transistors - Input resistance of small signal equivalent

I'm trying to calculate the input resistance of the depicted small signal equivalent circuit.

I was just wondering why I can't see the current source as an interrupt and then the input resistance will be: r_pi+R.

I recall that replacing current sources with interrupts and voltage sources with short circuits is done when calculating the resistance between two nodes in a circuit. Why isn't that allowed in the small signal equivalent?

^{simulate this circuit – Schematic created using CircuitLab}

Answer

Your final equation looks good, but you can simplify it further:

$$R_{IN} = R \left( 1 + r_\pi\left(gm + \frac{1}{R}\right)\right)$$

$$R_{IN} = R \left( 1 + r_\pi g_m + \frac{r_\pi}{R}\right)$$

$$R_{IN} = R+ r_\pi g_m R + \frac{r_\pi}{R}R $$

$$R_{IN} = R+ r_\pi g_m R + r_\pi $$

And now knowing that \$r_\pi \cdot g_m = \beta\$ because

\$ \beta= \frac{d(Ic)}{d(Ib)} \$ and \$ \textrm{d(Ib)} = \frac{d(v_{be})}{r_\pi} \$

Therefore:

\$ \frac{\beta}{r_\pi} = \frac{d(Ic)}{d(Vbe)}=g_m\$

So, we can write it like this:

$$R_{IN} = R+\beta R + r_\pi $$

$$R_{IN} = (\beta + 1) R + r_\pi $$

$$R_{IN} = r_\pi + (\beta + 1) R $$