# Thévenin's Theorem Example

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

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 $$6 k \Omega + 4.9 k \Omega = 10.9 k \Omega$$

• @jsotola What in particular? – The Pointer Mar 19 '19 at 2:23
• upvote for not walking away right after posting the question – jsotola Mar 19 '19 at 2:25
• 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. – Dwayne Reid Mar 19 '19 at 2:25
• @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. – The Pointer Mar 19 '19 at 2:29
• @DwayneReid thanks for the hint. Why do we disconnect the 4.9k one instead of the 4k one? – 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.

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

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