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Bipolar Junction Transistor Principles

Apr 03, 2018

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Sagar Shah
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    1

    Bipolar Junction Transistors (BJT)

    A Bipolar Transistoressentially consists of a pair ofPN Junction

    Diodes that are joined back-to-back.

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    3

    p n

    BJT operation

    Reverse polarity is appliedto the right diode

    There are still no electronson the p-side of the diode(they have all recombinedat the left junction).

    The current is nearly zero.

    Forward polarity isapplied to the left diode.

    There is a high current

    flowing through it.The electrons enteringp-region recombine with

    holes to form a currentloop.

    pn

    Battery 1 Battery 2

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    BJT operation

    Forward polarity is applied to the left diode. Electrons from

    the left n-region get injected into the p-layer driven by apositive terminal of the battery 1. The p-layer being very thin, most of the injected electrons

    have low chances to recombine; they reach the 2ndjunction

    and drift toward positive terminal of the battery 2. Therefore high current flows through 2nd p-n junction.

    nn

    Battery 1 Battery 2

    The P-layer is made VERY THIN

    p

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    nn p

    n-p-n bipolar transistor

    Device structure Circuit symbol

    Emitter Base Collector

    p-n-p bipolar transistorDevice structure Circuit symbol

    pp n

    Emitter Base Collector

    E

    B

    C

    E

    B

    C

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    6

    nn pe

    b

    c

    Electron and hole fluxes in n-p-n transistor:

    (1) electrons injected from emitter into the base and reaching the collector junction (as large as

    possible);

    (2) holes supplied by the base contact to initiate electron injection (as small as possible);

    (3) injected electrons lost to recombination in the base (as small as possible);

    (4) holes injected from the base into emitter recombining with electrons (as small as possible).

    1

    2

    3

    4

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    7

    pp ne

    b

    c

    Electron and hole fluxes in p-n-p transistor:

    (1) holes injected from emitter into the base and reaching the collector junction;

    (2) electrons supplied by the base contact to initiate hole injection;

    (3) injected holes lost to recombination in the base;

    (4) electrons injected from the base into emitter recombining with holes.

    1

    2

    3

    4

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    8

    pp ne

    b

    c

    Carrier concentrations and Terminal Currents

    (p-n-p transistor case):

    Hole concentration at the e-b junction:0 01

    beqV kT e b bp p e p

    /( ) = >>

    p(x)

    x

    pe

    Hole concentration at the b-c junction (Vcb

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    9

    The base width W is larger than that of the e-b and c-b depletion regions.

    Therefore, the electric field in most of the base, E 0. Therefore, the hole current (1) through the base is mainly the diffusion current:

    pp ne

    b

    c1

    2

    3

    4

    p(x)

    x

    The current (1) is nearly constant throughout the base because therecombination current (3) is very small.

    Therefore:

    dp/dx =const;The line that has a constant slope is a straight line.

    p pp

    I Aq Dx

    =

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    10

    pp ne

    b

    c

    Terminal Currents : Collector and Emitter currents

    (p-n-p transistor case):

    Assuming small recombination loss in the base,

    p(x)

    x

    pe

    pcW0

    ec e p

    ppI I Aq D Aq D

    x W

    W is the BJT base thickness

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    11

    pp ne

    b

    c

    Terminal Currents : Base recombination current

    (p-n-p transistor case):

    Total charge of the holes injected into the base (the area under the p(x) line:

    p(x)

    x

    pe

    pcW0

    2

    ep

    p WQ Aq

    =

    This charge reproduces due to recombination; the required time is the hole lifetime p

    Therefore, the base recombination current, Ibrec = Q/t:

    2e

    brecp

    p WI Aq

    =

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    12

    Current gain (recombination limited)

    Collector current:

    2

    ebrec

    p

    p WI Aq

    =

    ec p

    pI Aq D

    W

    Base recombination current:

    Collector-base current gain (limited by recombination only) rec = Ic/Ibrec:

    2

    22

    e

    p p p pcrec

    brec e

    pAq D

    DI WI Aq p W W

    = =

    The diffusion length: p p pL D = Therefore:2

    2

    2 p

    rec

    L

    W =

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    13

    pp ne

    b

    c

    Emitter injection efficiency

    Electrons injected from the base into emitter the current component (4), increase total base

    and emitter currents but DO NOT contribute into the collector current as they do not

    increase the injection.

    The ratio of the useful emitter current (the hole current in p-n-p device) over the total

    emitter current is called emitter injection efficiency; for a p-n-p transistor:

    1

    2

    3

    4

    ep

    ep en

    II I

    =+

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    Emitter injection efficiency (cont.)

    For a p-n-p transistor:

    1

    1

    ep

    enep en

    ep

    I

    II I

    I

    = =+

    +

    pp ne

    b

    c1

    2

    3

    4

    pp ne

    b

    cpp n

    e

    b

    c11

    22

    33

    44

    The electron and hole currents are mainly the diffusion currents:

    For the hole current in the base:Iep qA peb Dp/W;

    The electron current from the base into the emitter:Ien qA nbe Dn/Ln;The current ratio:

    en be n

    ep eb p n

    I n D W

    I p D L

    =

    must be as small as possible for the

    highest injection efficiency

    2

    0qV kT qV kT i

    be pAe

    nn n e e

    N

    / / = =

    2

    0qV kT qV kT i

    eb nDb

    n

    p p e eN

    / /

    = =

    1be Db

    eb Ae

    n N

    p N;

    =

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    Emitter injection efficiency (cont.)

    1b

    e

    N

    N;

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    BJT current gain

    Current gain limited by recombination in the base:

    2

    2

    2 prec

    L

    W =

    0pDe n

    Ab p

    LN DN D W

    =

    Current gain limited by injection efficiency:

    0 1

    = 0 is related to emitter injection efficiency as

    The overall current gain:

    0

    1

    1 1rec

    =

    +

    p-n-p BJT: n-p-n BJT:2

    2

    2 nrec

    L

    W =

    0pAe n

    Db n

    DN LN D W

    =p-n-p BJT: n-p-n BJT: