AC equivalent circuit of transistor
instead it offers a resistance called ac resistance. why it is so ? why dont it rectify ?


Let the voltage across the junction be of the following form

$$v_{BE} = V_{BE} + v_{be}$$

where \$V_{BE}\$ is positive and constant while \$v_{be}\$ is AC and is, in some sense, small (I'll later clarify what that means).

Now, recall the equation for the base current:

$$i_B = \frac{I_S}{\beta}e^{\frac{v_{BE}}{V_T}}$$

Substituting the form for \$v_{BE}\$ given earlier yields

$$i_B = \frac{I_S}{\beta}e^{\frac{V_{BE} + v_{be}}{V_T}}=\frac{I_S}{\beta}e^{\frac{V_{BE}}{V_T}}e^{\frac{v_{be}}{V_T}}$$ 

But, the quiescent (no signal) base current is just

$$I_B = \frac{I_S}{\beta}e^{\frac{V_{BE}}{V_T}}$$

Thus, the total base current is

$$i_B = I_Be^{\frac{v_{be}}{V_T}}$$

The crucial next step is to assume that \$v_{be}\$ is small enough \$^1\$ such that we can approximate the exponential term with

$$e^{\frac{v_{be}}{V_T}} \approx \left(1 + \frac{v_{be}}{V_T}\right)$$

In that case, we can write the base current in the following form:

$$i_B = I_B + i_b $$

where

$$i_b = \frac{I_B}{V_T}v_{be} = \frac{v_{be}}{r_{\pi}}$$

We call \$r_{\pi} = \frac{V_T}{I_B}\$ the small signal base-emitter resistance but it's not a true resistance, it's really just the (inverse) slope of the base-emitter IV curve at the point \$I_B, V_{BE}\$.

So, as long as the junction voltage is approximately the constant \$V_{BE}\$  with only a small variation \$v_{be}\$ added, the current is approximately the constant \$I_B\$ with a small variation \$i_b\$ added that is linearly proportional to \$v_{be}\$.



\$^1\$ e.g., \$v_{be}\lt \lt 4V_T \approx 100\mathrm{mV}\$