Thevenin's theorem BJT derivation question

Question

I'm trying to convert the left circuit to the one on the right using Thevenin's theorem, but I'm not understanding how this is done for this circuit.

From my understanding of Thevenin's theorem:
(1)You disconnect a load resistor R_{L} and calculate the open circuit voltage v_{oc} where the resistor would be.

(2)Then short the voltage source and find the thevenin resistance R_{TH}.

(3)Replace the original voltage source with v_{oc} in series with the thevenin resistance R_{TH} connected to the load resistor R_{L}.

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In the below picture, what is the load resistance that is removed and what are the terminals where V_{oc} is measured? How specifically is +5 volts found to be the open circuit voltage? How specifically is the thevenin resistance calculated in this image?

Answer 1

As I wrote in the comments, you should get into the mental practice of converting every single pair of resistors, those pairs spanning from one voltage source to another voltage source, into their Thevenin equivalent. You just "turn the crank." You should get to the point where you do it almost without thinking about it.

Let's take your first diagram and make it worse. It's still the same circuit, except that now this circuit requires three newly added power supply rail voltages: \$-7.5\:\text{V}\$, \$+7.5\:\text{V}\$, and \$+30\:\text{V}\$. But other than that annoyance, it's the same circuit.

^{simulate this circuit – Schematic created using CircuitLab}

See all those pairs in the circuit, itself, shown on the upper left area?

I've computed the Thevenin equivalents of each of them on the right upper side. From this, and some imagination, I think you can see that the circuits are the same.

Just get into the practice of it.

For complicated cases, such as my added collector and emitter pairs, you can look at the very bottom of the above schematic. The equation is simple:

$$\begin{align*} R_\text{TH} &= \frac{R_A\cdot R_B}{R_A+R_B}\\\\ V_\text{TH} &= \frac{V_B\cdot R_A + V_A\cdot R_B}{R_A+R_B} \end{align*}$$

Just memorize it.

You can see in the above schematic at the bottom section where I draw the locations of each of these voltages and resistors, along with the new equivalent. That's really all there is to it.

Answer 2

... you guys are making things way too complicated, the derivation itself has nothing to do with the BJT, we don't even care about the BJT, all that matters is the same i-v relationship is being sent to the BJT (or whatever device we are dealing with).

Proof of derivation: