This question is about the common emitter circuit, shown below, concerning the influence that the emitter capacitor (\$C_E\$) has on the gain of the amplifier.

Common-emitter amplifier

In the book "Transistor Circuit Techniques", by G. J. Ritchie, the author says that the common emitter has the following critical frequencies, due to the influence of capacitor \$C_E\$:

  1. \$f_0=\dfrac{1}{2\pi C_E(r_e\parallel R_E)}\$ is the frequency at which the voltage gain is 0.707 (\$\sqrt{2}/2\$) times its mid-band level,

  2. \$f_1=\dfrac{1}{2\pi C_ER_E}\$ is the frequency at which the voltage gain is 1.414 (\$\sqrt{2}\$) times the DC voltage gain.

He also says that the voltage gain of the emitter follower can be expressed in the following form, which he calls "standard form":

\$ A_v = A_{dc} \dfrac{ 1 + jf/f_1 }{ 1 + jf/f_0 } \$, where \$A_{dc}\$ is the DC voltage gain (that is, the gain for frequencies where \$C_E\$ can be approximated as an open circuit).

I'm trying to understand the author's statements that I outlined above, and would like some insight. Below are my attempts.

First, I drew the AC model:

Common-emitter amplifier AC model

(In the AC model, since I care about the effect of \$C_E\$'s reactance on the gain, I don't neglect it. However, I assume that the input capacitor \$C_C\$ is a short-circuit for all the frequencies of interest.)

Then, I calculated the gain of the amplifier. Here is what I did:

Calculating \$v_{in}\$ in terms of \$i_b\$:

\$v_{in} = i_b\left (r_\pi + (\beta+1)\left (R_E\parallel \dfrac{1}{j\omega C_E}\right )\right )\$

Calculating \$v_{out}\$ in terms of \$i_b\$:

\$v_{out}=-i_cR_C=-\beta i_bR_C\$

Calculating the gain (\$\dfrac{v_{out}}{v_{in}}\$):

\$A_v=\dfrac{-\beta R_C}{r_\pi + (\beta+1)\left (R_E\parallel \dfrac{1}{j\omega C_E}\right )}=\dfrac{-\beta R_C}{r_\pi + (\beta+1)\dfrac{R_E}{j\omega C_ER_E+1}}\$

Now, in order to express that in the "standard form" given by the book's author, I will find \$A_{dc}\$ (the DC gain). The DC gain can be found by making \$\omega=0\$ in the expression for \$A_v\$:

\$A_{dc}=\dfrac{-\beta R_C}{r_\pi + (\beta+1)R_E}\$

Now, \$A_v\$ can be expressed in tems of \$A_{dc}\$:

\$A_v=A_{dc}\dfrac{j\omega C_ER_E+1}{1+\dfrac{j\omega C_ER_Er_\pi}{r_\pi+(\beta+1)R_E}}\$

\$A_v=A_{dc}\dfrac{1+j\omega C_ER_E}{1+\dfrac{j\omega C_ER_Er_e}{r_e+R_E}}=A_{dc}\dfrac{1+j\omega C_ER_E}{1+j\omega C_E(R_E\parallel r_e)}\$

Comparing this with the standard form, I can see that \$f_1=\dfrac{1}{2\pi C_ER_E}\$ and \$f_0=\dfrac{1}{2\pi C_ER_E\parallel r_e}\$, just as stated by the author.

Now, I can also calculate these frequencies directly from their definition. For example, if calculate \$\omega_1\$ (the angular frequency corresponding to frequency \$f_1\$) by making \$\left | \dfrac{A_v}{A_{dc}} \right | = \sqrt{2}\$, I get:

\$\left | \dfrac{1+j\omega_1 C_ER_E}{1+j\omega_1 C_E(R_E\parallel r_e)} \right |=\sqrt{2}\$

Solving the above for \$\omega_1\$, I get:

\$\dfrac{1+\omega_1^2 (C_ER_E)^2}{1+\omega_1^2 C_E(R_E\parallel r_e)^2}=2\$

\$1+\omega_1^2 (C_ER_E)^2=2+2\omega_1^2 C_E(R_E\parallel r_e)^2\$

\$\omega_1=\dfrac{1}{C_E\sqrt{R_E^2-2(R_E\parallel r_e)^2}}\$

That gives me an \$f_1\$ of: \$f_1=\dfrac{\omega_1}{2\pi}=\dfrac{1}{2\pi C_E\sqrt{R_E^2-2(R_E\parallel r_e)^2}}\$

This is close to the result stated by the author. If I neglect \$r_e\$ as being very small compared to \$R_E\$, I get the exact same result stated by the author (\$f_1=\dfrac{1}{2\pi C_ER_E}\$).

Is my reasoning correct?

  • 1
    \$\begingroup\$ Knowing where to approximate is part of the art of electronics. Remember this was probably written before the wide spread availability of modelling software when engineers did calculations on the back of a brown envelope. Of course I could point out that you have failed to add into your calculation the tolerances of the capacitor (usually 20% for electrolytics) and resistances which will produce a spread of values. Now ask yourself the questions - was it worth all that extra time and effort? Will the circuit you build really be any better? The answer is a simple No. \$\endgroup\$ Nov 23, 2016 at 13:29
  • \$\begingroup\$ Actually $$ f_1=\frac{1}{2\pi C_\text{E} R_\text{E}} $$ is not an approximated relation but it's truly exact. Some mistake must have crept in your calcutations. \$\endgroup\$
    – carloc
    Nov 23, 2016 at 21:08

1 Answer 1


Your reasoning is correct. As pointed out in the comment, approximations are very useful in reducing the amount of calculations required to determine the performance of a circuit. In general, most transistor circuits are designed to make their performance independent of transistor parameters as much as possible because those parameters are quite variable in practical transistors. In this case, a good approximation, as you already know, is that the value of the emitter resistance is much smaller than the external resistor. This greatly simplifies the calculations. In any case, tolerances on components will probably swamp out the value of the exact calculation.

  • \$\begingroup\$ Barry, I suppose with "emitter resistance" you are referring to the inverse of the transconductance (1/gm), correct? For my opinion, we shouldn`t call this quantity "emitter resistance" because (a) it does not belong to the emitter and (b) it is a "transresistance": d(VBE)/d(IC). \$\endgroup\$
    – LvW
    Nov 23, 2016 at 14:44
  • \$\begingroup\$ no 1/gm is not equal to "re". "re" = d(VBE)/d(IE) = VT/IE = Alpha/gm; And "re" belongs to the emitter. \$\endgroup\$
    – G36
    Nov 23, 2016 at 16:14
  • \$\begingroup\$ G36, are you aware that we speak about a difference of 1% or less? (Difference between iE and IC). How can you say that re "belongs to the emitter"? I am sure you know that a resistance is defined by a current through a certain path and the corresponding voltage across both ends of this path. Here: VBE between B and E - and the current between B and E is IB but not IE. Don`t mix "resistance" with "transresistance". The transconductance gm=1/re does not "belong to the emitter". This is a wrong understanding. Rather, it is the slope of the curve IE=f(VBE). \$\endgroup\$
    – LvW
    Nov 23, 2016 at 18:18
  • \$\begingroup\$ @LvW re "belongs to the emitter" because for example the CE amplifier gain with emitter degeneration resistor RE is equal to $$ Av = - (Rc/(re + RE)) * hfe/(hfe +1) $$ And as you can see re is in series with RE. Also sometimes it is easier to visualize the behaviour of a circuit by treating the transconductance of the transistor as if it were a built in dynamic intrinsic emitter resistance re =d(VBE)/d(IE). And this is why re is so popular among the engineers. users.physics.harvard.edu/~horowitz/aoe/sm/smlitlre.htm \$\endgroup\$
    – G36
    Nov 24, 2016 at 17:52
  • \$\begingroup\$ G36, while saying "belongs to the emitter" means that re is the resistance of ...which part of the transistor? Which path between which points? Have you an answer? No, I don`t think so. Again: It is the voltage-to-current ratio of two quantities which do NOT belong to a common path - hence, it has the dimension of a resistor, but it is not a resistor. Instead, we call it "transresistance" (inverse of transconductance). To me, it is not important if it is "popular" or not - remember: Many engineers still think that the BJT would be current-controlled (also a "popular" but wrong view). \$\endgroup\$
    – LvW
    Nov 24, 2016 at 21:47

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