Why to calculate base voltage in given example differently?
You can use the divider in all cases of analysis. The following schematic illustrates the Thevenin equivalency:







simulate this circuit &ndash; Schematic created using CircuitLab

This Thevenin equivalence always works and it never fails. There aren't any tea leaves you need to read to figure out when a divider somehow isn't a divider in one case but is a divider in another. You can always replace the divider, as shown above, whether the transistor is saturated or isn't saturated. It's all the same thing.

Now, you can easily prepare the following equation using Kirchhoff's laws and then solve for \$I_B\$. Treating all terminal currents in \$Q_1\$ as positive and also treating a forward-biased base-emitter junction voltage as positive:

$$\begin{align*}
V_{TH}+I_B\cdot R_{TH}+V_{BE}+I_B\cdot\left(\beta+1\right)\cdot R_1&amp;=V_{CC}
\\\\
\therefore I_B&amp;=\frac{V_{CC}-V_{TH}-V_{BE}}{R_{TH}+\left(\beta+1\right)\cdot R_1}
\end{align*}$$

Of course, normally you can just plug in a reasonable estimate for \$\beta\$ and be done with it. From there, you can work out \$I_C=\beta\cdot I_B\$ and \$I_E=\left(\beta+1\right)\cdot I_B\$.

But what if you can't rely on the assumption of some a priori estimated value (or bracketed range of values) for \$\beta\$? Well, there's another Kirchhoff's law equation you can write down here, and solve it for \$V_{CE}\$:

$$\begin{align*}
0\:\textrm{V}+I_B\cdot\beta\cdot R_3+V_{CE}+I_B\cdot\left(\beta+1\right)\cdot R_1&amp;=V_{CC}
\\\\
\therefore V_{CE}&amp;=V_{CC}-I_B\cdot\left(\beta\cdot R_3+\left(\beta+1\right)\cdot R_1\right)
\end{align*}$$

Now, just plug in the equation for \$I_B\$ into that equation to get:

$$\begin{align*}
V_{CE}&amp;=V_{CC}-\big[V_{CC}-V_{TH}-V_{BE}\big]\cdot\frac{\beta\cdot R_3+\left(\beta+1\right)\cdot R_1}{R_{TH}+\left(\beta+1\right)\cdot R_1}
\end{align*}$$

If \$\vert V_{CE}\vert \ge \vert V_{BE}\vert\$, then your estimated \$\beta\$ value appears to work okay and this means that \$\beta\$ is an input and that \$V_{CE}\$ is an output of the above pair of equations.

But if not, then you know the opposite is true: That \$V_{CE}\$ is an input (or bracketed range of inputs) to the pair of equations (roughly, it's an estimated saturated value) and then \$\beta\$ is an output you need to solve. In this case, perhaps set \$V_{CE}\approx 300\:\textrm{mV}\pm 200\:\textrm{mV}\$ and solve for \$\beta\$:

$$\begin{align*}
\beta&amp;=\frac{R_{TH}\cdot\left( V_{CC} - V_{CE}\right) + R_1\cdot\left( V_{TH} + V_{BE} - V_{CE}\right) }{ R_3\cdot \left(V_{CC} - V_{BE} - V_{TH}\right) - R_1\cdot\left( V_{TH} + V_{BE} - V_{CE}\right)}
\end{align*}$$

Here, you will find the required low value for \$\beta\$ in order for the circuit to work at the given saturation level for \$V_{CE}\$.

But in all cases, your voltage divider concept remains workable. The issue is simply whether or not \$\beta\$ can be treated as constant (active, not saturated) or \$V_{CE}\$ is to be treated as nearly constant (in shallow or deep saturation.)



In your first case, assume \$\beta=200\$ and \$R_3=2\:\textrm{k}\Omega\$. Then you compute \$I_B=4.968\:\mu\textrm{A}\$ and \$V_{CE}\approx 1.724\:\textrm{V}\$. That looks like it is in active mode. So that's fine. So \$I_C\approx 994\:\mu\textrm{A}\$ and \$I_E\approx 999\:\mu\textrm{A}\$.

In your second case, assume \$\beta=200\$ and \$R_3=7\:\textrm{k}\Omega\$. The equation for \$I_B\$ doesn't care about the change in \$R_3\$, so we get the same \$I_B=4.968\:\mu\textrm{A}\$, again. But this time, \$V_{CE}\approx -3.245\:\textrm{V}\$. That tells you that it cannot be in active mode. So now you let go of \$\beta\$ and instead (let's say) choose \$V_{CE}=200\:\textrm{mV}\$. Now, using the last equation above, you get \$\beta\approx 1.914\$.

So that is your saturation \$\beta\$ and the circuit is saturated. Take that value for \$\beta\$ and go back and plug it into the first equation to compute \$I_B\approx 314\:\mu\textrm{A}\$ and \$I_C\approx 601\:\mu\textrm{A}\$ and \$I_E\approx 915\:\mu\textrm{A}\$.



Bottom line? The divider is still a divider. It still has a Thevenin equivalent. All is right with the world. The only question you really need to resolve is if \$\beta\$ may be treated as a constant or if \$V_{CE}\$ may be treated as a constant.



Final note: I wanted to include 3 places of significance in the calculations above so that the quantitative results from entering these expressions correctly can be verified. That is my entire intent in providing that level of precision -- a way of cross-checking the algebra.

I do not, in any way, mean to imply that the resulting quantities will be anywhere near as accurate given actual parts. All of the inputs to these calculations are, at best, approximate. The value of \$\beta\$, assuming that the BJT is in an active region, will vary significantly from part to part. So will the \$V_{BE}\$ for each part. The same is true for any estimation of the saturated \$V_{CE}\$, if the transistor is saturated.

For a practical design, reasonable variations in supply rails, variations around nominal resistor values, and BJT variations around \$V_{BE}\$ and forward \$\beta\$ values should be bracketed during the analysis and then verified by similar runs using Spice to help cross-check those estimates. This was not done here as this is just a question about "what happened to the divider?" Which can be addressed without having to be side-tracked into yet another chapter on a different (important) topic.