DC Bias

An important part of amplifier design is setting the DC bias point Q. In this section we look at a general bias network, Figure 107.

**Figure 107:** General DC bias network.
$\begin{figure} \begin{center} \epsfig{file=images/bjtimg15.eps}\end{center}\end{figure}$

Applying KVL to the left loop gives

V_BB = I_B R_B + V_BE + I_E R_E .

Now since $I_E = (\beta + 1) I_B$ we can eliminate I_E and solve for I_B:

$\begin{displaymath}I_B = \frac{V_{BB} - V_{BE}}{R_B + (\beta + 1) R_E} . \end{displaymath}$

(101)

Now $V_{BE} \approx 0.7$ V, and since V_BB and the resistances are known, we can determine I_B.

The right hand loop gives, using KVL,

V_CC = I_C R_C + V_CE + I_E R_E .

Now if $\alpha$ is close to 1, we assume $I_C \approx I_E$ , and so eliminating I_E and solving for I_C we get

$\begin{displaymath}I_C = \frac{V_{CC} - V_{CE}}{R_E+R_C} . \end{displaymath}$

(102)

This is the equation of the load line, Figure 108.

**Figure 108:** DC load line for BJT transistor circuit of Figure 107.
$\begin{figure} \begin{center} \epsfig{file=images/bjtimg16.eps}\end{center}\end{figure}$

The operating point Q=Q_DC is determined from the load line and the collector characteristic curves, or via $I_C = \beta I_B$ . Notice that in this analysis we have made (implicit) use of the DC model.

Example. Find the DC operating point for the voltage divider bias network shown in Figure 109. Use $R_{B1}=470k\Omega$ , $R_{B2}=18k\Omega$ , $R_C=9.8k\Omega$ , V_CC=20 V, V_BE=0.7 V, R_E=0, and $\beta = 500$ . Note that R_E=0 in this example.

**Figure 109:** Voltage divider bias network.
$\begin{figure} \begin{center} \epsfig{file=images/bjtimg17.eps}\end{center}\end{figure}$

To solve this problem, we first replace the voltage divider V_CC, R_B1, R_B2 by its Thevenin equivalent, which will be V_BB in series with R_B, as in Figure 107. Now

$\begin{displaymath}R_B = R_{B1} \parallel R_{B2} = 17.34 \ k \Omega \end{displaymath}$

and

$\begin{displaymath}V_{BB} = \frac{V_{CC}R_{B2}}{R_{B1} + R_{B2}} = 0.738 \ V \end{displaymath}$

So from (101) we get I_B = 2.2 $\mu$ A, and $I_C = \beta I_B = 1.1$ mA. Also from (102) we get V_CE=9.22 V. Thus the DC operating point Q is I_C = 1.1 mA, V_CE=9.22 V, which is not far from midway along the load line (draw it!).

Note that absence of R_E makes I_B critically dependent on exact V_BE. Using an emitter resistor R_E improves the bias stability, meaning robustness of the bias point with respect to parameter variations.

[ENGN2211 Home]

ANU Engineering - ENGN2211