Doubt related to impedance seen by the bypass capacitor:


What is the impedance seen by the bypass capacitor \$C_E\$

enter image description here

To calculate the impedance as seen by \$C_E\$, we attach a Thevenin volatge source as shown.

Applying Kirchoff's current law:

$$ \frac{V_T}{\beta r_e+R_S||R1||R_2}-\beta I_B+\frac{V_T}{R_E}=I_T$$ $$\frac{V_T}{\beta r_e+R_S||R1||R_2}+\beta \frac{V_T}{\beta r_e+R_S||R1||R_2} +\frac{V_T}{R_E}=I_T$$ $$V_T[\frac{(1+\beta)}{\beta r_e+R_S||R_1||R_2}+\frac{1}{R_E}]=I_T$$ $$V_T[\frac{1}{\beta r_e+R_S||R1||R_2}+\frac{1}{r_e+\frac{R_S||R_1||R_2}{\beta}}+\frac{1}{R_E}]=I_T$$

From here I get the resistance as $$\frac{1}{R_e}=\frac{1}{\beta r_e+R_S||R1||R_2}+\frac{1}{r_e+\frac{R_S||R_1||R_2}{\beta}}+\frac{1}{R_E}$$

However in book, the resistance has been given as:


It seems the first term in \$\frac{1}{R_e}\$ vanishes! Where might have I gone wrong?

(I have referred to the following text book: Electronic Devices and Circuit Theory, by Boylestad and Nashelsky.)

  • \$\begingroup\$ Do you know the voltage gain of a common-emitter stage with degeneration? If you know, then it would be much easier to figure out the gain with the capacitor added in parallel with Re \$\endgroup\$
    – dirac16
    Commented Feb 27, 2017 at 8:22
  • \$\begingroup\$ I'm not even sure what your question is! Also, you might get a better response if you learnt how to formally accept answers to questions you've raised. If you don't understand an answer then comment but, basically, the only fee that you are charged for receiving good information is pressing a button on the best answer. \$\endgroup\$
    – Andy aka
    Commented Feb 27, 2017 at 8:36
  • \$\begingroup\$ I think he wants to calculate the gain w/ the cap added \$\endgroup\$
    – dirac16
    Commented Feb 27, 2017 at 8:38
  • \$\begingroup\$ @Andyaka I want to calculate the impedance seen by the bypass capacitor \$\endgroup\$
    – Soumee
    Commented Feb 27, 2017 at 8:38
  • 1
    \$\begingroup\$ My wild guess would be assuming beeta*re value is too high and so the first part is negligible. \$\endgroup\$ Commented Feb 27, 2017 at 8:41

2 Answers 2


Soumee, I cannot identify any error in your calculation.

However, if you replace in the third line of your calculation the term (1+β) by β, your result will be identical to the expression as given in the book. As you know the current gain β is relatively large (mostly > 100), not a constant but dependent on Ic and - more important - equipped with large tolerances.

Therefore, we often simplify (1+β) to β (without expecting not acceptable errors) - and this seems to be the only reason for the discrepancy you have observed.


Here put It=0 because according to KCL algebraic sum of currents in the node is zero. \$ 1+b = b \$ (it doent make any change)


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