Voltage divider bias (BJT) leads to a huge voltage drop across the collector resistor. Why?

Question

I've already looked for similiar questions, but couldn't find any. Basically, I'm new in electronic circuits analysis and, after following my professor's lecture and looking on the internet I found out that the following circuit (please note that the voltage source on the top is meant to "replace" a voltage line, as I have no idea of how to draw it)

^{simulate this circuit – Schematic created using CircuitLab}

can be approximately solved, using Thévenin's equivalent, as

^{simulate this circuit}

with $$ R_{th}=R_1∥R_2\approx 2.3k\Omega, V_{BB}=\frac{R_2}{R_1+R_2}V_s\approx 9.23V.$$

Hence, assuming FAR operating conditions (with voltage drop between the base and the emitter of about 0.7V), according to my computations, $$ V_{BB}-I_BR_{th}-V_{BE}-I_ER_e=0\Rightarrow I_B=\frac{V_{BB}-V_{BE}}{R_{th}+(1+\beta)R_e}.$$

Let's further assume beta=100; then, $$I_B\approx 82\mu A, I_c\approx \beta I_B\approx 8.20 mA.$$ But, to me, this amount of current through a 2700 ohms resistor looks just insane, as it would bring about a 22V voltage drop across Rc, while the voltage source delivers only 12V. Moreover, $$V_s-R_cI_c-V_{CE}-I_ER_e=0\Rightarrow V_{CE}=V_s-R_cI_c-I_ER_e\approx -18.42V,$$ which looks totally wrong to me. Where am I wrong? Sorry if this may be a trivial question, but I've been trying for 3 days already and looking on the internet didn't help me at all. Thanks in advance!

Answer 1

If you getting such a strange result that's mean that your BJT is in the saturation region. In your circuit, the collector current cannot be larger than: $$I_{Cmax} = \frac{V_{CC}}{R_C+R_E} \approx 3.2mA $$ and you get \$8.2mA\$ hence your BJT is in the saturation region.

So, to be able to solve it without using the advanced math you are forced to assume some \$V_{CEsat}\$ value. Typically in hand calculations, we are assuming that the saturation voltage is equal to \$V_{CEsat} \approx 0.2V\$

Additional we use this equation \$I_E=I_B+I_C\$ which is always true is saturation and in the active region.

$$I_E = \frac{V_E}{R_E}$$

$$I_C = \frac{V_{CC} - (V_{CEsat}+V_E)}{R_C} $$

$$I_B= \frac{V_{BB}-(V_{BE}+ V_E)}{R_{th}}$$

So finally we have this equation with only one unknown \$V_E\$

$$\frac{V_E}{R_E}=\frac{V_{BB}-(V_{BE}+ V_E)}{R_{th}}+\frac{V_{CC} - (V_{CEsat}+V_E)}{R_C}$$

And the solution is

$$V_E = \left(\frac{V_{BB} - V_{BE}}{R_{th}} +\frac{V_{CC} -V_{CEsat}}{R_C}\right)\cdot R_E||R_{th}||R_C \approx 4.49V$$

And the currents are:

$$I_E \approx 4.49 \textrm{m}A$$

$$I_B \approx 1.79\textrm{m}A $$

$$I_C \approx 2.7\textrm{m}A $$

Or you could have written this two equations and solve for \$I_B\$ and \$I_C\$

$$V_{BB} - I_B R_B - V_{BE} - (I_B+I_C) R_E = 0$$

$$V_{CC} - I_C R_C - V_{CEsat} - (I_B+I_C) R_E = 0$$