bjt - Why the input resistance of a common emitter amplifier is like this?

I am trying to find the input resistance of this BJ common-emitter amplifier:

I replace the transistor with the hybrid-pi model

It appears clear to me that the input impedance will be

$ R_{in} = R_1 // R_2 // (R_E + r_\pi) $

but some authors say

$ R_{in} = R_1 // R_2 // h_{FE}(R_E + r_\pi) $

What is the correct value?

Answer

Draw this small-signal equivalent circuit:

$R_{IN} = \frac{V_X}{I_B}$

^{simulate this circuit – Schematic created using CircuitLab}

And we can see that $R_{IN} = \frac{V_X}{I_B}$

$$V_X = I_B\cdot r_\pi + I_ER_E = I_B\cdot r_\pi + (I_B + I_C )R_E = I_B\cdot r_\pi + (I_B + h_{FE}I_C )R_E$$ $$=I_B\cdot r_\pi +I_B(h_{FE}+1)R_E$$

Therefore

$R_{IN} = \frac{V_X}{I_B} =r_\pi + (h_{FE}+1)R_E$

Or simply think about emitter current

$I_E = I_B + I_C = I_B + \beta I_C = I_B(\beta+1) $

Try read this http://www.ittc.ku.edu/~jstiles/412/handouts/5.6%20Small%20Signal%20Operation%20and%20Models/section%205_6%20%20Small%20Signal%20Operation%20and%20Models%20lecture.pdf