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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
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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
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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
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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
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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
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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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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
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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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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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= 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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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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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
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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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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
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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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22 EDC UNIT-III22
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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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23 EDC UNIT-III23
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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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24 EDC UNIT-III24
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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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25 EDC UNIT-III25
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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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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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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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31 EDC UNIT-III31
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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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32 EDC UNIT-III32
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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