First, translate the specifications into constraint equations.
For the static power dissipation:
Assume, for now, that \$I_{R2} \ge 10 \cdot I_B = \dfrac{I_C}{10}\$ for the worst case \$\beta = 100 \$.
The supply current is then:
\$I_{PS} = I_C + 11 \cdot I_B = 1.11 \cdot I_C \$
The static power constraint then becomes:
\$\rightarrow I_C < \dfrac{25mW}{1.11 \cdot 10V} = 2.25mA\$
The bias equation:
The BJT bias equation is:
\$I_C = \dfrac{V_{BB} - V_{EE} - V_{BE}}{\frac{R_{BB}}{\beta} + \frac{R_{EE}}{\alpha}} \$
For this circuit, we have:
\$V_{BB} = 10V \dfrac{R_2}{R_1 + R_2}\$
\$V_{EE} = 0V\$
\$V_{BE} = 0.6V\$
\$R_{BB} = R_1||R_2\$
\$R_{EE} = R_E\$
So, the bias equation for this circuit is:
\$I_C = \dfrac{10V \frac{R_2}{R_1 + R_2} - 0.6V}{\frac{R_1||R_2}{\beta} + \frac{R_E}{\alpha}} \$
Now, you want less than 5% variation in \$I_C\$ for \$100 \le \beta \le 800\$. After a bit of algebra, find that this requires:
\$ \rightarrow R_E > 0.165 \cdot R_1||R_2 \$
Output swing:
The positive clipping level can be shown to be:
\$v^+_O = 3V = I_C \cdot R_C||R_L \$
The negative clipping level can be shown to be about:
\$v^-_O = -3V = I_C(R_C + R_E) - 9.8V \rightarrow 6.8V = I_C(R_E + R_C)\$
Put all this together:
Choose, for example, \$I_C = 1mA \$ then:
\$R_C||10k\Omega = 3k\Omega \rightarrow R_C = 4.3k\Omega\$
\$R_E + R_C = 6.8k\Omega \rightarrow R_E = 2.5k\Omega \$
Thus, \$V_E = 2.5V\$ and \$V_B = 3.1V\$
Then,
\$R_2 = \dfrac{V_B}{10 \cdot I_B} = \dfrac{3.1V}{100\mu A} = 31k\Omega \$
\$R_1 = \dfrac{10 - V_B}{11 \cdot I_B} = \dfrac{6.9}{110\mu A} = 62.7k\Omega \$
Now, check
\$0.165 \cdot R_1||R_2 = 3.42k \Omega > R_E \$
So, this doesn't meet the bias stability constraint equation we established earlier.
So run through this again (use a spreadsheet!) with larger \$I_C\$ until you've met the bias stability constraint equation.
If you can't meet the constraint with \$I_C < 2.25mA \$, you may need to increase current through the base voltage divider, e.g., \$I_{R2} = 20 \cdot I_B \$ and work through the static power constraint again.
As the the correctness of the clipping level calculations above has been questioned, I simulated the circuit using values calculated from the above except that \$I_C \$ was increased to \$2mA\$ for the calculation.
The DC solution:
Driving the amplifier with a 500mV 1kHz sine wave:
Note the clipping levels are precisely +3V and -3V as designed. The variation in \$I_C\$ is just over 5% over the range of \$\beta\$ so the next step would be to increase the multiple of base current through R2 to e.g., 20 and plug in the numbers (which does result in meeting all the constraints).
7 years late, but this was a fun one to do some forensic math on! And it actually may be a bit of a trick question! The trick comes down to that you're given an input current \$I_B\$, not an input voltage.
First: The error was on line 3 when you calculated \$I_E\$. You actually found \$I_C\$ instead. You have to add \$I_B\$ to that to get \$I_C\$.
Here's the shortcut solution. If the BJT is in the active mode:
$$
I_E=(\beta+1)I_B
$$
That means that it doesn't matter what \$R_E\$ is, the current will just be \$101\times I_B\$. We can explore this a little bit more thinking about Ebers Moll
simulate this circuit – Schematic created using CircuitLab
Let's say we start off with \$I_B=0\$. At this point CE is positive, reverse biasing D2 - so no current flows anywhere. This is in cutoff. It doesn't matter what \$R_E\$ is in this case either.
Let's turn on \$I_B\$ a little bit. D1 is forward biased and D2 is still reverse biased (so it's in the active mode). The current through D1 is \$I_{D1} = I_B+\alpha_F I_{D1}\$. Solving for the current through D1 is \$I_{D1}=I_E= I_B/ (1-\alpha_F)I_B = (\beta +1)I_B\$. Again, the value of \$R_E\$ does not matter.
Ok, so our BJT is in the active state. That means:
$$
V_BE=0.7\\
I_C=\beta I_B\\
V_o = V_+ - \beta I_B R_C=V_C\\
V_E = (\beta+1)I_B R_E-V_+\\
V_B \approx 0.7 - V_+ +(\beta+1)I_B R_E\\
V_{BE} \approx 0.7\\
V_{BC} \approx -2V_+ + 0.7 + (\beta+1)I_B R_E + \beta I_B R_C
$$
That means that at some point \$I_B\$ gets large enough that the \$V_{BC}\$ becomes positive, and our diode D1 becomes forward biased. We are now in saturation mode. At that point, no matter how much we increase IB, \$I_C\approx I_E\$. \$\beta\$ is effectively reduced, but it's still basically \$I_E=(\beta_{reduced}+1)I_B\$.
So it's not exactly a trick question, but it's a question about controlling a BJT with current instead of voltage.
Best Answer
The DC collector current is determined by \$R_E\$:
\$I_C = \alpha \dfrac{9.4V}{R_E} \approx \dfrac{9.4V}{R_E}\$
Since you require \$I_C < 1.25mA \$, the constraint equation is:
\$R_E > \dfrac{9.4V}{1.25mA} = 7.52k\Omega\$
The second requirement, maximum output voltage swing, without any other constraint, doesn't fix the collector resistor value.
We have:
\$ V_{o_{max}} = 19.8V - I_C(R_C + R_E)\$
But, the voltage across \$R_E\$ is fixed at 9.4V so:
\$V_{o_{max}} = 10.4V - I_C R_C\$
\$V_{o_{min}} = -I_C * R_C||R_L\$
If you stare at this a bit, you'll see that maximum output voltage swing is 10.4V but this requires that the product \$I_C R_C = 0\$* which is absurd.
Now, if we also require symmetric clipping, then, by inspection:
(1) \$V_{o_{max}} - V_{o_{min}} = 2 I_C (R_C||R_L)\$
(2) \$10.4V = I_C(R_C + R_C||R_L) \$
Looking at (1), note that, for maximum swing, we get more "bang for the buck" by increasing \$I_C \$ rather than \$R_C \$.
Since we have an upper limit on \$I_C\$, (2) becomes:
\$R_C + R_C||R_L = \dfrac{10.4V}{1.25mA} = 8.32k \Omega\$
which can be solved for \$R_C\$.
*unless \$R_L\$ is an open circuit