That depends of your circuit. Obviously you have to phase split the signal if you're planning to use two identical amp stages. If you're planning to bridge something it probably isn't a good idea to use one circuit as non-inverting and the other as inverting since the gains would have to be matched - though with this kind of circuit you would not need a phase splitter. I hope this answered your question.

Edit:
http://linkan.lewander.com/projects/amp/amp.htm
http://www.national.com/an/AN/AN-1192.pdf

No. You don't have to change bias. The idea is that bridged amps are as identical as they can be. They just have to work in an opposite phase.

I think it actually is. I consider the bridge circuit analogous to a PP output transformer coupled amplifier, in which the load connects between the two output nodes . However, one can not make a bridged amplifier by grounding the other side of the load and using only one node for load input: The opposite phase signals would cancel each other out. With PP transformer coupling you don't have to care about this as long as you retain the right winding polarity in the primary side. Note that tubes have no negative polarity so the amplified signal has to be phase splitted to acquire push-pull effect - then the OT's primary acts as a phase inverter for the other amp stage's output. As a result the signals are either "summed up correctly" or an (opposite) phase difference affects between the load nodes. This allows either amplifying both halfwaves (class A) or using each stage to amplify it's own portion of halfwave (class B). I hope this made some sense.

The circuit configuration defines whether you need an output transformer or a phase splitter. It is possible to build a push-pull circuit without either one too - just two opamp stages working identically but biased differently. What you're suggesting about class B/AB is possible by seriously altering the bias points of the opamps. i.e. bias voltage of Vcc to one opamp and bias voltage of Vee to the other. This allows more swing on either positive or negative side of the wave but clips the other halfwave completely. Finally the two signals would be summed (i.e. using output transformer or mixer resistors). To lower crossover distortion the bias point could be altered. This works at least in theory but I don't know how well real-life devices could cope with such an arrangement - Probably not very well and there seems to be so many things that could go wrong. Frankly, I see no point in the whole thing: Why beg for more crossover distortion and instability when you can gain more power by using parallel or bridged configuration.

This will be problematic only if you want to switch the Peavey´s speaker off while plugging to PA. But why use a transformer: Just put a high impedance voltage divider to the output (in parallel with the speaker) and buffer it's output. A high impedance (>1Meg) will be virtually "invisible" in parallel with the speaker load and the voltage divider is used to attenuate amplifier´s too high output voltage to 1-2V range - which should be suitable for most line level inputs. Then just feed the output signal from the buffer to the PA. This is the most basic form of the circuit you'll need. If you want to isolate the source from PA (to reduce hum, have balanced input etc.) you can put a 1:1 transformer to the buffer output.

Exactly. If you follow the layout you will see that VB voltage is formed by two 10k resistors and filtered with 22uF cap. The trace connects both non-inverting inputs of the opamp 1. I wonder where are the signal ground points for input and output though... Maybe the idea is to ground them via chassis with non-insulated jacks (0V point connects both battery - and chassis). It might be a good idea to add ground nodes for both input and output to the PCB layout - just for backup.

Edit: The second opamp doesn't need biasing because it's not AC coupled. There are no capacitors blocking DC and that's why the 4,5V bias voltage is already present at the inputs and outputs. Now that i think of it, the output will have a DC offset of 4,5V because of this. It might be a good idea to AC couple the output with a 100 uF capacitor (smaller values should do as well). Just solder the minus pin of it it to the output jack and connect the plus pin to the out terminal on the PCB. This is just an extra safety measure, probably the device input you'll be plugging into is already AC coupled. Anyway, some extra protection shouldn't do any harm.

No it's not. The Vcc corresponds to the positive voltage, Vee is either negative or zero (gnd) depending on whether the supply is dual or single. Single supply has only positive or negative voltage referenced to zero, dual supply has both positive and negative voltage referenced to zero. The VB pin is for "biasing" the opamp input and output into the middle of Vcc and Vee. Vbias=(vcc+Vee)/2. In single supply this must be 1/2 of Vcc since Vee is zero. In dual supply it's most often zero. ie. (+15-15)/2=0/2=0.

In short: Connect opamp's supply pins so that + goes to Vcc and - goes to Vee. Note: Supply pins, not input pins! If you have a dual supply just connect VB to ground and that's it. If you have a single supply it´s just a little more complicated: You have to connect Vee to ground and VB to potential which is half of Vcc. This is the most common way to do it: Connect two same size resistors from Vcc to Vee (gnd in this case) and take the VB from the middle of the two resistors. That should be 1/2 Vcc. If you want to stabilize the bias voltage at this point (recommended) just connect a 10-100uF potentiometer to ground from this point.

No. VB is in the halfway of Vcc and Vee. If you use single supply then Vee is zero and VB is half of Vcc. If dual supply then Vcc is Vcc, Vee is Vee and VB is 0 V.

Edit: I hope you are not confusing the opamp's inverting/non-inverting inputs with the +/- Vcc/Vee pins, which are not shown in the schematic.

http://focus.ti.com/lit/an/sloa058/sloa058.pdf

If the theoretical maximum efficiency of a class B amplifier is approximately 71% then efficiency of class AB has to be bit less than this. Then there's losses in the power supply too. I'd say around 1/2 of VA rating is a good rule of thumb for maximum output power of SS class AB amp.

I use this. Nothing fancy but simple enough to rat's nest inside a Hammond box.

I could be utterly wrong but i guess the basic topology is something similar to this: (Note the two feedback paths.)

Looks like Vox uses similar feature (mixed-mode or current drive) in the power amp section of Valvetronix line amplifiers:

http://www.korgmidwest.com/vox/tips/Valvetronix_Overview.htm
http://www.voxamps.co.uk/products/valvetronix/vtoverview_inside_story.htm

The following text is excerpt from the page under the first link:

"The Reactive Feedback technology used in the VariAmp Circuit "reads" the impedance curve of the speaker and then reports this reactive information back to the secondary side of the output transformer. This information is fed back to the primary side of the output transformer and therefore changes the loading on the tubes, another important part of the vital role an all-tube power amp plays in the creation of traditional tube tone.

Unlike most solid-state amplifiers used in guitar amps which are Constant Voltage designs, the Valvetronix Constant Current design used in the VariAmp allows the output voltage of Valvetronix amplifiers to vary along with the speaker's changing impedance, just like in a traditional tube amplifier."

The implementing of a transformer coupled, low power tube push-pull stage before the real SS power amp was quite clever but even more clever was the idea to make it sense the speaker impedance, this should give a realistic loading for the tube "power amp" section. My theory is that a huge part of the tube sound lies in the compression of the voltage peaks that occur during a high load resistance. I have been playing around with ideas to use mixed-mode feedback and compress the signal within the power amplifiers gain set/feedback loop to achieve similar results.

The "marketing statements" presented on the article have too many contradictions which makes it very hard to figure out what kind of circuit the Valvetronix power amp really is. I have been looking for the Valvetronix schematics but haven't found any yet - it would be really nice to see what kind of circuit the power amp section really hides inside.

I googled.  He reputedly used Roland's JC-120 and SS Acoustic Control Corp. amp, I believe it was 260 (http://acoustic360.homeunix.net/products/main/260.html, http://acoustic360.homeunix.net/images/schematics/260_160.jpg). The same amp was also used by numerous other artists including Chuck Berry (this guy used whatever he got) and The Doors guitarist Robbie Krieger.

A thought occured to my mind... If the switching arrangement gets too complex it might be a good idea to replace it with CMOS switches and design a control arrangement for them. This has some additional benefits: The signal path will be shortened and the switch's "life" will be hugely extended. Also, it allows implementing a good foot pedal control. This could be done with discrete components too but i consider using CMOS chips as much simpler solution.