Electronic – Designing yet another BJT amplifier given some constraints

amplifierbjtdesign

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.
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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}}|\$])?

Best Answer

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