Thevenin equivalent for a transistor circuit
Call the emitter resistor, \$R_E\$. Just go around the loop:

$$5.6\:\textrm{V} - I_B\cdot R_B - V_{BE} - I_E\cdot R_E = 0$$

Assuming active region (reasonable here, though you should check the results when you are done to be sure about it) and knowing that \$I_E=\left(\beta+1\right)\cdot I_B\$, you can work out:

$$\begin{align*}
5.6\:\textrm{V} - I_B\cdot R_B - V_{BE} - \left(\beta+1\right)\cdot I_B\cdot R_E &amp;= 0\\\\
5.6\:\textrm{V} - V_{BE}&amp;= I_B\cdot R_B - \left(\beta+1\right)\cdot I_B\cdot R_E \\\\
5.6\:\textrm{V} - V_{BE}&amp;= I_B\cdot\left( R_B - \left(\beta+1\right)\cdot R_E\right) \\\\
&amp;\therefore\\\\
I_B=\frac{5.6\:\textrm{V} - V_{BE}}{ R_B - \left(\beta+1\right)\cdot R_E}
\end{align*}$$

From that, you should be able to calculate anything else what you want, I think. However, it ignores the Early effect. But I don't know if you care about it, or not.

If you want, you can re-estimate \$V_{BE}\$ from the currents you work out and then work through the calculations another time to narrow it down some.



If you use \$V_{BE}\approx V_T\cdot\operatorname{ln}\left(\frac{I_C}{I_{SAT}}\right)\$, removing the \$+1\$ in the Shockley equation to simplify the result, and solve it this way:

$$\begin{align*}
I_B&amp;=\frac{5.6\:\textrm{V} - V_T\cdot\operatorname{ln}\left(\frac{\beta\cdot I_B}{I_{SAT}}\right)}{ R_B + \left(\beta+1\right)\cdot R_E}, \quad\textrm{setting: }I_T=\frac{V_T}{R_B + \left(\beta+1\right)\cdot R_E}\\\\
&amp;\textrm{then,}\\\\
I_B&amp;= I_T\cdot \operatorname{LambertW}\left(\frac{I_{SAT}}{\beta\cdot I_T}\cdot e^{\frac{5.6\:\textrm{V}}{V_T}}\right)
\end{align*}$$

Using branch 0 of the LambertW function, I think. Then you don't have to iterate the computation. But most calculators don't include that function key.

(If you are interested in what the LambertW function is [how it is defined] and in seeing a fully worked example on how to apply it to solving problems like these, then see: Differential and Multistage Amplifiers(BJT).)