How do we derive the values for these Thévenin equivalent circuits?

Question

I am currently studying the textbook The Art of Electronics, third edition, by Horowitz and Hill. Appendix D Thévenin's Theorem says the following:

In Chapter 1 we stated (but did not "prove") Thévenin's Theorem, namely that any two-terminal network whose internal circuitry consists solely of resistors, batteries, and current sources, interconnected in any manner whatsoever, is equivalent (and indistinguishable) from the two-terminal network consisting of a single battery \$V_\text{TH}\$ in series with a single resistor \$R_\text{TH}\$; see Figure D.1. We did not prove it, because, in the spirit of this book, we don't prove anything; we show you how to design circuits, instead. We make an exception here, because it's nice to see something proved, right? D.1 The proof
For linear circuit elements (here resistors), the "nodal equations" (Kirchhoff's voltage law, KVL, and Kirchhoff's current law, KCL) are a set of linear equations. So we can find any circuit quantity (a voltage or a current), which depends on all the "independent sources" (batteries, current sources), by turning on each source in turn, and adding the partial contributions. (This is exactly analogous to using superposition to find, say, the electric field from a set of charges.) This technique is often useful in circuit analysis.
Here we wish to mimic the \$V\$ versus \$I\$ of the actual circuit with the (simpler) Thévenin equivalent of a single battery in series with a single resistor. Imagine we determine that \$V\$ versus \$I\$ function by applying an external current \$I_\text{ext}\$ that flows through the two-terminal circuit, and observing the resultant \$V\$ across those same two terminals. \$V\$ depends on \$I_\text{ext}\$ and on all the internal batteries (\$V_\text{int}\$) and the current sources (\$I_\text{int}\$).

1. Set all \$V_\text{int} = 0\$ and all \$I_\text{int} = 0\$; that is, replace all internal batteries with short circuits and all current sources with open circuits. Now, with a given applied \$I_\text{ext}\$, observe \$V_1\$.
2. Define \$R_\text{T} = V_1 / I_\text{ext}\$. (They must be proportional, by linearity.)
3. Now set \$I_\text{ext} = 0\$, and turn on the internal batteries and current sources. Observe \$V_2\$, which we will call \$V_\text{T}\$.
4. Finally, by superposition it must be the case that $$V(\text{actual}) = V_1 + V_2 = I_\text{ext} R_\text{T} + V_\text{T}.$$ This is true for all \$I_\text{ext}\$, and is exactly what you get with the Thévenin equivalent circuit, when connected to any load (which need not be linear); see Figure D.2.
   To summarize: (a) you determine \$R_\text{T}\$ and \$V_\text{T}\$ by first finding the open-circuit voltage, which equals \$V_\text{T}\$; then (b) you find the short-circuit current, \$I_\text{SC}\$, which equals the ratio of \$V_\text{T}\$ to \$R_\text{T}\$. In other words, \$V_\text{T} = V_\text{OC}\$ and \$R_\text{T} = V_\text{OC}/I_\text{SC}\$. You do this by analysis, if you know the "black-box" circuit; or by measurement, if you don't. D.1.1 Two examples – voltage dividers
   Figures D.3. and D.4. show two simple examples, variations on the resistive divider. Interestingly, their Thévenin equivalent circuits are different, even though the resistor values and the open-circuit voltages are the same.

I'm attempting to apply the sequence of steps 1., 2., 3., 4. to re-derive the values shown in figure D.3. and figure D.4., but I cannot seem to make sense of how this is supposed to work. How does one apply the sequence of steps 1., 2., 3., 4. to re-derive the values \$V_\text{OC}\$, \$I_\text{SC}\$, \$V_\text{T}\$, and \$R_\text{T}\$ in figure D.3. and figure D.4.? And why does figure D.3. have a Thévenin equivalent resistance equal to the parallel resistance, whereas figure D.4. does not? I would appreciate it if someone would please demonstrate this, and also provide explanations/justifications for each step, so that I can clearly understand what my misunderstanding is. Please do not use any concepts that are not covered in this section/explanation of the textbook.

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EDIT

I was under the impression that this is a simple question, but, so far, no one has been able to offer a clear answer. This really does seem like a simple question: Why is the Thévenin equivalent resistance for D.3. calculated in parallel as \$R_\text{T} = \dfrac{R_1 \times R_2}{R_1 + R_2} = \dfrac{10 \times 10}{10 + 10} = 5 \text{k} \$, whereas the Thévenin equivalent resistance for D.4. is calculated in series as \$R_\text{T} = \dfrac{V}{I} = \dfrac{10 \text{V}}{1 \text{mA}} = 10 \text{k} \$?

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EDIT 2

After reading the answers and comments, and referring back to the textbook, it has become clear that the textbook has not actually properly explained this concept, and has simply introduced it with the implicit assumption that the reader somehow already understands the details. This is despite the fact that this is, explicitly, an introductory textbook to electronics, with the assumption that the readers have no prior study of the subject. I have no idea how a textbook referred to as "the bible of electronics" has such poor explanations – especially when it's in its third edition!

Answer 1

An ideal current source has an infinite series impedance. If you put a resistor in series with that current source, you're adding the value of your resistor to an already infinite resistance (open circuit), which makes no change. So the series resistor is doing nothing in D.4.

^{simulate this circuit – Schematic created using CircuitLab}

On the other hand, in D.3 the series resistor does do something, because an ideal voltage source has zero series impedance (equivalent to a short circuit for impedance calculations). Adding your series resistor to this gives an equivalent resistance equal to that of the series resistor, so it does impact the impedance seen at the output nodes.

^{simulate this circuit}

This is what they mean when they say that you should open-circuit all current sources and short-circuit all voltage sources when calculating the Thevenin or Norton impedance of a circuit; you're replacing them with their equivalent impedances of infinity and zero, respectively.

Answer 2

The circuits are completely different, one has voltage source and the other has current source. As explained in the text you quoted, voltage supplies have zero impedance, and current supplies have infinite impedance. So, the voltage supply circuit has two 10k in parallel because the voltage supply has zero impedance, and the current supply has only 10k in the circuit because the other 10k has no effect because it is in series with a current source which already has infinite impedance.

Answer 3

To find the Thevenin resistance by looking through the terminals, you have to turn off all independent sources and find the resistance seen by the terminals. So

1. Voltage sources become short circuits
2. Current sources become open circuits

In figure D.3 we have a voltage source, and two resistors. To find the Thevenin resistance we turn off (short circuit) the voltage source. Clearly, \$R_\text{th} = \frac{R_3R_4}{R_3+R_4} = 5 \; \text{k}\Omega \$.

^{simulate this circuit – Schematic created using CircuitLab}

In figure D.4 we have a current source, and two resistors. To find the Thevenin resistance we turn off (open circuit) the current source.

^{simulate this circuit}

Now, the top resistor \$ R_3\$ is effectively disconnected from the circuit (no current can run through it). Therefore the Thevenin resistance (the resistance seen by the terminals) simply becomes \$R_\text{th} = R_4 =10\text{k}\Omega\$.

Answer 4

^{simulate this circuit – Schematic created using CircuitLab}

Answer 5

To get the Thevenin values, you can measure the open-circuit V and the short-circuit I. If you analyze you can see the short-circuit I is different for each circuit (2 mA for the V source one; 1 mA for the other).

An alternative (but equivalent) way of getting the Thevenin resistance is to make all (independent) voltage sources 0 (i.e. a short) and the current sources also 0 (an open circuit). Then you can see why the R's are going to be different.

"Why does figure D.3. have a Thévenin equivalent resistance equal to the parallel resistance, whereas figure D.4. does not?" -- while they both would include the parallel resistance, in the 2nd case that resistance is in series with the (infinite) resistance of the current source -- so it has no effect; you have 10k//infinite == 10k.

Answer 6

If I added this resistor (in red) in parallel with the voltage source, do you see that it has no effect on the outcome: -

That added red-boxed resistor is equivalent to a resistor in series with a current source.

Answer 7

Calculate what happens if you stick a 10 kohm load resistor to the right side of the circuit in each diagram, between the terminals. Check the voltage over that load (and current and power, if you like).

For the first two circuits, in D.3, it's easy to calculate that the voltage over the load is 20 V * 5 kohm / 15 ohm = 6.67 V, and 10 V * 10 ohm / 15 ohm = 6.67 V. So indeed the circuits are equivalent.

Looking at the circuit with the current source, there's a 1 mA current through the top resistor, and then that splits in two: 0.5 mA through both the 10 kohm shunt and the 10 kohm load resistor. That's 0.5 mA * 10 kohm = 5 V over the load. For the equivalent circuit on the right, we get 10 V * 10 kohm / 20 kohm = 5 V over the load. Indeed the two lower circuits in D.4 are also equivalent with each other, but different from the ones at the top in D.3.

That's because of voltage and current sources are different: a voltage source keeps the voltage the same, so adding parallel loads increases the current produced by the load. (And with voltage constant, also the power increases.) With a current source, adding parallel loads splits the same current in more parts, while the voltage stays the same.