Rectifiers&Resonance

ANALOG ELECTRONICS LAB

DEPARTMENT OF TELECOMMUNICATION ENGINEERING,CMRIT 1

EXPERIMENT-03

HALF WAVE RECTIFIER

AIM: To study Half Wave Rectifier and to calculate ripple factor, efficiency and

regulation with filter and without filter.

COMPONENTS REQUIRED:

Sl. No. Components Details Specification Qty

1. Diodes BY127 1 No.

2. Capacitor 0.1f, 470f Each 1 No.

3. Power Resistance Board 1 No.

4. Step down Transformer 12 V 1 No.

5. CRO, Multimeter, Milliammeter, Connecting Board

THEORY:

Half wave rectifier circuit consists of resistive load, a diode and source of ac

voltage, all connected in series. In half wave rectifier, rectifying element conducts only

during positive half cycle of input ac supply. The negative half cycles of ac supply are

eliminated from the output. The dc output waveform is expected to be a straight line

but the half wave rectifier gives output in the form of positive sinusoidal pulses. Thus

the output is called pulsating dc.

CIRCUIT DIAGRAM:

HALF WAVE RECTIFIER WITHOUT FILTER CAPACITOR

C2

0.1UF BY127

A K

RL

AC (230V/50HZ)

12V

12V

0

Step down Transformer

A

Ammeter(0-250mA)

+ -

VODC VOAC



HALF WAVE RECTIFIER WITH FILTER CAPACITOR

DESIGN:

V12VrmsIN

V97.16V2VrmsINmIN

V4.5/VV mDCO

Given V5V DCO

mA100I DCO

50I/VR DCODCOL

Ripple = r = Vo rms / VO DC = 1.21

Design for the filter capacitor

Ripple = 1/(43 f C RL)

Given r = 0.25

C = 1/(43 f r RL)

RL = 50

f = 50Hz

= 461.88F 470F Efficiency = PDC /PAC (I

2DC * RL) / [(Irms)

2 * (RL + RF)]

Regulation % Regulation = 100

FL

FLNL

V

VV

C2

0.1UF BY127

A K

RL

AC (230V/50HZ)

12V

12V

0


A

Ammeter(0-250mA)

+ -

470UF

+

- C1

VODC VOAC



PROCEDURE:

1. Connections are made as shown in the circuit diagram

2. Switch on the AC power supply

3. Observe the wave form on CRO across the load resistor and measure the o/p

amplitude and frequency.

4. Note down RL, IDC, VODC, VINAC, and VOAC in the tabular column for different load

resistances.

5. Calculate the ripple and efficiency and Regulation for each load resistance.

6. Repeat the above procedure with filter capacitor.

TABULAR COLUMN:

Sl.

No. RL IDC VO (DC) VIN (AC) VO (AC) Ripple Efficiency Regulation

WAVEFORMS:

12

0 t

-12

0 t Vo (Without Filter)

Vo (with filter)

t

VC

VIN

VO



FULL WAVE RECTIFIER

AIM:To study the full wave rectifier and to calculate ripple factor and

efficiency and Regulation with filter and without filter.



1. Diodes BY127 2 Nos.





THEORY:

The center tapped full wave rectifier circuit is similar to a half wave rectifier

circuit, using two diodes and a center tapped transformer. Both the input half cycles

are converted into unidirectional pulsating DC.

CIRCUIT DIAGRAM:

FULL WAVE RECTIFIER WITHOUT FILTER CAPACITOR

Step down

C2

0.1UF BY127

A K

RL

AC (230V/50HZ)

12V

12V

0

Transformer

A

Ammeter(0-250mA)

+ -

VO(DC)

BY127

A K

VO (AC)



FULL WAVE RECTIFIER WITH FILTER CAPACITOR

DESIGN:

Vin rms = 12V

Vin m = 2Vin rms = 16.97V

VO DC = 2Vm/ = 10.8V

Given VO DC = 10V

IO DC = 100mA

RL = VO DC / IO DC = 100




Given r = .06

C = 1/(43 f r RL)

RL = 100

f = 50Hz

= 470UF

Efficiency = PDC /PAC (I2DC * RL) / [(Irms)2 * (RL + RF)]


FL

FLNL

V

VV

(230V/50HZ)

C2

0.1UF BY127

A K

RL

AC

12V

12V

0


A

Ammeter(0-250mA)

+ -

470UF

+

- C1

VO(DC)

BY127

A K

VO(AC)



PROCEDURE:





4. Note down RL, IDC, VODC , Vinac, Voac in the tabular column for different load

resistances.

5. Calculate the ripple and efficiency and regulation for each load resistance.


TABULAR COLUMN:

Sl. No.

RL IDC VO (DC) VIN (AC) VO (AC) Ripple Efficiency Regulation

WAVEFORMS:

t

0

-

0 Vo (Without Filter)

t

Vo (with filter)

t

VC

VIN

VO



BRIDGE RECTIFIER

AIM:

To study the bridge rectifier and to calculate ripple factor and efficiency

and regulation with filter and without filter.



1. Diodes BY127 4 Nos.





THEORY:

The bridge rectifier circuit is essentially a full wave rectifier circuit, using four

diodes, forming the four arms of an electrical bridge. To one diagonal of the bridge,

the ac voltage is applied through a transformer and the rectified dc voltage is taken

from the other diagonal of the bridge. The main advantage of this circuit is that it does

not require a center tap on the secondary winding of the transformer; ac voltage can

be directly applied to the bridge.

The bridge rectifier circuit is mainly used as a power rectifier circuit for

converting ac power to dc power, and a rectifying system in rectifier type ac meters,

such as ac voltmeter in which the ac voltage under measurement is first converted

into dc and measured with conventional meter.

CIRCUIT DIAGRAM:

BRIDGE RECTIFIER WITHOUT FILTER CAPACITOR

RL

- +

BRIDGE

1

4

3

2

C2

0.1UF

AC (230V/50HZ)

12V

12V

0


Vo

A

Ammeter(0-250mA)

+ -



BRIDGE RECTIFIER WITH FILTER CAPACITOR

DESIGN:

Vin rms = 12V

Vin m = 2Vin rms = 16.97V

VO DC = 2Vm/ = 10.8V

Given VO DC = 10V

IO DC = 100mA

RL = VO DC / IO DC = 100




Given r = .06

C = 1/(43 f r RL)

RL = 100

f = 50Hz

= 470UF

Efficiency

= PDC /PAC

= (I2DC * RL) / [(Irms)2 * (RL + RF)]


FL

FLNL

V

VV

C1

470UF

RL

- +

BRIDGE

1

4

3

2

C2

0.1UF

AC(230V/50HZ)

12V

12V

0

Step downTransformer

Vo

A

Ammeter(0-250mA)

+ -

+ -



PROCEDURE:





4. Note down RL, IDC, VODC , Vinac, Voac in the tabular column for different load

resistances.

5. Calculate the ripple factor, efficiency and regulation for each load resistance.


TABULAR COLUMN:

Sl.

No.

RL IDC VO (DC) VIN (AC) VO (AC) Ripple Efficiency Regulation

WAVEFORMS:

Vin

12

t

0

-12

Vo

Vo (Without Filter)

0

t

VC Vo (with filter)

t



EXPERIMENT-04

SERIES AND PARALLEL RESONANCE CIRCUITS

SERIES RESONANCE CIRCUIT

Aim : To obtain the frequency response of an RLC series circuit and hence to

determine

a) Resonance frequency fo b) Band width ,Upper and Lower half power frequency

c) Q-factor.



1. Resistors 47 1 No.

2. Capacitor 0.22f 1 No.

3. Inductor 4.7 mH 1 No.

Signal generator, Multimeter

Circuit diagram:

4.7mH 0.22F 10Vp-p

47

DESIGN:

Let = , R=47 Assume ,C= 0.22F

0 =1

2

L=4.61mH

L

C

R VO



Procedure: 1. Connections are made as shown in the circuit diagram.

2. AC Supply is switched on. oscillator output voltage is adjusted to about

maximum i.e 10V P-P 3. The frequency is gradually varied from zero hertz and for different value of f,

voltage is noted down. The results are tabulated in the tabular column.

4. Frequency response i.e a graph of frequency versus voltage is drawn.

5. From the graph , resonant frequency fo is noted down at which voltage is maximum(Vo).

6. Lower half power frequency f1 and upper half power frequency f2 are noted corresponding to a voltage of Vo/ 2

Band width=f2-f1=_____________ hertz

7. The Q-factor =fo/f2-f1

Tabular column

FREQUENCY RESPONSE CURVE ( in Semilog )

f in hz V in VOLTS

0

BW

f1 f0 f2 f in Hz

HHz f, Hz

VO

VOmax

VOmax/2



PARALLEL RESONANCE CIRCUIT:

Aim : To obtain the frequency response of an RLC series circuit and hence to

determine

a) Resonance frequency fo

b) Band width ,upper and lower half power frequency

c) Q-factor.



1. Resistors 47 1 No.

2. Capacitor 0.22f 1 No.

3. Inductor 4.7 mH 1 No.

Signal generator, Multimeter

Circuit diagram:

Frequency response

Vo

Vomin

Vomin

f1 fO f2 fin Hz

BW

0

L C

R VO



Procedure:

1. Connections are made as shown in the circuit diagram. 2. AC Supply is switched on. oscillator output voltage is adjusted to about

maximum i.e 10V P-P

3. The frequency is gradually varied from zero hertz and for different value of f, voltage is noted down. The results are tabulated in the tabular column.

4. Frequency response i.e a graph of frequency versus voltage is drawn. 5. From the graph , resonant frequency fo is noted down at which voltage is

minimum (Vo).

6. Lower half power frequency f1 and upper half power frequency f2 are noted corresponding to a voltage of VOmin x 2.

a. Band width = f2 - f1=_____________ Hz

7. The Q-factor =fo/f2-f1

Result:

Rectifiers&Resonance

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