Consider the following circuit:

enter image description here

It is required to calculate the small signal voltage gain (i.e \$\frac{v_o}{v_i}\$) based on \$R_s, R_c, I_c, V_A\$ and β.

So I draw the small signal model as below:


simulate this circuit – Schematic created using CircuitLab

So, for \$A_v\$ we have:

\$v_o = g_m v_{be} (r_o || R_c)\$

\$v_{be} = v_i \frac{rπ}{(rπ + Rs)}\$


\$A_v = \frac{v_o}{v_i} = g_m \frac{r_π}{R_s + r_π} (r_o || R_c)\$

I think everything is okay so far! Am I right?

In the next part of question, I have been asked to find the value of \$I_c\$ which we have maximum \$A_v\$ for it. I have been stuck in this point.

Would you please give me some hints?

  • \$\begingroup\$ Hint: Think about when does the circuit become non-linear? \$\endgroup\$ – The Photon Aug 4 '17 at 15:04
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    \$\begingroup\$ Abraham, try using \\$\frac{A}{B}\\$ in the future, it will look like this: \$\frac{A}{B}\$. It's much easier to read. Also, it will make people not give up because of poor formatting. Just look at this: vo = gm vbe (ro || Rc) => \$V_O = gm×V_{BE}×(R_O||R_C)\$ \$\endgroup\$ – Harry Svensson Aug 4 '17 at 15:07
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    \$\begingroup\$ \$R_C\$ => R_C, \$R_C^2\$ => R_C^2, \$V_{BE}\$ => V_{BE}, \$1.3×10^5\$ => 1.3×10^5, alt gr + shift + * = ×, alt gr + shift + Q = Ω. at least those shortcuts works on my nordic Swedish keyboard. \$\endgroup\$ – Harry Svensson Aug 4 '17 at 15:23
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    \$\begingroup\$ MathJax tutorial is here. Unlike other SE sites, EE uses \$ instead of just $ to start and end inline math, since prices of things are often on-topic here. \$\endgroup\$ – The Photon Aug 4 '17 at 15:43
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    \$\begingroup\$ The gain is missing a negative sign, min Ic is 0 max Ic is Vcc/RC which value is best between the two \$\endgroup\$ – sstobbe Aug 4 '17 at 15:43

Max gain for a CE topology is VDD / 0.026 volts. Assuming the Vsource drives the base directly.

  • \$\begingroup\$ I really do not understand it, this does not give me any sense... (btw., neither you specify what VDD means). I have never, either nowhere, seen anything like that. Would you be so kind and provide a reference proving your claims? \$\endgroup\$ – Eric Best Jun 25 '18 at 16:04
  • \$\begingroup\$ Maximim gain (without feedback) - and assuming DIRECT base drive - is Gmax=gmRc=IcRc/Vt=(Vcc-Vce)/Vt=0.5*Vcc/0.026. Here I have assumed Vce=0.5*Vcc (which is best for symmetric operation). \$\endgroup\$ – LvW Dec 4 '18 at 14:01

Quick estimate.

Ib is vi/Rs

Ic is beta x Ib or beta x vi / Rs

Vout is Ic x Rc = beta x Rc / Rs x vi

You can get the gain from that quickly.

  • \$\begingroup\$ The first equation assumes that the base node is at ground potential. (A very quick and very rough estimate). \$\endgroup\$ – LvW Dec 4 '18 at 13:52

The task is to calculate the small-signal voltage gain based on \$R_s\$, \$R_c\$, \$I_c\$, \$V_A\$, and \$ \beta \$ for the given schematic diagram. Therefore, the small-signal model you have drawn does not correspond to the task because you have used small-signal parameters (\$r_\pi \$, \$g_m\$, and \$r_o\$) different from the required ones. Also, you do not specify anywhere what \$V_A\$ means, neither you use it and I must confess I do not have any idea as to its meaning here (any suggestion?). However, the designation \$ \beta \$ is normally used as a synonym for the hybrid small-signal parameter \$h_{21e}\$ (the suffix e indicates the common emitter topology), so it is obvious the small-signal model based on the \$h_e\$-parameters was assumed to be used for the calculations. The explored circuit including the mentioned full small-signal model then looks like this:


simulate this circuit – Schematic created using CircuitLab

The corresponding equations of that model are as follows:

\$ v_1 = h_{11e} \cdot i_1 + h_{12e} \cdot v_2 \$

\$ i_2 = h_{21e} \cdot i_1 + h_{22e} \cdot v_2 \$

and the equations for the whole explored circuit can obviously be written as:

\$ V_i = (R_s + h_{11e}) \cdot i_1 + h_{12e} \cdot V_o \$

\$ -\frac{V_o}{R_c} = h_{21e} \cdot i_1 + h_{22e} \cdot V_o \$

From the above equations there can be derived the small-signal voltage gain \$ A_v \$ of the circuit:

\$ A_v = \frac{V_o}{V_i} = \frac{1}{h_{12e} - \frac{(R_s + h_{11e}) \cdot (h_{22e} + \frac{1}{R_c})}{h_{21e}}} = \frac{-h_{21e} \cdot R_c}{(R_s + h_{11e}) \cdot (h_{22e} \cdot R_c + 1) - h_{12e} \cdot h_{21e}} = \frac{-\beta \cdot R_c}{(R_s + h_{11e}) \cdot (h_{22e} \cdot R_c + 1) - h_{12e} \cdot \beta} \$

In the first approximation, let's presume both \$ h_{12e} \rightarrow 0 \$ and \$ h_{22e} \rightarrow 0 \$, then the formula would simplify to:

\$ A_v = \frac{-h_{21e} \cdot R_c}{R_s + h_{11e}} = \frac{-\beta \cdot R_c}{R_s + h_{11e}} \$

To find the \$I_c \$ at which the voltage gain has its maximum, the h-parameters as functions of \$I_c \$ have to be known (they are nonlinear functions, of course; W. Shockley transistor equations). Then the first derivative \$ \frac{\partial A_v}{\partial I_c}\$ of the function \$ A_v = f(I_c) \$ has to be set to 0 and the corresponding \$I_c(\frac{\partial A_v}{\partial I_c}=0) \$ expressed out of it. The second condition for the gain to be a maximum is that the second derivative of the same function at that calculated \$I_c \$ has to be less than 0. In fact, if the function in charge is known, it can also be plotted and its maximum found this way.

Another view: the maximum theoretical accessible amplitude of the amplified signal is \$ \frac{V_{cc}}{2} \$, it means the operating point must be set to \$ I_c = \frac{V_{cc}}{2R_c} \$.


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