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 1


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}\$.

  • \$\begingroup\$ 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). \$\endgroup\$
    – mHo2
    Commented Aug 16, 2014 at 13:06

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