Sorry for asking a question about the same subject as my last question, but I am once again stuck on a BJT Amplifier design problem. enter image description here Where the beta parameter may vary from 100 to 800, the voltage between the emitter and the base equals 0.6V (active mode), Vt=25mV and the Early Effect may be ignored.

It can also be supposed that the bypass capacitors simply act as a short circuit for AC and open circuit for DC.

There are two constraints:

  • Input impedance > \$2k\Omega\$
  • Maximum possible output signal swing

What have I already done (\$i_C\$ is the polarization current which runs through the collector):

I found the signal swing equations:

\$ V_{o_{max}} = 19.8 - i_C(R_C + R_E)\\ V_{o_{min}} = -i_C * R_C//R_L\$

I also found out that the imput impedance will be \$r_\pi = \frac{\beta V_T}{i_C}\$ from the small signal model. One can infer that if the input impedance > \$2k\Omega\$ for \$\beta = 100\$, then it will continue > \$2k\Omega\$ for \$\beta = 800\$. So we can work with \$\beta = 100\$, which yields:

\$R_i = r_\pi = \frac{\beta V_T}{i_C} = \frac{100 * 0.025}{i_C} \rightarrow \frac{2.5}{i_C} > 2000 \rightarrow i_c < 1.25mA\$

From now, I don't know what to do. I have already tried some values for \$i_c\$, being able to calculate the resistances (only supposing symmetrical output) and I noticed that bigger \$i_c\$ gives biggers signal swing. How can prove that? Also, how can I solve the problem without supposing symmetrical output (having one less equation [\$ |V_{o_{max}}| = |V_{o_{min}}|\$])?

  • \$\begingroup\$ To the best of my knowledge your small signal input impedance r(pi) is a function of the large signal collector current I(c) not the small signal collector current i(c). \$\endgroup\$
    – jippie
    Commented Sep 29, 2013 at 2:44
  • \$\begingroup\$ That's what I meant. Added a note. \$\endgroup\$
    – Thiago
    Commented Sep 29, 2013 at 5:57

2 Answers 2


The DC collector current is determined by \$R_E\$:

\$I_C = \alpha \dfrac{9.4V}{R_E} \approx \dfrac{9.4V}{R_E}\$

Since you require \$I_C < 1.25mA \$, the constraint equation is:

\$R_E > \dfrac{9.4V}{1.25mA} = 7.52k\Omega\$

The second requirement, maximum output voltage swing, without any other constraint, doesn't fix the collector resistor value.

We have:

\$ V_{o_{max}} = 19.8V - I_C(R_C + R_E)\$

But, the voltage across \$R_E\$ is fixed at 9.4V so:

\$V_{o_{max}} = 10.4V - I_C R_C\$

\$V_{o_{min}} = -I_C * R_C||R_L\$

If you stare at this a bit, you'll see that maximum output voltage swing is 10.4V but this requires that the product \$I_C R_C = 0\$* which is absurd.

Now, if we also require symmetric clipping, then, by inspection:

(1) \$V_{o_{max}} - V_{o_{min}} = 2 I_C (R_C||R_L)\$

(2) \$10.4V = I_C(R_C + R_C||R_L) \$

Looking at (1), note that, for maximum swing, we get more "bang for the buck" by increasing \$I_C \$ rather than \$R_C \$.

Since we have an upper limit on \$I_C\$, (2) becomes:

\$R_C + R_C||R_L = \dfrac{10.4V}{1.25mA} = 8.32k \Omega\$

which can be solved for \$R_C\$.

*unless \$R_L\$ is an open circuit


I have the impression that this is a homework question, so I'll answer with some simplifications/neglections. Up to you to do more accurate calculations.

Let's say Ic = Ie (which isn't exactly true, but good enough for me as it is you homework, not mine ;o)).

Ie = Ic = 1.25mA

Also you know that the base voltage is 0V because the input signal is to be replaced with a short for the large signal analysis.

You know the emitter current; you know the base voltage and with that you know the emitter voltage; therefore you can calculate the voltage across the emitter resistor; and finally you can calculate the emitter resistor by dividing the voltage across it by the current through it.

Order of magnitude 7500 ohm. Try the exact calculations for yourself. Consider adding those to your initial question by clearly marking your progress.

Next step is to calculate Rc, but I leave that for someone else to answer.

  • \$\begingroup\$ For I_E = 1.25mA, R_E, with 9.4V across, should be about 7500 ohms. \$\endgroup\$ Commented Sep 30, 2013 at 2:29
  • \$\begingroup\$ @AlfredCentauri yes you are right, I made a type there. Corrected it. \$\endgroup\$
    – jippie
    Commented Sep 30, 2013 at 5:17

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