Powering a 2 watt 6 ohm speaker with a 9V battery?
The following schematic uses fewer parts than the one shown at this link as Fig 5.5.3.

So here's my simpler version:







simulate this circuit &ndash; Schematic created using CircuitLab

Speaker/Bootstrap

Let me start at the speaker and bootstrap, without getting into details here. You may usually see \$C_1\$ with a speaker tied to ground. But you can also tie the speaker to the plus rail. Either way, \$C_1\$ will wind up with a nearly constant voltage across it. But that voltage is more useful to you if you direct it upwards towards the plus rail, instead. So that's what this circuit does.

The useful trick in making this choice is that since \$Q_1\$ "looks like" an emitter follower here, there is a fixed voltage across the BE junction of \$Q_1\$. With \$C_1\$ also having a relatively fixed voltage, too, this means there is an almost fixed voltage across \$R_3\$. And that means that \$R_3\$ "looks like" a current source. Which is perfect for this application and saves parts while at the same time providing a better way to handle biasing for \$Q_1\$ and \$Q_2\$.

BJT emitter degeneration

I've added \$R_1\$ and \$R_2\$ so that they provide, at maximum output, voltage drops of about \$100\:\textrm{mV}\$. This helps a little with variations between the transistor parameters.

You could eliminate them, if you want to. But if you have parts of those values (or as much as an Ohm, or so) then it might be nice to experiment with and without them to see if you can tell a difference or not.

The main reason I want them there for now, other than what I already said about it, is that you will need them for later circuit adjustments. So for now, keep them and find something near to those values if possible.

The \$R_3\$ current source

If things are tweaked right (and that's coming up), then there should be approximately \$2.9-3.7\:\textrm{V}\$ across \$C_1\$ and therefore about \$2.2-3.0\:\textrm{V}\$ across \$R_3\$. (The large variation here is due to the fact that a \$9\:\textrm{V}\$ battery varies over its lifetime something like \$7-9\:\textrm{V}\$.) So I set this to source about \$8-11\:\textrm{mA}\$ over the operating life of the battery. The reason I picked that range is that I expect the base currents for \$Q_1\$ and \$Q_2\$ to be around \$1\:\textrm{mA}\$, or so, and I'd like to have nearer 10X that much available near the bases so that variations in base currents can be tolerated well.

\$Q_1\$ and \$Q_2\$ AB biasing

Start with the \$R_5\$ and \$R_6\$ pair, without any input signal. Temporarily replace both of those resistors with a single resistor and don't include \$C_2\$, to start. Just focus on the resistance of a single resistor here. Adjust the value until you see \$\frac{1}{2}\:\textrm{V}\$ more than half your battery voltage at the point shown by the red arrow. So if you have a fresh battery, this means you want to hit about \$5\:\textrm{V}\$. Adjust until you get close. Then record that resistance value.

To complete the biasing, place a voltmeter across \$R_1\$ and plan to make adjustments to \$R_4\$ (blue arrow.)

I've slightly differently arranged biasing of the two output BJTs, \$Q_1\$ and \$Q_2\$, using a series resistor \$R_4\$ here rather than the arrangement used at the link at the top, above. The idea of \$R_4\$, \$D_1\$, and \$D_2\$ is to operate \$Q_1\$ and \$Q_2\$ so that they are both active, but not overly so, when there is no input signal applied. I'd recommend trying to hit about \$5\:\textrm{mA}\$ with no signal applied.

So with the voltmeter in place, adjust the value of \$R_4\$ until you see a voltage drop across \$R_1\$ that is predicted by \$5\:\textrm{mA}\cdot R_1\$ (whatever the value for \$R_1\$ in your case might be.) If you used the values I show, then this would be around \$1.5-2.0\:\textrm{mV}\$.

If it turns out that the current is still higher, even with \$R_4\$ shorted, then just short out \$R_4\$ (remove it and short the nodes.) If that still has the measured voltage too high, then put \$R_4\$ in parallel with the two diodes instead as shown in Fig 5.5.3, start perhaps with \$1\:\textrm{k}\Omega\$, and bring the value downward until that goal is reached.

Now go back and reverify that the node at the red arrow is still where I said you wanted to target, above. Again, adjust your temporary resistor value until that's true. Then record that value, again (if needed.)

Now, you must make arrangements so that the sum of \$R_5\$ and \$R_6\$ (green arrows) is approximately this temporary resistor value, but you need to split the difference between them. You can work that part out later. For now, just divide it about in half and re-add \$R_5\$ and \$R_6\$ and \$C_2\$.

Done?

Well, that's about it. Until you apply a signal. That's when you get to worry about how you divided out that temporary resistance value into \$R_5\$ and \$R_6\$. I show an uneven distribution in the circuit because, well, it's likely that the best setting will be uneven.

Putting a higher percentage into \$R_6\$ will means more negative feedback at AC, which may be required. But I'll leave the exact distribution as a task for you to experiment with. Keep a significant portion in each of them, but feel free to play a bit.

Also, experiment with the circuit using a somewhat drained battery, as well. Make sure it all still works reasonably well.

And of course, certainly try out Fig 5.5.3. If it is all equal to you, that one is probably better to go with. I just wanted to give you a flavor of how to approach a case with even fewer parts.