I made a CE amplifier as the first stage with a gain of around 10 with Rc = 10K, Re = 910, and biased the base voltage using Rb1 = 270K and Rb2 = 30K (to GND).

I want to preserve my signal of -5V to 5V with a gain of 10 at the collector output of the CE amplifier and I read that I have to use a voltage follower using a CC configuration however because of the high output impedance of my CE amp at 10K, I can hardly get any output signal. I don't know how to match the impedance of my CE and CC amplifier to preserve my signal of 10Vpp. I want my CC amplifier to be a voltage follower that also outputs around the same 10Vpp signal as my CE amplifier. I also have trouble choosing Re for the CC configuration. Choosing Vcc = 12V, I assumed that Vre~0.5Vcc and Icq = Ieq = 0.2A such that Vre = 6V and Re = 30 so then I used a voltage divider to bias the base at around this same voltage.

I also tried using a push-pull configuration as suggested by YouTube videos for the second stage and it works in preserving a raw 10Vpp signal without source impedance through it but as soon as I connect my two transistor stages, the output voltage drops significantly because of I presume the CE output impedance. I will attach my circuit here.

 

From the 8Ω output load, I assume you are designing an audio speaker amp.
Instead of reinventing the wheel, why not use one of the circuits already designed.

Below is an interesting, fairly simple design I like:
It has a differential input which provides both AC and DC negative feedback to stabilize the AC gain at 10, output stage DC bias point to 1/2 Vcc, and gives output low harmonic distortion (a little over 0.5% simulated).

The AC gain is determined by the feedback attenuation ratio of R4 to R5 (here 0.1 giving a gain of 10).

D1 and C6 bootstrap Q5's bias voltage to raise the positive output clipping voltage to be close to the negative (about ±6.5V with the single 15V supply).

 

Why is the CE collector resistance much too high at 10k? If it is much lower then its output level will not be reduced as much.
Why does the output use 4 little overloaded transistors instead of only 2 medium power transistors?

 

Hello, thank you for providing an example circuit but I was already able to make an update to my previous circuit with just the CC emitter follower and discarded the push-pull configuration which solved some problems and introduced another problem which is negative clipping where my output waveform got clipped for the negative cycles. I apologize for forgetting to mention that we were constrained to using a CE and or CC configuration to amplify our signal. I was able to preserve half of the 10vpp signal with only its positive cycle. I am not sure if it is a biasing problem for the base of my CC amplifier. Since changing my CC Rb1 and Rb2 resistors didn't change the output (Rb5 and Rb6 in this case). I assumed that Rb6 should be much greater for a higher Vbq but again there wasn't much change. Edit: I also forgot to mention that the output impedance was supposed to be 10K so I thought Rc should be 10K.

Below is the updated circuit and its output waveform.


From the output of the CE amplifier, I was getting a 10Vpp signal from -5V to 5V since I need a voltage gain of about 10. However after connecting the CC stage, the output signal has its negative cycles clipped. How do I choose Re2 or R3, the CC emitter resistor? Do I have to do a push pull configuration again to add the negative portions of my signal?

 

Your emitter-follower is severely overloaded and is operating in class-A which needs to use a lot of current to drive an 8 ohms load. Your 8 ohms speaker is shorting the emitter to ground so of course its output cannot go negative.

You must use a coupling capacitor to feed the speaker AC, not DC. Audio is AC and a speaker with DC in it does not work.
You need capacitor coupling to a push-pull pair of output transistors.