While we're hijacking this thread, can I ask please, does it really need 2 capacitors to couple the output stage?

Of course not. Amplifiers today are completely DC-coupled.

The very old circuits on the internet used coupling capacitors like that.

Why is the driver transistor an emitter-follower with no voltage gain instead of common-emitter that has plenty of voltage gain?
Oh yeah, maybe because there is no overall negative feedback like ALL amplifiers have.

Why don't teachers in his country today teach about circuits that are used today??
Oh, I guess teachers in his country do not teach anything today.

EDIT: Lazy teachers in his country are very different from teachers in most countries.

 

Umm...same question: In the circuit you posted as #19, AG, there are 2 capacitors, one for each base. Is that really necessary for that circuit? Seems to me you only need one to couple the signal to the bases.

 

You'll still need two capacitors. The DC voltages at the base of each transistor are different. Plus you need to decouple from the input voltage.

 

Misunderstood MrChips. I could almost swear I've seen that circuit with only one coupling capacitor because the diodes keep the base voltages apart and the audio signal travels right through the diodes.

but that was back when I "partied" too much on the weekends. Maybe I'm wrong.

 

Ah, you're right. You can remove one of the capacitors (don't know which one).
The other base will be coupled to the input signal via the two-diode voltage drop.

 

The way it works in my head, either capacitor can be removed and get the same result. Doesn't matter which one.

anyway, thanks for double checking this for me.

 

Misunderstood MrChips. I could almost swear I've seen that circuit with only one coupling capacitor because the diodes keep the base voltages apart and the audio signal travels right through the diodes.

but that was back when I "partied" too much on the weekends. Maybe I'm wrong.

Yes, a single coupling capacitor can be used or DC-coupling can be used:

 

Wow, you must have a time lag in Argentina! The problem is fixed and we had time to make derisive remarks about the simulator program that failed to give the correct answers.

I guess I'm a little late coming to the party, but I gather you think the simulator was wrong. You shouldn't be so hasty.
As Agustin suggested, the problem is with Q4. Specifically, it is with Q4's emitter resistor, R14. During the negative half of the signal, current must flow from right to left through C1||C4. If that current exceeds the emitter bias current, Q4 will cut off, causing clipping.
Given the crappy circuit topology (I don't recommend it), the easiest solution is to return R14 to -25V, and reduce the value to the point where Q4 doesn't cut off.
To calculate the value, recognize that the maximum voltage at the base of Q4 is about 22V p-p. Assume that the Darlington betas are 2000. With a 3 ohm load, this makes their input impedance about 6kΩ. This is in parallel with the two 50kΩ bias resistors, making the load on Q4 approximately 5kΩ. Therefore, the current through C1||C4 is about 11v/5kΩ=2.2mA peak. The quiescent voltage at the emitter of Q4 is ≈12.5V. So, when the Q4 emitter is (12.5V-11V) 1.5V, the voltage across R14 will be (1.5V+25V)=26.5V.R14 needs to be less than (26.5V/2.2mA) 12k to avoid cutoff on the negative half cycle. Keep in mind that the emitter current at this point (the negative signal peak) will be zero. We need to make R14 lower, for design margin and to reduce distortion.
I hope this makes sense.

 

Took me a while but it seems that Q4 only delivers 1 ma to start with and it's trying to drive a 12k resistor in parallel with a capacitively coupled 6k load. Sum=4k

Not enough drive current.
Got it.

I was only seeing an emitter follower and expecting a sine wave driving a load that is high enough impedance to be irrelevant. Caught me assuming.

 

Took me a while but it seems that Q4 only delivers 1 ma to start with and it's trying to drive a 12k resistor in parallel with a capacitively coupled 6k load. Sum=4k

Not enough drive current.
Got it.

I was only seeing an emitter follower and expecting a sine wave driving a load that is high enough impedance to be irrelevant. Caught me assuming.

The transistor can source plenty of current for the positive half cycles, but it can't sink current. R14 has to sink the current required for the negative half cycles.

 

Yes. I thought of phrasing it like that after a while.

 

You guys are absolutly right I am student trying to learn more stuff by myself because teachers only teach basic amplifier. What I'd like to learn is to build an amplifier that works similar to the one I just attached if anybody could help me with this it would be greatly appreciated.

 

You want to learn how to build an amplifier? What help do you need?
You found one that uses extremely old 2N2219A output transistors that are difficult to cool.
Simply replace them with more modern transistors that are easy to cool.
Design a pcb for the circuit and etch one.
Then solder the parts to the pcb and power it.

 

I mean designing one such as the one I attached putting one together is alot simpler.

 

You must learn the basics of electronics first so that you can design a simple audio amplifier. There are many IC audio power amplifiers available that work very well.

Do you know how much output power you need? 1W? 10W? 100W?
Do you know how to calculate the output power of an amplifier schematic?

The schematic you found produces an output power of only about 4W which is fairly low.

 

I don't see a feedback mechanism....You will need some for stability.

 

I don't see a feedback mechanism....You will need some for stability.

Look again. It's there.

 

Ok so I;ve done some reading and came up with a new circuit, can one of you guys look at it and comment on it and then help find a way to adding a class ab amp to drive it to get my current gain?

 

1) Why does your new amplifier have R6 that does nothing?
2) Why does it have coupling capacitor C1 and bias resistors R3 and R4 when amplifiers have been DC-coupled for a long time?
Its resistor values are high so it will not have enough output current to drive an output stage unless it uses darlington transistors.
3) It will probably oscillate at a high frequency because it is missing a frequency compensation capacitor.

 

Ok so I have taken R6 out of the circuit and I have coupling capacitors because it is the way i was taught and dont know how to get out of it?
And what can I do to fix the resistor values?