# Need help in designing a BJT Amplifier with Avnl = 100

For our final project, we need to design a "single-stage BJT-based highpass amplifier using voltage divider configuration".

The amplifier has the following requirements: $$\V_{CC} = 12V, f_L = 300 Hz, Vs = 300 m V_{pk-pk}, and A_{VNL} = 100 to 300 \$$.

I decided to use a 2N3904. And the following is what I have so far

I set the voltage at R2 to 6.35 to have the maximum swing for the output. I set R2 to 10k ohms and R1 needs to be 8.87k ohms to set the voltage divider.

$$\ 6.35 = 12 \times \frac{10k}{10k+R1} \$$

Next, I set $$\I_C = 10 mA\$$. Then $$\R_E = \frac{6.35-0.7}{10mA}=565\Omega\$$.

For the gain to be 100, $$\R_C = 100\times re = 100\times\frac{26mV}{10mA} =260\Omega\$$

For the capacitors, I use the formula $$\1/(2\pi\times R\times300) \$$ and then increase it a bit more.

The problem is then I can only get a maximum of $$\2.39V\$$ and a minimum of $$\ -3.439V\$$. I don't what else I can do.

Thank you for the help.

EDIT: Here is the result of the DC point analysis:

Based on my understanding, the DC point analysis looks good to me. That being said, I still do not know what I am missing to make it work.

• Run an operating point simulation and see if all your design choices gave you the results you expected. Are you really biased at the point that allows maximum swing? Nov 16 '19 at 4:53
• I just added the DC Analysis point and it looks good to me. Nov 16 '19 at 5:20
• That V(c) result doesn't look like the optimum for output amplitude. Nov 16 '19 at 5:32
• Do you have any idea about the proper value for that? Do I need to set it to 6V too? Nov 16 '19 at 5:42
• @JakeAvila Does this only have to work in simulation at only one simulated temperature? Or do you have to be able to randomly grab BJTs from a box and make it work on an actual construction and at various temperatures, whatever it happens to be on the day? (If actual, then I'd assume you have a 50 Ohm source that drives it.) Also, how do you expect to get Av=100 with an input of 300 mVp-p? Wouldn't that imply 30 Vp-p on a 12 V rail? I don't see how that ever works!
– jonk
Nov 16 '19 at 7:27

Your decoupling capacitor is too low for 300 Hz. The 50 uF capacitor has an impedance $$\X \approx 10\ \Omega\$$. This will decrease your gain significantly. Even worse, the impedance is frequency dependent. So the gain will change with frequency. If you decrease the biasing current, it will have less of an effect, as the resistors will all increase in value.