We shall follow the rule of thumb mentioned above and allocate one-third of the supply voltage to the voltage drop across $R_2$ and another one-third to the voltage drop across $R_C$, leaving one-third for possible signal swing at the collector. Thus,

$$
\begin{aligned}
& V_B=+4 \mathrm{~V} \\
& V_E=4-V_{B E} \simeq 3.3 \mathrm{~V}
\end{aligned}
$$

and $R_E$ is determined from

$$
R_E=\frac{V_E}{T_E}=\frac{3.3}{1}=3.3 \mathrm{k} \Omega
$$


From the discussion above we select a voltage-divider current of $0.1 I_E=0.1 \times 1=0.1 \mathrm{~mA}$. Neglecting the base current, we find

$$
R_1+R_2=\frac{12}{0.1}=120 \mathrm{k} \Omega
$$

and

$$
\frac{R_2}{R_1+R_2} V_{C C}=4 \mathrm{~V}
$$


Thus $R_2=40 \mathrm{k} \Omega$ and $R_1=80 \mathrm{k} \Omega$.
At this point, it is desirable to find a more accurate estimate for $I_E$, taking into account the nonzero base current. Using Eq. (5.70),

$$
I_E=\frac{4-0.7}{3.3(\mathrm{k} \Omega)+\frac{(80 / / 40)(\mathrm{k} \Omega)}{101}}=0.93 \mathrm{~mA}
$$


This is quite a bit lower than the value we are aining for of 1 mA . It is casy to see from the above equation that a simple way to restore $I_E$ to its nominal value would be to reduce $R_E$ from $3.3 \mathrm{k} \Omega$ by the magnitude of the second term in the denominator ( $0.267 \mathrm{k} \Omega$ ). Thus a more suitable value for $R_E$ in this case would be $R_E=3 \mathrm{k} \Omega$, which results in $I_E=1.01 \mathrm{~mA} \simeq 1 \mathrm{~mA}$.

It should be noted that if we are willing to draw a higher current from the power supply and to accept a lower input resistance for the amplifier, then we may use a voltage-divider current equal, say, to $I_E$ (i.c., 1 mA ), resulting in $R_1=8 \mathrm{k} \Omega$ and $R_2=4 \mathrm{k} \Omega$. We shall refer to the circuit using these latter values , for which the actual value of $I_E$ using the initial value of $R_E$ of $3.3 \mathrm{k} \Omega$ will be

$$
I_E=\frac{4-0.7}{3.3+0.027}=0.99 \simeq 1 \mathrm{~mA}
$$


In this case,  we need not change the value of $R_E$.
Finally, the value of $R_C$ can be determined from

$$
R_C=\frac{12-V_C}{I_C}
$$


Substituting $I_C=\alpha I_E=0.99 \times 1=0.99 \mathrm{~mA} \simeq 1 \mathrm{~mA}$ results, for both designs, in

$$
R_C=\frac{12-8}{1}=4 \mathrm{k} \Omega
$$