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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 farBJT Amplifier

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: enter image description here

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.

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    \$\begingroup\$ 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? \$\endgroup\$
    – The Photon
    Commented Nov 16, 2019 at 4:53
  • \$\begingroup\$ I just added the DC Analysis point and it looks good to me. \$\endgroup\$
    – Jhaendar
    Commented Nov 16, 2019 at 5:20
  • \$\begingroup\$ That V(c) result doesn't look like the optimum for output amplitude. \$\endgroup\$
    – The Photon
    Commented Nov 16, 2019 at 5:32
  • \$\begingroup\$ Do you have any idea about the proper value for that? Do I need to set it to 6V too? \$\endgroup\$
    – Jhaendar
    Commented Nov 16, 2019 at 5:42
  • \$\begingroup\$ @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! \$\endgroup\$
    – jonk
    Commented Nov 16, 2019 at 7:27

1 Answer 1

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There are a bunch of problems with your circuit. Your collector is at 9.5 V and your emitter is at 5.3 V. I would expect the output to swing between -3.2 V and 2.5 V having 1 V across the collector emitter to keep the transistor biased. You need to bring the emitter voltage down to about 1 V. Put your Q point at half the supply voltage.

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.

The standard method for controlling the gain is to put a resistor in series with the emitter capacitor. Then the AC signal will see a resistor as well as a capacitor at the emitter.

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  • \$\begingroup\$ Hi @user110971! I have taken into account your recommendations. How can I control the gain if I set the RC to be at 6V drop? My current approach is to set IC at 10mA, and then set RC to be 6V/10mA == 600 Ohms. But if do that, the gain will be at about 230 since the gain depends only on RC and IE (IC). I can't control the gain then/ \$\endgroup\$
    – Jhaendar
    Commented Nov 16, 2019 at 10:52
  • \$\begingroup\$ One possible method for controlling the gain is to use different feedback resistances for DC and AC in the emitter path (CE shunts a part of the DC resistor only). \$\endgroup\$
    – LvW
    Commented Nov 16, 2019 at 14:27
  • \$\begingroup\$ @JakeAvila As LvW has suggested, the standard approach is to put a resistor in series with the decoupling capacitor. I would also decrease the current to at least 1 mA, if I were you. The capacitor impedance will cause your gain to vary too much at low frequencies (you said 300 Hz). Don’t forget to take into account re when calculating the resistor value. \$\endgroup\$
    – user110971
    Commented Nov 16, 2019 at 18:14
  • \$\begingroup\$ @LvW Good point. I forgot to mention that. \$\endgroup\$
    – user110971
    Commented Nov 16, 2019 at 18:21

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