# BJT small signal analysis, ignoring current sources

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?

## 1 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 $\beta_2i_{b2}$.

• EDIT: Nevermind, obviously in the case of this transistor the first base/emitter node is grounded and thus no base reflection occurs. Thanks! ------------------------------------- Interesting, as base reflection still applies in other cases when that current source is not present. Why is this? (e.g. with the first transistor base/emitter node, the current source goes straight to ground, yet base reflection still applies).
– mHo2
Aug 16, 2014 at 13:06