bjttransistors
Here is the hybrid-\$\pi\$ model:
with \$I_c\$ and \$I_b\$ going in, and \$I_e\$ going out.
The same circuit can be used for both NPN and PNP, the only thing changing being \$V_{be}\$ for NPN and \$V_{eb}\$ for PNP.
The currents direction does not change, which is what I don't understand.
We know that for a PNP in active mode, \$i_c = i_e+i_b\$.
For a NPN in active mode, \$i_e = i_b+i_c\$, which corresponds to the current flow shown on the hybrid-\$\pi\$ model.
Why is it the same for PNP?
simulate this circuit – Schematic created using CircuitLab
The model of each transistor is not dependent on what it will be connected.
Yes it is possible, but you must have into account the sign of the currents and which voltage controls the collector source current, Vbe or Veb.
You can take the NPN model and apply Vbe = -Veb (that's by definition). Then, on the equations for the NPN model you'll have ib = -Veb/rb, ic = -gm Veb and ie = ib + ie;
But now all currents are negative and the arrows pointing against the flow. Since that's not intuitive and we're free to choose the signs, we consider positive the currents that flow outwards and turn around the current source. That translates to the equations to multiply the right members of the equations by -1. This way, you have the PNP model.
As you see, It's a matter of avoiding minus signs. Just like in large signal analysis we say VEB = 0.6V instead VBE = -0.6V for a PNP transistor.
The original equations are: $$I_C = \alpha I_E + I_{CBO} \tag{a}$$ $$I_C = \beta I_B + (\beta +1)I_{CBO} \tag{b}$$
Before equation 4, you mentioned that \$\alpha I_E = I_C\$. But this is valid only when \$I_{CBO} = 0\$ and that is what equation 4 says.
In the same way, \$I_C = \beta I_B\$ is valid only if \$I_{CBO} = 0\$ and that is what equation 5 says.
Both equation 4 and 5 are valid only under the approximation that \$I_{CBO} = 0\$.
Best Answer
It's a small signal AC model, so the DC currents don't matter. Since \$ \frac{dI_{C}}{dV_{BE}}\$ is the same polarity for both, we can use the same model (i.e. a larger b-e voltage results in a larger load current, just they are both negative for the PNP version (which equates to the same result as the NPN)