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Phase splitting--how does it work?

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Hi all!

Sorry if this is a dumb question, but I promise that I *have* searched the forums, and I am new here :)

Basically, I'm trying to figure out how a push-pull circuit works. Not from the point of view of what it is, nor even how to wire up a circuit that can accomplish it, but rather how the valve being used is able to "split up" the signal to be sent out to the power tubes before being "joined" back together.

For example, let's say I put a (sinusoidal) signal into a 12AX7 valve. I know that this is a dual triode valve, and by coupling the plate of the first triode to the grid of the second one, I get the phase splitting I'm after. Of course the wiring would be more complicated than what I've indicated, but I don't have any diagrams handy at the moment...

So, basically, what I want to know is how the phase splitting is able to be achieved? Many thanks in advance!
I think the term "phase splitter" is misleading. A phase splitter stage produces 2 outputs: one of the original phase and one of the same signal but reverse (opposite) phased.
Imagine banging a drum and placing a microphone on each side of the membrane. The signals of those mics are of opposite phase.

In a push-pull circuit the 2 signals are recombined in the output stage to produce a single phase signal similar to the original.
As dhaen stated above, the term is misleading, as the stage doesn't exactly split anything. What it does is to create two nearly identical outputs that are complete but are out of phase with one another. When some signal is applied to the grid of a common cathode stage, let's assume a triode, the currents act in such a way as to produce an amplified, but inverted signal at the anode. This principle is used to end up with two signals, 180 degrees out of phase with one another, to apply to two (or more) output valves who are most often driving the opposing sides of an output transformer. One output valve's current is increasing just as the other one's is decreasing. The idea of two microphones each on opposite sides of a drum membrane is an excellent illustration.

Does this make sense?

Another way to think of it is simply a change in reference point. Think of a sine wave of amplitude 1 referenced to one extreme. Make the reference point 0. Then the signal goes from 0 to 1. But you don't have to have the reference point at one end, you can put it in the middle. Then, the signal goes from -.5 to +.5.

A single ended interstage transformer has the common at one end and takes the signal from the other, call it positive (for transistor amps with a negative rail, you can make it negative). But you can put a tap in the middle and take a signal from either end. One end will be positive with respect to the common, the other negative. Same exact wave form, just a different point used for reference.

In the single ended case, you have one output - single ended. In the second case you have two outputs. Looked at from the center reference point, one end is swinging away from the 0 point, the other end towards 0, so they are considered in opposite phase.

Wow, thanks for all the replies--this has got to be one of the friendliest forums I've ever used! As well, thanks also to tubewade for the link--in fact I had not seen that page.

However, I'm still a bit stumped--this is so much easier with pictures!

Basically, assuming I'm understanding you all correctly, is that you are describing scenario much like the one illustrated here:


correct? What I am stumped with then is how each triode actually manges to achieve the class-B amplification without the use of a transformer (I've seen designs that use them, of course) but rather a creative coupling of triodes. As an example--and this one is used only because it's the first thing that appeared on my search, not playing favorites here--there is the circuit


which has the dual triode configuration I've mentioned. The basic thing I'm stumped with is how does a configuration like that one correspond to the illustration of class-AB push pull illustrated in the first link?

Thanks again!
Joined 2004
The basic thing I'm stumped with is how does a configuration like that one correspond to the illustration of class-AB push pull illustrated in the first link?

The example in your first link (Wikipædia) shows a pair of complementary power transistors (a PNP and an NPN), which do their own phase splitting (or phase inversion). No separate splitter is needed. Unfortunately, we don't have the luxury of complementary tubes (or at least not yet, but I'm working on it! :D ).

To split the phase with tubes, we need a splitter that will, as others have said, produce two signals that are the same shape and amplitude but of opposite phase. One of the confusing things is that there are several different splitter designs and they work in different ways. Another confusing thing is that the same splitter design can be known by several different names.

The example in your second link shows a 'concerrtina' or 'caythodyne' or 'split load' phase splitter. The way it works is by taking the outputs from both plate and cathode, which are opposite in phase. The plate and cathode resistors are made equal, so that the output signals will be of equal amplitiude. This splitter is very accurate but it has no gain and is limited in the amplitude of signal it can provide. That's why Williamson used a differential driver stage after the splitter in his amplifier design.

Another method, called the 'paraphase', uses two common-cathode triodes, one after the other, and takes the outputs from the plates of both. These will obviously be opposite in phase but they also need to be of equal amplitude. In the simple paraphase splitter, this is achieved by using a voltage divider to counteract precisely the gain of the second triode (e.g. if its gain would be 20, the voltage divider attenuates the signal going into it by a factor of 1/20). A better and more consistent way of doing this attenuation is by negative feedback - the 'floating paraphase' or 'seesaw' splitter. This splitter has high gain and can tolerate high signal amplitude but is unpopular these days bacause it is considered not very accurate over the full audio spectrum.

Yet a third method of splitting phase is the 'long tail pair' or 'differential' or 'cathode coupled' or 'Schmitt' phase splitter. This uses two triodes which are coupled together at the cathodes and use as the common cathode load a high value of cathode resistor, or better still a constant current sink (CCS). The signal goes in via the grid of the first triode. The grid of the second triode is grounded. The triodes have equal plate resistors and each has the same gain (if there is a CCS in the cathode). The signals at the plates are of equal amplitude and are also of opposite phase, because the first triode is acting as a phase-inverting grounded cathode amplifier and the second as a non-inverting grounded grid amplifier. This splitter is accurate if a CCS is used, it has gain of about 50% of a single cathode-coupled triode and it can tolerate a high signal amplitude. It is perhaps the most popular tube-based splitter.

There is also a transformer splitter, which others have explained.

Ray Moth and I were writing at the same time and he posted before I did and maybe did a better job of answering your question, but here is my response to your last post...

The first image gives an idea of how this works, although what they are showing is "Class B" operation. Notice how the top half of the waveform and the bottom half of the waveform are amplified by separate devices and the re-assembled at the emitters? Class B means that with no signal there is essentially no current flowing in either device. Each one "switches on" as needed and as called for by an applied signal. This can be done with valves as well but is less common than in transistor amplifiers. The second circuit you provided could not be a "Class B" amplifier but rather a "Class A" because of the drive requirements of Class B. In a "Class A" amplifier both output valves are conducting 100% of the time, even when there is no signal, and are never driven to cutoff. The grids are never driven so hard that they require any current, only voltage swing. A Class B amplifier has current flowing in the grid circuit when a signal is applied and that current has to be provided by the previous stage. Therefore, any Class B stage has to be driven by another power amplifier and the cathodyne inverter on that second amplifier cannot provide any significant current.

As for how that inverter/splitter works... do you understand ohms law and understand that the current flowing through the 45k3 anode resistor, the triode, and the 45k3 cathode resistor is the same at any point? If this is true then the voltage drop across each resistor will be the same. The valve is essentially a voltage controlled variable resistance. As the internal resistance of the valve changes to follow the signal the voltage drop across each resistor is modulated accordingly. If the resistance of the valve is less, the drop across both resistors will be greater. The anode will now appear to have become more negative and the cathode will appear to have become more positive. If the resistance of the valve becomes greater, the voltage drop across the resistors will be less and the anode will appear more positive while the cathode will appear more negative. Since the two resistances, anode and cathode, are the same the change in voltage across them will essentially be the same, but opposite in phase.

Does that help?
Ah! I think I understand, but let just make sure I got it right...

If I'm reading this correctly--and let's just focus on the concertina design here--my confusion arose from the fact that the triodes are not actually performing a class-B amplification, but rather sending two signals that have a phase lag of \pi to the power valves. It's then the power valves that do the class-B amplification by way of biasing the grids on each, and the signal is then "joined" again after the power valves in the OPT.

If, however, the signal was sent straight from the phase splitter triodes to the OPT (ignoring the impracticalities of the small signal not being able to drive any sort of load) the resulting output would be zero owing to the two signals being out of phase by \pi.

Did I get it right? :)
Joined 2004
No, that's not right. The power tubes in push-pull feed opposite sides of the OP transformer primary and are in anti-phase, so they alternately push and pull the signal through the primary. If they were in the same phase, they would indeed cancel one another. It would be just the same if it were possible to connect the phase splitter directly to the OP transformer (which, in fact, can be done in some low power amps, where the power tubes can be used in an LTP splitter arrangement).

This has nothing to do with the class of amplification, by the way, except to say that Class A can work in both single ended and push-pull amplifiers, while Class AB and B need push-pull, otherwise part of the waveform will be missing and there will be horrible distortion.
OK, so I am close--sorry for being such a pest but I this has been very helpful for me and I really appreciate it:)

The good news is that I think I get it. I hope. What you're saying is that after the signal has been split by the triode pair, it can then be fed to the OPT for "joining" by way of the fact that signal(s) are fed to opposite sides of the OPT. OK, so if this is the case then what is going on with the concertina phase splitter is that one output is like the input (possibly with a little gain), while the other one is an inverted copy of the same output signal, as opposed to an out of phase version of it. By doing this, the two signals not actually "joined" back together (as in the wikipedia link I referenced earlier), but rather used, like you say, to alternatively push and pull the OPT.

Please tell me I got it this time, this is starting to get embarassing :blush:

wally.gribbles said:
If, however, the signal was sent straight from the phase splitter triodes to the OPT (ignoring the impracticalities of the small signal not being able to drive any sort of load) the resulting output would be zero owing to the two signals being out of phase by \pi.

Did I get it right? :)

Not really. The whole business really isn't all that difficult. A push-pull amp works in the same manner as AC power distribution. The overhead power lines carry single phase AC at 4.0KVrms to the "pole pig" transformers. The secondary of the pole pig is center tapped, with the center tap being the AC neutral. If you take your DMM and measure the voltage from each "hot" side to the neutral, you get a reading of 120Vrms (nominal). Measure from hot-to-hot, you get 240Vrms. This is possible since the two voltages are 180 deg out of phase, and:

Vtotal= 120/_0 - (120/_180) (by the KVL)
= 120/_0 + 120/_0= 240/_0Vrms

So your 240V line is really a two-phase system.

The main differences between that and a VT amp are: the two phases are on the supply side, not the load side, and the wave form is more complex than a simple sine wave. Otherwise, it's no different, and Class A operates just like the AC distribution system in reverse.

Of course, you aren't limited to Class A operation, since the xfmr itself can "put together" two halves of a partial signal so long as the electrical balance is perfect (close, but not absolutely perfect in the real world) and there is at least half the signal waveform correctly reproduced at each end. You can come pretty close to the ideal with practical xfmrs used today.

For PP operation, you don't necessarily have to have a xfmr. In the case of output transformerless PP, you need an active current source/current sink. Being that solid state isn't limited to just electron conduction, it's NBD to make use of NPN/PNP complimentary pairs to get balanced source and sink currents. Not so easy with VTs since these just come in the one flavour (N-Channel), so that's why you need phase splitters
Sweet! I can feel the enlightenment washing over me :)

But yeah, inverted = out of phase, but only when the phase angle is an odd multiple of \pi. My background is more mathematical than electronics, so I tend to be kind of pedantic about these sorts of things. ;)

Many thanks again to all of you, this has been a great help.

EDIT: Sorry, I see a few replies have appeared while I was typing a post and it looks like I've ignored you. I haven't really, simply the timing is all...
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