bloodhawk, yes.
Back to tube theory:
So we've gone thru stages using cathode as output and anode as output. I've also said that the tube doesn't know which output it's supposed to use. Let's look at an extreme case of this:
This is called a concertina phase splitter, or alternatively a cathodyne phase splitter. (I think both of these come from HIFI manufacturer brand names.)
There is equal amount of resistance in series both above the tube and below the tube. Nothing is bypassed with a cap. Knowing what we just went thru in my previous message, what kind of an output do you now predict this produces?
Think about it for a second before looking at the next picture.
Did you get it right? This is exactly as theory would predict. Anode output is inverted, cathode non-inverted, and equal resistances produce equal amplitudes.
So it really is just a matter of Ohm's law, and the ratio between the anode and cathode resistances. The tube changes transconductance (meaning alters the amount of current that goes thru it), and the rest is just Ohm's law.
The concertina has very good properties on paper, (pretty much perfect balance), but it has some shortcomings. As you can see, the amplitude is compressed; 2VPP at grid produces 1.6VPP outputs. This reduces dynamics.
Also the anode and the cathode don't have equal output impedances (meaning they don't have equal ability to output current to load). This sim has 470k load on the other side of the capacitors, but in many real life cases there is a much harsher load there (especially on transients), and then bad things happen to the balance.
So this type of phase splitter is not ideal for transparent high fidelity sound reproduction. It's very commonly used in guitar amps though.
We know that triodes amplify the difference between grid and cathode voltage. In the previous part I stated right in the beginning that a triode has two inputs; grid and cathode.
We looked at a situation where cathode is at AC ground, so essentially grounded for signal purposes. What about this:
Would it work? There is a difference between grid and cathode voltage after all!
Let's look at the waves:
Grid is grounded of course, green is cathode at 2VPP and blue is anode at 36VPP! This gives a gain of about 18.
Works just as theory predicted. Triode amplifies difference between cathode and grid voltage. I keep banging on this particular drum because it is essential to the LTP operation.
As an aside, this kind of grounded grid amplifier was very common in radio frequency amplification as the first stage amplifying input from an antenna. The cathode resistor was replaced with a small choke, giving high impedance at very high frequency while having a low DC resistance to bias the tube right. The point of it was that capacitance between cathode and anode was almost nonexistent; this is important because the antenna produces pretty much no current at all to drive this parasitic capacitance. So while a grid input would have a high Miller capacitance, cathode input doesn't.
We noted earlier that the cathode OUTPUT is non-inverting. This is also true of the cathode INPUT. If you look at the waves above, they are in phase.
So, let's move closer to the goal:
Here we have a pair of single ended gain stages. They are not connected to each other in any way except via the B+, which here is pure DC. Left side receives a grid input of 2VPP, and right side receives no input with grounded grid.
No surprise what the outputs look like:
I'm sure you can guess that was the right side.
Left side has some cathode degeneration (green wave), but there is a decent output still.
This type of circuit is of course quite pointless. Let's see what happens when we connect the cathodes:
Now this is starting to look a lot more familiar now! Based on the previous examples, predict what happens here after the cathodes are connected together.
Since they now share a cathode node, there is only one voltage there:
It is of course produced by the left tube; it is in phase with the left tube grid signal.
Now this is where it might get a bit tricky, keeping all the phase in your brains.
So left tube grid gets signal. Let's call it's phase as 0. There is an unbypassed resistor at the cathode, so there appears "cathode output" voltage. It also has phase 0.
Now left tube anode output is at phase 180 since grid to anode output is inverted phase.
Right tube receives a cathode input, when using cathode to anode input phase is not inverted. So right tube anode output signal has phase 0.
So what do we end up with?
Green wave is left anode, and blue wave is right anode. Left output is about 24VPP, right is about 7VPP.
Of course left output is much bigger; left tube input (at the grid) is 2VPP. There is some cathode degeneration, but not that much, about 600mVPP as was seen above.
This 600mVPP is the whole of the input signal for the right side tube, so naturally it's output is much smaller.
Why is the cathode signal (parasitic output for left side tube, input signal for the right side tube) so small?
Well because of Ohm's law; the 750 ohm cathode resistor (the tail) is not that big. So what happens if we make it a lot bigger?
Now if we make the tail resistor bigger, we must make the bottom of the tail voltage much lower, negative. I tried to keep the current about the same, so the bias points are about the same; if the same current flows thru a resistor, there needs to be a voltage across it. Ohm's law again.
So about -1000V B- there. Easy to build, at least in Spice!
But now the tail is 120k instead of 0.75k. Based on what we've gone thru here, predict what happens and why.
Well, first of all, the left tube experiences a lot more cathode degeneration.
Because of that, the right side receives a much bigger input signal via cathode input.
So, outputs should be much closer to amplitude.
Let's look at the outputs then:
Almost same in amplitude! With only one input into the whole circuit!
Anode output amplitudes are now about 16VPP. But, since the signal is now balanced, we don't have to think in terms of SE signals anymore; the DIFFERENTIAL output amplitude is both waves added together, so 32VPP.
When we look at the previous example with the 750 ohm tail, the outputs were about 24 and 7 VPP, together about 31 VPP, this tells us that the sum of the output amplitudes is pretty much always the same (with the same anode resistors).
If we were to increase anode resistance, we'd get an overall larger differential output (increased gain).
Let's look at the left grid voltage and the cathode node voltage:
Blue is left grid input, and green is cathode node, which has of course some DC component also (bias voltage for the tubes).
You can see the cathode wave is 50% amplitude of the left grid wave. This causes two things:
1) it suppresses left tube gain (cathode degeneration); we can see this by looking at the individual outputs and comparing to the previous scheme with the 750 ohm tail
2) increases right tube input signal via cathode input; this increases right side anode output amplitude.
Balance achieved!
But how does the circuit 'know' how to stop the cathode degeneration at 50% of left grid amplitude? After all, there is now 120k beneath the left side cathode, and only 33k above the anode. Shouldn't the left side be much more of a cathode follower, yet there is a big anode output?
Think about it and give your suggestions. We'll go thru this at a later time.
This is the essentials; the same way that this topology can compensate for missing input signal on one side, it can compensate for tube differences.
