Thevenin equivalent in a diode circuit with ground between resistors

Question

I'm trying to find the Thevenin equivalent of this circuit with respect to the diode.

I know the steps needed to find Vth and Rth and I was able to find Rth already (2.2222k) but I'm having a hard time finding Vth mainly because of the ground between between the lower 2k resisters and not next to the negative terminal of the source.

Can someone advise me of what I'm missing here or what I need to do?

Answer 1

First, I will present a method that uses Mathematica to solve this problem. I know that this approach is not 'smart' but this method will work all the time, even when the circuit is way complicated than this one.

Well, we are trying to analyze the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

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

$$ \begin{cases} \text{I}_0=\text{I}_1+\text{I}_2\\ \\ \text{I}_3=\text{I}_1+\text{I}_5\\ \\ \text{I}_2=\text{I}_4+\text{I}_5\\ \\ \text{I}_0=\text{I}_3+\text{I}_4 \end{cases}\tag1 $$

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

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

Using \$(2)\$ we can rewrite \$(1)\$ as follows:

$$ \begin{cases} \text{I}_0=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_\text{i}-\text{V}_2}{\text{R}_2}\\ \\ \frac{\text{V}_1}{\text{R}_3}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_2-\text{V}_1}{\text{R}_5}\\ \\ \frac{\text{V}_\text{i}-\text{V}_2}{\text{R}_2}=\frac{\text{V}_2}{\text{R}_4}+\frac{\text{V}_2-\text{V}_1}{\text{R}_5}\\ \\ \text{I}_0=\frac{\text{V}_1}{\text{R}_3}+\frac{\text{V}_2}{\text{R}_4} \end{cases}\tag3 $$

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

In[1]:=Clear["Global`*"];
FullSimplify[
 Solve[{I0 == I1 + I2, I3 == I1 + I5, I2 == I4 + I5, I0 == I3 + I4, 
   I1 == (Vi - V1)/R1, I2 == (Vi - V2)/R2, I3 == V1/R3, I4 == V2/R4, 
   I5 == (V2 - V1)/R5}, {I0, I1, I2, I3, I4, I5, V1, V2}]]

Out[1]={{I0 -> (((R1 + R2) (R3 + R4) + (R1 + R2 + R3 + R4) R5) Vi)/(
   R3 R4 R5 + R1 R4 (R3 + R5) + R2 R3 (R4 + R5) + 
    R1 R2 (R3 + R4 + R5)), 
  I1 -> ((R4 R5 + R2 (R3 + R4 + R5)) Vi)/(
   R3 R4 R5 + R1 R4 (R3 + R5) + R2 R3 (R4 + R5) + 
    R1 R2 (R3 + R4 + R5)), 
  I2 -> ((R3 R5 + R1 (R3 + R4 + R5)) Vi)/(
   R3 R4 R5 + R1 R4 (R3 + R5) + R2 R3 (R4 + R5) + 
    R1 R2 (R3 + R4 + R5)), 
  I3 -> (((R1 + R2) R4 + (R2 + R4) R5) Vi)/(
   R3 R4 R5 + R1 R4 (R3 + R5) + R2 R3 (R4 + R5) + 
    R1 R2 (R3 + R4 + R5)), 
  I4 -> ((R3 (R2 + R5) + R1 (R3 + R5)) Vi)/(
   R3 R4 R5 + R1 R4 (R3 + R5) + R2 R3 (R4 + R5) + 
    R1 R2 (R3 + R4 + R5)), 
  I5 -> (-R2 R3 Vi + R1 R4 Vi)/(
   R3 R4 R5 + R1 R4 (R3 + R5) + R2 R3 (R4 + R5) + 
    R1 R2 (R3 + R4 + R5)), 
  V1 -> (R3 ((R1 + R2) R4 + (R2 + R4) R5) Vi)/(
   R3 R4 R5 + R1 R4 (R3 + R5) + R2 R3 (R4 + R5) + 
    R1 R2 (R3 + R4 + R5)), 
  V2 -> (R4 (R3 (R2 + R5) + R1 (R3 + R5)) Vi)/(
   R3 R4 R5 + R1 R4 (R3 + R5) + R2 R3 (R4 + R5) + 
    R1 R2 (R3 + R4 + R5))}}

Now, we can find:

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

Where I used the following Mathematica-codes:

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

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

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

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

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

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

So, using your values we get:

• $$\text{V}_\text{th}=\frac{3}{10}=0.3\space\text{V}\tag7$$
• $$\text{I}_\text{th}=\frac{3}{22000}\approx0.000136364\space\text{A}\tag8$$
• $$\text{R}_\text{th}=2200\space\Omega\tag9$$