JNTUK_EDC-Unit-IV_Notes.pdf

8/11/2019 JNTUK_EDC-Unit-IV_Notes.pdf

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1 EDC UNIT-III1

1

UNIT III

HALF WAVE RECTIFIER: -

VS = Vmsin t

= VmSin

Imaxm

s f L

V

R R R=

+ +

(Assuming voltage drop across diode is zero)

Where RSis source resistanceRfis Diode resistance in forward bias condition

RLis load resistance.

Current i = 0 for 2

2) AVERAGE CURRENT IDC.

The current shown by DC ammeter, is the average current.2 2

max

max

0 0

1 . sin sin

2 2

II Sin d d d

= = +

[ ] [ ]

( )

max max

0

max max

cos cos 02 2

2 _________________(1)

2 2

dc

I IC os

I II

= =

= =

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2 EDC UNIT-III2

2

3) R.M.S (ROOT MEAN SQUARE) CURRENT1 2

2

2 2

max

0

1sin

2rms

i I

=

1 2

2

max

1sin

2I d

=

; max

2rms

II = (2)

max

2RMS

VV = (When RS& Rfare negligible) (3)

4) DC OUTPUT VOLTAGE: VRL

max

max

max.

.

(When R & R are negligable)

L

L

R dc L L

S f L

L

R s f

iV i R R

V

R R R

R

VV

= =

+ +

=

=

(4)

5) RIPPLE FACTOR

rms value of alternate current componentr

average value=

Total current = DC component + AC. component1 1 ( )dc dci I i or i i I = + = . (1)

i rms =2

2

0

1( ) .

2i

=2

2

0

1 ( )

2 dc

i I d

=2

2 2

0

1( 2 )

2 dc dci I I d

+

=

2 2 2

2 2

0 0 0

1 1 1 - .2 + .

2 2 2

dci d I id I dc d

I II III

a) 1stTerm: -

2

2 2

0

1.

2 rms rmsi d i i

= =

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3 EDC UNIT-III3

3

b) 2nd

Terms: -2 2

2

0 0

1 1.2 2 . = 2 I I = 2 I

2 2dc dc dc dc dc

I id I id

=

c) 3rdTerm: -2

2 2 2

. 0

0

1 1. [ ]2 2

dc dcI d I

=

= 2 2.

1.2

2dc dcI I

=

Combining 3 terms

i rms = 2 2 22rms dc dci I I +

= 2 2rms dc

i I

The ripple factor r =1

rms

dc

I

I

2 2=

rms dc

dc

I I

I

=2 2

2 2

rms dc

dc dc

I I

I I

Ripple Factor r =

2

1rms

dc

I

I

_______________ - (5)

The above expression is independent of current wave shape and therefore notrestricted to half wave rectifier alone.

Similarly it can be expressed in terms of Voltage under: -

r =

2

1

DC

rmsV

V

- (6)

6) RIPPLE FACTOR OF HALF WAVE RECTIFIER

Irms = 2

mI

- eqn. No. 2

IDC =mI

- eqn. No. 1

Putting these values in equation No. 5

Ripple factor of half wave rectifier.

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4 EDC UNIT-III4

4

HW =2

m mI I

/ 12

HW =

2

12

HW = 2.467 1 1.21 121% = =

7) VOLTAGE REGULATION.

Voltage regulation Variation of Output voltageFull load voltage

=

Voltage regulation = 100DCNL DCFL

DCFL

V V

V

- (1)

VDCFL= VDCNL IDC(Rf+ RS)

8) FOR HALF WAVE RECTIFIER

( )mDCNL DC f S L

VV I R R R

= = + + - (2)

.DCFL DC L

V I R= - (3)

( ) ( )DCNL DC f S DC L

V I R R I R= + + from equation No. 2

= ( )DC f S DCFLI R R V+ +

( )DCN DCFL DC f SV V I R R = + - (4)

Voltage regulation (HWR) =( )

( )

DC f s

m

dc S F

I R R

VI R R

+

+ - (5)

For good rectifier voltage regulation must be as less as possible.

9) RECTIFICATION EFFICIENCY: -

= 100%

DC power delivered to the load

AC power input

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5 EDC UNIT-III5

5

= 100%dc

ac

P

P

2

2

2

2

( )

( )

dc dc L

ac rms f s L

dc L

rms f s L

p I R

P I R R R

I R

I R R R

=

= + +

=+ +

For a good rectifier voltage regulation must be as less as possible.

10) OF HWR

( ) ( )

( )

( )

2

2dc dc L L

2

2ac

22

L

2 2

2

2

P =I R = R

P2

4 R100 100

.( )4

4 = 100 40.5%

m

mrms f s L f S L

m L

m f s Lf s

f L L

I

II R R R R R R

I R

I R R RR R R

If R R R

then

= + + = + +

= = + +

+ +

+


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6 EDC UNIT-III6

6

11) TRANSFORMER UTILIZATION FACTOR (TUF)

DC power delivered to load

AC power rating of transformer secondary

( )

dc

ac

TUF

P

P rated

=

=

Where Pacrated(rms) = Vac(rms) x Irms

Pdc =2 .DC LI R

12) FOR HALF WAVE RECTIFIER: -

mI

( ) I22

m

ac rms

VV rms and = = ; Idc m

I

=

2

.

=

.22

m

L

m m

IR

TUFV I

= We know ( )m m f s LV I R R R= + + ;( )

2L

2

R

.22

m

m f s L m

I

TUFI R R R I

=+ +

2

2 2L

f S L

RTUF

R R R

=

+ +

If (Rf+ RS)


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7 EDC UNIT-III7

7

In HWR PIV = VmWhen a fitter with capacitor is used with HWR then PIV = 2 Vm.

Disadvantages of HWR.

a)

Very high ripple factor (r = 1.21)

b)

Low rectification efficiency (=40.5%)c) TUF is low (0.287)d)

There is a DC current component through the winding of the transformer,

which can lead to saturation of transformer core, which is un desirable.

Full Wave Rectifier: -

1) Working Principle1)

Being a center tapped transformer VAB= VBC

When A is positive with respect to B (or C)D1 conducts (say for +Ve halfcycle)When C is positive with respect to B (or A) D2 conducts (say for Ve half

cycle)2)

Voltage VABis applied to D1& VBCis applied to D2.

3) Direction of currents i1 & i2are same through RL.

2) AVERAGE CURRENT IDC

Idc

[ ]

2 2

m

0 0

2

m

0

m

m m

m

1 1 I sin

2 2

Isin sin

2

I( 2)( 2)

2

I 2I.4 0.637 I .

2

id d

d d

=

=

=

= = =

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8 EDC UNIT-III8

8

Idc = 0.637Im.

IDCFWR = 2 IDCHWR.

3) RMS LOAD CURRENT IRMSm

m .

I0.707 I

2

axrms axI = =

4) DC Output voltage VDC.

m2I. . .dc dc L LV I R R

= = We know mI

m

S f L

V

R R R=

+ +

.

.

.

2.

(

( )

20.637

0.637

m Ldc

S f L

s f L

mdc m

dc m

V RV

R R R

If R R R

VV V

V V

=

+ ++


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9 EDC UNIT-III9

9

6) VOLTAGE REGULATION: -

Voltage Regulation =

100

( )

DCNL DCNL

DCFL

DCFL DCN L DC f S

V VV

V V I R R

= +

For full wave Rectifier.

m2 2.I &mDCNL DC

VV I

= =

or2. 2

.m mDC

f S L

I VI

R R R = =

+ +

2( ) I ( + R +R ) = .

( ) I ( + R ) + I .R = V

( ) I ( + R ) + V = V

( ) V - V = ( + R )

DC f s L m

DC f s DC L DCNL

DC f s DCFL DCNL

DCNL DCFL f s

or R V

or R

or R

or R

Voltage regulation =

( + R )

2( )

dc s f

mDC f S

I R

VI R R

= +

7) RECTIFICATION EFFICIENCY: -

100%dc

ac

p

p=

For FWR,

2

2

m

2. .I .

dc dc L Lp I R R

= =

Pac= I2

rms(Rf+ RS+ RL)2

m

2

2

2 2

2 2

I( )

2

4.

.( )2

( )

4 2 8. 0.812 81.2%1

f S L

mL

f s L

f S L

R R R

IR

I mR R R

If R R R

+ +

=

+ +

+


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10 EDC UNIT-III10

10

8) TRANSFORMER UTILIZATION FACTOR (TUF)

a) TUF (Secondary) =sec

dcP delivered to load

AC power rating of transformer ondary

b)

Since both the windings are used TUF FWR= 2 TUF HWR= 2 x 0.287 = 0.574

c) TUF primary = Rated efficiency = 100 81.2%dc

ac

P

P =

d) Average =0.812 0.574

0.6932

+=

9) PEAK INVERSE VOLTAGE (PIV): -

PIV = 2 Vm

10) CORE SATURATION IN FWR.

Currents i1, i2 flow in opposite direction in the secondary winding. Therefore no

saturation.

11) Advantages

1) Ripple factor = 0.482 (against 1.21 for HWR)

2) Rectification efficiency is 0.812 (against 0.405 for HWR)

3) Better TUF (secondary) is 0.574 (0.287 for HWR)

4)

No core saturation problem

Disadvantages:

1) Requires center tapped transformer

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11 EDC UNIT-III11

11

Bridge Rectifier: -

1) Diode D1, D2, D2, D4are forming four arms of bridge

* During positive half cycle (as shown) diodes D2& D3conduct through RL.

* During negative half cycle diodes D4 & D1 conduct through RL.* Direction of current flow through RLis same in both half cycles and we get

the same wave forms as that of full wave rectifier.*Therefore, the following expressions are same as that of full wave rectifier.

a)

Average current Idc=2

mI

b) RMS current Irms=2

mI

c) DC output voltage (no.load) VDC2

mV

d)

Ripple factor =0.482

e) Rectification efficiency = = 0.812

f)

DC output voltage full load.

=2

( 2 )mDCFL dc S fV

V I R R

= + ; i.e., less by one diode loss.

-

TUF of both primary & secondary are 0.812 therefore TUF overall is 0.812

(better than FWR with 0.693)

Comparison:

Sl.No.

Parameter HWR FWR BR

1 No. of diodes 1 2 42 PIV of diodes Vm 2 Vm Vm

3 Secondary voltage (rms) V V-0-V V

4 DC output voltage at no

loadmV

=0.318 Vm

2mV

=0.636 Vm

2mV

=0.636 Vm

5 Ripple factor 1.21 0.482 0.482

6 Ripple frequency f 2f 2f

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12 EDC UNIT-III12

12

7 Rectification efficiency 0.406 0.812 0.812

8 TUF 0.287 0.693 0.812

1 Q) A diode whose interval resistance is 20 is to supply power to a 1000 load

from a 110V (rms) source of supply calculate.i) Peak load currentii)

DC load current

iii) AC load currentiv) DC load voltage

v)

Total power input to the circuit% regulation from no load to full load. (Nov. 2000, June 2006)

solution: -

i) Peak load current = Im = ?

max maxmI ;

2ax RMS

f L

V VV

R R= =

+

Or Vmax 2 2 110 155.56RMSV V= = =

m

155.56 155.56I 152.5

1000 20 1020ax mA= = =+

ii) DC load current IDC= ?

mI 152.5 48.54axDC

I mA

= = =

iii) AC load current = IRMS=max

2

I=

152.5= 76.25 mA

2

Or AC load current .idc

( )1.21HWR = = 1.21 x 68.66 = 83.07mA.

iv) DC load Voltage .dc dc LV i R=

= 48.54 x 10-3

x 1000 = 48.54 Volts

v) Total input power to the circuit( )2RMS L FI R R+ = (76.25 X 10

-3)

2(1000 + 20) = 5.93 W

vi) % regulation20

100 100 2%1000

f s

L

R R

R

+ = = =

2 Q) A half wave rectifier uses was a diode with forward resistance of 100. If the ACinput voltage is 220V (rms) and the load resistance is 2 k, determine.

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13 EDC UNIT-III13

13

i) Imax, IDC& IRMS.ii) PIV of the diode when the diode is ideal.

iii)

Load output voltageiv) DC output power & AC output

v) Ripple factor

vi)

Transformer utilization factorvii) Rectification efficiency. (May, 2001)

Solution:

Case (i)

max max

mI &

2ax rms

f L

V VV

R R= =

+

Vmax= Vrmsx 2 = 220V x 2 = 311V

m

m

m

311 311I 148

100 2000 2100

I 14847.1

I 148

742 2

ax

ax

DC

ax

RMS

mA

I mA

I mA

= = =+

= = =

= = =

Case (ii) PIV = Vmax= 311 V.

Case (iii) Load output voltage VDC= IDC.RL

= 47.1 x 10-3

x 2000 = 94.2 Volts.Case (iv) DC output power PDC= IDC.VDC

= 47.1 x 10-3

x 94.2 = 4.4 watts.

AC power output = ( )2

2 3. R = 74 10 2000 = 10.9 WRMS L

I

Case (v) Ripple factor (HWR) = 1.21

Case (vi) 2 22 2 2 2 2000

2100

L

f S L

R

TUF R R R

= = + +

= 0.287 (0.952) = 0.273.

Case (vii) Rectification efficiency

= DC power delivered to the load / AC power input.

2 2 6

4.4 4.4

( ) (74) 10 (200 2000)RMS f L

I R R = =

+ +

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14 EDC UNIT-III14

14

4.40.383 38.3%

11.5= = =

3 Q) Show that the maximum power output Pdc= VDCx Idcin a half wave single phase

circuit when load resistance equals diode resistance Rf. ( Sep.06, Dec.04)

SOLUTION:IdcRf

Pdc = VDC.IDC = IDC.RL. IDC = IDC2

RL.Where IDC= Vs /( Rf+ RL)

Or

2

.DC L

f L

VsP R

R R

= +

22

2 2 2 2

.

2 2

SL

f L S L f L f L

L

VVs R

R R R R R R R R

R

= =+ + + +

2

2

2

DC

f

L f

L

VsP

R

R RR

=

+ +

PDC will be maximum if denominator is minimum according to Maxima and Minima,denominator (D) will be minimum.

2

0

2

L

fLf

L L L L L

dDIf

dR

RdRdD d d R

dR dR dR R dR

=

= + +

2 1

2

2 1 1 2

2 2

1 ( ) 0

11 ( 1) 0 1 0 1

f

L

f

f L f

L L

dD dR RL

dR dL

RR R R

R R

= + +

= + + = + + =

If

2

2= 0 . .,1 0

f

L L

RdDi e

dR R =

Or Rf2= RL

2

Or Rf= RL

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15 EDC UNIT-III15

15

PDCwill be maximum when Rf= RL

4 Q) A 230V, 60 Hz voltage is applied to the primary of 5:1 step down center tapped

transformer used in a FWR having a load of 900. If the diode resistance and

secondary coil resistance together has a resistance of 100, determine:a)

DC voltage across the load

b) DC current flowing through the loadc) DC power delivered to the load

d)

PIV across each diodee)

Ripple voltage and frequency.

(May, 2006)

900

Given:AC input 230V, 60 Hz/

RL= 900RS+ Rf= 100

Part (a)

DC voltage across load = ?Voltage secondary of transformer = 230/5 = 46 V.

Each of half = 23 volts,(rms); Vrms=23V; Vm= ? Vrms= Vmax(0.707)

Or Vmax= 23 32.530.707 0.707

rmsV Volts= =

m

32.53I 32.53

900 100

m

L S f

V

R R R= = =

+ + +mA

Part (b) DC current IDC= Imax(0.636)

= 32.53 (0.636) = 20.69 mA.VRL= IDC.RL= 20.69 x 10

-3x 900 = 18.62 volts

Part (c) DC power Pdc= Vdc. Idc= 18.62 x 20.69 x 10-3

= 3.85 milliwatt.

Part (d) PIV across each diode = Vmaxx 2 = 32.53 x 2 = 65.06 volts

Part (e) Ripple voltage = ? Ripple factor =Ripple Voltage

Load Voltage

V

= .VRL = 0.483 x 18.62 = 8.99 volts

Ripple frequence = 2 x Input source frequency

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16 EDC UNIT-III16

16

= 2 x 60 Hz= 120 Hz.

5 Q) In a full wave rectifier the required DC voltage is 9V and diode drop is 0.8V.Calculate ac rms input voltage required in case of bridge rectifier and full wave

rectifier. (Dec. 2005)

Solution: Given VDCrequired = 9V.Vd = 0.8V

Case IBRIDGE RECTIFIER: - Input DC voltage required = 9+2(0.8) = 10.6V

maxmax

max

2or V V .

2

1 10.6. . 11.7722 2 2 2

DC DC

rms DC

VV

VV V volts

= =

= = = =

Case II FWR.Input DC voltage required = 9+0.8 = 9.8

Input RMS voltage required =. 9.8

10.882 2 2 2

DCV volts

= =

In two half windings Vrms= 2 x 10.88 = 21.76 Volts

6 Q) A full wave rectifier has a center tap transformer of 100 0 100V and each oneof the diodes is rated at Imax = 400mA and Iar = 150mA. Neglecting the

voltagedrop across the diode calculate (June,2005)a) The value of load resistor that gives largest DC power output.

b)

DC load voltagec) DC load current

d) PIV of each diode.

Diode Ratings:

Ian= 150 mAImax= 400 mA

Solution: -

a) RL= ? for max Pdc. Pdc= Idc2.RL

Max. Idcthat can be handled by the diode = 150 mA.

Max. Peak current that can be delivered to the load = ?

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17 EDC UNIT-III17

17

m

m

3

m

max

m max

max

3

m

I 2

I . 2

I 150 10 235.6

2I ; 2 100 2 141.4

141.4600

I 235.6 10

ax

dc

ax dc

ax

ax rms

L

L

ax

I

I

mA

VV V volts

R

VR

=

=

= =

= = = =

= = =

b) Load voltage VDC= IDC.RL= 150 x 10-3

x 600 = 90 voltsc) DC load current Idc= 150 mA

d) PIV of each diode = 2Vmax= 2 x 141.4 = 282.8 volts

7 Q) Draw the circuit diagram of a FWR using center tapped transformer to obtain anoutput DC voltage of 18V at 200 mA and VDCno load equals 20V. Find the

transformer ratings. (Dec. 2004)Solution:

VDC= 18V

VDCNL= 20V.Idc= 200 mA

VDCNL VDCFL= IDC(RS+ Rf)

20 18 = 200 x 10-3

(RS+Rf) or RS+ Rf= 2 / 200 x 10-3

= 10Transformer rating voltage (Vrms) = ?

max

max

. 20

31.442 2

31.4422.24

2 2

DC

rms

V

V volts

VV volts

= = =

= = =

Transformer rating is Input 220 V Ac.Output 22 0 22V (RMS)DC current 200 mA.

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18 EDC UNIT-III18

18

CAPACITOR FILTER WITH HWR

Cut In angle wt2 (a) Capacitor charging through diode

(Wt2 Wt1)Cut out angle = Wt1 (b) Capacitor discharging through RL

Wt1= - tan1

WCRL (Wt1to Wt2)(c) Average (DC) voltage with fitter

(d) Average (DC) voltage without fitter.

CAPACITOR FITTER WITH FWR.

Ripple factor1

4 3L

rfCR

= Ripple freqFWR= 2 ripple freqHWR.

Inductor Filter with HWR.

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19 EDC UNIT-III19

19

INDUCTOR FILTER WITH FWR.

Ripple factor1

3 2

L

L

R

W =

-

Ripple magnitude reduces

with increase of L-

L Section Filter Section Filter

r = 0.83 / LC Critical Inductance = 3300/C1C2L.RL

LC= RL/1130COMPARISON OF FILTERS: -

1)

A capacitor filter provides Vmvolts at less load current. But regulation is poor.2) An Inductor filter gives wire ripple voltage for low load currents. It is used for

high load currents3)

L Section filter gives a ripple factor independent of load current. Voltage

regulation can be improved by use of bleeder resistance

4) Multiple L Section filter or filters give much less ripple the single L

Section.BLEEDER RESISTANCES: -

1) Vo of L Section filter at no load = Vm.

Voof L Section filter at any load2 mV

The output voltage falls sharply from no load to same load.

By adding bleeder resistance RB, I min. is increased and thereby avoiding suddenfall of output voltage Vo.

RB< 800 Lc

2) For L section filter to work properly LC 3

RL

.

In the absence of RL (When no load ES connected keeps the filter functioning

properly.

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20 EDC UNIT-III20

20

SUMMARY OF FILTER INFORMATION: -

Sl.

No

Parameter None L C L-

Sectionsection

1 VDCNL 0.636 Vm 0.436 Vm Vm Vm Vm

2 VDCNL 0.636 Vm 0.636 Vm Vm 4170

Idc.CRL

0.636Vm 4170 dcm

IV

C

3 Ripple

factor

0.48

16000

LR

L

2410

LCR

0.83

LC

1 1

3330

LCC L R

Where C in F and f = 60 Hz.L om Henms

R ohmsVmVolts

EMITTER FOLLOWER REGULATIOR

Un Regulated power supply unit consists of transformer, rectifier and filter.

PROBLEMS:

Output voltage varies as the load current varies

-

Output voltage varies as the input voltage varies- Output voltage varies with the temperature

REMEDY:

-Voltage stabilization ratio Sv = Vo / Vi Rz / (RZ + R.)

Distance of Emitter follower regulator.

1) i.e., Svrequires increase of R VCEPower dissipation .2) Output voltage cannot be varied i.e., variable power supply is not possible.3)

Changes in VBEand VRdue to temperature variations appear at the output.

4) An electronic control (feed back) is used to counter above problem.

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21 EDC UNIT-III21

21

SERIES VOLTAGE REGULATOR.

Vo = 12

1 2

R BE

RV R Vo

R R+ +

+

Or 1 21 2

o o R BE

RV V V V

R R = +

+

( )

12

1 2

1 2 1

2

1 2

2

2

1 2

2 12

2

1

2

2

1

( )

1

o R BE

o R BE

o R BE

o R BE

o R BE

RV V VR R

R R RV V V

R R

RV V V

R R

R RV V V

R

RV V V

R

= + +

+ += +

+

= +

+

+= +

= + +

11 Q) If = 0.8, VBB= 15V and VD= 0.7V. Find the value of VP.

Vp = VBB + VD

= 0.8 x 15 + 0.7 = 12.7 Volts

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RELAXATION

OSCILLATOR

PNPN DIODE (SHOCK LEY DIODE)

Initially UJT is cut off and capacitor

is charging slowly.

*Anode is applied with +Ve voltage (Wrt

to cathode)- When Ve = VP, UJT conducts

heavily and capacitor dischargescompletely,

*J1is forward biased J2is reverse biased

- UJT cut off and capacitor chargesslowly again.

*Anode current is very small (leakagecurrent of J2(equivalent to switch off)

- Output across capacitor is a saw

tooth wave.

*When you increase anode volt everse bias of J2

increased and it breaksdown (Avalanche BD)

*Heavy anode current flows (and equivalent to onswitch)

*To make it switch off,wehave to reduce the

anode voltage such that anode current is verysmall less than holding Current IH

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6 a) Q)TUS-IIIDefine the terms as referred to FWR circuit.

i) PIVii) Average DC voltage

iii) RMS current

iv)

Ripple factor. ( May 2007)

A. i) PIV

Peal Inverse voltage (PIV) is the maximum voltage across the reverse biased(diode during the half cycle when diode is not conducting).

It is equal to 2 Vmfor full wave rectifier.ii) Average DC voltage (VDC)

It is the voltage measured by a DCvoltmeter across the load resistance RL.VDC= IDL. RL. & VDC= 0.636 Vmax.for FWR.

iii) RMS current ( IRMS)

If the root of mean of square of instantaneous current and given by

( )

1/22

2

m m

0

1I sin 0.707 I

2RMS

I

= =

for FWR.

iv) Ripple Factor (r)

It is defined as the ratio of RMS value of alternate current component and averagecurrent component flowing through the load resistance.

2

1rms

dc

I

I

=

and = 0.483 for FWR.

1 b) Q) (TUS-III)

A full wave rectifier circuit is fed from a transformer having a center tappedsecondary winding. The rms voltage from either end of secondary to center tap is

30V. If diode forward resistance is 5and that of secondary is 10for a load of900calculate.

i) Power delivered to the load

ii) % regulation at full loadiii) efficiency at full load

iv)

TUF of secondary (August, 2007)

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SOLUTION: -

Part (i)

Power delivered Pdc2 .

DC LI R= Idc=Imx 0.636

mI m

S f L

V

R R R=

+ + Vm = ? Vrms= 0.707 Vm or

0.707

rmsm

VV =

or30

40.43707

mV = = volts

m

2 3 2

40.43I 44.18

10 5 900

0.636; Im 0.636 44.18 28.09

(28.09 10 ) .900 710 .

m

S f L

dc

dc dc L

VmA

R R R

I mA

P I R mw

= = =+ + + +

= = =

= = =

Part (ii)

% Regulation =5 10

100 1.67%900

f S

L

R R

R

+ += =

Parr (iii)

Efficiency at full load

0.812

9000.812 79.86%

900 5 10

L

f S L

R

R R R=

+ +

= =+ +

Part (iv)TUF of Primary = 81.2% = 0.812

TUF of Secondary = 2 x 0.287 = 0.574

Average TUF =0.812 0.574

0.693

2

+=

3 b) Q) TUS-IIIA 3 K resistive load is to be supplied with a DC voltage of 300V from ac voltage

of adequate magnitude and 50Hz frequency by full wave rectification. The LCfilter is used along the rectifier. Design the Header resistance, turns ratio of the

transformer, VA rating of transformer and PIV rating of the diodes.

(August 2007)

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Solution: -VDC= 300V

RL= 3KTo find out

i) RB= ? (ii) N1:N2= ? (iii) VA rating of transformer (iv) PIV of diode(Vrms, Idc)

Part (iii) Vrms= ? Idc= ? Idc=300

1000.636

dc

L

VmA

R= =

With LC filter (FWR)2

.mdc dc

VV I R

=

Where R = rs+rf+rL where rLis the internal resistance of L

Let us assume R = 0

Then

2

0.636m

dc m

V

V V= =

Or300

471.690.636 0.636

m

m

VV volts= = =

Vrms= 0.707, Vm= 0.707 x 471.69 = 333.5 volts 330V VA rating of transformer is 220V/330 0 330V, 100 mA.PART-II

Input is 220V Ac.

Then turns ratio N1:N2 = 220 : 660 = 1:3iv) PIV of each diode in FWR = 2Vm= 2 x 471.69 = 943.38 Volts.

Part I

RB < 800 LC Where Lc = RL/ 3w = 3000/3 x 2 x x 50 = 10/= 3.18RB < 800 x 3.18 = 2544 2500.

4.b) Q)TUS-III

A HWR circuit has a filter capacitor of 1200F and is connected to a load of 400.The rectifier is connected to a 50Hz, 120 vrms source. It takes 2 m.sec. for thecapacitor to recharge during each cycle. Calculate the mini value of the repetitive

surge current for which diode should be rated. (August, 2007)

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26 EDC UNIT-III26

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Solution: -

Given:

Frequency = 50Hz. Time T = 1/f = 1/50Hz = 100050

m/sec. = 20 m/sec.

Capacitor charges to Vmand discharges through RL.

Surge current = ? m cS

S f

V VI

R R

=

+

Where Vc is voltage across the capacitor at the end of previous cycle.

Vrms =2

mV

or Vm= 2 x Vrms= 2 x 120 = 240 Volts

Vc = V(e-t/cR

) Capacitor discharges through RL.

Here V = 240 V 1/= Vm.t = T discharging = Time T charging

= 20 2 = 18 msec.CR = 1200 x 10

-6x 400 = 480 m/sec.3

3

18 10 18

480 10 480240 240

240 0.963 231.12 231

Vc e e

volts

= =

= =

Surge current = m cs

S f

V VI

R R

=

+

Where Rs& Rfare not given. Let us assume RS= 0& Rf= 25

Then240 231 9

360 .25 25

sI mA

= = =

5 b) Q)TUS-III

A FWR circuit uses two silicon diodes with a forward resistance of 20each. A DC volt meter connected across the load of 1 Kreads 55.4 volts.Calculate

i) Irms (ii) Average voltage across each diode

iii) Ripple factor (iv) Transformer secondary voltage ratings.

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27 EDC UNIT-III27

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Solution: -

Given:

Rf= 20VDC= 55.4 Volts

RL= 1 KPart I

Irms= Imx 0.707 IDC= 0.636 Im. Or Im = IDC/0.636

IDC= VDC/ RL= 55.4 / 1000 = 55.4 mA.

Im = IDC/ 0.636 = 55.4/0.636 = 87.1mA.

Irms= 0.707 x Im = 87.1 x 0.707 = 61.58 mA,Part II

Average voltage (VDC) across each diode Vd(DC) = ?

Vd(DC) = IDC.Rf= 55.4 x 10-3

x 20 = 1.1 volts

Part III

Ripple factor for FWR is 0.482.

Part IV

Transformer secondary ratings voltage are in rms volts

Vrms= 0.707 of Vm & Vm 0.636

DCV=

VDCacross one half of secondary = VRL+Vd= 55.4 + 1.1 = 56.5 volts

88.840.636

DCm

VV Volts= =

Vrms= 0.707 x 88.84 = 62.8 63 voltsTransformer secondary voltage rating is 63 0 63 Volts (rms).

3 a)Q) Derive the expression for ripple factor for FWR with L section filter.

(August, 2007)

Solution:

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28 EDC UNIT-III28

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-

Ripple factorrms

DC

V

V

=

- Object of LC filter here is to suppress the ripple frequency components i.e.,

2f.

-

To suppress 2f, the reactance offered by L must be very very high comparedto reactance offered by C. i.e., XL>> XC and R=(rf+rs+rL) is also negligible

-

Then alternating current through the filter circuit is given by

4 1.

3 2

2 2 1 2 1. . .

3 3

m

rms

L

mdc

L L

VI

X

VV

X X

=

= =

-

Ripple voltage across RL = Ripple voltage across XC=1

rmsV

-

Or

1 1

2

2 1

. . .3rms ms C dc C V I x V xx= =

-

Ripple factor1 2 2

. . .3 3

rms dc c C

dc dc L L

V V X X

V V X X = = =

-

Or2 1 1

. .3 2 2wc wL

=

-

8 a)Q) TUS-III

In a full wave single phase bridge rectifier circuit, can the transformer andload be interchanged? Justify your statement.

Solution: -Circuit Diagram of bridge rectifier with transformer and load interchanged

is given below.

If Transformer and load are interchanged as shown in the figure (b) above, for

positive half cycle, i.e., point D is positive wrt point C, current flows through D1 andD2. At the same time current also flows through D4 and D3. The potential at point A

is equal to point B and there is no current through RL and no output voltage available.

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29 EDC UNIT-III29

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The current through the diodes will be very high as diode resistance is very small.Therefore diodes may be burnt.

For negative half cycle, when point D is negative not C, all the diodes will bereverse biased and only reverse saturation current flows. Again points A and B

will be at same potential and no current flows through RL. No output voltage

available.Therefore, there will not be any output voltage if we interchange the load and thetransformer.

5 a) Q) TUS-IIIExplain about regulation characteristics of a zener diode with a circuit and

wave forms.

Characteristics of Zener Diode.

-

As shown in the graph zenerdiode works as a normal

diode when it is forwardbiased.

-

In reverse bias, when it

breaks down,voltage across itwill be constant.

-

- This phenomenon is used for

oltage regulation.Regulated out put.

As shown in the graph above, a regulated output voltage (10V) can be obtained

being a zener regulation. If the input voltage is less than zener voltage output voltage isapproximate is equal to input voltage and will be varying with input voltage. Once input

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30 EDC UNIT-III30

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voltage is more than the zener voltage (V2) the zener diode conducts heavily and voltageacross it will be constant at VZ.

Remaining voltage i.e., V in VZwill be developed across the series resistanceRs. The value of Rs is such selected that the current flowing if (which is equal to the

current flow through zener when no load) should be less than the rated (maximum

current) of zener diode.

6 b) Q. TUS-III

A full wave rectifier (FWR) supplies a load requiring 300 V at 200 mA, Calculatethe transformer secondary voltage for?

(i) a capacitor input filter using a capacitor of 10 F

RL= 300/200x10-7

= 1.5 K

300V1 200mA

10F

Where R = XS+Xf+XL

; 504

dcdc m

IV V f Hz

fc= =

2.

dc dc

VmV I R

=

3

5

200 10300

4 10 10mV

=

2300 200.0m

V

=

= Vm 100 Vm= 300 x /2 = 471Vrms= 0.707 Vm Vrms= 0.707 x 471

= 0.707 x 400V = 282.8 283 V. = 333 Volts

Transformer sec voltage = 2x283 = 566 volts Tr. rating 333 0 333 V.

7(b) Q. A full wave rectified voltage of 18 V peak is applied across a 500 F filter

capacitor. Calculate the ripple and D.C.voltages if the load takes a current of 100mA

Vm= 18V Vr= ? Vr= Idc/ Zfc.

C = 500F Vdc= ? Vdc= Vm Idc/4fc.Idc= 100 mA

3

6

100 102

2 50 500 10V V

= =

3

3

100 1018

4 50 500 10dc

V

=

= 18 1 = 17V.IV

15V Idc= ?

100 10000 Vdc= ?

F4

dcm

IV

fc

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Vrms = 0.707 Vm .4

dcdc L m

c

II R V

f=

1521.2

0.707 0.707

rms

m

VV V= = =

6.100 20.2

4 50 10000 10

dc

dc

II

=

Vdc= Idc.RL 100 Idc= 20.2 Idc/2

= 212 x 103

x 100 100 Idc=(Idc)/2 = 21.2= 21.2 Volts 100 Idc= 20.2

Idc= 21.2/100.5 = 212 mA.

9b) Q. Draw the circuit diagram of a full wave rectifier using center tapped transformerobtain an output DC voltage Vdc= 18 V at 200 mA and Vdcno load = 20V also

mention transformer rating and sketch the input and output wave forms.Solution:

VDCNL= 20V Vrms= 0.707Vm

18V1 Vrms= ? VDC= 636 Vm.200 mA. Vrms= 0.707 Vm= 0.707 x 31.45

= 22.24

11C)Q.TUS-III

In a full wave rectifier using an LC filter L = 10H, C= 100F and RL =

500 . Calculate I dc, Vdc,for an input Vi= 30 sin (100 t )Solution:

10H

IDC = ?100F Vdc= ?500

Vi= 30 sin (100 t) Vm= 30 V

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f = 50 Hz.

Vdc= 2 Vm/- Idc. R. When R = rs+ rf + rL Say R = 02 30 2

19.1

19.1

38.2 .500

mdc

dc

dcL

VV Volts

V

I mAR

= = =

= = =

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