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  • 2/7/2011

    1

    1

    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

    2

    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

  • 2/7/2011

    2

    EE3601-06Electronics Circuit Design

    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

    EE3601-06Electronics Circuit Design

    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

  • 2/7/2011

    3

    EE3601-06Electronics Circuit Design

    5

    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

    EE3601-06Electronics Circuit Design

    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

  • 2/7/2011

    4

    EE3601-06Electronics Circuit Design

    7

    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

    Peak Inverse Voltage (PIV) of the diode

    n:1+1

    220V50Hz

    Vsm

    RS

    RS

    RL

    IrmsIavIm

    Vsm

    n:1+1

    220V50Hz

    Vsm

    RS

    RS

    RL

    IrmsIavIm+

    +

    +

    -

    -

    + -

    PIV

    Vsm+- -

    neglected

    Vm=ImxRL

    EE3601-06Electronics Circuit Design

    8

    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

    Peak Inverse Voltage (PIV) of the diode

    VP

    n :1

    220V50Hz Vsm

    RL

    IrmsIdc

    RS

    Im

    n :1

    220V50Hz Vsm

    RL

    IrmsIdc

    RS

    Im+

    -

    PIV

    PIV

    -+

    +

    -

    neglected

  • 2/7/2011

    5

    EE3601-06Electronics Circuit Design

    9

    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

    EE3601-06Electronics Circuit Design

    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

  • 2/7/2011

    6

    EE3601-06Electronics Circuit Design

    11

    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

    EE3601-06Electronics Circuit Design

    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

  • 2/7/2011

    7

    EE3601-06Electronics Circuit Design

    13

    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

    EE3601-06Electronics Circuit Design

    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 seco