0
\$\begingroup\$

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}.

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?

enter image description here

|improve this question
\$\endgroup\$
  • 1
    \$\begingroup\$ The base of the transistor is the load. Can you solve it now? (The load doesn't have to be a resistor.) \$\endgroup\$ – Jonathan S. Mar 15 '18 at 21:10
  • \$\begingroup\$ @JonathanS. So the positive (open) terminal would be the base and the negative (open) terminal would be ground? \$\endgroup\$ – Oliver G Mar 15 '18 at 21:12
  • 1
    \$\begingroup\$ No, you'd just disconnect the base, leaving a voltage divider between +15V and GND. Then calculate the thevenin equivalent of that voltage divider. \$\endgroup\$ – Jonathan S. Mar 15 '18 at 21:31
  • \$\begingroup\$ @JonathanS. I guess I mean: for the +open terminal and -open terminal resulting from disconnecting a load and trying to calculate V_{oc} where the load would have been, is the +V_{oc} located at the terminal point where the base was disconnected and the -V{oc} at ground? \$\endgroup\$ – Oliver G Mar 15 '18 at 21:41
  • 1
    \$\begingroup\$ Yes, the resulting thevenin equivalent voltage (at the base) is relative to ground. (Everything is relative to ground unless explicitly stated otherwise, so the "+15V" markers in your circuit diagrams mean "+15V between here and ground", same for the 5V ones) \$\endgroup\$ – Jonathan S. Mar 15 '18 at 21:52
2
\$\begingroup\$

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.

schematic

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.

|improve this answer
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.