Electronic – How to approach the circuit analysis of a BJT in saturation

You can solve this circuit by more or less the same method you've given in the question; however you need to plug in one more equation (\$V_1=V_2+12\$) into the system and introduce an unknown current variable. So I'm not sure if we can call it a 'pure' nodal analysis.

This is what you've got to do:

• Write KCL on the left node (\$I_s\$ is the current through the voltage source): $$6A=\frac { { V }_{ 1 }-{ V }_{ 2 } }{ { R }_{ 3 } } +\frac { { V }_{ 1 } }{ { R }_{ 3 } } +{ I }_{ s }$$

• Do it again on the node on the right side: $$4A={ I }_{ s }-\frac { { V }_{ 2 } }{ { R }_{ 2 } } +\frac { { V }_{ 1 }-{ V }_{ 2 } }{ { R }_{ 3 } } $$

• So far, we're in line with the method described in the question. As a last step, write down this one: $$V_1=V_2+12$$

Now we're left with 3 equations in three unknowns, which you can easily solve to get \$V_1, V_2\$ and \$I_s\$

I think you are asking how beta can be less in saturation than in active mode when it appears from a calculation of Ic that Ic is highest in saturation. If that is your question, the answer is that in saturation, if you increase the base current this fails to further increase the collector current. So in betasat = Ic/Ib, Ib increases but Ic does not, hence betasat reduces.