jueves, 28 de enero de 2010

Bipolar Junction Transistor (BJT)

 Bipolar Junction Transistor (BJT) has three terminals connected to three doped semiconductor regions. In an npn transistor, a thin and lightly doped p-type material is sandwiched between two thicker n-type materials; while in a pnp transistor, a thin and lightly doped n-type material is sandwiched between two thicker p-type materials. In the following we will only consider npn BJTs.
transistors1.gif
transistorBJT1.gif
In many schematics of transistor circuits (especially when there exist a large number of transistors in the circuit), the circle in the symbol of a transistor is omitted.
transistorBJT2a.gif
transistorBJT2b.gif
The three terminals of a transistor are typically used as the input, output and the common terminal of both input and output. Depending on which of the three terminals is used as common terminal, there are three different configurations: common emitter (CE), common base (CB) and common collector (CC). The common emitter (CE) is the most typical configuration:
transistors2.gif

  • Common-Base (CB)Two voltages $V_{BE}$ and $V_{CB}$ are applied to the emitter $E$ and collector $C$ of the transistor with respect to the common base $B$. Te BE junction is forward biased while the CB junction is reverse biased.
    CB.gif
    CBnpn.gif
    The behavior of the npn-transistor is determined by its two pn-junctions:


    • The forward biased base-emitter (BE) junction allows the free electrons to flow from the emitter through the PN junction to form the emiiter current $I_E$.
    • As the p-type base is thin and lightly doped, most electrons from the emitter $\alpha I_E$ (e.g. $\alpha \approx 0.99$) go through the base to reach the collector-base junction, only a small number of the electrons are combined with the holes in base to form the base current $I_B=(1-\alpha)I_E$.
    • The reverse biased collector-base junction blocks the majority carriers (holes in the p-type base, electrons in n-type collector), but lets through the minority carriers, electrons in base and holes in collector, including most of the electrons from the emitter $I_{CN}=\alpha I_E$, and the reverse saturate current of the CB junction $I_{CP}=I_{CB0}$,
    The relationship between the output $I_C$ and the input $I_E$ can be found as:


    \begin{displaymath}I_C=I_{CN}+I_{CP}=\alpha I_E+I_{CB0}\approx \alpha I_E \end{displaymath}


    The base current $I_B$ is the small difference between two nearly equal currents $I_E$ and $I_C$:


    \begin{displaymath}I_B=I_E-I_C\approx I_E-\alpha I_E=(1-\alpha)I_E \end{displaymath}


  • Common-Emitter (CE)Two voltages $V_{BE}$ and $V_{CE}$ are applied to the base $B$ and collector $C$ of the transistor with respect to the common emitter $E$. The BE junction is forward biased while the CB junction is reverse biased. The voltages of CB and CE configurations are related by:


    \begin{displaymath}V_{CE}=V_{CB}+V_{BE} \end{displaymath}


    CE.gif
    CEnpn.gif
    The input current is $I_B$, $I_E=I_B+I_C$, and the output current is


    \begin{displaymath}I_C=\alpha I_E+I_{CB0}=\alpha (I_C+I_B) + I_{CB0} \approx \alpha (I_C+I_B) \end{displaymath}


    Solving for $I_C$, we get the relationship between the output $I_C$ and the input $I_B$:


    \begin{displaymath}I_C=\frac{\alpha}{1-\alpha} I_B+\frac{1}{1-\alpha} I_{CB0} =... ...a I_B +(\beta+1)I_{CB0}=\beta I_B + I_{CE0} \approx \beta I_B \end{displaymath}


    Here $\beta\stackrel{\triangle}{=}\alpha/(1-\alpha)$ is the current-transfer ratio for CE (e.g., $\alpha=0.99$ and $\beta=99$), and $I_{CE0}=(\beta+1) I_{CB0}$ is the reverse saturation current between collector and emitter. In summary:


    \begin{displaymath}\left\{ \begin{array}{l} I_C=\beta I_B\ I_E=I_C+I_B=(\beta+1) I_B \end{array} \right. \end{displaymath}


    Source: http://fourier.eng.hmc.edu/e84/lectures/ch4/node3.html
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