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When analyzing this transistor circuit it was suggested that I simplify the left side by using Thevenin's theorem with the points A and B as follow. Before I jump in let me apologize the lenght of the question and my english. But I would really appreciate any insight because my professor told me Thevenin is important in the electronics course, I would like to learn it well. The actual questions are in bold (1) and (2).

schematic

simulate this circuit – Schematic created using CircuitLab

The simpified circuit after using Thevenin's theorem is the following:

schematic

simulate this circuit

What I couldn't understand is why the voltage source of 22V still appears in the simplified circit. By my what I understand of Thevenin, the 22V source should not appear in the final circuit, since Vth's calculations involved the 22V source. This is what it would look if I were to do it (the simplified circuit below):

schematic

simulate this circuit

Now, calculating Vth and Rth I get the same results, but when plugging again the dead network, here's what mine would look like:

schematic

simulate this circuit

I tried to make the most of the "name nodes" feature to explain myself better, especially because my questions are messy sometimes. Back to business. Obviously, I would not know where to "plug" "Wire X" terminals. It could not be at point A because R3 was connected between R1 and the 22V source, and it could not be between Vth and Rth because it also doesn't make sense and is probably wrong.

(1) I think that, maybe, the 22V source was plugged in the final circuit because this is the voltage in R3 even before we simplify it and it is like it is there, is this correct? But in case there is a resistance between V1 and R1 in the original I would run into trouble because I would first need to known what is the voltage drop between that resistor. (2) How do we deal when our points of analysis A and B are in the middle of the circuit?

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  • \$\begingroup\$ You have everything ok but remember that the source conversion is separate from the load so your output [?]tag is still 22V ...Point [A] is slightly loaded by input imepdance= hFE*R4 value in parallel with [A] which is higher resistance so Vb [A] actually controls the voltage on Ve across R4 which determines the current thru R3 and thus Vc . \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 8 '16 at 0:47
  • \$\begingroup\$ Thanks Tony for helping understand more about this circuit. I guess I'm just confused about the process of applying Thevening when the points AB are in the middle of the circuit, but I'm starting to think that in some cases is just not worth it. \$\endgroup\$ – João Pedro Nov 8 '16 at 1:35
  • \$\begingroup\$ you have done it correctly for AB ..... emitter Q1-e is typically 0.65 V below Q1-b then Q1-c is easy ratio of Rc/Re for ac gain while Vdc is same ratio but drop from 20V thus Ve=2-0.65=1.35 (aprox) and Vc= 20- 10/1.5*1.35 = 20-9=11V \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 8 '16 at 1:54
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Perhaps all you need is some visual aid. Here's your circuit and its equivalent version once I cut the top 22V rail in the middle. (You have to ask yourself why I can cut the circuit between those upper nodes when they both are at the same voltage).

cutting the rail

And now I can redraw it so that I have a black box with one port. You can apply Thevenin to that one port and get your final circuit

enter image description here

It's as easy as that.

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    \$\begingroup\$ Hi Sredni. I have a doubt exact in this point: "why I can cut the circuit between those upper nodes when they both are at the same voltage", Why can I do that since, although the nodes are at the same voltage, the current between then isn't zero? \$\endgroup\$ – Vinicius ACP Jul 25 '20 at 6:34
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    \$\begingroup\$ The current is zero. That branch has zero voltage across it and zero current through it. It's as if it weren't there at all. Even if you put a resistor in the middle of it, you have exactly 22V on one terminal and exactly 22V on the other (this is important). The presence of the resistor does not alter the voltage at the nodes and the current in the rest of the circuit. \$\endgroup\$ – Sredni Vashtar Jul 25 '20 at 6:55
  • \$\begingroup\$ I simulated the circuit in the question using CircuitLab. I placed an ammeter between these two points and the measured current wasn't zero, it was ~900µA. \$\endgroup\$ – Vinicius ACP Jul 25 '20 at 7:20
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    \$\begingroup\$ @ViniciusACP did you put two batteries in the circuit? Like in the the second picture at the top (the one with the scissors)? \$\endgroup\$ – Sredni Vashtar Jul 25 '20 at 7:36
  • \$\begingroup\$ Oh, I see it now. It's not just cut the circuit...you have to add the voltage source to the other side too. Now I understand, thank you Sredni! \$\endgroup\$ – Vinicius ACP Jul 25 '20 at 8:18
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Thevenin's transformation is applied only to the part of the circuit consisting of the 22V source and the voltage divider (R1 and R2).

Anything else is left as it was; especially the fact that R3 is connected to a 22V source. Therefore a 22V source must still be present in the simplified circuit.

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