Suppose you have the schematic:

enter image description here

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. enter image description here

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\$?

  • \$\begingroup\$ 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. ;) \$\endgroup\$
    – jonk
    Commented Jan 11, 2019 at 0:55
  • \$\begingroup\$ 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/… \$\endgroup\$
    – Bidon
    Commented Jan 11, 2019 at 1:01
  • \$\begingroup\$ Can you explain to us how the RB resistor can affect the DC operating point? \$\endgroup\$
    – G36
    Commented Jan 11, 2019 at 16:48

1 Answer 1


For the dc analysis, I would replace the bias by its equivalent Thévenin generator and have the following circuit:

enter image description here

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:

enter image description here

Looks like your dc expression has a dimension issue and Mathcad flags it.

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.

enter image description here


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