1
\$\begingroup\$

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

enter image description here

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?

\$\endgroup\$
10
  • 1
    \$\begingroup\$ convert each pair to Vth, Rth then solve. Assume Vpn=0.6 \$\endgroup\$ Jan 23, 2022 at 7:38
  • 2
    \$\begingroup\$ What did you get for Rth? Vth is quite easy without even redrawing the circuit. Two voltage dividers when the diode is removed. \$\endgroup\$ Jan 23, 2022 at 7:49
  • 2
    \$\begingroup\$ Remove the diode and then find the voltage on the right (Vr) side and next, find the voltage on the left side (VL). And the Vth is Vr - VL. How do you get Rth = 2.2222k? \$\endgroup\$
    – G36
    Jan 23, 2022 at 8:07
  • 1
    \$\begingroup\$ @Amanda What others have said: You can treat any pair of resistors, used as a divider pair and an ideal voltage source across them, as a Thevenin equivalent of a different ideal voltage source having a series resistance. As you have two such divider pairs, they can each be treated separately and then the analytic schematic is much simpler, allowing you to combine the two equivalent series resistances and take the difference of the two voltage sources. That's simpler still. And \$2\:\text{k}\Omega\mid\mid 2\:\text{k}\Omega+3\:\text{k}\Omega\mid\mid 2\:\text{k}\Omega=2.2\:\text{k}\Omega\$. \$\endgroup\$
    – jonk
    Jan 23, 2022 at 8:33
  • 1
    \$\begingroup\$ @Amanda You should find a rather small voltage difference. If your mental model of the diode is a fixed but non-zero voltage difference, for example, then you likely will "find" that the diode doesn't conduct. If your mental model of the diode is still simpler, having a zero voltage difference, then the diode conducts. But this situation is such that only the Shockley diode model is sufficient to compute a specific quantity for the resulting diode current. Just by way of seeing how complicated that can be for this situation, see here. \$\endgroup\$
    – jonk
    Jan 23, 2022 at 8:43

1 Answer 1

3
\$\begingroup\$

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:

schematic

simulate this circuit – Schematic created using CircuitLab

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

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

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:

  • $$\text{V}_\text{th}=\frac{3}{10}=0.3\space\text{V}\tag7$$
  • $$\text{I}_\text{th}=\frac{3}{22000}\approx0.000136364\space\text{A}\tag8$$
  • $$\text{R}_\text{th}=2200\space\Omega\tag9$$
\$\endgroup\$
2
  • \$\begingroup\$ Thank you so much @J.W.L. Jan Eerland ! This is extremely helpful. I guess I just had no idea how to move from my circuit in the question to the circuit you have in the very beginning with a common ground! Once I did that (and learned the logic behind it) I was able to get those answers! Thanks again for all your help! \$\endgroup\$
    – Amanda
    Jan 24, 2022 at 2:32
  • \$\begingroup\$ @Amanda I am glad that I could help you, you are welcome! \$\endgroup\$ Jan 24, 2022 at 8:46

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