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I am hoping someone can help me understand Open Circuit voltage and Thevenin Resistance.

To test my understanding I created the attached circuit and ran a transient analysis using LTSpice.

With a large RLoad resistance of 9e9 Ohms, the node voltage at node N001 is 8.36V, the node voltage at node N002 is about 1.407V.

My understanding is that with an Open Circuit, there can be no current, so there can be no voltage drop across the RLoad resistor.

Does that mean that the Open Circuit voltage is 8.36V?

For the Thevenin Resistance, I changed the RLoad resistor to have a tiny resistance, 1e-9, and the reported current was 1.233 milliamps. From that, I thought rth = 8.36/1.233e-3 = 6780.21 Ohms.

My understanding is there is another way to derive rth. Replace the current source with an open circuit, the voltage source with a short circuit and then determine what the resistance should be. When I try to do that, I don't get the same 6780.21 value. (R3+R4) || (R1+R2) = 6446.69.

If someone could help point out my mistakes I would appreciate it.

Simple Circuit

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    \$\begingroup\$ Watch out for some simulators that don’t with open switches connected to 1 nohm but Ok with 1mohm . ;). tinyurl.com/yea8mk2m \$\endgroup\$ Commented Jul 31, 2021 at 0:20
  • \$\begingroup\$ resolutions can lead to a singularity in matrix \$\endgroup\$ Commented Jul 31, 2021 at 4:16

2 Answers 2

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Open circuit

With an open-circuit, there is no current. That's true. But that doesn't mean there cannot be any voltage difference. That's not true. The mistake here is that you are multiplying a zero current by an infinite resistance. Any finite value can be justified when you multiply infinity by zero. It's not sound reasoning to say the result must be zero.

So, it's best to say that open-circuits permit any finite voltage difference between the points. An open-circuit has no impact on the difference.

Thevenin approach using very high and very low impedance in LTspice

Your approach is one reasoned way to find the Thevenin impedance between two points. In your case, \$N_1\$ and \$N_2\$, using \$1000\:\text{M}\Omega\$ and \$1\:\text{n}\Omega\$ and measuring the current in both cases. You could also just leave it open-loop for one measurement, picking out the node voltages like you did, and then using a \$0\:\text{V}\$ voltage source between the two points and measuring the current through it. Either way, you'll get similar results.

Here, again, you made a mistake. You need to use the voltage difference when measuring the node voltages using the \$1000\:\text{M}\Omega\$ resistance. Here, I get an open loop voltage difference of \$8.36\:\text{V} - 1.4072\:\text{V}\approx 6.95\:\text{V}\$. Note that you don't just take the voltage at one side. Voltages are always measured between two points.

Never, ever, do you consider an absolute voltage value for anything useful. It's just not done. Don't ever do it. It's never useful taken by itself.

I also get your current of about \$1.233\:\text{mA}\$, when shorted with the \$1\:\text{n}\Omega\$ resistance.

I then get a Thevenin resistance of \$\frac{6.95\:\text{V}}{1.233\:\text{mA}}\approx 5.634\:\text{k}\Omega\$.

Verification

Your circuit can be trivially analyzed for its Thevenin resistance without resorting to the above steps.

As seen by node \$N_1\$, \$I_1\$ has an infinite impedance so it can be tossed out and ignored. \$R_1\$ goes to ground and \$R_2\$ goes to a voltage source. So the impedance seen by \$N_1\$ is \$R_1\mid\mid R_2\$.

As seen by node \$N_2\$, \$R_4\$ goes to ground and \$R_3\$ goes to a voltage source. So the impedance seen by \$N_2\$ is \$R_3\mid\mid R_4\$.

Clearly, the impedance seen looking through N1 towards N2 will just be the sum of the above results, or \$\left(R_1\mid\mid R_2\right)+\left(R_3\mid\mid R_4\right)\approx 5.64\:\text{k}\Omega\$.

Which is close enough to the results using your method, when properly applied (taking the voltage difference instead of the absolute value that picked.)

Note, here again, your approach wasn't correct. Compare it to how I just did it and you may see why.

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    \$\begingroup\$ I did not expect such a thorough answer. I very much appreciate it! \$\endgroup\$
    – pgk
    Commented Jul 31, 2021 at 1:09
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First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course).

Well, we are trying to analyze the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \text{I}_\text{a}=\text{I}_1+\text{I}_2+\text{I}_5\\ \\ \text{I}_3=\text{I}_2+\text{I}_6\\ \\ \text{I}_4=\text{I}_3+\text{I}_5\\ \\ \text{I}_6=\text{I}_4+\text{I}_7\\ \\ \text{I}_1=\text{I}_\text{a}+\text{I}_7 \end{cases}\tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1-\text{V}_\text{i}}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_\text{i}-\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_2}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_1-\text{V}_2}{\text{R}_5} \end{cases}\tag2 $$

We can subsitute \$(2)\$ into \$(1)\$, to get:

$$ \begin{cases} \text{I}_\text{a}=\frac{\text{V}_1}{\text{R}_1}+\frac{\text{V}_1-\text{V}_\text{i}}{\text{R}_2}+\frac{\text{V}_1-\text{V}_2}{\text{R}_5}\\ \\ \frac{\text{V}_\text{i}-\text{V}_2}{\text{R}_3}=\frac{\text{V}_1-\text{V}_\text{i}}{\text{R}_2}+\text{I}_6\\ \\ \frac{\text{V}_2}{\text{R}_4}=\frac{\text{V}_\text{i}-\text{V}_2}{\text{R}_3}+\frac{\text{V}_1-\text{V}_2}{\text{R}_5}\\ \\ \text{I}_6=\frac{\text{V}_2}{\text{R}_4}+\text{I}_7\\ \\ \frac{\text{V}_1}{\text{R}_1}=\text{I}_\text{a}+\text{I}_7 \end{cases}\tag3 $$

Now, we can set up a Mathematica-code to solve for all the voltages and currents:

In[1]:=Clear["Global`*"];
FullSimplify[
 Solve[{Ia == I1 + I2 + I5, I3 == I2 + I6, I4 == I3 + I5, 
   I6 == I4 + I7, I1 == Ia + I7, I1 == V1/R1, I2 == (V1 - Vi)/R2, 
   I3 == (Vi - V2)/R3, I4 == V2/R4, I5 == (V1 - V2)/R5}, {I1, I2, I3, 
   I4, I5, I6, I7, V1, V2}]]

Out[1]={{I1 -> (Ia R2 R4 R5 + 
    Ia R2 R3 (R4 + R5) + (R2 + R3) R4 Vi + (R3 + R4) R5 Vi)/(
   R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + 
    R2 R3 R4 + (R1 + R2) (R3 + R4) R5), 
  I2 -> (Ia R1 (R4 R5 + 
       R3 (R4 + R5)) - (R3 (R1 + R4) + (R3 + R4) R5) Vi)/(
   R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + 
    R2 R3 R4 + (R1 + R2) (R3 + R4) R5), 
  I3 -> (-Ia R1 R2 R4 + R1 (R2 + R5) Vi + R2 (R4 + R5) Vi)/(
   R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + 
    R2 R3 R4 + (R1 + R2) (R3 + R4) R5), 
  I4 -> (Ia R1 R2 R3 + R2 R5 Vi + R1 (R2 + R3 + R5) Vi)/(
   R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + 
    R2 R3 R4 + (R1 + R2) (R3 + R4) R5), 
  I5 -> (Ia R1 R2 (R3 + R4) + (R1 R3 - R2 R4) Vi)/(
   R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + 
    R2 R3 R4 + (R1 + R2) (R3 + R4) R5), 
  I6 -> (-Ia R1 ((R2 + R3) R4 + (R3 + R4) R5) + ((R2 + R3) (R1 + 
          R4) + (R1 + R2 + R3 + R4) R5) Vi)/(
   R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + 
    R2 R3 R4 + (R1 + R2) (R3 + R4) R5), 
  I7 -> (-Ia R1 (R3 R4 + 
       R2 (R3 + R4) + (R3 + R4) R5) + ((R2 + R3) R4 + (R3 + 
          R4) R5) Vi)/(
   R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + 
    R2 R3 R4 + (R1 + R2) (R3 + R4) R5), 
  V1 -> (Ia R1 R2 (R4 R5 + R3 (R4 + R5)) + 
    R1 ((R2 + R3) R4 + (R3 + R4) R5) Vi)/(
   R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + 
    R2 R3 R4 + (R1 + R2) (R3 + R4) R5), 
  V2 -> (R4 (Ia R1 R2 R3 + R2 R5 Vi + R1 (R2 + R3 + R5) Vi))/(
   R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + 
    R2 R3 R4 + (R1 + R2) (R3 + R4) R5)}}

Now, we can find:

  • \$\text{V}_\text{th}\$ we get by finding \$\text{V}_1-\text{V}_2\$ and letting \$\text{R}_5\to\infty\$: $$\text{V}_\text{th}=\frac{\text{I}_\text{a}\text{R}_1\text{R}_2\left(\text{R}_3+\text{R}_4\right)+\left(\text{R}_1\text{R}_3-\text{R}_2\text{R}_4\right)\text{V}_\text{i}}{\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_3+\text{R}_4\right)}\tag4$$
  • \$\text{I}_\text{th}\$ we get by finding \$\text{I}_5\$ and letting \$\text{R}_5\to0\$: $$\text{I}_\text{th}=\frac{\text{I}_\text{a}\text{R}_1\text{R}_2\left(\text{R}_3+\text{R}_4\right)+\left(\text{R}_1\text{R}_3-\text{R}_2\text{R}_4\right)\text{V}_\text{i}}{\text{R}_2\text{R}_3\left(\text{R}_1+\text{R}_4\right)+\text{R}_1\text{R}_4\left(\text{R}_2+\text{R}_3\right)}\tag5$$
  • \$\text{R}_\text{th}\$ we get by finding: $$\text{R}_\text{th}=\frac{\text{V}_\text{th}}{\text{I}_\text{th}}=\frac{\text{R}_2\text{R}_3\left(\text{R}_1+\text{R}_4\right)+\text{R}_1\text{R}_4\left(\text{R}_2+\text{R}_3\right)}{\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_3+\text{R}_4\right)}\tag6$$

Where I used the following Mathematica-codes:

In[2]:=FullSimplify[
 Limit[(((Ia R1 R2 (R4 R5 + R3 (R4 + R5)) + 
      R1 ((R2 + R3) R4 + (R3 + R4) R5) Vi)/(
     R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + 
      R2 R3 R4 + (R1 + R2) (R3 + R4) R5)) - ((
     R4 (Ia R1 R2 R3 + R2 R5 Vi + R1 (R2 + R3 + R5) Vi))/(
     R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + 
      R2 R3 R4 + (R1 + R2) (R3 + R4) R5))), R5 -> Infinity]]

Out[2]=(Ia R1 R2 (R3 + R4) + (R1 R3 - R2 R4) Vi)/((R1 + R2) (R3 + R4))

In[3]:=FullSimplify[
 Limit[(Ia R1 R2 (R3 + R4) + (R1 R3 - R2 R4) Vi)/(
  R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4 + (R1 + R2) (R3 + R4) R5),
   R5 -> 0]]

Out[3]=(Ia R1 R2 (R3 + R4) + (R1 R3 - R2 R4) Vi)/(
R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4)

In[4]:=FullSimplify[%2/%3]

Out[4]=(R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4)/((R1 + R2) (R3 + R4))

So, using your values we get:

  • $$\text{V}_\text{th}=\frac{29028}{4175}\approx6.95281\space\text{V}\tag7$$
  • $$\text{I}_\text{th}=\frac{16933}{13736000}\approx0.00123275\space\text{A}\tag8$$
  • $$\text{R}_\text{th}=\frac{6593280}{1169}\approx5640.1\space\Omega\tag9$$
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