http://www.ecse.rpi.edu/~schubert/Course-ECSE-6290%20SDM-2/1%20BJT-2%20Basics.pdf (More Basics about Transistors)

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

Bipolar Junction Transistor (BJT)

A 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.

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.

All previously considered components (resistor, capacitor, inductor, and diode) have two terminals (leads) and can therefore be characterized by the single relationship between the current going through and the voltage across the two leads. Differently, a transistor is a three-terminal component, which could be considered as a two-port network with an input-port and an output-port, each formed by two of the three terminals, and characterized by the relationships of both input and output currents and voltages.
Depending on which of the three terminals is used as common terminal, there can be three possible configurations for the two-port network formed by a transistor: common emitter (CE), common base (CB), and common collector (CC). Here we only consider CE and CB, as CC is not widely used.

Common-Base (CB)Two voltages $V_{BE}$ and $V_{CB}$ are applied to the emitter

and collector

of the transistor with respect to the common base

. The BE junction is forward biased while the CB junction is reverse biased.

The behavior of the NPN-transistor is determined by its two PN-junctions:

The forward biased base-emitter (BE) PN-junction allows the free electrons to flow from the emitter through the PN-junction to form the emitter current .
As the P-type base is thin and lightly doped, only a small number of the electrons from the emitter are combined with the holes in base to form the base current , while most of the electrons go through the base to reach the collector-base junction to form the collector current .The ratio of the output current and the input current is defined as the CB current gain or current transfer ratio:

$\begin{displaymath}\alpha =\frac{I_C}{I_E}<1 \end{displaymath}$

For example, $\alpha=0.99$ indicates that $99\%$ of the electrons from the emitter arrive at the collector to form while only $1\%$ combined with the hole in the base to form ,
The reverse biased collector-base (CB) PN-junction blocks the majority carriers (holes in the P-type base and electrons in N-type collector), but lets the minority carriers to go through, including the free electrons in the base coming from the emitter $\alpha I_E$ , and the reverse saturation current of the collector-base PN-junction $I_{CB0}$ (much smaller than $\alpha I_E$ .

The relationship between the output

and the input

can be found as:

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

The base current

is:

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

In summary:

$\begin{displaymath}\left\{ \begin{array}{l} I_C=\alpha I_E I_B=I_E-I_C=(1-\alpha)I_E \end{array} \right. \end{displaymath}$

Input characteristics:This is the relationship between and $V_{BE}$ both associated with the input port:

$\begin{displaymath}I_E=f(V_{BE}, V_{CB})\approx f(V_{BE})=\frac{1}{1-\alpha} I_B =\frac{1}{1-\alpha} I_0 ( e^{V_{BE}/V_T}-1 ) \end{displaymath}$

Here

$\begin{displaymath}I_B=I_0 ( e^{V_{BE}/V_T}-1 ) \end{displaymath}$

as the EB junction is essentially the same as a forward biased diode. Also, higher $V_{CB}>0$ can slightly increase .
Output characteristics:This is the relationship between and $V_{CB}$ both associated with the output port:

$\begin{displaymath}I_C=f(I_E,V_{CB})\approx f(I_E)=\alpha I_E,\;\;\;\;\;\;\;\mbox{(in linear region)} \end{displaymath}$

When $V_{CB}>0$ , i.e., the CB junction is reverse biased, the current depends totally on . When , $I_C=I_{CB0}$ is the current caused by the minority carriers crossing the PN-junction. This is similar to the diode current-voltage characteristics seen before, except both axes are reversed (as both $I_{CB}$ and are defined in the opposite directions). When is increased, $I_C=\alpha I_E+I_{CB0}$ is increased correspondingly. Higher $V_{CB}$ can slightly increase .As $I_C=\alpha I_E<I_E$ , CB configuration does not have current-amplification effect. However, if $V_{BE}$ is held constant, and therefore will also be held constant, i.e., CB transistor circuit can be used as a current source.

Common-Emitter (CE)Two voltages $V_{BE}$ and $V_{CE}$ are applied to the base

and collector

of the transistor with respect to the common emitter

. The BE junction is forward biased while the CB junction is reverse biased. The voltages of CB and CE configurations are related by:

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

The input current is

, 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 this equation for

, we get the relationship between the output

and the input

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

Here the ratio of the output current

and the input current

is defined as the CE current gain or current transfer ratio:

$\begin{displaymath}\beta=\frac{I_C}{I_B}=\frac{I_E-I_B}{I_B}=\frac{I_E}{I_B}-1 =\frac{1}{1-\alpha}-1=\frac{\alpha}{1-\alpha} \end{displaymath}$

and $I_{CE0}=(\beta+1) I_{CB0}$ is the reverse saturation current between collector and emitter. In summary:

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

The CB and CE current gains are related by:

$\begin{displaymath}\beta=\frac{\alpha}{1-\alpha},\;\;\;\;\;\;\alpha=\frac{\beta}... ...+\beta=\frac{1}{1-\alpha},\;\;\;\;\;1-\alpha=\frac{1}{1+\beta} \end{displaymath}$

For example, if $\alpha=0.99$ , then $\beta=0.99/(1-0.99)=99$ .

Input characteristics:Same as in the case of common-base configuration, the EB junction of the common-emitter configuration can also be considered as a forward biased diode, the current-voltage characteristics is similar to that of a diode:

$\begin{displaymath}I_B=f(V_{BE},V_{CE})\approx f(V_{BE})=I_0 ( e^{V_{BE}/V_T}-1 ) \end{displaymath}$

$V_{CE}$ has little effect on .
Output characteristics:

$\begin{displaymath}I_C=f(I_B,V_{CE})\approx f(I_B)=\beta I_B\;\;\;\;\;\;\;\;\;\;\mbox{(in linear region)} \end{displaymath}$

Higher $V_{CE}>0$ can slightly increase .The CB junction is reverse biased, the current $I_C=\beta I_B+I_{CE0}= \beta I_B+(\beta+1)I_{CB0}$ depends on the current . When , $I_C=I_{CE0}$ , the current caused by the minority carriers crossing the PN-junctions. When is increased, is correspondingly increased by $\beta$ fold.

The collector characteristics of the common-base (CB) and common-emitter (CE) configurations have the following differences:

In CB circuit $I_C=\alpha I_E$ is slightly less than

, while in CE circuit $I_C=\beta I_B$ is much larger than

In CB circuit,

when $V_{CB}=0$ ; while in CE circuit

when $V_{CE}=0$ . This is because in the CE circuit, when $V_{CE}=0$ , $V_{CB}=-V_{BE}$ has the effect of suppressing

Increased $V_{CE}$ will slightly increase $\alpha$ but more greatly increase $\beta=\alpha/(1-\alpha)$ , thereby causing more significantly increased

is determined by two variables $V_{CE}$ and

. When $V_{CE}$ is small ( $\approx 0.3V$ ), its slight increase will cause significant increase of

. But when $V_{CE}>0.3V$ , its further increase will not cause much change in

due to saturation (all available charge carriers arrive at collector C), and $I_C=\beta I_B$ is mostly determined by

Various parameters of a transistor change as functions of temperature. For example, $\beta$ increases along with temperature.
Load line and DC operating point
The DC operating point (also known as bias point, quiescent point, or Q-point) is the steady-state operating condition of a transistor with no AC input signal applied.
A typical CE circuit is shown in the figure below, where

, $V_{in}=V_{BE}=V_B$ , and $V_{out}=V_{CE}=V_C$ .

The input current

and voltage $V_{BE}$ can be determined by the base characteristics of the PN-junction between base B and emitter E, together with the external circuit including the voltage source $V_{CC}$ and resistor

satisfying the equation $V_{BE}=V_{CC}-I_C R_B$ , represented by the load line as shown in the figure below. This load line can be found as the straight line that passes through two special points corresponding to the open-circuit voltage and short-circuit current:

$\begin{displaymath} (I_B=0,\;V_{BE}=V_{CC})\;\;\;\;\;\mbox{and}\;\;\;\;\;(I_B=V_{CC}/R_B,\;V_{BE}=0) \end{displaymath}$

The actual current

and $V_{BE}$ can be found at the intersection of the two curves, so that both the internal current-voltage characteristics of the transistor and the external circuit parameters are both satisfied. The voltage $V_{BE}$ can also be typically approximated to be

when

is within a certain range of typical

values in practice.

Similarly, the output current

and voltage $V_{CE}$ can also be determined by both the output characteristics of the transistor and the external circuit including the voltage source $V_{CC}$ and resistor

, satisfying the equation $V_{CE}=V_{CC}-I_C R_C$ , represented by the straight line that passes through two special points corresponding to the open-circuit voltage and short-circuit current:

$\begin{displaymath} (I_C=0,\;V_{CE}=V_{CC})\;\;\;\;\;\mbox{and}\;\;\;\;\;(I_C=V_{CC}/R_C,\;V_{CE}=0) \end{displaymath}$

The actual current

and voltage $V_{CE}$ , called the DC operating point or -point, can be obtained as the intersection of the load line and the curve of the current-voltage characteristics, corresponding to a given base current

The output characteristic plot of the transistor can be divided into three regions:

cut-off region: $V_{BE}<0.7V$ , , $I_C=I_{CE0} \approx 0$ , $V_C=V_{CC}$ , i.e., the transistor (between collector and emitter) is cut off (immediate above the horizontal axis of the output plot).
linear region: $V_{BE}\approx 0.7V$ , , $I_C=\beta I_B$ , i.e., the output current is proportional to the input current , $V_C=V_{CC}-R_C I_C$ .
saturation region: $V_{BE}$ is further increased and so is , $I_C=\beta I_B$ approaches its maximum $V_{CC}/R_C$ . As can never exceed this value, it is no longer proportional to , i.e., $I_C<\beta I_B$ , and $V_{CE} \approx 0.2V$ independent of (to the immediate right of the vertical axis of the output plot).

Example: In the CE circuit shown above, $V_{CC}=12V$ , $R_B=6 K\Omega$ , $R_C=2 K\Omega$ , $\beta=60$ . The load line can be determined by two points: $(V_{CE}=0,\;I_C=V_{CC}/R_C=6\;mA)$ and $(I_C=0,\;V_{CE}=V_{CC}=12\;V)$ . Find output voltage $V_{out}=V_{CE}$ when $V_{in}$ takes the following values:

$V_{BE}=V_{in}=0<0.7V$ , and $I_C=\beta I_B=0$ , $V_C=V_{CC}=12\;V$ , the transistor is cut off.
$V_{in}=1V$ , $V_{BE}=0.7\;V$ , and $I_B=(V_{in}-V_{BE})/R_B=(1-0.7)/6=0.05\;mA$ . $I_C=\beta I_B=60\times 0.05\;mA=3\; mA$ . $V_{CE}=V_{CC}-I_C R_C=12\;V-3\;mA \times 2\;K\Omega=6\;V$ .The transistor is working in linear region.
$V_{in}=2V$ . , $I_C=\beta I_B=13\;mA$ , and $V_{CE}=12V-13mA\times 2K\Omega=-14\;V$ .We get this unreasonable negative voltage $V_{CE}=-14\;V$ because the base current is so high that the transistor is working in its saturation region where the linear relationship $I_C=\beta I_B$ is no longer applicable (It is only valid in linear region). The actual output voltage can be estimated to be about $V_{CE}=0.2V$ , and the actual can be found to be $(V_{CC}-V_{CE})/R_C=(12-0.2)/2=5.9\;mA$ .

Jana

Tuesday 8 January 2013

Bipolar Junction Transistor (BJT)

Bipolar Junction Transistor (BJT)

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