The most important thing to recognize in this circuit, is that the current through R1 and R2 is not equal and that everything can be solved algebraically. Let's just write down everything we know and see where we land. In this notation, \$I_x\$ means the current through x
- \$(I_{R_2}+I_B)R_1 + I_{R_2}R_2 = 20 V\$
- \$(I_B+I_C)R_E + V_{CE} + I_CR_C = 20 V\$
- \$I_C = \beta * I_B\$
- \$(I_B + I_C)Re + V_{BE} + (I_{R_2} + I_B)R1 = 20 V\$
This system of equations has too many unknowns: \$I_{R_2}, I_B, I_C, V_{BE}\$ and \$V_{CE}\$. We need to either find a fourth independent equation or guess one of them. I assume that you are being expected to assume that \$V_{BE}\$ is constant, as per your statement that it is 0.7V. This allows you to solve this system of equations:
- \$I_B = (20 - 0.7 - {R_1* 0.7\over R_2})/(R_1 + R_E*(1+\beta) + {Re*R_1*(1 + \beta)\over R_2}) = 28.75904 µA\$
- \$I_C = 50*I_B = 1.43795245 mA\$
- \$I_E = I_B + I_E = 1.46671149 mA\$
- \$V_{CE} = 20 - I_ER_E - I_CR_C = 20 - 1.613382639 - 8.91530519 = 9.471312171 V \$
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
Best Answer
You can calculate back to the base current.
\$I_e = \beta * I_b +I_b\$
-> \$I_e = I_b * (\beta + 1)\$
-> \$I_b =I_e/(\beta +1) = 2/101 = .0198mA\$ (Formula 1)
\$I_c = I_e - I_b = 2 -.0198 = 1.98mA\$
You can also do it this way
\$I_c = \beta * I_b\$
Substituting (Formula 1) from above
\$I_c = I_e * \beta/(\beta + 1)\ = 2 * 100 /101 = 1.98mA\$