Question for you. Currently 'm in the middle of studying for an analog II exam. I've got the following midband small signal model:
When analyzing for midband gain, more specifically VB2/VPi1, the solution just considers the one current source (GmVpi1) to be running across the resistance RC1//Ri2.
$$\large\frac{V_{b2}}{V_{\pi1}} = -g_m \left[R_{C1} \| \left(r_{e2} + R_{E2} \| R_L \right)\left(1 + \beta_2\right) \right]$$
In reality there are two current sources! What gives? Is it because we know that the current being supplied by Gm2Vpi2 is much much smaller than Gm1Vpi1 that we ignore it?
Best Answer
The current supplied 2nd source \$g_{m2}v_{\pi 2}\$ is not negligible and is not ignored here.
The current entering to node \$v_{b2}\$ from left is the base current to second transistor (let it be \$i_{b2}\$). Then the collector current will be \$\beta_2i_{b2}\$ hence the emitter current will be \$(\beta_2 + 1)i_{b2}\$. That is why the final equation has a multiplication term of \$(\beta_2 + 1)\$ in it.
OR the 2nd source provides a current of \$\beta_2i_{b2}\$.