Dioda Penyearah - ELDAS

2/7/2011

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6 Unregulated Power Supply Design

6.1 Rectifier Circuit Classifications

6.2 Half-wave Rectifier Circuits Design

6.3 Full-wave Rectifier Circuits Design

6.4 Bridge Rectifier Circuits Design

6.5 Power Supply Classifications

6.6 Design equation of Power Supply with filter capacitor

6.7 Power Supply Circuits Design

EE3601-06Electronics Circuit Design

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6.1 Rectifier Circuit Classifications

Im

D

Im

D1 D2

Im

D1D2

D3D4

n:1

220V

50Hz RLVSm

RS Im Idc Irms

D

half-wave rectifier circuit

n:1+1

220V50Hz

Vsm

RS

RS

RL

IrmsIavImD1

D2Vsm

full-wave rectifier circuit

bridge rectifier circuit

n :1

220V50Hz Vsm

RL

IrmsIdc

RS

ImD1

D2 D3

D4

rectifier = diode is used to convert input ac into output dc

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3

rms value of half-wave rectified sine wave

average (dc) value of half-wave rectified sine wave

mm0

m

0

mavdcV

112

Vcos

2

VdsinV

2

1VV

2

V

2

V02sin

2

10

2

V

0sin2

12sin

2

10

2

V2sin

2

1

2

V

d2cosd12

Vd

2

2cos1

2

V

dsin2

VdsinV

2

1V

mm2

1

m

2

1

m2

1

00

m

2

1

00

m2

1

0

m

2

1

0

2m

0

22mrms

Vm=ImxRL

half-wave rectified sine wave

Ls

LsmLmm

Ls

smm RR

R7.0VRIV

RR

7.0VI

n:1

220V

50Hz RLVSm

RS Im Idc Irms

D

rms and average (dc) values of rectified sine wave


4

2

V

2

V02sin

2

10

2

V

0sin2

12sin

2

10

2

V2sin

2

1

2

V

d2cosd12

Vd

2

2cos1V

dsinV

dsinV1

V

mm2

1

m

2

1

m2

1

00

m

2

1

00

m2

1

0

m

2

1

0

2m

0

22mrms

rms value of full-wave rectified sine wave

average (dc) value of full-wave rectified sine wave

mm0

P

0

mavdcV2

11V

cosV

dsinV1

VV

n :1

220V50Hz Vsm

RL

IrmsIdc

RS

Im

Ls

LsmLmm

Ls

smm RR

R4.1VRIV

RR

7.02VI

Vm=ImxRL

full-wave and bridge rectified sine wave

n:1+1

220V50Hz

Vsm

RS

RS

RL

IrmsIavIm

Vsm

Ls

LsmLmm

Ls

smm RR

R7.0VRIV

RR

7.0VI

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Besarnyategangan DC yang dayanyasetara dengantegangan AC

Nilai teganganDC hasil

penyearah diodapada sebuah

tegangan AC.

Vm

Vm

Vm

t

t

t

rms and average (dc) values of rectified sine wave

2

Vm0Sine wave

2

Vm mVHalf-wave

2

Vm mV2Full-wave

rms dcwave


6

6.2 Half-wave Rectifier Circuits Design

half-wave rectifier circuits

n:1

220V50Hz RLVsm

RS Im=0

-

+

- +PIV

n:1

220V50Hz RLVsm

RS Im

+

- PIV of the diode is found across the diode when the diode is not conducting here in half-wave rectifier, PIV = Vsm

Peak Inverse Voltage (PIV) of the diode

Vm=ImxRLn:1

220V50Hz RL

RS Im

Vsm sin =Vsm sin t

IdcIrms

Ls

LsmLmm

Ls

smm RR

R7.0VRIV

RR

7.0VI

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6.3 Full-wave Rectifier Circuits Design

full-wave rectifier circuits

PIV of the diode is found across the diode when the diode is not conducting here in full-wave rectifier, if the voltage drop due to Rs and diode are neglected, PIV = 2Vsm


n:1+1

220V50Hz

Vsm

RS

RS

RL

IrmsIavIm

Vsm

n:1+1

220V50Hz

Vsm

RS

RS

RL

IrmsIavIm+

+

+

-

-

+ -

PIV

Vsm+- -

neglected

Vm=ImxRL


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6.4 Bridge Rectifier Circuits Design

bridge rectifier circuits

PIV of the diode is found across the diode when the diode is not conducting here in bridge rectifier, if the voltage drop due to Rs and diode are neglected, PIV = Vsm


VP

n :1

220V50Hz Vsm

RL

IrmsIdc

RS

Im

n :1

220V50Hz Vsm

RL

IrmsIdc

RS

Im+

-

PIV

PIV

-+

+

-

neglected

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Summary of Design EquationsRectifier Circuits

n:1

220V

50Hz RLVSm

RS Im Idc Irms

D

half-wave rectifier circuit

n:1+1

220V50Hz

Vsm

RS

RS

RL

IrmsIavImD1

D2Vsm

full-wave rectifier circuit

bridge rectifier circuit

n :1

220V50Hz Vsm

RL

IrmsIdc

RS

ImD1

D2 D3

D4

Ls

LsmLmm

Ls

smm RR

R4.1VRIV

RR

7.02VI

Ls

LsmLmm

Ls

smm RR

R7.0VRIV

RR

7.0VI

PIV = Vsm PIV = 2Vsm

PIV = Vsm

Vm

Vm

Vm

t

t

t

2

Vm0Sine wave

2

Vm mVHalf-wave

2

Vm0Sine wave

2

Vm mVHalf-wave

2

Vm mV2Full-wave

Ls

LsmLmm

Ls

smm RR

R7.0VRIV

RR

7.0VI


10

Design of half-wave rectifier-1

V4.141VPIV

299.1n7.141

2220

V

2220

2

2

)rms(V

)rms(V220

1

n

V7.1417.0)9010(41.1V9010

V7.0VA41.1I

A41.145.0II

A45.0I

A45.090

V5.40IV5.4090IV

sm

sms

smsm

m

mm

av

avavav

A half-wave rectifier is to deliver an average voltage of 40.5V to a dc load of RL=90 from an ac supply of 220V, 50Hz. Design the transformer turn ratio n:1 , PIV of the diode and average current rating of the diode. Assume that the secondary winding resistance of the transformer is 10

n:1

220V

50Hz RLVsm

RS Im Idc Irms10

90

Vdc=

40.5V

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Design of half-wave rectifier-2

A half-wave rectifier is to heat a resistive load of RL=90 with a power of 44.1W from an ac supply of 220V, 50Hz. Design the transformer turn ratio n:1 , PIV of the diode and average current rating of the diode. Assume that the secondary winding resistance of the transformer is 10

V7.140VPIV

21.2n7.140

2220

V

2220

)rms(V

)rms(V220

1

n

V7.1407.0)9010(4.1V9010

V7.0VA4.1I

A446.04.1I

I

A4.17.02IA7.02

II

A7.090

1.44IW1.4490IP

sm

sms

smsm

m

mav

mm

rms

rms2rmsL

n:1

220V

50Hz RLVsm

RS Im Idc Irms10

90

PRL=

44.1W


12

Design of full-wave rectifier-1

A full-wave rectifier is to heat a resistive load of RL=90 with a power of 44.1W from an ac supply of 220V, 50Hz. Design the transformer turn ratio n:1+1 , PIV of the diode and average current rating of the diode. Assume that the secondary winding resistance of the transformer is 10

V4.2017.1002V2PIV

09.3n7.100

2220

V

2220

)rms(V

)rms(V220

1

n

V7.1007.0)9010(1V9010

V7.0VA1I

A64.02I2

I

A17.02IA7.02

II

A7.090

1.44IW1.4490IP

sm

sms

smsm

m

mav

mm

rms

rms2rmsL

n:1+1

220V50Hz

Vsm

RS

RS

RL

IrmsIavIm

Vsm

10

1090

PRL=44.1W

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Design of full-wave rectifier-2

A full-wave rectifier using a full average current rating of the diode of 5A is to deliver a heating power of 1KW to a resistive load of RL from an ac supply of 220V, 50Hz. Design the value of RL , transformer turn ratio n:1+1 , PIV of the diode. Assume that the secondary winding resistance of the transformer is 1

V72.52636.2632V2PIV

18.1n46.263

2220

V

2220

)rms(V

)rms(V220

1

n

V36.2637.046.3385.7V46.321

V7.0VA85.7I

46.3255.5

1000RRIPKW1

A55.52

85.7

2

IIA85.7

2

5I

I2I5

sm

sms

smsm

m

2LL2rmsL

mrmsm

mav

n:1+1

220V50Hz

Vsm

RS

RS

RL

Irms

Idc=5A

Im

Vsm

1

1

PRL=1kW


14

Design of bridge rectifier

A bridge rectifier using a full average current rating of the diode of 5A is to deliver a heating power of 1KW to a resistive load of RL from an ac supply of 220V, 50Hz. Design the value of RL , transformer turn ratio n:1 , PIV of the diode. Assume that the secondary winding resistance of the transformer is 1

V264VPIV

178.1n264

2220

V

2220

)rms(V

)rms(V220

1

n

V2644.146.3385.7V46.321

V4.1VA85.7I

46.3255.5

1000RRIPKW1

A55.52

85.7

2

IIA85.7

2

5I

I2I5

sm

sms

smsm

m

2LL2rmsL

mrmsm

mav

n :1

220V50Hz Vsm

RL

Irms

Idc=5ARS

Im

1

1kW

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6.5 Power Supply Classifications

Vm

n:1

220V50Hz RLVm

Vm Vdc V

C

+

-

half-wave power supply circuit

n:1+1

220V50Hz

Vm

RL

VVdcVm

Vm

+

-

C

full-wave power supply circuit

bridge power supply circuit

n :1

220V50Hz Vm

RL

VVdcVm

C

DC Power Supply = filter C is used to smooth the output dc from the rectifier

Vm

Vm


16

6.6 Design equation of Power Supply with filter capacitor

n:1

220V

50Hz

VSm

VSm

RLC

VO

t1t2

Vmin

Vsm=Vmax

V

T

V= - m t V= m tVO

CRV

t

v)m(isslopeeargDisch

)CRtif(CRtVeVvisequationeargDisch

L

maxc1

LL

maxCR

t

maxcL

max

L11

L

max111

L

max1 V

CVRtt

CR

VtmVttatBut

CR

Vm

2T

V

Vt

2T

tVtmVand

2T

Vm

max2

2max22

max2

21 tTtmaxmaxmax

L

V

V2

2

T

V2

TVT

V

CVR

maxpmaxpmaxmax

L V2Vf

1

V2

V1

f

1

V

V2

2

T

V

CVRor

equationdesignCapacitorRVf

VC

Lp

max

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Ripple factor of the dc output voltage

T/2

t1 t2

Vmin

Vmax

V

T/2

t

VO

VdcRippledc

Ripple factor of the dc output voltage is the amount of rms. voltage of the ripple with respect to the dc output voltage

VV23

V

VV22

32

V

V

Vr

maxmaxdc

rms

32

VV

32

V

3

VV

isvalue.rmswhose,wavetriangularaisripple

minmaxprms

2

VV2

2

VVV

2

VVV maxmaxmaxminmaxdc

dc

rmsV

)ripple(Vr


18

Summary of Design EquationsPower Supply Circuits

n:1

220V50Hz RLVm

Vm Vdc V

C

+

-

half-wave power supply circuit

n:1+1

220V50Hz

Vm

RL

VVdcVm

Vm

+

-

C

full-wave power supply circuit

bridge power supply circuit

n :1

220V50Hz Vm

RL

VVdcVm

C

equationdesignCapacitorRVf

VC

Lp

max

)wavehalf(f)wavefull(ff ssp 2

dc

rms

V

)ripple(Vr

32323

minmaxprms

VVVVV

isvalue.rmswhose,wavetriangularaisripple

2

2

22

VVVVVVVV maxmaxmaxminmaxdc

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6.7 Power Supply Circuits Design

Draw the full-wave power supply with filter capacitor. Design the size of the capacitor for the following conditions. Find Ripple factor of the dc output voltage. Transformer turn ratio is 1:2 (1:1+1) with 100V, 60Hz. ac at the primary winding. Load resistor RL=2k Required minimum output voltage is 70V.

Design Example 1

1:2

100V

60Hz

VS

VS

RL

C

VO

100V

100V

F23.81021204.71

4.141RVf

VC

3Lp

max

)wavefull(Hz120602f2f sp

%8.19198.05.10588.20

ripple

V88.2032

4.71rmsrippleV4.71V

V5.105270141

Vdc

V4.71704.141V

,V70V,V4.1412100V

V2

2100

V2

V100

2

1

1

n

minsm

sms


20

Draw the half-wave power supply with filter capacitor. Design the size of the capacitor for the following conditions. Transformer turn ratio is 1:1 with 100V, 60Hz. ac at the primary winding. Load resistor RL=2k The required ripple factor is 10% what is the dc voltage output?

Design Example 2

1:1

100V

60Hz RL100V C

1.0VV23

V

VV22

32

V

V

Vr

V4.1412100V

maxmaxdc

rms

max

76.41V1.0V4.14123

V

F2.281026076.41

4.141RVf

VC

3Lp

max

V52.1202

76.414.1414.1412

VVV minmaxdc

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