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

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!

• If you are getting strange results you should question your assumption. What if the transistor is actually saturated? – Spehro Pefhany Apr 8 '18 at 18:23
• related – The Photon Apr 8 '18 at 18:34
• Ohhh, I gave FAR operation for granted because my professor gave this exercise with the usual FAR constraints (VBE and that stuff)... Thanks! – Maurizio Carcassona Apr 8 '18 at 21:48

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 $I_C$

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

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

• Thanks a lot dude! Tomorrow I'll have the practical lab about this, so I will feel safer about this. Thanks again! – Maurizio Carcassona Apr 8 '18 at 21:49
• Surprisingly, in practice, the lab measurements showed closer-to-active results for VBE, Ib and VCE, while Ic was just as expected from the above results. It's so fascinating how reality can totally change the theoretical approximations! – Maurizio Carcassona Apr 9 '18 at 22:33
• @MaurizioCarcassona Can you explain more? You want to say that in your lab the BJT was is the active region? What resistor values and BJT type you have used in your lab measurement? And Vcc was? – G36 Apr 10 '18 at 13:51
• Sorry for not having answered you - I saw the comment only now. I consider the matter solved as of now, but for the sake of completeness, here's the data: we used a 2N2222A BJT, and we obtained the following results; Vbe=6.32mV, Vce=4.38V, Ib=7.5µA, Ic=2.08mA. After having re-looked at the data I provided, there's a datum, R1, which isn't correct, as it was actually 33kOhm. Nevertheless, by looking at Vbe, it seems like it wasn't operating in FAR anyway. – Maurizio Carcassona Sep 20 '18 at 12:44
• @MaurizioCarcassona For R1 = 33kΩ the situation is completely different. We have Vbb = 2.79V therefore the emitter current is now equal to Ie = Ic = (2.97V - 0.7)/1kΩ = 2mA. As you can see this time Ic current is smaller than Ic_max (3.2mA). Hence our BJT will be working in the active region with Vce around Vce = 12V - 2mA*(1kΩ + 2.7kΩ) = 4.6V. Quite a big change don't you think? – G36 Sep 20 '18 at 14:25