pn juction characteristics

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    Graphical

    PN-Junction Diode V-I Characteristic

    Reverse

    breakdown

    Forward Bias Region

    Reverse Bias Region

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    Mathematical Approximation

    D

    T

    V

    V

    D sI = I (e -1)

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    Ideal PN Junction Diode V-I Characteristic

    Forward BiasShort Circuit

    Reverse BiasOpen Circuit

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    Diode Reverse Recovery Time

    tais the time to remove the charge stored in the

    depletion region of the junction

    tbis the time to remove the charge stored in the bulk

    semiconductor material

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    Reverse Recovery Characteristics

    Soft Recovery

    Reverse recovery time = trr= ta+tb

    Peak Reverse Current =IRR= ta(di/dt)

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    Reverse Recovery Characteristics

    Abrupt Recovery

    Reverse recovery time = trr= ta+tb

    Peak Reverse Current = IRR= ta(di/dt)

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    Diode Characteristics

    Due to differences

    between devices,

    each diode has a

    different voltageacross it.

    Would like to

    Equalize the

    voltages.

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    Series-Connected Diodes with

    Voltage Sharing Resistors

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    Series-Connected Diodes with

    Voltage Sharing Resistors

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    Example 2.3

    Is1= 30mA, Is2= 35mA

    VD= 5kV

    (a)R1=R2=R=100k,

    find VD1and VD2

    (b)Find R1and R2for VD1=VD2=VD/2

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    Example 2.3 (a)

    s1

    s2

    1 2

    D D1 D2

    D2 D D1

    D1 D2s1 s2

    DD1 S2 S1

    -3 -3

    D1

    D2 D D1

    I = 30mA

    I = 35mA

    R = R = R = 100k

    -V = -V - V

    V = V - V

    V VI + = I +

    R R

    V RV = + (I - I )2 2

    5kV 100kV = + (3510 - 3010 ) = 2750Volts

    2 2

    V = V - V = 5kV - 2750 = 2250Volts

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    Example 2.3 (a) simulation

    D1

    DIODE_VIRTUAL*

    D2

    DIODE_VIRTUAL**

    R1

    100kOhm

    R2

    100kOhm

    U1

    DC 1MOhm-2.727k V+

    -

    U2

    DC 1MOhm-2.273k V+

    -

    V1

    5000 V

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    Example 2.3 (b)

    s1

    s2

    DD1 D2

    D1 D2

    s1 s2

    1 2

    D2 12

    D1 1 s2 s1

    1

    2 -3 -3

    2

    I = 30mA

    I = 35mA

    VV = V = = 2.5kV

    2

    V V

    I + = I +R R

    V RR =

    V - R (I -I )

    R =100k

    2.5kV100kR =

    2.5kV -100k(3510 - 3010 )

    R =125k

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