Thank you! This is perfect

DIP32 socket needed. Already looked for just the socket, couldn’t find a 32 pin DIP anywhere… maybe I’ve missed it though. And I know I could make my own I just thought I’d ask first

Thank you very much. I understand that now. Again, much appreciated!

Guessing about 700mV for the base-emitter voltage, this means the base voltage for 𝑄1 is −15V+460mV+1.19V+700mV=−12.65V. - Please can you clarify where the 1.19V has come from? Maybe it is very simple, but I should ask rather than guess.

Thank you so much for the help! It’s really appreciated that you’ve done so much. This has really given me some extra understanding of the circuit, which I have not been as clear previously

@James does this mean you think the 12x gain is not possible with the circuit I am driving?

@James Great, so I have some great results from the information you have provided. I appreciate that greatly. When I reduce the load impedance the gain reduces. Can you advise on a way to increase it again? I presume the CE amp isn't providing enough current throug, is that a correct statement? I can see the current doubles through capacitor C2 from using a 100K load to simulating with the push pull amp I have to use as the load

@James Thanks for the info - I will give this a try. In terms of R3 being 3.3K, I have been told a rule of 10 should apply between R3 and R2 to avoid any significant loading. Is this true? Can you please explain whether this is valid?

@Audioguru unfortunately not

@Bimpelrekkie The signal generator is 50 ohms.

@jonk Yes, apparently 24V pk-pk is achievable (my lecturers words).

@jonk The output load will be removed and connected directly to a push pull amplifier. I can provide the schematic if that is needed. The input signal is 1Vpk (2Vpk-pk) @ 1KHz (these are both in the question). The output impedance of the source signal in real life is 50 ohms. I understand how the gain is achieved yes.