Question 13.4: Analysis of the Fixed Base Bias Circuit The dc bias circuit ...
Analysis of the Fixed Base Bias Circuit
The dc bias circuit shown in Figure 13.18(a) has RB = 200 kΩ, RC = 1 kΩ, and VCC = 15 V. The transistor has β = 100. Solve for IC and VCE.
We will eventually see that the transistor is in the active region, but we start by assuming that the transistor is cut off (to illustrate how to test the initial guess of operating region). Since we assume operation in cutoff, the model for the transistor is shown in Figure 13.16(c), and the equivalent circuit is shown in Figure 13.18(b).We reason that IB = 0 and that there is no voltage drop across RB. Hence, we conclude that VBE = 15 V. However, in cutoff, we must have VBE < 0.5 for an npn transistor. Therefore, we conclude that the cutoff assumption is invalid.
Next, let us assume that the transistor is in saturation. The transistor model is shown in Figure 13.16(b). Then, the equivalent circuit is shown in Figure 13.18(c). Solving, we find that
I_C=\frac{V_{CC}-0.2}{R_C}=14.8\mathrm{~mA}
and
I_B=\frac{V_{CC}-0.7}{R_B}=71.5~\mu\mathrm{A}
Checking the conditions required for saturation, we find that IB > 0 is met, but βIB > IC is not met. Therefore, we conclude that the transistor is not in saturation.
Finally, if we assume that the transistor operates in the active region, we use the BJT model of Figure 13.16(a), and the equivalent circuit is shown in Figure 13.18(d). Solving, we find that
I_B=\frac{V_{CC}-0.7}{R_B}=71.5~\mu\mathrm{A}
(We have assumed a forward bias of 0.7 V for the base emitter junction. Some authors assume 0.6 V for small-signal silicon devices at room temperature; others assume 0.7 V. In reality, the value depends on the particular device and the current level. Usually, the difference is not significant.) Now, we have
I_C=\beta I_B=7.15\mathrm{~mA}
Finally,
V_{CE}=V_{CC}-R_CI_C=7.85\mathrm{~V}
The requirements for the active region are VCE > 0.2 V and IB > 0, which are met. Thus, the transistor operates in the active region.
