Mesh analysis in BJT circuits
Sure! It's easier if you redraw the circuit to make the three-terminal BJT a pair of two-terminal components. You can do a large-signal analysis; it doesn't have to be small-signal.







simulate this circuit &ndash; Schematic created using CircuitLab

Note that there's an extra unknown -- the voltage across the dependent current source. This is the collector-emitter voltage \$V_{CE}\$. To complete the analysis, you need an extra equation:

$$I_C = \alpha I_E = \frac {\beta}{\beta + 1} I_E$$

When \$\beta\$ is large,

$$I_C \approx I_E$$ (to within a few percent)

You might be tempted to go one step further and assume that \$I_B \approx 0\$. This would let you use a quick-and-dirty method for finding the bias point:


Calculate \$V_B\$ by treating the base bias resistors as a simple voltage divider.
Subtract 0.6V to get \$V_E\$.
Use \$V_E\$, \$R_E\$, and Ohm's Law to get \$I_E\$.
\$I_C\$ and \$I_E\$ are approximately the same, so now you have \$I_C\$ too.
Use \$I_C\$, \$R_C\$, and \$V_{CC}\$ to get \$V_C\$.


But this method only gives good results when the base resistors are small. (They usually aren't.) So it's better to include the base current in your analysis.

UPDATE:

In your schematic, you don't have loops for \$R_{C1}\$ and \$R_{C2}\$. You should have five loops total.

Let's go through the analysis of this circuit. I'll give the resistors some arbitrary values and assume \$\beta = 100\$.

$$-V_{CC} + (I_{CC} - I_C)R_{B1} + (I_{CC} - I_E)R_{B2} = 0$$
$$(I_E - I_{CC})R_{B2} + 0.6\mathrm{V} + I_E R_E = 0$$
$$(I_C - I_{CC})R_{B1} + I_C R_C + V_{CE} - 0.6\mathrm{V} = 0$$
$$I_C = \frac {\beta} {\beta + 1} I_E$$

Solving this gives:

$$I_{CC} = 5.272 \mathrm{mA}$$
$$I_E = 4.294 \mathrm{mA}$$
$$I_C = 4.251 \mathrm{mA}$$
$$V_{CE} = 1.455 \mathrm V$$

and from there, we can get:

$$I_B = I_E - I_C = 43 \mathrm {\mu A}$$
$$I_{RB1} = I_{CC} - I_C = 1.021 \mathrm{mA}$$
$$I_{RB2} = I_{CC} - I_E = 0.978 \mathrm{mA}$$
$$V_C = V_{CC} - I_C R_C = 5.749 \mathrm V$$

The only hard part is solving the system of equations. I use a TI-89 for this. On a networked computer, there are many other options.

Let's try this in the CircuitLab simulator with a \$\beta = 100\$ transistor and see what we get:







simulate this circuit

With this transistor (2N2222), the simulation results for \$I_C\$ and \$V_C\$ are within ~1% of our calculated values. Not bad!