Maximum load resistor for a given single transistor circuit
The following two schematics are equivalent:







simulate this circuit &ndash; Schematic created using CircuitLab

Where \$V_{TH}=V_{CC}\frac{R_2}{R_1+R_2}\$ and \$R_{TH}=\frac{R_1\cdot R_2}{R_1+R_2}\$.

From the above, and assuming \$I_B^{'}=\mid I_B\mid\$ and \$V_{BE}^{'}=\mid V_{BE}\mid\$, you can compute:

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

Given your values, I get \$I_B^{'}\approx 9.6\:\mu\textrm{A}\$.

The start of shallow entry into saturation occurs right when \$V_{BC}=0\:\textrm{V}\$ or when \$V_B=V_C\$:

$$\begin{align*}
V_C&amp;=V_B\\\\
I_C\cdot R_L &amp;= V_{TH}+I_B^{'}\cdot R_{TH}\\\\
\beta\: I_B^{'}\cdot R_L &amp;=V_{TH}+I_B^{'}\cdot R_{TH}\\\\
\therefore \quad R_L &amp;=\frac{V_{TH}+I_B^{'}\cdot R_{TH}}{\beta\: I_B^{'}}=\frac{1}{\beta}\left(R_{TH}+\frac{V_{TH}}{I_B^{'}}\right)
\end{align*}$$

From which I get \$R_L\approx 2130\:\Omega\$.