# Electrical – DC and AC analysis of a Common Emitter Amplifier Circuit

transistors

What we know about the circuit is that in DC-analysis (when $$\V_{in}=0\$$), $$\V_{out}=10V\$$.

We also know that the circuit (even outside of DC analysis) will have $$\I_{C}=2mA\$$ and $$\V_{CE}=5V\$$. $$\R_{1}\$$ and $$\R_{2}\$$ should be so that $$\I_{B(max)}=\frac{1}{10}I_{2}\$$ where $$\I_{2}\$$ is the current over $$\R_{2}\$$.

I'm assuming

$$\V_{out}-V_{B}=0.7V\$$

and

$$\V_{B}=V_{CC}(\frac{R_2}{R_2+R_1})\$$

And in DC-analysis if $$\V_{out}-V_{B}=0.7V\$$ holds, I can conclude that

$$\V_{B}=V_{out}-0.7V\$$, so that $$\V_{B}=10V-0.7V=9.3V\$$

But I'm not 100% certain on what I should do to figure out the formulas to calculate the resistors to match the assumptions and how to proceed overall.

Edit 1

If $$\V_B=9.3V\$$, $$\I_2*R_2=9.3V\$$ and $$\I_1*R_1=20V-9.3V=10.7V\$$

$$\I_1=I_2+I_B\$$

assuming $$\I_B=max\$$

$$\I_1=I_2+\frac{1}{10}I_2=1.1I_2\$$