I just started learning Thévenin's Theorem. My textbook gives the following example:

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I'm referring to online sources such as this to complete this problem. However, all of the circuit diagram examples that I find online are highly similar and unlike the textbook example.

Using my naive understanding, I removed the 4 kilo-ohm resistor and shorted out the 20V and 0.7V (diode) voltage sources. We then have 6 kilo-ohm resistor and 4.9 kilo-ohm resistor in series. Calculating the equivalent resistance, we get $$\frac{(6 k \Omega) (4.9 k \Omega)}{6 k \Omega + 4.9 k \Omega} \approx 2.7 k \Omega$$

But I suspect that I'm doing this incorrectly.

I would appreciate it if people could please take the time to explain this example.

EDIT: I misread the source material, which uses a parallel circuit as an example. Since we have a series circuit, the equivalent resistance would be $$6 k \Omega + 4.9 k \Omega = 10.9 k \Omega$$

  • \$\begingroup\$ @jsotola What in particular? \$\endgroup\$ – The Pointer Mar 19 '19 at 2:23
  • \$\begingroup\$ upvote for not walking away right after posting the question \$\endgroup\$ – jsotola Mar 19 '19 at 2:25
  • \$\begingroup\$ Hint: completely disconnect the 4.9k resistor and calculate the Thevenin Equivalent for the voltage source and the two resistors 6k & 4k. Then re-connect that equivalent circuit to the 4k9 resistor / diode combination and do your calculations. \$\endgroup\$ – Dwayne Reid Mar 19 '19 at 2:25
  • \$\begingroup\$ @jsotola Thanks. Are you saying that I should be simply summing the resistances to find the equivalent resistance? The calculation I did was copied from the specified source. (see electronics-tutorials.ws/dccircuits/dcp_7.html) EDIT: Oh, wait, that calculation is for resistors in parallel, not series. \$\endgroup\$ – The Pointer Mar 19 '19 at 2:29
  • \$\begingroup\$ @DwayneReid thanks for the hint. Why do we disconnect the 4.9k one instead of the 4k one? \$\endgroup\$ – The Pointer Mar 19 '19 at 2:30

Since you are just learning about Thevenin, it's probably easiest to see things in the following way:

  1. Notice that the \$6\:\text{k}\Omega\$ resistor and the \$4\:\text{k}\Omega\$ resistor span between two voltage sources (at \$0\:\text{V}\$ and \$20\:\text{V}\$.)
  2. Convert that pair of resistors and voltage sources to their Thevenin equivalent (which will be a new voltage source and a series resistance.)
  3. Now you should have a Thevenin voltage, \$V_\text{TH}\$, followed by its Thevenin resistance, \$R_\text{TH}\$, followed by a \$4.9\:\text{k}\Omega\$ resistor (in series), followed by the diode.
  4. Since you know the diode drop (given to you, a priori), you just subtract it from the Thevenin voltage value. This will be the remaining voltage that is across the remaining Thevenin resistance in series with the \$4.9\:\text{k}\Omega\$ resistor.
  5. The problem is now reduced to \$I_D=\frac{V_\text{TH}-V_D}{R_\text{TH}+4.9\:\text{k}\Omega}\$.

That's all there is to it.

The primary insight is recognizing step (1) above. Those two resistors and their voltage sources can be converted readily into \$V_\text{TH}\$ and \$R_\text{TH}\$. The rest is just basic machinery steps.

  • \$\begingroup\$ Thanks for the answer. I'm going to study this further in my textbook and come back to your answer. \$\endgroup\$ – The Pointer Mar 20 '19 at 2:53
  • \$\begingroup\$ Insightful and well written. Thanks. \$\endgroup\$ – K H Mar 23 '19 at 2:22

The circuit will simplify to the following:

Disconnect R2 from the power source voltage divider. Calculate Vth and Rth. These are easy to calculate.

Now you have a voltage source with two resistors in series feeding a diode. Calculate the total current. Use that current to calculate the voltage across R2.

Takes only a minute or two, total.


  • \$\begingroup\$ Please explain how this works. I am unsure of how to do the necessary calculations. \$\endgroup\$ – The Pointer Mar 19 '19 at 4:19

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