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

enter image description here

I replace the transistor with the hybrid-pi model

enter image description here

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?

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    \$\begingroup\$ Have you tried to do the analysis yourself? As a first approximation use this $$R_{IN} \approx R_1||R_2||(r_\pi + (h_{FE}+1)\cdot R_E) $$ \$\endgroup\$ – G36 Nov 20 '18 at 17:34
  • \$\begingroup\$ Both are wrong. (G36 was some seconds earlier than me). \$\endgroup\$ – LvW Nov 20 '18 at 17:34
  • \$\begingroup\$ My recommendation: Do not (blindly) rely on some obscure internet contributions. Instead, use a good text book and/or do your own calculations. \$\endgroup\$ – LvW Nov 20 '18 at 17:37
  • \$\begingroup\$ @G36 - where do I find the explanation about this formula? \$\endgroup\$ – SpaceDog Nov 20 '18 at 17:48
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    \$\begingroup\$ SpaceDog_you can derive the result by simple inspection of the circuit - if you consider the fact, that the current through RE is larger by the factor (1+beta) if compared with the current into the base node. \$\endgroup\$ – LvW Nov 20 '18 at 17:52

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}\$

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


\$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

  • \$\begingroup\$ Now I see. THANKS. Just one question: why is the upper part of the current source connected to ground? \$\endgroup\$ – SpaceDog Nov 20 '18 at 18:17
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    \$\begingroup\$ Because we are interested in AC dynamic resistance only we use the superposition principle and we turn-off all DC-voltage sources ( set all DC sources to zero). Look at this answer electronics.stackexchange.com/questions/298560/… \$\endgroup\$ – G36 Nov 20 '18 at 18:24
  • \$\begingroup\$ OK, I was suspecting that. Just checking... 😃 Thanks again. \$\endgroup\$ – SpaceDog Nov 20 '18 at 18:35

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