Hello, its me once again.

I came up with another circuit, which im trying to simulate in order to keep learning.

I made this circuit that has 4 transistors, 2 npn and 2 pnp.

What i expect from this circuit is the the input sine wave.
From what i understood the circuit below should work like this: i apply a sine wave to a npn and pnp transistors, what happens is that the transistors npn ( which is called Q1 in the schematic ) will shift down the signal by 0.7 ( this is what i see on emitter )
the same sine wave gets shifted up by the pnp transistor on its emitter ( which is called Q2 on the schematic )

At this point, the signal at the emitter of the npn transistor is given into the base of the second PNP ( which is called Q4 ), what the second pnp does it shifting it up again by 0.7 ( 0.7 is an approximation, lets say thats the drop on the diode), in this way u cancel the previous offset given by the npn, same thing work for the pair Q2 / Q3.

Now my questions are:
1)Is this the correct way to analyze this circuit? i got it well or i misunderstood? and if so then how it works?
2)Why the simulation shows that on its output, the signal goes from -4.5 to 4.5V? ( sine wave is 9V amplitude)

Thanks.

 

Why the simulation shows that on its output, the signal goes from -4.5 to 4.5V? ( sine wave is 9V amplitude)

Because R3/4/5 form a divide-by-two voltage divider.

 

Because R3/4/5 form a divide-by-two voltage divider.

ahh right... im stupid lol...
another thing, during positive cycle there is the voltage divider R4/R5 and during the negative cycle there is R3/R5?
secondly, what value should i choose to make this circuit work for the voltage divider? is like 10 ohm sufficient ( so in this way i get 9v in output )

 

during positive cycle there is the voltage divider R4/R5 and during the negative cycle there is R3/R5?

For positive it's R3/R5. For negative it's R4/R5.

[Edit] Correction. It's R3/R4/R5 for both positive and negative.

what value should i choose to make this circuit work for the voltage divider?

I don't know what you mean.

 

For positive it's R3/R5. For negative it's R4/R5.

I don't know what you mean.

For positive it's R3/R5. For negative it's R4/R5.

I don't know what you mean.

why for positive R3 R5?, during positive cycle the transistor Q1 is conducting and thats the path i see no?:

 

Yes, that’s called a “diamond buffer”.
It doesn’t need R3 and R4, you can short them out, but a value of a couple of ohms or so would stop large standing currents if your transistors are mismatched.
The circuit works better with A big capacitor between the emitters of the input pair.
SPICE generally works better if you tell it some real transistor part numbers rather than using its generic NPN and PNP.
There’s a good article here:
https://www.tubecad.com/2022/06/blog0558.htm

 

during positive cycle the transistor Q1 is conducting and thats the path i see no?:

During both the positive and negative parts of the input voltage all four transistors are conducting. All four emitters follow the input, albeit with different DC offsets. The signal doesn't follow the path you drew in post #5, because the voltage at the bottom of R2 is clamped at -10V by the negative supply voltage.
Btw, you might want to increase R1 and R2. At the moment, the output current available from the buffer is less than the current from the signal source.