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Bipolar Junction Transistor

A bipolar junction transistor (BJT) has three operating regions:
  1. Cut off ($ V_B<V_E$ for NPN BJT)
  2. Active region ( $ V_E<V_B<V_C$ for NPN)
  3. Saturated ($ V_B>V_C$ for NPN)

In active region, $ V_{BE}\approx 0.7V$ for silicon BJT, and $ \approx
0.3$ for Germanium BJT.

In saturated region, $ V_{CE}\approx 0.2 V$.

Common realizations of BJT are shown in 10.1

Figure 10.1: Realizations of BJT

In active region:
Three configurations in the active region are shown in figure 10.2. For active region, the specified biasing condition is satisfied.

Figure 10.2: Configurations of BJT

When transistor is used for switching purposes, it works in either cut-off or saturation mode.

In active region, the base and collector currents satisfy the condition $ \frac{I_C}{I_B}=\beta$ (DC Current gain. Ratio of absolute values). $ \beta$ is a constant for a particular transistor, which varies from $ 50$ to $ 400$ for different transistors. Note that this condition does NOT hold for saturation and cut-off operations of the BJT.

Figure 10.3: Sample circuit for design problem

Now we address the problem of circuit design, in which we find appropriate values of resistances and voltages in figure 10.3 to ensure BJT in active region. The problem assumes importance as many transistor applications are those in which it is in active region.

In cut-off, $ I_B=0$, as $ V_{BE}<0.7V$. If $ \frac{I_C}{I_B}$ becomes less than $ \beta$, the transistor is in saturation. We need to ensure that the BJT is not in these states.

In active region, as $ V_{BE}>0.7$,

$\displaystyle I_B=\frac{6-0.7}{R_B}$      
$\displaystyle I_C=\beta_{dc}I_B$      
$\displaystyle \Delta I_C=\beta_{dc}\Delta I_B$      

The last equation shows that the transistor, in this mode (active), is basically a current amplifier.

Let $ R_B=1k\Omega$. Then, $ I_B=\frac{5.3}{1k\Omega}=5.3\;mA$. Suppose the BJT has $ \beta=50$. $ I_C=5.3mA\times 50=265mA$.

Also, we need to ensure $ V_{CE}\geq 0.2V$, so that BJT is not in saturation. In the limiting case, $ V_{CE}=0.2V$, just when the BJT is entering saturation from active region. (In active region, $ V_{CE}\geq 0.2V$).

Thus, $ 10-265mA\times R_C\geq 0.2$. That is, $ R_C\leq
\frac{9.8}{265}k\Omega\approx 37 \Omega$ for ensuring BJT in active region.

Suppose we increase $ R_C$ to $ R_C=500\Omega$. Then, $ I_C<\frac{10-0.2}{500}=19.6mA$. Thus, the current gain $ \frac{I_C}{I_B}<\frac{19.6}{5.3}=3.7<\beta_{active}=50$.

Cut off and saturation are used in switching application. For the circuit shown in figure 10.4, we find conditions for operating BJT as a switch.

Figure 10.4: BJT as a switch

When $ V_B<0.7V$, $ I_B=0$, $ I_C=0$, and $ V_{CE}=V_{CC}$, since BJT is in cut-off.

Now find $ V_B$ such that the BJT is in saturation.

$\displaystyle I_B=\frac{V_C-0.2}{R_B}$      
$\displaystyle I_C=\beta I_B$      
$\displaystyle \frac{V_{CC}-0.2}{R_L}=I_C$      
$\displaystyle \frac{I_C}{I_B}=\frac{(V_{CC}-0.2)R_B}{R_L(V_B-0.7)}<50$      
$\displaystyle \frac{(V_{CC}-0.2)R_B}{R_C\times 50}< V_C-0.7$      
$\displaystyle V_E\geq 0.7+\frac{(V_{CC}-0.2)R_B}{R_C\times 50}$      
Thus, we get: $\displaystyle V_B\geq 4.62 V$      

Thus, for $ 0.7<V_B<4.62V$, the BJT is in active region.

Figure 10.5: $ V_{CE}$ Vs $ V_B$ for the design of BJT as a switch

Two different biasing strategies are shown in figure 10.6 and 10.7.

Figure 10.6: Fixed bias circuit

Figure 10.7: Voltage divider bias

next up previous contents
Next: CE Characteristics Up: Introduction to Electronics Previous: LM 317: Regulator   Contents
ynsingh 2007-07-25