Question 13.6: Analysis of a BJT Bias Circuit Solve for IC and VCE in the c...
Analysis of a BJT Bias Circuit
Solve for IC and VCE in the circuit of Figure 13.21(a) given that VCC = 15V, VBB = 5V, RC = 2 kΩ, RE = 2 kΩ, and β = 100. Repeat for β = 300.
We assume that the transistor is in the active region and use the equivalent circuit shown in Figure 13.21(b). Writing a voltage equation through VBB, the base emitter junction, and RE, we have
V_{BB}=0.7+I_ER_E
This can be solved for the emitter current:
I_E=\frac{V_{BB}-0.7}{R_E} =2.15\mathrm{~mA}
Notice that the emitter current does not depend on the value of .
Next, we can compute the base current and collector current using Equations 13.2 and 13.10.
i_E=i_C+i_B (13.2)
i_C=\beta i_B (13.10)
I_C=\beta I_B
I_E=I_B+I_C
Using the first equation to substitute for IC in the second equation, we have
I_E=I_B+\beta I_B=(\beta +1) I_B
Solving for the base current, we obtain
I_B=\frac{I_E}{\beta+1}
Substituting values, we obtain the results given in Table 13.1. Notice that IB is lower for the higher β transistor, and IC is nearly constant.
Now, we can write a voltage equation around the collector loop to find VCE:
V_{CC}=R_CI_C+V_{CE}+R_EI_E
Substituting values found previously, we find that VCE = 6.44 V for β = 100 and VCE = 6.42 V for β = 300.
Table 13.1. Results for the Circuit of Example 13.6
| 0 | I_B(\mu\mathrm{A}) | I_C(\mathrm{mA}) | V_{CE}(\mathrm{V}) |
| β | |||
| 100 | 21.3 | 2.13 | 6.44 |
| 300 | 7.14 | 2.14 | 6.42 |