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## Q. 12.7

Analysis of the Four-Resistor Bias Circuit
Find the values of $I_C \text{ and } V_{CE}$ in the circuit of Figure 12.23 for β = 100 and β = 300.
Assume that $V_{BE} = 0.7 \text{ V}$. ## Verified Solution

Substituting into Equations 12.20 and 12.21, we find that

$R_B=\frac{1}{1/R_1+1/R_2} =3.33 \text{ k}\Omega \\ V_B=V_{CC}\frac{R_2}{R_1+R_2}=5 \text{ V}$

Then, substituting into Equation 12.23 and using β = 100, we have

$I_B=\frac{V_B-V_{BE}}{R_B+(\beta +1)R_E}=41.2 \ \mu \text{A}$

For β = 300, we find that $I_B = 14.1 \ \mu\text{A}$. Notice that the base current is significantly smaller for the higher β.
Now, we can compute the collector current by using $I_C = \beta I_B$. For β = 100, we find that $I_C = 4.12 \text{ mA}$, and for β = 300, we have $I_C = 4.24 \text{ mA}$. For a 3:1 change in β, the collector current changes by less than 3 percent. The emitter current is given by $I_E = I_C + I_B$. The results are $I_E = 4.16 \text{ mA}$ for β = 100 and $I_E = 4.25 \text{ mA}$ for β = 300.
Finally, Equation 12.24 can be used to find $V_{CE}$. The results are $V_{CE} = 6.72$ for β = 100 and $V_{CE} = 6.51$ for β = 300.

$V_{CE}=V_{CC}-R_CI_C-R_EI_E \quad \quad \quad (12.24)$