I am trying to design a Common-Emitter amplifier to the following specifications:

  1. Midband voltage gain of 50
  2. Frequency range 100Hz to 20kHz
  3. Load \$5k\Omega\$ with a coupling capacitor (not shown below)
  4. 12V supply lines
  5. Input source resistance \$100 \Omega\$

Using a 2N2222 BJT transistor and the following CE configuration:

Circuit configuration

I've drawn the small-signal model as follows, assuming that in the midband coupling capacitors are treated as shorts and bypass and load capacitors are treated as open circuits:

Small Signal Model

I began my design by picking a maximum current. I want to keep that low so I choose \$2mA\$ and plot the IC vs VCE curves for this specific transistor in a spice simulator:

IV curves

Choosing a point half way on the load line for symmetrical swing, I obtain:

  • \$V_{CE} = 6V\$
  • \$I_C = 1mA\$
  • \$I_B = 5.5\mu A\$

I calculate \$ \beta = \frac{I_C}{I_B} = \frac{1mA}{5.5\mu A} = 182\$

\$R_c = \frac{V_{cc}}{I_c}=\frac{12}{2mA}=6000 \Omega\$

I pick \$R_E = 0.1R_c = 600 \Omega\$ based on a rule of thumb for beta stability.

Now I would like to design my bias network \$R_1, R_2\$ to enforce the bias conditions above and also give a gain of 50. The voltage gain expression is given as:

\$A_v = - \frac{\beta R_c || R_l}{r_\pi + (1+\beta)R_E}(\frac{R_i}{R_i+R_s})\$

I first calculate

\$r_\pi = \frac{V_T}{I_B} = \frac{0.026V}{5.5\mu A} = 4727 \Omega\$

\$R_{ib} = r_\pi + (1+\beta)R_E = 4727 + (183)(600) = 114527 \Omega\$

Solving for the input resistance:

\$A_v = 50 = \frac{182(6000) || (5000)}{4727 + (183)(600)}(\frac{R_i}{R_i+100})\$

Giving \$R_i=92.02 \Omega\$

\$R_i = R_{thev} || R_{ib}\$

Some algebraic manipulation results in

\$R_{thev} = \frac{-R_i R_{ib}}{R_i - R_{ib}} = 92 \Omega\$

I write a loop around the Emitter-Base loop as:

\$ -Vcc(\frac{R_2}{R_2 + R_1}) + R_{thev} + 0.7 + I_E R_E = 0 \$

Selecting \$R_2 = 6000 \Omega\$, \$R_1\$ is solved for \$50 000 \Omega\$

Simulating the circuit and running a dynamic DC analysis shows that the bias conditions are enforced:

Bias conditions in Simulation

However I am confused because when I work out \$R_i\$ as:

\$ R_{i} = R_{thev} || R_{ib} = \frac{1}{\frac{1}{6000} + \frac{1}{50 000} + \frac{1}{114527.27}} = 5117 \Omega \$

And furthermore the thevenin resistance of \$R_1, R_2\$

\$R_{Thev} = \frac{R_1 R_2}{R_1 + R_2} = 5357 \Omega\$

I would expect these values to be the same as what I worked them out to be previously from the voltage gain equation (\$92 \Omega \$)? In simulation this method seems to work however I can't understand how the input and thevenin resistances have seemingly changed value. Could anyone explain what is happening here?

I'd also be interested to know how more experienced designers would approach this simple design problem. I find that my courses at university are very theoretical and so I tend to over complicate the designs. In practice I suspect that designers tend to take more of a heuristic approach to circuit design rather than solve equations like this.

  • 1
    \$\begingroup\$ one observation: the load line you traced is the static load line. When the frequency is high enough for the capacitors to be considered irrelevant, you will find a completely different slope. The dynamic load line will still be centered at the quiescent point, but the slope will be steeper - and out of the window goes your symmetric swing. It is no coincidence that analogsystemsrf suggested you to use a higher Ic. \$\endgroup\$ Commented Jul 17, 2020 at 18:04
  • \$\begingroup\$ @SredniVashtar Are there some rules of thumb for selection of Ic? The max value for the device is 800mA so what trade offs are there to selecting say 400mA vs 10mA, I would imagine that it depends on how much power should be delivered to the load? \$\endgroup\$
    – Blargian
    Commented Jul 18, 2020 at 7:49
  • \$\begingroup\$ In general you choose Ic in order to set a desired value for gm = Ic/Vt. This will set the 'device gain', but in a circuit with feedback, like the one you use, the gain is set by the resistors in the feedback network. In this case you are choosing Ic by choosing the value of Ie that will give you a Ve low enough to accomodate a reasonable swing. Some might choose Ve as 1V, or 2V, other might like to pick it as a fraction of Vcc. I suggest you trace the static and dynamic load lines for different choices to see what is going on and why, for example, your output might clip in certain cases. \$\endgroup\$ Commented Jul 18, 2020 at 16:54

1 Answer 1


[For best design, add a 2nd Re in series with first R, with no bypassing. These 2 Rs allow stabile biasing and stable Gain. And thanks to LvW for error detection and system thinking.

Input Resistance will be parallel combination of R1, R2, and the transistor Rin.

The transistor Rin, assuming excellent bypassing of Re, will be Beta/gm where gm is the transistor transconductance.

If you have NO emitter bypassing, then Rin is beta * (1/gm + Re). [note: had been, in error, beta * (1/gm * Re); thank you, LVW]

[by the way, the method I use here took me years to "invent". There is no magic in biasing Emitter to 1/3 VDD and biasing collector to 2/3 VDD; and setting Rb1 and RB2 to 10X/20X the Re and Rc, just that you can mostly ignore the beta and the beta variations. In starting out, I used the single_resistor from base_to_VDD as do most newbies; then I learned about Re stabilization for Ie and for gain. I think the GE Transistor manual I was given did show Re stabilization, but I had to slowly learn all the tradeoffs. And even later I still had to learn/visualize the energy sucking o C_Miller_Effect.]

Your instructor is letting you immerse, and learn quickly. Hang in there.


For a gain of 50X, using a single transistor, with resistive load, first let us be sure we can achieve that 50X.

The maximum gain a single bipolar can provide, with resistive load, is

  • VDD/ 0.026 volts.

Since 12/0.026 is about 480, then yes, we can move ahead.

Bias the transistor emitter to VDD/3. Bias the collector to 2/3 VDD.

This is very stable biasing.

You have no required Vout_peakPeak, despite that Rload.

Bias the transistor to 10mA (why not? power is free).

The transconductance "gm" is Ie/0.026 volts = 0.010 / 0.026 = 1/2.6 ohms.

The gm is 1/2.6 ohms. For a gain of 50, set Rc = 50 * 2.6 ohms, or 130 ohms.

With 130 ohms as Rcollector, we will mostly ignore the Rload value in computing the Voltage Gain.

Set the Re to 130 ohms also. Pick a very large Ce, to ensure Zc << 2.6 ohms.

Make top base bias be 20 * 130 ohms

Make bottom base bias be 10 * 130 ohms.


For high bandwidth, you might worry about C_Miller_Effect.

At high frequencies, that charge consumption will suck up lots of input energy.

I ignored any source impedance.


If you must have a high Rin, then consider a Darlington. Which also gives you some flexibility about handling C_miller_effect.

  • \$\begingroup\$ Thank you for your detailed answer. Reading your approach to designing this circuit really makes me question the usefulness of an academic qualification, but I guess that's a whole different topic. I guess this comes from experience and actually building many many circuits, not just analysing them on paper. Cheers! \$\endgroup\$
    – Blargian
    Commented Jul 17, 2020 at 8:44
  • \$\begingroup\$ @analogsystermsrf...you wrote: "Rin is beta * (1/gm * Re)".....typing error? Without cap bypassing the input resistance is rin=hie(1+gmRe) which is app. beta*Re \$\endgroup\$
    – LvW
    Commented Jul 17, 2020 at 9:04
  • 1
    \$\begingroup\$ @Biargian...Concerning academic qualification: I think, at least the knowledge about the role (pros and cons) of negative feedback can help to solve the problem under discussion.(removing Ce stabilizes the gain against parameter uncertainties and parts tolerances). More than that, knowing that the BJT is a voltage controlled device, the base voltage divider should be made as low-resistive as allowed (to reduce the influence of the base current with its large tolerances) \$\endgroup\$
    – LvW
    Commented Jul 17, 2020 at 9:08
  • \$\begingroup\$ @LvW ...or, since the BJT requires a current to operate, you need to make sure that such current is a negligible portion of that required by the voltage divider in order not to load it. And to do that you can only act on the divider, not Ib - which has to have the value required to give the set Ic (one of the first design parameters to be chosen). (to the OP: no, I am not stating that the BJT is solely current controlled, but that you can see it both ways, according to what is more useful to you). \$\endgroup\$ Commented Jul 17, 2020 at 22:07

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