Finding current through an LED in an amplifier using a BJT

Suppose you have the schematic:

My job is to find $$\I_{LED} \$$ knowing that the impedance of the capacitor in negligible against $$\V_s\$$. Here's my attempt:

I intend to do an AC analysis followed by a DC analysis, and get both components of the current that goes by the LED.

Where $$\R_A=R_1||R_2\$$

For the AC analysis: I did the schematic in the figure and with the following set of equations I get to the final result: $$i_1=i_2+i_B \\v_s-i_1R_B-i_2R_A=0 \\ v_s-i_1R_B-i_ER_E=0 \\ i_C=\beta i_B \\ i_C\approx i_E \\ \therefore i_C=\frac{\frac{v_s}{R_B}}{\frac{1}{\beta}+R_E\bigg(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_B}\bigg)}$$

For the DC analysis I use the following set, reaching the result: $$V_cc-I_1R_1-I_2R_2=0 \\ V_cc-I_1R_1-V_{BE}-I_ER_E=0 \\ I_1-I_2=I_B\\ I_B=\frac{I_C}{\beta} \\ I_C \approx I_E \\ \therefore I_C=\frac{\frac{V_{cc}}{R_1}-V_{BE}\bigg(\frac{1}{R_1}+\frac{1}{R_2}\bigg)}{\frac{1}{\beta}+R_E\bigg(\frac{1}{R_1}+\frac{1}{R_2}\bigg)}$$

Finally I get to:

$$I_{LED}=I_C+i_c=\frac{\frac{V_{cc}}{R_1}-V_{BE}\bigg(\frac{1}{R_1}+\frac{1}{R_2}\bigg)}{\frac{1}{\beta}+R_E\bigg(\frac{1}{R_1}+\frac{1}{R_2}\bigg)}+\frac{\frac{v_s}{R_B}}{\frac{1}{\beta}+R_E\bigg(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_B}\bigg)}$$

However the correct solution is: $$I_{LED}=\frac{\frac{V_{cc}}{R_1}-V_{BE}\bigg(\frac{1}{R_B}+\frac{1}{R_1}+\frac{1}{R_2}\bigg)}{\frac{1}{\beta}+R_E\bigg(\frac{1}{R_B}+\frac{1}{R_1}+\frac{1}{R_2}\bigg)}+\frac{\frac{v_s}{R_B}}{\frac{1}{\beta}+R_E\bigg(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_B}\bigg)}$$

I know that this is a very long question and I don't ask for a solution to the problem, because I have one solution that leads to the correct result, however they don't do the DC and AC analysis separately and I was trying to do it because it should work, but I don't know where the mistake is, the answers are really similar and the AC analysis appears to be correct but the DC does not, and I can't find the mistake.

EDIT: I have found the mistake, In the DC analysis schematics I left out the resistance $$\R_B\$$, and if I leave it in the schematics connected to the ground then I can substitute, in the DC analysis only, all the $$\R_2\$$´s by the paralel resistance of $$\R_B\$$ and $$\R_2\$$ and the answers match. But a new question comes up, why do I have to leave $$\R_B\$$ connected to the ground? I thought that with a DC analisys capacitors would be replaced by an open circuit, therefore, no current should go through $$\R_B\$$, and that's why I ignored it. And this is just th case where I have a resistance $$\R_B\$$ there, if I didn't, but left that branch connected to the ground, then wouldn't it act as a short and we could ignore $$\R_2\$$?

• I pretty much agree with your DC analysis, just glancing over it. The DC operating point is not affected by $R_B$. And the DC operating point sets the quiescent DC collector current driving the LED. So I can't find a way to include $R_B$ in the DC term of the correct solution you show. I suspect they made a mistake. But then, I've been known to wrong on occasion. ;)
– jonk
Commented Jan 11, 2019 at 0:55
• I'm really new to this so I'm not familiar nor confortable with technical terms so I'm sorry I could not follow your reasoning, however in case you have the patience here's a link to the correct answer (The text is in portuguese but it is not that relevant to the solution) :drive.google.com/file/d/14pjoPtAmpJ6Jm1OPxklDsEf7UMGVE2SM/… Commented Jan 11, 2019 at 1:01
• Can you explain to us how the RB resistor can affect the DC operating point?
– G36
Commented Jan 11, 2019 at 16:48

The base current is defined as the voltage drop across the equivalent output resistance divided by $$\R_{eq}\$$. The voltage across the emitter resistor is the collector current plus the base current times $$\R_E\$$. Extract $$\i_b\$$ from this expression and substitute it in the first base current definition, multiply the result by the transistor gain and you have your collector current:
For the ac analysis, combine $$\R_A\$$ and $$\R_b\$$ into a Thévenin generator again then scale $$\V_{in}\$$ accordingly and it should be easier: Thévenin and Norton are your friends in all these analyses : ) These are the first steps to the FACTs. By the way, I don't see $$\r_{\pi}\$$ or $$\h_{11}\$$ in your equivalent circuit.