Consider what happens when you have an NPN bipolar transistor with a resistor from its collector to base (let's call this R1) and another from base to emitter (call this one R2). We'll also assume that we have a battery and resistor in series feeding current into the collector and out the emitter, with several volts of voltage and the current limited by the external, unnamed resistor to some value that won't cause excess heating of the NPN.
Initially, there is voltage across the NPN, but no currrent through it. The voltage causes a voltage divider effect across the two resistors, so if we take the emitter as a 0V reference point, the voltage at the base tries to go up to Vcollector times R2 divided by R1+R2. At some point the voltage reaches the turn-on voltage for the NPN, 0.5 to 0.7V depending on the device, and the base starts conducting.
If the NPN gain is reasonably high, say 49, then 49 times that much current flows in the collector. The current is limited by the external resistor though, so when it reaches the current limit, the collector is eating 49 parts of the current while 1 part is going through the resistors and base. (The sharp guys are now figuring out why I hand-waved a gain of 49 instead of 50... 8-) )
So what's the voltage on the collector where this magic point occurs?
OK... I picked a current gain high enough to ignore the base current - trust me on this one - so we have the voltage across R2 being Vbe for silicon, and to get it up to there, the voltage at the collector has to be enough to get it to Vbe. Since Vbe= Vcoll * (R2/(R1+R2)), then we can get out our high school algebra texts and solve for Vcoll.
Vcollector turns out to be Vbe*(R1+R2)/R2, or Vbe*(1+R1/R2), independent of everything else.
Looked at another way, this is a feedback amplifier, with feeedback from the collector to the base. Same equations come out.
So for your biasing you remove the diodes and paralleled resistors, etc. and substitute in a medium power transistor (we'll get to why medium power) with a couple of resistors, one from collector to base, one from base to emitter. Now let's count junctions. The Randall output stage is quasi-complementary, so there are two base-emitters in series that have to be turned on in the top half, Q11 and Q13. There is only one in the bottom half, Q12. We know that the correct bias voltage is where these base-emitterses are just barely on, so the nominally correct voltage is three times about 0.6, or 1.8V.
To get to that, you need R1 and R2 so that 1+R1/R2 is about 3, or R1=2R2. But what are the absolute resistor values? Let's calculate a bit. Get out your Ohm's law text.
How much current is going through there? Well, at no signal conditions from the schematic, there is 40V on the - power rail, and R48+R49 is 2.5K. The specified bias point is at -0.6, so there's 40- 0.6V across 2.5K, or 39.4/2500 = 15.7ma of standing bias current. Our Vbe multiplier needs to have Vce of 1.8V at 15.7ma.
So we can assign how much current we want where. We're free to have the split of current between the collector/emitter path in the Vbe multiplier and the R1/R2 path be almost anything as long as the collector can overwhelm the R1/R2 path in operation and the base current sucked out of R1/R2 does NOT overwhelm the R1/R2 current.
With a target gain of 50, let's pick the R1R2 current to be ten times the base current, so just counting fractions we have one part base current, ten parts R1/R2 current and 50 parts collector current, and a total of 61 parts. One 1/61th of 15.7ma is 259uA, so the resistor string current is 2.58ma.
Back to Ohm plate. the voltage across R1/R2 is 1.8V, the current is 2.58ma, so R1+R2 is 1.8/2.56ma = 697ohms. R2 is 1/3 of that, or 232 ohms and R1 is 2/3 or 465 ohms.
But we actually wanted to vary that a bit, maybe from 1.7 to 1.9V or even a little more.
So we need to put a pot in to vary the Vbe ratio.
First, simple choice is to use a 750 ohm (or 1K) pot, stringing the outer lugs from collector to emitter and connecting the wiper to the base. Oly problem is range. The pot can now cause Vbe ratios from one (that is, R1/R2 is about 0) to substantially infinity by shorting the base to the emitter. Bad juju indeed.
So let's only vary one resistor. Which one? If we use the pot to vary R1, and the wiper goes open, the transistor turns off, the output transistors are turned maximally off, and the amp pours out prodigous amounts of smoke and flame. Ooops, can't do that one.
Let's use the pot for R2. Now if the pot opens, we get transistor fully on and the bias goes to one Vbe, which sounds terrible, but does not kill the map... better juju.
So we want a fixed resistor for R1, variable one for R2. Pick 470 ohms for R1. That resistor dissipates 0.00258A**2 * 470ohms, or 3.128milliwatts; a 1/4 W will do nicely.
We put all our variation into R2. Again, we could use the whole pot range by using a 500 ohm pot, but that gives us an output range of 0.6V (R2=0) to 1+470/500 = 1.94 Vbes. We can do better.
Ideally, we'd like the entire range of the pot to go from just a little too little voltage to a little too much. The output devices might start turning on at as little as 0.45V, so three of those is 1.35V. They might be hard cases up at 0.7 or 2.1V. The Vbe of our multiplier transistor may be the same range (although chances heavily favor it being close to 0.55V for a variety of reasons), so our multiplier needs to cover the case where we put out a minimum of 1.35V when our multiplier Vbe is 0.7V and where we put out at least 2.1V when our multiplier is 0.45V.
So we need a multiplier range from 1.35V/0.7V = 1.928 to 2.1/0.45 = 4.67.
We're going to use 470 for R1, and the range on R2 is that R1/R2 is 0.928 to 3.67. That makes R2 be 128 to 506 ohms. So we use a 120 ohm resistor in series with a 380 ohm pot, and we're golden.
Errr... there aren't any 380 ohm pots...
OK, pick a 500 ohm pot. Scale up the series resistor to (500/380)*128=168 ohms and R1 to 618 ohms. So we get a 620 ohm 1/4W resistor for R1, a 160 or 170 ohm resistor in series with a 500ohm pot for R2. We connect only to one lug and the wiper of R2 pot because if the wiper fails we want that pot to go open, not to the full value of R2.
Now we need a transistor.
The transistor needs to conduct 16ma with say 2V across it. That's 32milliwatts. But there is another requirement. This transistor also does the thermal compensation for the output stage. That's why there is that thermistor on the schematic. The transistor MUST be bolted to the heat sink right where that thermistor is so it heats up like the thermistor. That's what gives you thermal compensation so as the output devices get hot, the bias keeps them out of thermal runaway.
One path is to pick a TO-126 (weenie-pack) device, another is to get a TO-220. Both of these require insulation from the heat sink because the full signal voltage is on their collector, which is tied to the heat sink. My favorite is to get full-pack to-220s which are encased in epoxy and need no insulating. If you get an uninsulated package, also get an insulator and some thermal grease or one of those sil-pad things and be sure you insulate the collector from the heat sink. Check it with an ohmmeter when you're done.
I got the following out of the Digi-key catalog.
ZTX455 - $0.63 Super E-Line -no insulation needed, bolt it on.
2SC1567or - $0.70 TO126 package, needs insulated
2SC3063 - $0.66 TO126 also needs insulated
With the range of the pot restricted, you can probably use a single turn pot instead of a ten turn.