My textbook says that biasing current through Q8 and Q9 is about 20mA and through Q10 and Q11 about 100mA. How can I analyze the circuit to get these values with back-of-the-envelope calculations?
So far I only calculated that biasing voltage provided by Q5, R8 and R7 is 2V, but cannot get any further.
The truth is that it is impossible to tell/predict what current will flow without making a "big" error. There are far too many unknowns.
The wild guess for the drive stage will be \$ \frac{V_{BEQ10}}{R_{11}} \approx 20mA \$
I used Bob's Cordell models in LTspice and get these results: \$I_{CQ8} = 18.1mA\$ and the output stage current is \$20.5mA\$ and \$\textrm{Vbe}\$ multiplier (Q5) bias voltage equal to \$2.4V\$.
Of course, when you build the circuit in real life you will get completely different results. Because the output transistors will get hot and after some time they will reach the thermal equilibrium and the bias current will settle on the new value.
This is why in practice we adjust on the benchtop the \$R_7\$ resistor value to set the output stage bias current to the desired value after warming-up the amplifier.
I'll assume we're calculating bias current in the same setup we'd use to measure it: the amp is idle, 0V input, 0V output (neglecting offset), no output load. All transistors are ON, linear mode (not saturated), Vbe=0.6V. I'll start by neglecting base currents.
Q6/Q7 is a current source, voltage on R9 is Q6 Vbe so current in R9 and Q7 is 9.7 mA, a common value.
Q5 is wired as a Vbe multiplier, which is a shunt regulator with feedback. There is one Vbe across R7, so Q5 Vce = Vbe/(R7/(R7+R8)) = 2V as you said. Note Q5 should be mounted on the heat sink with power transistors so its Vbe tracks Q8-9-10-11 Vbe to ensure bias does not drift (or go into thermal runaway) as the output transistors heat up.
Now there is 2V between the bases of Q8 and Q9, but we have to fit the sum of Vbe's of Q8-9-10-11 inside those 2V... and 4*Vbe is 2.4V.
If we assume Vbe=0.5V instead of 0.6V then the Vbe multiplier bias voltage shrinks in the same proportion, and it doesn't work.
This means there is a problem somewhere. Either the output transistors Vbe is quite lower than 0.6V, or the resistor values on the schematic are wrong and the Vbe multiplier actually sets a higher bias voltage than 2V.
Also the schematic introduces unnecessary crossover distortion. If you build the amp, DO NOT CONNECT the midpoint of R11-R12 to the midpoint of R13-R14. You want your drivers to run in class A push-pull so they are able to suck charge out of Q10/Q11 bases quickly when Q10/Q11 have to turn off, instead of simply relying on a 33R resistor. On the schematic as shown, when say the top transistor has to turn off, the top driver also turns off, and there is only Vbe/33R = 18mA current available to suck charges out of Q10 base. So it will take extra delay before it turns off, while Q11 is driven by Q9, and turns on much quicker. This can lead to cross-conduction and crossover distortion.
It is impossible to calculate the exact current through the resistors although it is possible to get a ball park figure for the current through R11 & R12.
That is not a practical design for the Vbe multiplier. A practical Vbe multiplier would include a pot in series with R7. When setting up the amp the pot would then be adjusted to give a few mVs quiescent voltage across the output resistors, R13 & R14. I designed and built a power amp recently with 0.22R output resistors and adjusted the pot to give 8mV across each output resistor (measured with a volt meter) giving a quiescent bias current of about 36mA.
There are 4 Vbe drops between the Vbe multiplier and the amp's output. Therefore the Vbe multiplier should be adjusted to produce a bias voltage of about 4 x 0.7V = 2.8V between the output stage transistor bases Q8 & Q9 but it is not possible to say exactly what the Vbe voltage values are. You can say that there should be approximately 2 x 0.7 across R11 and R12 giving about 21mA through them.
Incidentally, the Vbe multiplier adjustment pot should be in series with R7 not R8 and then if the wiper goes open circuit the bias voltage is removed rather than being massively increased.
The output stage has two Darlington pairs (Q8 and Q10, and Q9 and Q11). Q8 and Q9 are connected as emitter follower while Q10 and Q11 as common emitter. Usually, the transistors in the pair are fabricated so that their Beta is equal. The total gain of the pair will roughly be in that case Beta squared
To find the bias current you can initially ignore the resistor R11 (R12), and note that the base current of the second transistor is the emitter current of the first (Ib + Ic). The calculation is pretty straightforward.
It's interesting to notice that R11 and R12 are there to help discharge the charge accumulated in the base capacitance when the output is switching.
EDIT:
Let's go through Q8 and Q10 only and ignoring R11. the total collector current for Q8 and Q10:
\$\ Ic = Ic(Q8) + Ic(Q10) = \beta\ (Q8)Ib(Q8) + \beta\ (Q10)Ib(Q10)\$
Q10 base current is Q8 emitter current:
\$\ Ib(Q10) = Ie(Q8) = [ \beta\ (Q8) + 1] Ib(Q8) \$
The total collector current will then be:
\$\ Ic = \beta\ (Q8)Ib(Q8) + \beta\ (Q10) [ \beta\ (Q8) + 1] Ib(Q8)\$
\$\ Ic = [\beta\ (Q8) + \beta\ (Q10) \beta\ (Q8) + \beta\ (Q10)]Ib(Q8)\$
assuming \$\ \beta\ (Q10) = \beta\ (Q8) = \beta\ \$
\$\ Ic = (\beta\ ^2+2\beta\ )Ib(Q8) ≈ \beta\ ^2 Ib(Q8)\$
if you now consider the pair Q8 and Q10 as one transistor with collector current Ic, base current Ib(Q8) and emitter current Ie(Q10), you can say that the pair has beta squared as beta.
