________________________________________________________________________________

Oscillators are waveform generators. Usually, it is their frequency of

oscillation and wave shape that is important. Oscillators can be classified as:

• sinusoidal

• square wave

• others, e.g. triangular, staircase, trapezoidal, etc.

We have found the general condition for oscillation in Lesson 1 of this topic.

Remember the condition for oscillation is a gain of at least +1 when the phase

shift round a feedback loop is 0° at some frequency. We shall first investigate

some oscillators giving sinusoidal output.

________________________________________________________________________________________

SINE WAVE OSCILLATORS________________________________________________________________________________________

wien bridge oscillator

FIG. 1

V2V1

R

RC

Z2

0 V

C

Z1

2



Consider the network of FIGURE 1.

and

Now, when does this expression have either 0° or 180° phase shift?

When the above expression is purely real! The only way this can apply in this

equation is if the real term in the denominator is 0; then the 'j' factors in the

numerator and denominator cancel.

Hence,j

j

j

j j

j

V

V

Z

Z Z

R

2

1

2

1 2

1

ωω

ωω ω

( )( ) =

( )( ) + ( )

=+ ωω ω

ω

ω ω

CR R CR

CR

R

R CR C

( ) + ++

⎛⎝⎜

⎞⎠⎟

=+ + +

11

12

/

/

jj

j j RR R

CR

CR CR

+

=( ) +

j

1 j ..................

ωω ω– 2 3

............. 1( )

Z R C

Z

1

2

1j Series R and C j

j Paralle

ω ω

ω

( ) = = +

( ) =

/

ll R and C

j1/j

1 j

=+

=+

R C

R C

R

CR

/ ωω

ω

3



i.e.

So the circuit will oscillate at a frequency given by Equation (2):

provided that V2 is fed back to V1 with a gain of +3 (to make the loop gain 1 or

more).

Let us look at how we design an amplifier with a gain of +3. See FIGURE 2.

FIG. 2

What should R2/R1 be? Suggest suitable values, bearing in mind that resistor

tolerances could bring the ratio below the critical value.

....................................................................................................................................................

....................................................................................................................................................

....................................................................................................................................................

....................................................................................................................................................

________________________________________________________________________________________

0 V

–+

R1

R2

ω ω π π= = = ( )⎡⎣ ⎤⎦1 2 1 2/ / /CR f CRor Hz

1 02– ωCR( ) = ........................................................

j

j1

2

2

( )

( )( ) =

V

V

ωω

jj0 3 j

..........................ω

ωCR

CR+= 1

3..... 3( )

4



The gain of this amplifier is 1 + R2/R1, so R2/R1 must be greater than 2. R2 =

22 kΩ and R1 = 10 kΩ would be suitable, unless the resistor tolerance is very

high.

What determines the amplitude of the oscillation?

....................................................................................................................................................

....................................................................................................................................................

....................................................................................................................................................

....................................................................................................................................................

....................................................................................................................................................

....................................................................................................................................................

________________________________________________________________________________________

Only the output range of the op amp limits it. Since the characteristic of the

amplifier has a hard saturation, the circuit above would generate a distorted

sine wave as illustrated in FIGURE 3. This is because there is a sudden gain

reduction to almost 0. To minimize distortion, we should build in a soft

saturation. FIGURE 4 shows a possible modification, and shows the complete

oscillator.

FIG. 3

0

+ve saturation

–ve saturation

5



FIG. 4

We arrange for R2/R1 > 2 (say 2.2) and < 2 (say 1.5). Then,

when the Zener diodes conduct, the gain falls below the crucial level required

to maintain oscillation. This damps the oscillation, preventing it from building

up.

Suppose we want a ±10 V peak-peak output. If V1 = 10, V2 = 10/3 (since the

gain is nominally 3). Hence, there is about (10 - 10/3) = 6.7 V across R2 and

this is the value of the Zener diode we should choose (say 6.3).

Another way of controlling the amplitude of oscillation is by using a voltage

dependent resistor instead of R2.

( ) /R R R3 2 1

0 V

–+R1

R3

R2

C R

0 V

R C

V2

V1

6



COLPITTS OSCILLATOR

FIG. 5

We want to find Vo/Vi in FIGURE 5.

Also, Z (jω) is the complex impedance obtained from

ZL C C

L C Cj

j j j

j j jω

ω ω ωω ω ω

( ) = +( )+ +

=

1 1

1 1

1

2 1

1 2

/ /

/ /

––

–

ωω ω

22

31 2 1 2

5LC

LC C C Cj j ...........

+ +( ) ( ))

1 1

1 2j j

j ωω

ωCL

C+

⎛⎝⎜

⎞⎠⎟

V

V

C

L C LCo

x

j

j

j

j j ...

ωω

ωω ω ω

( )( ) = +

=1

11

12

22

2

/

/ –....... 4( )

0 V

Vi

R VxVo

C1 C2

L

Z

7



We are now able to derive an expression for Vo/Vi.

Make sure you can derive this equation from the previous line.

This is real when the imaginary terms of the denominator sum to 0.

At this frequency,

The negative sign in this equation indicates that there is a 180° phase shift. So,

to oscillate at the frequency defined by Equation (7), we require an amplifier

with a gain of –C2/C1 as shown in FIGURE 6. Note that the resistor plays no

direct part in the circuit; it merely provides a convenient impedance between

the amplifier and the frequency dependent elements.

The above analysis is only one way of finding Vo/Vi; you could use Thevenin

or nodal analysis. It comes to the same thing.

FIGURE 6 shows the complete oscillator.

ω 2 1 2

1 2

=+C C

LC C .............................................

.........

7

1

2

( )

=V

V

C

Co

i

– .......................................... 8( )

– j jω ω31 2 1 2 0LC C R C C R+ +( ) =

V

V

V

V

V

V LC

Z

R Z

LC LC C

o

i

o

x

x

i

= =+

=

11

1

1

22

22

31

–

– –

ω

ω ωj 22 1 2

6R C C R+ +( ) ( )

j ..........

ω

8



FIG. 6

SUMMARY OF SINE-WAVE OSCILLATORS

The analysis of all sine-wave oscillators follows more or less the same pattern.

The analysis can be fairly complex, and there are plenty of opportunities for

making algebraic errors. The process steps are as follows.

1. Analyse a complex network.

2. Find at what ω the gain of the network is purely real.

3. Find the magnitude of the gain at this ω.

4. Design an amplifier to make the overall gain of +1.

5. Include a soft saturation characteristic.

The ordinary 741 op amp is only useful over a low frequency range, say up to

about 3 to 4 kHz. Above these frequencies, slew rate limitations start to show

up. So you need a higher frequency op amp, or you can use a simple transistor

amplifier!

0 V

–+R1

R3

R2

R

C1 C2

L

9



________________________________________________________________________________________

THE OPERATIONAL AMPLIFIER AS A SWITCH________________________________________________________________________________________

FIG. 7

With a high gain op amp, any small difference between the inputs will send the

output into saturation. If, in the circuit above, V2 is positive then, if V1 > V2,

the output will go to negative saturation; on the other hand, a signal less than

V2 will cause the amplifier to go to positive saturation.

Remember that Vo = –A (V1 – V2).

Hence, the circuit acts as a switch; Vo changes when V1 = V2.

If we want to avoid the uncertain values of amplifier saturation, we can add the

Zener diode network as shown above. The output is then clamped at voltages

of VZ + VD as shown above. The value of the resistor is not critical, but it has

to supply the load current and the Zener diode current.

V V V

Ro Z Dsat

Load current Zener current( ) +( )

> +– (( )

–+

0 V

V1

V2

Vo Vo

–Vo

V1

Vo

0

–ve Saturation

+ve Saturation

V2

Vo

10



________________________________________________________________________________________

LEVEL DETECTOR WITH HYSTERESIS________________________________________________________________________________________

FIG. 8

FIG. 9

In this circuit, we have positive feedback from the output to the non-inverting

input. So the output can only be stable in the two saturation limits. There are

therefore only two stable levels for V2. Switching of the output occurs when

V1 = V2 and so there are two different input voltages at which the output will

switch. This is similar in concept to a thermostat which is used to switch on at

a low temperature and off at a higher temperature to control an oven.

Vh

0

Vo

Vin

+(VZ + VD)

–(VZ + VD)

Vin1Vin2

Vh

–VR

–+

0 V

V1

V2

R

R3

R2

R1

R1

Vin

VR(VZ + VD)

11



Since there are two stable output levels, another name for this circuit is a

bistable, or sometimes bistable multivibrator. It also performs the same

function as a Schmitt trigger circuit. You will find all these names in

textbooks.

Switching occurs when V1 = V2.

Note that V2 can be either positive or negative, depending on the amplifier

output.

So we can write down the equations for the two values of Vin at which the

circuit output switches.

where Vh is half the hysteresis voltage as shown in FIGURE 9.

V V Vo Z D= ± +( ) .................................................. 13( )

V VR V V

R RV Vin R

Z DR h1

2

2 3

2= +

+( )+

= +– – .......... ....

– – – –

11

22

2

2 3

( )

=+( )

+=V V

R V V

R RV Vin R

Z DR h .............. 12( )

Also,

Hence,

V V

R

V V

R

V V V

in R

in R

– –

–

1

1

1

1

12

=

= + ...................................................... 10( )

But ................VR V V

R RZ D

22

2 3

=+( )

+................................... 9( )

12



NOTES

1. Note that –VR is the mid-voltage between the two input switching levels.

(e.g. if we want to design a level detector that switches at +4 V and –8 V,

then VR = +2 V).

2. The voltage difference between the two input switching levels is the

hysteresis 2Vh.

3. Many other forms of level detector are possible. They can all be analysed

by considering the conditions when V1 = V2.

4. In practice, the slew rate of the common op amps limits the switching

speed of the circuit. It is therefore better to use a comparator, an

integrated circuit which much resembles the op amp, but is specifically

designed for high slew rates.

WORKED EXAMPLE 1

Design a level detector which will switch at ± 5 V. Take (VZ + VD) as 7 V.

It is clear that, by adding equations (11) and (12), we can eliminate Vh.

V V V

V

V

in in R

in

in

1 2

1

2

2

5

5

+ =

= +

=

–

–

So, if

and

then VVR = 0 .

13



Now substitute the figures into equation (11).

We should choose R3 = 18 kΩ and R2 = 10 kΩ.

The value of R1 is not critical; choose R1 = 10 kΩ.

In designing level detectors, the significant errors will clearly be:

• accuracy and stability of the reference voltage VR

• stability and matching of the Zener diodes

• resistor tolerance

• errors due to op amp bias current and offset voltage.

+ = +× ×

+

+ =

=

=

5 02 7

5 5 14

5 9

1

2

2 3

2 3 2

3 2

3 2

R

R R

R R R

R R

R R/ .88.

14



________________________________________________________________________________________

ASTABLE MULTIVIBRATOR________________________________________________________________________________________

FIG. 10

This circuit is a modification of the level detector and provides a continuous

square-wave output. Typical waveforms are shown in FIGURE 11. A

capacitor C is charged and discharged via a resistor R1. Switching takes place

when V1 = V2. Immediately after the output switches, the polarity of V2 also

changes, and the voltage across the capacitor charges towards this new level

exponentially.

–+

0 V

V1

V2

R

R3

R2

(VZ + VD)

R1

C

15



FIG. 11 Typical Waveforms

As in the level detector,

The equation for V1 can be written by considering the initial and final voltages

of an exponential with a time constant of R1C.

You may find this a little hard to see. Compare this equation with that for the

simple circuit shown below. We have to consider an initial condition; the

exponential starts from –V2 and would continue until the asymptotic value

[VZ + VD], if switching did not intervene.

V V V V t R C VZ D1 2 1 21= +⎡⎣ ⎤⎦ +( ) ( )( )– – / – ......exp .. 15( )

VR V V

R RZ D

22

2 3

=+( )

+ ..................................................... 14( )

0t

–V2

+V2

+(VZ + VD)

–(VZ + VD)

T Tt = 0

V1

16



As a check, put t = 0 in equation (15) and remember e0 = 1.

Then put t = infinity.

Switching occurs when V1 = V2, and t = T = half the period of the cycle.

R V V

R RV V

R V V

R RZ D

Z DZ D2

2 3

2

2 3

+⎡⎣ ⎤⎦+

= +⎡⎣ ⎤⎦ ++⎡⎣ ⎤⎦

+⎧⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪( )( )

+⎡⎣ ⎤⎦+

1 1

2

2 3

– – /

–

exp T R C

R V V

R RZ D

V V V V V V VZ D Z D2 2 2= +⎡⎣ ⎤⎦ +( ) = +⎡⎣ ⎤⎦–

V V1 2=

t

Finalvoltage

Initial capacitor voltage

VB

VC

VC0

(VB + VC0)

t = 0

Switch closes

R

CVC

+VB

17



Simplify by dividing by [VZ + VD] and multiplying by (R2 + R3):

The full cycle period is 2T, since the circuit is symmetrical. It is interesting

that the frequency does not depend on the Zener voltages.

What errors would you expect between predicted and actual performance?

....................................................................................................................................................

....................................................................................................................................................

....................................................................................................................................................

....................................................................................................................................................

....................................................................................................................................................

....................................................................................................................................................

....................................................................................................................................................

________________________________________________________________________________________

The errors would be the same as for the level detector, plus the limitation of

the amplifier slew rate, which means that the waveform will not be square, but

a trapezoid.

R R R T R C R

R

R R

2 2 3 1 2

2

2 3

2 1

2

21

= +( ) ( )( )

+=

– – / –

– –

exp

exxp

exp

So exp

– /

– /

/

T R C

R

R RT R C

T R C

1

3

2 31

1

2

( )

+= ( )

( ) == +1 2 2 3R R/

18



WORKED EXAMPLE 2

What is the oscillation frequency of the astable multivibrator if R1 = R2 = R3

= 10 kΩ and C = 10 nF?

Substitute the values into equation (16) to obtain the half period.

As with the level detector, there are many similar square-wave oscillator

circuits. A very popular integrated circuit which can be used for oscillators or

timers (as in the next section) is the type 555.

________________________________________________________________________________________

MONOSTABLE MULTIVIBRATOR________________________________________________________________________________________

This is the third circuit in this family of oscillators. The level detector had two

stable states, while the astable had none; the monostable has one stable state.

In other words, the monostable stays in a stable condition until it is triggered

externally. It then produces a single pulse of defined length before recovering

to its initial condition.

T

f

= × × × +( ) =

=

10 10 10 1 2 109 86

1

4 9– .ln s

Frequency,

μ

224551

T= Hz.

19



FIG. 12

FIG. 13 Illustrative Waveforms

The diode in FIGURE 12 clamps the capacitor voltage at about +0.7 V, thus

preventing the voltage rising to +V2; the circuit is in a stable condition. A

negative trigger pulse, VT, pulls down V2 to the level of the capacitor voltage

and initiates a negative output pulse of defined width. Immediately after the

trigger pulse, the waveform is defined by the following equation, obtained by

the same sort of considerations that generated the astable exponential equation.

0t

–V2

+V2

+(VZ + VD)

–(VZ + VD)

Tmt = 0

+VD

Tr

V1V1

–+

0 V

V1

V2

R

R3

R2

VZ + VD

R1

C

CTVT

20



The capacitor starts to charge from +VD to – (VZ + VD).

Reset occurs when V1 = –V2, and t = Tm = the monostable period.

Re-arrange this equation in the same way as we did for the astable circuit.

Prove for yourself that this is true.

NOTES

1. To ensure triggering, VT must be greater than (V2 – VD).

2. To reverse the output and trigger sense, change the polarity of the diode

connection.

3. An expression for the recovery time can also be derived using a similar

analysis. The recovery time is less than the monostable time, but can still

be too long for convenience. If we demand another pulse before recovery

T R CV V R R

V VmD Z

Z D

=+( ) +( )

+( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

12 32 1

ln ./

..................... 19( )

–– – – /

R V V

R RV V V T RZ D

D Z D m2

2 3

1+⎡⎣ ⎤⎦

+= + +⎡⎣ ⎤⎦( ) exp 11 C VD( )( ) +

V V V V t R C VD Z D D1 11= + +⎡⎣ ⎤⎦( ) + ( )( ) +– – /exp .....

Also ..........

17

22

2 3

( )

= ±+⎡⎣ ⎤⎦

+V

R V V

R RZ D ........................................... 18( )

21



is complete, the resultant period Tm will be shorter. FIGURE 14 shows a

modification to reduce the recovery time whilst leaving Tm unaffected.

4. The monostable suffers from the same errors as the astable, and you will

note that in this case Tm depends on [VZ + VD].

FIG. 14

WORKED EXAMPLE 3

Design a monostable with a Tm of 10 seconds.

As in all design problems there are choices to be made. The student finds it

much easier to calculate Tm if all the parameters of equation (19) are known. It

is much harder to choose sensible values to give a desired answer!

Let us choose VZ = 5.6 V, a readily available device, and assume that VD =

0.7 V. It also seems sensible to try making R2 = R3 (say 100 kΩ). Hence the

'ln' term yields the following value.

V1R1

R4

(VZ + VD)

R4 << R1

22



It is best to use a high value of R1 to keep the value of C low and cheap. A

reasonable solution would be to make C = 10 μF and R1 = 1.2 MΩ.

ln ln2 1 2 0 72 3V V R R

V VD Z

Z D

+( ) +( )+( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

× +/ . 55 6 1 1

5 6 0 7

0 80

.

. .

.

,

( ) +( )+( )

⎡

⎣⎢

⎤

⎦⎥

=

Hence R C Tm1 0 8 12 5= =/ . . .

23



________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS________________________________________________________________________________________

1. Find the formulae for oscillation frequency and gain required for a Wien

bridge oscillator where the two resistors are not equal as shown below.

2. Derive the formula for the frequency of oscillation of the Hartley

oscillator based on the circuit below. What sort of amplifier is needed

and what gain will be required if L1 = 1 mH and L2 = 0.5 mH?

3. A phase-shift oscillator has the circuit shown opposite.

Prove thatj

j j

V

V

CR

CR CR CR2

1

3

21 6 5= ( )

( ) +–

– –

ωω ω ω(( )3

V2V1

0 V

C

Z

R

L2L1

V

V2V1

CZ2

0 V

C

Z1

R1

R2

24



Remember that the ‘R’ into the amplifier loads the point V2.

Hence find the ratio RF/R to cause oscillation.

Choose components to give an oscillation frequency of 2 kHz.

4. Design a level detector to operate at input signal levels of +6 V and –2 V.

Assume [VZ + VD] = 7.5 V.

5. Design a square-wave generator to oscillate at a frequency of 1 kHz.

6. Show that the recovery time for the monostable circuit of FIGURE 12 is

given by the equation below.

T R C V VR

R Rr D Z= +{ } ++

⎧⎨⎩

⎫⎬⎭

⎡

⎣⎢

⎤

⎦⎥1

2

3 2

1 1ln /

–+

R

RF

0CCC

R R

V2VyVx

V1

0 V

25



7. Find the input voltages at which the circuit below switches.

8. In the level detector circuit below, the two R1 resistors no longer have the

same value. BUT the VR is defined to be +15 V or –15 V (depending on

the design requirement). Derive new formulae for the input switching

levels and design a circuit to switch at +6 V and –2 V. Assume (VZ +

VD) = 10 V.

–+

0 V

Vin

V2

R3

R2

VZ + VD

R1a V1

R1b

VR

0 V–+

Vz = 8.2 V

Vin

10 kΩ

22 kΩ

0 V

26



________________________________________________________________________________________

NOTES________________________________________________________________________________________

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

27



________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS________________________________________________________________________________________

1.

Multiply above and below by jωC (jω CR2 + 1).

This is real when [1 – ω2 C2 (R1 R2)] = 0

At this frequency,j

j

V

V

CR

C R R

R

R2

1

2

1 2

2

2=

+( ) =ω

ω 11 22+ R.

i.e. ω = ⎡⎣ ⎤⎦C R R1 2

1–

V

V

CR

CR CR CR

CR

2

1

2

2 1 2

2

1 1

1

=+( ) +( ) +

=

j

j j j

j

ωω ω ω

ω

ω– CC R R C R R[ ]( ) + +( )22 1 1 22jω

Put jj

j

and jj

Z R CCR

C

Z R CR

1 11

2 22

11

1

= + =+

= =

/

//

ωω

ω

ωωωω ω

ω

C

R C

R

CR

V

V

Z

Z Z

R

CR

2

2

2

2

1

2

1 2

2

1 1+=

+

=+

=

/ j j

Nowj 22

1 2

2

11

1

++

++

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

j

j j

ωω ω

CR

C

R

CR

28



Therefore, to ensure oscillation, the gain of an accompanying amplifier

must be greater than

Note: If we make R1 = R2 = R, then ω = 1/CR and the gain

requirement is +3 as in the lesson.

2. We will follow the same steps as in the Colpitts oscillator.

This simplifies to the expression below.

V

V

L L C

R C L L L L C2

1

31 2

21 2 1

221 1

=+⎡⎣ ⎤⎦( ) +–

– –

j

j

ωω ω ω(( )

V

V

L

C L

L C

L C

Z L

2 2

2

22

22

1

1 1

1

=+

=

= ( )

j

j j

j j

ωω ω

ωω

ω

/

–

–

/ ωω ωω ω ωω ω ω

C LL C L

L L+( ) = ( ) +( )

+ +j

j j j

j j j21 2

1 2

1

1

/

/ CC

L L C

C L L

V

V

V

V

V

V

=( )( )

+( )

= =

jω ωω1

22

21 2

2

1

2

1

1

1

–

–

–ωωω

ωω

ω

22

22

22

22

1

1

1

L C

L C

Z

R Z

L C

L C

L

–

–

–

×+

=⎡

⎣⎢

⎤

⎦⎥

( j ))( )+( )

+( )( )

1

1

1

1

22

21 2

12

2

–

–

–

–

ωωω ω

ω

L C

C L L

RL L Cj

221 2C L L+( )

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

.

+ +( ) = +R R R R R1 2 2 1 22 2/ / .

29



To make this real, the real part of the denominator must be zero.

Now, using the frequency condition, substitute for ω2 C.

To obtain an overall gain of +1, we need an amplifier gain of –L1/L2. If

L1 = 1 mH and L2 = 0.5 mH, the required gain is –2.

3. Algebraically, this is probably one of the hardest problems which has

been set. Consider the currents at the nodes Vx, Vy and V2. Note that, at

V2, the current flows to the virtual earth via the right-hand resistance R.

Multiply each equation by R and put jωCR = K.

V V C V V C V Rx x y x1 – – /( ) = ( ) +j j ...............ω ω A

j j .........

( )

( ) = ( ) +V V C V V C V Rx y y y– – /ω ω2 ....... B

j ..............

( )

( ) =V V C V Ry – /2 2ω .............................. C( )

V

V

L

L2

1

2

1

= –

1 1 2/ L L+( )

At this frequency,j

j

V

V

L L C

L L C2

1

31 2

12

21=

–

–

ωω ω(( ) =

–

–

ωω

22

221

L C

L C

1 021 2

1 2

– ω

ω

C L L

C L L

+( ) =

= +( )⎡⎣ ⎤i.e. ⎦⎦– 1

2

30



Arrange each of these three equations so that terms are gathered.

We must now eliminate Vx and Vy from these simultaneous equations.

(A2) – (C2):

K (B2) + (2K + 1) (C2):

This eliminates Vy. Now we need to remove Vx using equations (D1) and

(D2).

K V K V K K Vx2 2

2 22 1 1 0+ +( ) +( ) =– ............................. D2( )

KV K V K Vx1 22 1 1 0– +( ) + +( ) = ............................. D1( )

KV K V KVx y1 2 1 0– +( ) + = ............................. A2

..........

( )

+ +( ) + =KV K V KVx y– 2 1 02 ................... B2

..

( )

+ +( ) =KV K Vy – 1 02 ........................... C2( )

V V K V V K Vx x y x1 – –( ) = ( ) + ......................... A1

...........

( )

( ) = ( ) +V V K V V K Vx y y y– – 2 .............. B1

..........

( )

( ) =V V K Vy – 2 2 ....................................... C1(( )

31



K2 (D1) + (2K + 1) (D2) gives equation (E1) below.

Gather all the terms in V2, multiplying out the brackets. An equation of

the following form results.

where

The coefficient F can be considerably simplified.

The voltage gain may then be obtained by rearrangement.

Now replace K with jωCR.

V

V

CR

CR CR CR2

1

3

2 31 6 5= ( )

( ) + ( )–

– –.

j

j j

ωω ω ω

V

V

K

K K K2

1

3

3 26 5 1=

+ + +

K V V K K K31 2

3 26 5 1 0+ ( ) =– – – –

K K K K K K K K K F3 2 3 2 3 2 22 4 4 4 4 1+ + +( ) =– – – – – –

K V FV31 2 0+ =

K V K K V K K V K K V31

22

22

221 2 1 2 1 1 0+ +( ) + +( ) +( ) +( ) =–

E11( )

32



This has been hard going; you need a clear head and a logical procedure

to be successful. If you refer back to Lesson 1 of this topic, you will find

that the required gain is –29, and the frequency is defined as below.

From this, we can derive the required value of CR.

If we choose C = 1 nF, then R = 32.5 kΩ.

Select R = 33 kΩ as the nearest value in the E12 series.

4. Use equations (11) and (12) of the lesson text and substitute in the values

given.

Add these equations to obtain the value of VR.

4 2

2

=

=

–

–

V

V

R

R V

+ = +×+

=×+

62 7 5

22 7 5

2

2 3

2

2 3

–.

– – –.

VR

R R

VR

R R

R

R

CR = × × ×⎡⎣ ⎤⎦ = ×2 6 2 10 32 5 1031

6π–

–.

ω

ω π

= ⎡⎣ ⎤⎦

= = ×

6

2 2 10

1

3

CR

f

–.

/

33



Substitute this value of VR into the first equation.

Rearrange to give a relationship bewtween R2 and R3.

Choosing R2 = 10 kΩ and R3 = 27 kΩ would be close.

Choose R1 = 10 kΩ (or any other convenient value).

5. f = 1/2T where T = half the period of oscillation.

So

Use equation (16) and substitute this value of T.

Choose R2 = R3 (this is almost always a sensible choice).

Some trial and error with standard component values gives C = 56 nF

and R1 = 8.2 kΩ as a close fit.

R C13 30 5 10 3 0 455 10= × = ×. / .– –ln

0 5 10 1 231 2 3. /–× = +( )R C R Rln

T f= = ×1 2 0 5 10 3/ . –

4 4 15

2 75

2 3 2

3 2

R R R

R R

+ =

= .

6 215 2

2 3

= + ++R

R R

34



6. The exponential equation for the recovery time, Tr, is as follows.

When t = Tr, V1 = VD. Substitute in these values.

Substitute for V2.

Rearrange this equation to isolate the exponential term.

Finally,

A few intermediate steps have been missed out here. Make sure you can

fill them in.

T R C V V R R Rr D Z= +( ) + +⎡⎣ ⎤⎦( )1 2 2 31 1ln / /

VR V V

R R

V VR

R R

DZ D

Z D

++( )

+

+( ) ++

⎛⎝⎜

⎞⎠⎟

⎡

⎣

⎢⎢

2

2 3

2

2 3

1⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

= ( )

( ) =

1 1

1

– – /

– /

exp

exp

T R C

T R CV

V

r

rZ

Z ++( ) + +⎡⎣ ⎤⎦( )V R R RD 1 2 2 3/

V V VR V V

R RT RD Z D

Z Dr= +( ) + +( )

+⎡

⎣⎢⎢

⎤

⎦⎥⎥

2

2 3

1 – – /exp 11

2

2 3

C

R V V

R RZ D

( )⎡⎣ ⎤⎦

+( )+

–

V V V V T R C VD Z D r= +( ) +⎡⎣ ⎤⎦ ( )⎡⎣ ⎤⎦2 1 21 – – / –exp

V V V V t R C VZ D1 2 1 21= +( ) +⎡⎣ ⎤⎦ ( )⎡⎣ ⎤⎦– – / –exp

35



7. With 7.5 V Zener diodes, the output voltages are

Since V1 = 0 V, switching occurs when V2 = 0 V. Input current must

flow via the 10 kΩ and 22 kΩ resistors, giving the following relationship.

Hence, the voltage at which the circuit switches is determined as follows.

This is an example of a different comparator, one that gives a positive

output transition for a rising input voltage.

8. In this level detector, switching still takes place when V1 = V2, and V2 is

still defined in the same way; the only change is the relationship between

the input voltage, VR and V1.

V V

R

V V

R

V

RV

R R

in

a

R

b

in

a a b

– –1

1

1

1

11

1 1

1 1

=

= +⎧⎨⎪

⎩⎪

⎫⎬⎪⎪

⎭⎪

= +( )

–

/ – /

V

R

V V R R V R R

R

b

in a b R a b

1

1 1 1 1 11

Vin = + × = +/ –.

/ – .10 8 2

223 73V

Vin

10 108 2

22 103 3×= +

×/ –

.

+ +( ) = + +( )

= +

/ – / – . .

/ – .

V VZ D 7 5 0 7

8 2V.

36



So we now have the two equations for the two input switching levels.

Substitute the values Vin1 = +6 V and Vin2 = +2 V and add equations

(A) and (B).

VR is either +15 V or –15 V, but we must choose –15 V in this case.

Choose R1b = 56 kΩ and R1a = 15 kΩ as appropriate E12 resistances to

yield this ratio.

Now substitute in equation (A), remembering that (VZ + VD) = 10.

It is not possible to get very close to this ratio with just two resistors.

R3 = 120 kΩ and R2 = 22 kΩ is about the best.

+ = +( ) +( )

+( ) +

+ =

610 1 1 3 75

15 3 75

2 2

2

2 3

2 3

R

R R

R R

/ ./ .

112 67

5 33

2

3 2

.

/ .

R

R R =

R Rb a1 1 30 8 3 75/ / .= =

6 2 2 1 1+ = – /V R RR a b

VR V V R R

R RV R Rin

Z D a bR a b1

2 1 1

2 31 1

1= +

+( ) +( )+

/– / ......... A( )

=+( ) +( )

+V

R V V R R

RinZ D a b

22 1 1

2

1–

/

RRV R RR a b

31 1– / ........ B( )

37



________________________________________________________________________________________

SUMMARY________________________________________________________________________________________

In this lesson, we have learnt how to design sine-wave oscillators, by using the

Barkhausen criterion of seeking the condition when the loop gain is ≥ 1 when

the phase shift round a feedback loop is 0°. We have also investigated a family

of circuits defined as follows:

• Bistable multivibrator or Level detector or Schmitt trigger or

comparator

• Astable multivibrator or square-wave oscillator

• Monostable multivibrator or timer.

All these circuits have been treated as op amp applications, but there are many

other ways of obtaining the same result.

38



at the end of the lesson, you should be able to: • analyse

Documents