I could figure out how the small signal pi model is obtained from the large signal model and below shows an example circuit where I also draw the pi model:

enter image description here

But from this point does KVL hold? How can we write the KVL equations? The change in small signal vin is not the same thing with the change in vpi. What kind of KVL would be meaningful to form in this case? Or we should use KCL?


1 Answer 1


Why do you think that KVL and KCL do not hold anymore?

For your circuit we can write using KVL and KCL and because \$ v_\pi = i_b \times r_\pi\$ we can write:

$$i_e = i_b + i_c = i_b + g_m i_b r_\pi = i_b(1 +g_mr_\pi) $$

$$v_{in} = i_b r_\pi + i_e R_E = i_b r_\pi + i_b(1 +g_mr_\pi)R_E $$

And if we treat the voltage across the emitter as an output wee will have:

$$v_o = i_e R_E = i_b(1 +g_mr_\pi)R_E$$

And the voltage gain is:

$$\frac{v_o}{v_{in}} = \frac{i_b(1 +g_mr_\pi)R_E}{i_b r_\pi + i_b(1 +g_mr_\pi)R_E } = \frac{(1+g_mr_\pi)R_E}{ r_\pi + (1 +g_mr_\pi)R_E }$$

Also you should remember that :

\$g_m r_\pi = \beta\$


Because the BJT transconductance is equal to \$\large g_m = \frac{dI_C}{d V_{be}}\$ and \$\large r_\pi =\frac{d V_{be}}{dI_B} \$ therefore:

$$ g_m \times r_\pi = \frac{dI_C}{d V_{be}} \times \frac{d V_{be}}{dI_B} = \frac{d I_C}{d I_B} = \beta$$

  • \$\begingroup\$ Thanks a lot! I wasnt sure about it. \$\endgroup\$
    – GNZ
    Commented Jan 17, 2020 at 22:42
  • \$\begingroup\$ @Genzo Also remember that we "invented" the small-signal model to be able to analyze the transistor circuit using only linear equations without involving nonlinear equations. But in both of these cases, the KVL and KCL hold true (can be used). \$\endgroup\$
    – G36
    Commented Jan 18, 2020 at 12:01

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