circuit analysis - 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

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

Answer

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.