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

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.

## 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} \$$?

## 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!

• You already know why Rt= Voc/Isc by definition. It is NOT because series vs parallel method , it's only because the R next to the current source has no effect so there is no "equivalent R value" .In fact there is only one effective R in the circuit D.4 and that is the 10k on the right. Vt=IR = 10V. Perhaps you are not trusting the method of Voc/Isc=Rt Oct 10 '21 at 4:26
• Maybe this answer will help you electronics.stackexchange.com/questions/377467/…
– G36
Oct 10 '21 at 8:42
• But the book explains that "as if the current source were replaced with an open circuit" And if you have a voltage source "as if the voltage source were replaced with a short circuit." And this is the answer to your question. In fig D.3 we have a voltage source, thus Rth = R1||R2 (replaces the voltage source with a short circuit). And in fig D.4 we have a current source thus, to find Rth we replace the current source with a open circuit. And kept in mind that AOE is not a book about circuit theory.
– G36
Oct 10 '21 at 19:43
• @G36 Yes, I understand the part about replacing the current source with an open circuit and replacing the voltage source with a short circuit. The part I don't understand is how things work from here. My first edit states clearly what I'm not understanding. Oct 11 '21 at 3:17
• So, you don't understand what "open circuit/"short circuit" means? Open circuit - conductive path for electric current is broken ( cut in a wire I = 0A) and therefore no current may pass between the open circuit (current cannot flow). On the other hand short circuit - joining together of two points by a conductor of negligible resistance, making those two points electrically common with each other and therefore no voltage may exist between them. But the current can flow easily.
– G36
Oct 11 '21 at 13:49

## 7 Answers

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.

• Again, this answer is not clear and understandable in the context of my question. Notice that I have made no mention of impedance – and neither has the textbook – so I don't understand why you're using it in your answer. Oct 10 '21 at 19:27
• Resistance is the real component of impedance--I'm not sure where your understanding is flawed, but if this isn't adequate, could you elaborate on what you're not understanding? Oct 10 '21 at 20:10
• @ThePointer That's the voltage source necessary to make the voltage between the two output nodes the same as in the original circuit. It's always a voltage source in the Thevenin equivalent circuit; if you do it with a current source instead, that's the Norton equivalent circuit. The resistance works out to be the same in either case, but the Norton equivalent has it in parallel with a current source, rather than in series with a voltage source as in the Thevenin equivalent. Oct 11 '21 at 4:24
• @ThePointer Just your standard circuit analysis. In the case of the voltage source, you can use the voltage divider principle to see that the voltage across R2 must be half of the supply voltage, or 10 V. In the case of the current source, you know that 1 mA through 10 kΩ gives you 10 V by Ohm's law. Oct 11 '21 at 4:49
• @ThePointer You're welcome! Sorry it took so long to figure out where the difficulty was. Oct 11 '21 at 19:08

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.

• I don't understand what you're saying regarding the impedance; there is nothing here that discusses impedance, as you are claiming. I don't see how this answers my question at all. Oct 10 '21 at 1:43
• @The_Pointer: That's something you need to know from your studies about voltage sources and current sources. Voltage sources can be replaced by shorts while current sources can be replaced by open circuits. Do that, then calculate the resistance the two-pole offers to the outside viewer. Oct 10 '21 at 3:17
• @Janka There has been no mention of such a thing so far in the textbook, which is why I found it to be so confusing. Oct 10 '21 at 3:33
• Since this is a DC circuit, in steady state, you can say "Resistance" instead of "Impedance". Does that help? Also, in simplest terms, voltage sources are "shorts" and current sources are "opens". Oct 10 '21 at 19:18
• @ThePointer The book is supposed to be practical how to do stuff with little proofs as possible. This may suit some people better than a dry mathematics or physics oriented electronics book with no practical examples. This chapter D is actually a chapter that proves something not proven elsewhere in the book. You are looking at excercises in chapter D.1.1, and how to solve them was explained on previous page in chapter "D1 The Proof". A quote says "replace all internal batteries with short circuits and all current sources with open circuits". It also teaches impedances of sources. Oct 10 '21 at 19:45

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. It's obvious that $$\R_{th} = \frac{R_3R_4}{R_3+R_4} \$$ (if not I can elaborate).

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_{th} = R_4 =10\text{k}\Omega\$$

• How is the top resistor $R_3$ effectively disconnected from the circuit? It is clearly still connected to the circuit. And, when you say that "no current can run through it", isn't that also true for every other component, including $R_4$, since we got rid of the current source? Oct 10 '21 at 18:54
• It's probably best if you include a full explanation of each aspect, because the textbook has clearly skipped over the details of Thévenin's theorem and just assumes that the reader knows what it's talking about. Oct 10 '21 at 19:04
• @ThePointer Looking from the perspective of the terminals A and B, $R_3$ is clearly disconnected since the other end of the resistor is connected to thin air (open circuit) and no current can run through it. I can't say it clearer than this, I think you have to think about it for some days. With regards to your request of me explanation Thevenin's theorem, I will maybe do it tomorrow, if I get the inclination.
– Carl
Oct 10 '21 at 20:28
• My understanding is that an electronic circuit does not necessarily have to be a loop, so the fact that there is a gap / thin air shouldn't matter. Or am I misunderstanding this? Oct 10 '21 at 21:27
• @ThePointer Yes, in circuit theory that is very wrong. Current flows in loops. I would be interested if you could make a circuit that wasn't a loop and still had current running.
– Carl
Oct 11 '21 at 16:15

simulate this circuit – Schematic created using CircuitLab

• I don't see how this answers my question at all. Oct 10 '21 at 1:43
• "D.4. does not? " It does not use both R's for the simplified circuit because R1 above is redundant and has no effect on current in series with a current source. ( I thought this was the most intuitive way to explain it) If you understand this logic, then you will know it is correct. LIkewise putting a resistor across a voltage source should not affect voltage because in theory a voltage source is 0 ohms Oct 10 '21 at 1:52
• Yet you will know that is not true in reality when you excessively load a battery , the voltage will drop from internal resistance by the load , resistor, RL ratio RL/(RL+Rs). Likewise if you remove all loads to a current source it cannot go to infinite voltage. So for now assume ideal properties. Oct 10 '21 at 1:59
• What do you mean putting a resistor across a voltage source should not affect voltage? Ohm's law says that $V = IR$, so I don't see how changing resistance would not affect voltage. Oct 10 '21 at 2:03
• Remember the true voltage source has 0 Ohms source impedance so it will not drop in voltage with any load R. But a Li Ion cell might be 50 mohms on a good day. Oct 10 '21 at 2:22

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.

• I don't see how this answers my question. Oct 10 '21 at 1:44

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.

• My understanding is that adding a resistor in the manner you have shown is adding another resistor in parallel, and so it would have an effect (how does it make sense to say that adding an entire resistor would have no effect?). You state that adding the resistor is equivalent to a resistor in series with a current source, but you have not offered any explanation at all to justify this, so I don't understand how this can be clear. Oct 10 '21 at 19:32
• A voltage source can provide infinite current and hence it has a zero output impedance. Therefore adding a parallel resistor has no overall effect to the bigger picture. If you think about a current source, it has infinite impedance and therefore, adding a series resistor doesn’t change a thing. Oct 10 '21 at 20:03
• I don't understand this; how can it make sense to say that a voltage source can provide infinite current? Current is calculated by Ohm's law $V = IR \Rightarrow I = \dfrac{V}{R}$. Oct 11 '21 at 3:08
• It can provide infinite current into a short circuit. I didn't say that it always delivers infinite current. A current source can provide infinite voltage into an open circuit in order to sustain the current it wants to deliver. Think also of the most perfect voltage source of 0 volts. Yes, it's a voltage source but produces 0 volts. Any amount of current you try and force into that voltage source has no effect whatsoever; i.e. it remains at 0 volts. What is its impedance; it has to be zero ohms (just like any other voltage source. Try thinking about a current source yourself. Oct 11 '21 at 7:02

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.