tutorial questions – oscillator with the aid of a suitable diagram, explain briefly what are meant...

TUTORIAL QUESTIONS – OSCILLATOR

With the aid of a suitable diagram, explain

briefly what are meant by the following terms;

Q1

a) open-loop gain;

b) Loop gain

c) closed-loop gain


SolutionA

b

V e

V f

V sV o

+The open-loop gain is the gain of the operational amplifier without feedback. Referring to the figure, the open-loop gain is A and is expressed as;

e

o

V

VA

a)


Solution (cont’d)

A

b

V e

V f

V sV o

+The loop gain is the amplification (or attenuation) experienced by the signal as it travels from the input to the operational amplifier to the output of the feedback network. From the figure, the loop gain is Ab and is expressed as;

b)

e

f

V

VA b


A

b

V e

V f

V sV o

+

The closed-loop gain is the amplification of the input signal by the amplifier with feedback. From the figure, the closed-loop gain is Af and is

expressed as;

c)

s

of V

VA

Solution (cont’d)


Q2 With the aid of suitable figures, describe the

term Barkhausen criterion as applied to

sinusoidal oscillators


Solution

In the figure, it can be shown that the closed-loop gain, Af, is given by the expression;

A

b

V e

V f

V sV o

+

AA

V

VA

s

of b

1


Solution (cont’d)

When the magnitude of the loop gain is unity and the phase is zero i.e. when;

A

b

V e

V f

V sV o

+

01Abthe system will produce an output for zero input.


A

b

V e

V f

V sV o

+

Barkhausen criterion states that in order to start and sustain an oscillation, the loop gain must be unity and the phase shift through the loop must be 0

Solution (cont’d)


The following figure

shows a Wien-Bridge

oscillator employing

an ideal operational

amplifier A. Derive

an expression for the

frequency of

oscillation o in

terms of R and C.

Q3

.

RRC

C

R 2

R 1

+

v OA


Solution

.

RRC

C

R 2

R 1

+

v OA

Z sZ p

v 1

v 2

1

1

11

sCR

RZ

R

sCR

sCRZ

p

p


Solution (cont’d)

.

RRC

C

R 2

R 1

+

v OA

Z sZ p

v 1

v 2

sCRZ s

1

sC

sCR 1


Solution (cont’d)

.

RRC

C

R 2

R 1

+

v OA

Z sZ p

v 1

v 2

ov

v2b

sp

p

ZZ

Z


Solution (cont’d)

sp

p

ZZ

Z

b

sCsCRsCRR

sCRR

/11/

1/

sCRsCR

13

1


Solution (cont’d)

1

21R

RA

.

RRC

C

R 2

R 1

+

v OA

Z sZ p

v 1

v 2


Solution (cont’d)

sCRsCR /13

1

b

1

21R

RA and

The loop gain;

1

21/13

1

R

R

sCRsCRAb


Solution (cont’d)

Substituting for s;

1

21/13

1

R

R

CRjCRjA

b

Since bA must be real at the oscillation frequency o, it follows that;

01

CRj

CRjo

o


Solution (cont’d)

or;RCo

1


Q4For the relaxation oscillator shown in the figure, sketch and label the waveforms of vC and vR2 and

indicate in your sketch, the relevant mathematical equations describing various sections of the waveform.


Solution


Solution (cont’d)


Q5 Design a phase shift oscillator in the following figure, to obtain a sinusoidal wave of 3 kHz. Use C = 22 nF and R1 = 50 k.


Solution

RC

6oFrom the expression;

we obtain;C

Ro6

Substituting values;

kΩ 5.910221032

69-3

π

R

Q1 A single-pole low-pass filter as shown in Fig.5-1 has a RC network with the resistance R of 2 kΩ and the capacitance C of 0.04μF. If the resistances of resistor R1 and R2 are 20 and 4 Ω respectively, determine:

i) cutoff frequency, fC

ii) pass band voltage gain or the gain of non-inverting amplifier, Acl iii) expression of output voltage, Vo

Figure 5-1.

a

Sol1

kHzHzx

FxxRCfc

21099.1

)104)(102)(14.3(2

1

2

1

3

83

614

201

2

1)( R

RA NIcl

(a)

(b)

The cutoff frequency for a single-pole low-pass filter is

The pass band voltage gain or the gain of non-inverting amplifier is calculated as follows:

fRCjv

RCjv

XR

jv

jRX

Xvv

inin

C

inC

Cina

21

1

1

1

1

1

C

Cin

C

Cina jXR

jXv

XR

Xvv

Applying the voltage divider rule, the input voltage, which is the voltage across the capacitor C, is determined as follows:

Sol1_Cont’d

Multiplying the numerator and the denominator by j and substituting for XC results in the following:

Sol1_Cont’d

RCf

orRCf cc

21

2

1

At the cutoff frequency fc, the magnitude of the capacitive reactance XC equals the resistance of the resistor R.

Substituting for 2πRC in Eq. results in the following equation for va:

c

ina

ff

jvv

1

1

Where f is the operating frequency and fc is the cutoff frequency.

Sol1_Cont’d

c

NIclinNIclao

ff

j

AvAvv

1

)()(

The output vo is the amplified version of vi.

where Acl(NI) is the pass-band voltage gain or the gain of the non-inverting amplifier.

Q2

Determine the critical frequency critical frequency, pass-band voltage pass-band voltage gaingain, and damping factordamping factor for the second-order low-pass active filter in Fig. 5-2 with the following circuit components:

RA = 5 kΩ, RB = 8 kΩ, CA = 0.02 μF, CB = 0.05 μF, R1 = 10 kΩ, R2 = 20 kΩ

Figure 5-2.

Sol2

Hz

xxxx

CCRRf

BABA

c

18.796

)105)(102)(108)(105()14.3(2

1

2

1

8833

The critical frequency for the second-order low-pass active filter is

Sol2_Cont’d

5.1120

101

2

1)( R

RA NIcl

The pass-band voltage gain set by the values of R1 and R2 is

Q3

Design a multiple-feedback band-pass filter with the maximum gain, Ao = 8, quality factor, Q = 25 and center frequency, fo =10 kHz. Assume that C1 = C2 = 0.01µF.

Draw the circuit design of the active band-pass

kkCAf

QR

oo

97.4)8)(01.0)(10(2

25

21

kkCf

QR

o

58.79)01.0)(10(

252

32)8)25(2)(01.0)(10(2

25

)2(2 223 kAQCf

QR

oo

Sol3

Sol3_Con’t

R 1

R 2

R 3

C 1

C 2

V inV out

The drawing of the circuit diagram :

Q4Determine the center frequency, fc, quality factor, Q and bandwidth, BW for the band-pass output of the state-variable filter in following figure. Given that R1 = R2 = R3 = 25 kΩ, R4 = R7 =2.5 kΩ, R5 = 150 kΩ, R6 = 2.0 kΩ, C1 = C2 = 0.003µF

Sol4_Con’t

i) Center frequency, fc

kHzkCRCR

f c 22.21)03.0)(5.2(2

1

2

1

2

1

2714

ii) Quality factor, Q

67.2510.2

150

3

11

3

1

6

5

k

k

R

RQ

Sol4_Con’t

iii) Bandwidth, BW

The critical frequency, fo of the integrators usually made equal

to the critical frequency, fc

Hzk

Q

fBW

ff

o

co

65.82667.25

22.21

tutorial questions – oscillator with the aid of a suitable diagram, explain briefly what are meant...

Documents

c solution contd slide

relaxation oscillator

expression slide

q3 slide

b slide

closedloop gain

closed loop gain

openloop gain