How the Long Tail Pair Phase Inverter Works

[robrob]

This is from my How Tube Amps Work webpage. As always I'd appreciate corrections and feedback.

Signal flow shown with red arrows. The Signal enters the phase inverter at V3A's grid and flows out both its plate (inverted signal) and cathode (non-inverted signal). The cathode signal flows to V3B's cathode where it is amplified.

The Long Tail Pair (LTP) Phase Inverter is the most popular phase inverter in guitar amplifiers. Unlike the non-amplifying cathodyne phase inverter it not only creates a dual mirror image signal stream but it also acts as a gain stage boosting the signal by about half of what a normal triode gain stage would do. The LTP is a true differential amplifier and uses both halves of a triode (usually a 12AX7).

Input signal goes to V3A's grid, V3B's grid is held at a constant voltage and capacitor C20 removes all AC signal. Since the grid is held constant voltage fluctuations on the cathode create an amplified signal on V3B's plate.

V3A acts as a normal gain stage (outputs inverted signal at the plate). V3A also acts as a cathode follower (outputs non-inverted signal at its cathode).

V3A and V3B's cathodes are tied together--all of V3B's input signal comes from V3A's cathode.

R34 is the bias resistor and creates the voltage difference between both tubes' grid and cathode.

R36 is the tail resistor that creates the relatively high voltage needed for the cathode follower function of V3A.

R37 and R38 are simply grid leak resistors.

The Presence control (R35 and C21) removes a variable amount of high frequency from Negative Feedback therefore boosting high frequency output (C21 shunts AC signal to ground).

Function Detail: When a positive voltage signal arrives at V3A's grid the reduction of blocking negative electrons on the grid allows electrons to flow from its cathode, through the grid, to its plate. The electrons flowing onto the plate decrease the voltage in the wire between the load resistor and plate--this is the inverted and amplified output signal. As electrons leave V3A's cathode a positive voltage is created on the cathode (scarcity of electrons = positive voltage). This positive signal voltage is also present in V3B's cathode because the cathodes are directly connected. Since V3B's grid is held constant at ground or 0 volts any change in its cathode voltage will create an amplified signal at its plate. As V3B's cathode goes positive (scarcity of electrons) fewer electrons will flow from it through the grid to the plate. The reduction of electrons flowing onto the plate increase the voltage in the wire between the load resistor and plate--this is the non-inverted and amplified output signal.

Bassman 5F6-A LTP PI Voltages

Note the 1.5v bias difference between the tail resistor junction of 32.5v and the cathodes at 34v.

 

[tubeswell]

R36 is the tail resistor that creates the relatively high voltage needed for the cathode follower function of V3A.

The reason the tail is exists in the LTP is to provide a constant current source which drives the non-inverting stage in the LTP.

The bigger you make the tail, the more even the gain in both the inverting stage and the non-inverting stage becomes (ultimately permitting use of same-sized plate resistors if the tail resistance is high enough (like 47k or so)), however this even-ness is at the expense of overall reduction in gain all other things being equal (because if you keep the HT supply voltage the same, then the plate-to-cathode voltage decreases as the cathode-to-ground voltage increases).

 

[Fred Mertz]

Tubeswell is 100% correct. Another consideration is the symmetry of the input signal to the output tubes. Asymmetry leads to higher levels of 2nd order harmonic distortion.

 

[printer2]

V3A also acts as a cathode follower (outputs non-inverted signal at its cathode).

I am assuming this statement will disappear.

The reason the tail is exists in the LTP is to provide a constant current source which drives the non-inverting stage in the LTP.

The bigger you make the tail, the more even the gain in both the inverting stage and the non-inverting stage becomes (ultimately permitting use of same-sized plate resistors if the tail resistance is high enough (like 47k or so)), however this even-ness is at the expense of overall reduction in gain all other things being equal (because if you keep the HT supply voltage the same, then the plate-to-cathode voltage decreases as the cathode-to-ground voltage increases).

I don't see it as gain reduction but I see it more as limiting the output swing.

 

[robrob]

The reason the tail is exists in the LTP is to provide a constant current source which drives the non-inverting stage in the LTP.

Both V3A and B's cathode equally use the constant current--it's constant because of the inverted and non-inverted signal being created simultaneously--as V3A's grid goes positive and allows electrons to flow from its cathode to plate, at the same time V3B's cathode goes positive and cuts the flow of electrons to its plate so the current through the tail and cathode resistors stays constant.

A cathode follower needs an elevated cathode voltage to function and the tail resistor provides it.

Classic Cathode Follower
image removed
Note how RL functions like the LTP tail resistor.

V3A is acting as a normal gain stage (inverted signal plate output) and as a cathode follower (non-inverted signal cathode output).

Quote:
V3A also acts as a cathode follower (outputs non-inverted signal at its cathode).

I am assuming this statement will disappear.

Why?

 

[printer2]

Both V3A and B's cathode equally use the constant current--it's constant because of the inverted and non-inverted signal being created simultaneously--as V3A's grid goes positive and allows electrons to flow from its cathode to plate, at the same time V3B's cathode goes positive and cuts the flow of electrons to its plate so the current through the tail and cathode resistors stays constant.

A cathode follower needs an elevated cathode voltage to function and the tail resistor provides it.

Classic Cathode Follower
image removed
Note how RL functions like the LTP tail resistor.

V3A is acting as a normal gain stage (inverted signal plate output) and as a cathode follower (non-inverted signal cathode output).



Why?

You are wrong. How can a stage be a cathode follower that has a gain of less than one and then be used as a gain stage? You left out the plate resistor that is in the LTP circuit, the classic cathode follower diagram does not have it, where is it?

The tail resistor is the constant current source not the two triodes. The second triode gets its signal from the common cathode resistor but the gain is not unity so it has a lower output than the first triode. That is why you see a lot of LTPI's with 82k 100k values on the plate to compensate for the difference in signal. In the 18 Watt or many other EL84 designs they get rid of some signal by using a large CCS resistor. This balances the stages better and the two plate resistors are normally the same value.

 

[robrob]

You are wrong. How can a stage be a cathode follower that has a gain of less than one and then be used as a gain stage?

Designing Tube Preamps for Guitar and Bass by Merlin Glencowe (the tube wizard)

page 155 "An alternative way to view the circuit is to say that V1 acts both as an ordinary gain stage and as a cathode follower."

Valve Amplifiers by Morgan Jones

page 457 "V2 can be considered to be a grounded grid amplifier, fed from the cathode of V1. It is this use of the first valve as a cathode follower to feed the second valve that results in the apparent loss of gain of the second valve, since for a cathode follower, Av <1. By inspection, we see that 2 Vgk is required to drive the stage, so the gain of the compound stage [input triode] to each output is half what we would expect from an individual valve."

In other words it's the cathode follower function of the first valve in an LTP that cuts the gain of the two valves in half.

 

[printer2]

No, it is not a cathode follower as the plate voltage on that triode varies with the signal. The second triode is getting its signal from the cathode of the first tube. That is the extent of it being a cathode follower.

The reason the gain is cut in half is because the input signal is shared between the two grids, one that is virtually grounded, the other has the signal on it and the mid point is the cathodes. Same circuit as a self split output stage.

 

[robrob]

The reason the gain is cut in half is because the input signal is shared between the two grids, one that is virtually grounded, the other has the signal on it and the mid point is the cathodes. Same circuit as a self split output stage.

Your statement is directly contradicted by Jones's quote, ". . . the gain of the compound stage [input triode] to each output is half what we would expect from an individual valve."

He's saying the cut in gain is caused by the dual output [plate & cathode] of the first triode.

But Merlin agrees with you, page 157 "Notice that we are using two triodes but the gain is the same as that of one equivalent gain stage, because each triode only sees half the total differential input signal applied between the two grids."

Jones makes more sense to me because the first triode has to split its output between the plate and cathode. Also 100% of the input signal goes to the first triode grid and while the second triode grid is at AC ground so 0% goes to the second triode grid.

No, it is not a cathode follower as the plate voltage on that triode varies with the signal.

Merlin, Morgan and I are not saying the first triode IS a cathode follower, only that is ACTS as a cathode follower. Anyway you may want to write to Merlin and Morgan and give them your corrections.

 

[Mhat]

I honestly find this very interesting. I'll be watching this tread. I wish I could add some input.

 

[printer2]

Your statement is directly contradicted by Jones's quote, ". . . the gain of the compound stage [input triode] to each output is half what we would expect from an individual valve."

He's saying the cut in gain is caused by the dual output [plate & cathode] of the first triode.

But Merlin agrees with you, page 157 "Notice that we are using two triodes but the gain is the same as that of one equivalent gain stage, because each triode only sees half the total differential input signal applied between the two grids."



Merlin, Morgan and I are not saying the first triode IS a cathode follower, only that is ACTS as a cathode follower. Anyway you may want to write to Merlin and Morgan and give them your corrections.

Think about it, if you had a regular gain stage with a 1.5k cathode resistor and a 100k resistor you get a certain gain. You have a large signal out of the plate, the cathode voltage changes due to the input signal. If the tube is a 12AX7 and you are biased about -1.5V the most the voltage will swing at the cathode is 1.5V.

Now look at the LTP circuit ignoring the triodes. You have the signal coming from C19 in your diagram. The signal is coming from the treble pot and it is referenced from ground potential. So say you are getting 2V out of the tone stack and into the PI. Ignoring the triodes as I mentioned before as the signal does not see them as they are a high impedance as far as the signal is concerned.

So this 2V has to go through R28 (1M) and through R37 (1M) in order to get to ground potential. It has to go through C21 and through the presence pot but 5k is of no consequence as the load on the signal is 2.005M, so we can just call that 2M. So now we have two volts into the 1M voltage dividers and one volt is across R28 and 1V is across R37.

The mid point of the two 1M resistors has the shared cathode resistor terminated there. So adding the top triode now, at the grid (referenced from the cathode) it will see the one volt difference across the top 1M resistor. This triode will see 1V rather than the 2V coming out of the tone stack.

I hope you can see where this is going. The first triode has not had its gain magically cut and as you said, Merlin agrees with me. So far I have said the first stage has normal gain and it is seeing half the applied signal resulting in half the signal you would normally see on a normal gain stage.

Now onto the other triode. The other triode will see the other half of the 2V signal being sent out of the tone stack. But rather than the signal going into the grid, the grid is tied to ground. But we still have the 1V signal across the 1M grid leak resistor referenced to the cathode through the shared cathode resistor.

And this is the important part that does the phase inversion. But instead of the signal coming in positive (sine wave starting from zero and rising) to the grid (since the grid is effectively tied to ground) the sine wave increases the bias voltage. The triode doing what any good triode would do amplifies this and the plate signal is inverted from the driving signal.

The top triode grid sees a signal going up and the bottom triode sees the signal going down as the cathode is referenced higher than the grid. Therefore you get two signals out of the PI, one inverted from the other.

So really the inversion is being driven by the two 1M resistors. The two triode gain stages are in their own little worlds and have no idea how they are hooked up in the grand scheme of things. They are just performing their doing their job in relation to their plate, cathode and grid leak resistors.

The circuit will operate with or without a constant current source, in our case the resistor. Fender had it down to somewhere around 10k in one amp in order to not loose the PS voltage as they wanted to give more signal driving the output tubes. The first tube is not acting as a cathode follower. Once you leave the confines of its plate cathode and grid leak resistor it does not care what the world is doing. Saying it is acting like a cathode follower does not change that.

 

[printer2]

I honestly find this very interesting. I'll be watching this tread. I wish I could add some input.

And you take that smile off your face.

 

[KCStratMan]

And you take that smile off your face.

I agree this is fascinating, the Royals are playing in The Show and I'm listening to youse guys. Recently followed a similar discussion of whether or not the fourth or fifth tube stage in a premp was a cathode follower and/or a gain stage. Are you guys both correct and just stating this differently or describing slightly different aspects?

 

[robrob]

So this 2V has to go through R28 (1M) and through R37 (1M) in order to get to ground potential. It has to go through C21 and through the presence pot but 5k is of no consequence as the load on the signal is 2.005M, so we can just call that 2M. So now we have two volts into the 1M voltage dividers and one volt is across R28 and 1V is across R37.

The first tube is not acting as a cathode follower. Once you leave the confines of its plate cathode and grid leak resistor it does not care what the world is doing. Saying it is acting like a cathode follower does not change that.

I can't find R28 and I can't follow your analysis. I can follow Glencowe and Jones description of the LTP function. They both say the first triode is acting as a cathode follower and I can see that clearly and intuitively. You saying the first triode does not act as a cathode follower does not change that.

 

[printer2]

I can't find R28 and I can't follow your analysis. I can follow Glencowe and Jones description of the LTP function. They both say the first triode is acting as a cathode follower and I can see that clearly and intuitively. You saying the first triode does not act as a cathode follower does not change that.

Sorry, R38.

A cathode follower has the plate at the power supply rail. What is screwing you up is the tail resistor. The phase inverter can work without it but it works better with a CCS. Really all you have to be able to see is the signal gets split between the two 1M resistors. Because of this the output is only half of what it would be if you fed the signal into a ordinary gain stage.

As far as the inverting function, the top triode acts like a normal gain stage, the only difference is the cathode resistor is half the normal value as it is being used by two triodes. It is no different than the two channels of a 5E3 using the same resistor. The second triode inverts the signal as the cathode is connected to the top of the 1M resistor and the grid is connected to the ground through the capacitor.

Split the two triodes up and give them each their own cathode resistor. The diagram you used makes it real easy for me to see how the circuit works. There was another inverter that I never could really understand until I had to explain the LTP with this diagram. Now the other one seems simple. But back to this one. I drew up a series of schematics outlining how to look at it but my photo software is hanging up so I can not change the schematic into a picture.

I think where we diverge is the cathode follower thing. You see the first triode acting like a CF because the first triode is swinging a current through it and it is effecting the second triode. This is wrong, if only the first triode was using the tail resistor it still would not be acting as a CF but rather a cathodyne splitter. Think of the plate resistor as 100k and the tail resistor (or CF cathode resistor), then you would see a voltage that followed the signal at the cathode. Actually a cathodyne and CF actually are the same circuit except their plate and cathode resistors do not match.

Now the reason the first triode does not act as a CF is the current does not change through this resistor. Both triodes feed current through it. When one goes up the other goes down. So no AC voltage is developed across this resistor. So effectively it is out of the AC circuit.

Anyway, without the tail resistor and without the shared cathode resistor you can see what the circuit is doing better. You just need to realize that the top triode works as a normal gain stage. The second triode works the same except rather than the grid being attached to the high side of the signal and the cathode grounder the two are reversed. And since they are reversed the action of the signal is reversed. You get an inverted signal. Again, it is because the grid and cathode are reversed as far as the signal on the second 1M resistor goes. Up is down and down is up.

 

[printer2]

I agree this is fascinating, the Royals are playing in The Show and I'm listening to youse guys. Recently followed a similar discussion of whether or not the fourth or fifth tube stage in a premp was a cathode follower and/or a gain stage. Are you guys both correct and just stating this differently or describing slightly different aspects?

No, not stating the same thing. You have to be able to split what the dc characteristics are doing and what the AC characteristics are doing. Once you do that it is easier to analyze.

 

[robrob]

No, not stating the same thing. You have to be able to split what the dc characteristics are doing and what the AC characteristics are doing. Once you do that it is easier to analyze.

I agree this is the key to understanding the LTP. We 'see' this circuit very differently. You see the AC signal being split as it flows through the two 1M grid leak resistors but and I don't see it that way. The grid leak resistor 'leaks' electrons that are unintentionally captured by the grid to ground to keep the DC bias from drifting negative. You could remove the grid leak resistor and the valve would continue to amplify but the DC bias would drift positive. Only a tiny bit of the AC signal goes to ground through the grid leak--that's what is meant by 'high impedance.' I see almost all of the input signal making it onto V3A's grid, not being split (what I do see being split is V3A's output between the plate and cathode).

The AC signal on the grid wire doesn't need to flow to ground, it just needs to pack the grid with electrons (negative signal voltage swing) so they block the flow of electrons from V3A's cathode to the plate.

The tail resistor reduces the expected gain from two triodes because both triodes' plates have to pull its electron flow through the tail resistor to generate their amplified voltage swing. Increasing the resistance of the tail resistor increases the signal output balance of the LTP which is desirable but it also reduces the LTP's gain.

With no AC signal on V3A's grid there's 34v DC at the cathode generated by the 'electron drag' of the tail and bias resistors. That drag reduces the gain of V3A and V3B.

 

[printer2]

I agree this is the key to understanding the LTP. We 'see' this circuit very differently. You see the AC signal being split as it flows through the two 1M grid leak resistors but and I don't see it that way.

Then you will never be good at electronics. The signal does not split, the two 1M resistors are a voltage divider. You put 2V AC out of the tone stack, two volts will be on the top triode's gridreferenced from ground. As far as the 2V signal is concerned it leaves the tone stack, goes through the top 1M, goes through the bottom 1M, through C20 and the presence pot. At this point you are at the same potential (ground) as you are referencing to measure the output of the tone stack.


Think of the tone stack as a power supply and ignore the capacitor. The two volts the power supply is putting out will be developed across the two 1M resistors. With your volt meter negative lead at ground you will measure two volts at the top of the top 1M and 1V at the node in between the two resistors.

If you put your meter lead across the top resistors with the negative lead at the node where the two 1M resistors meet you will read 1V. Now if you take your positive meter lead and put it on the bottom of the bottom 1M your meter would read -1V as the negative lead is referenced from a point with a higher potential.

The grid leak resistor 'leaks' electrons that are unintentionally captured by the grid to ground to keep the DC bias from drifting negative. You could remove the grid leak resistor and the valve would continue to amplify but the DC bias would drift positive. Only a tiny bit of the AC signal goes to ground through the grid leak--that's what is meant by 'high impedance.' I see almost all of the input signal making it onto V3A's grid, not being split (what I do see being split is V3A's output between the plate and cathode).

The signal makes it to the top grid, while the two 1M resistors may seem like high impedance, they are a low impedance as compared to the impedance between the triode's grid and cathode. So the 1M resistors determine that the 2V is shared between them.

Now since you see V3A's output split between the plate and cathode how much voltage is developed across each resistor? Not actual values but as a rough percentage? I see most of the voltage swing across the 82k resistor and about 5% across the 470R resistor.

The AC signal on the grid wire doesn't need to flow to ground, it just needs to pack the grid with electrons (negative signal voltage swing) so they block the flow of electrons from V3A's cathode to the plate.

The electrons need to leave the treble pot's wiper, go through the blocking capacitor, go through the 1M resistors, the other pot, up through the tone stack, and back to the wiper. If you do not have this closed circuit you have an open circuit. It is like measuring a voltage with one lead of your volt meter.

The tail resistor reduces the expected gain from two triodes because both triodes' plates have to pull its electron flow through the tail resistor to generate their amplified voltage swing. Increasing the resistance of the tail resistor increases the signal output balance of the LTP which is desirable but it also reduces the LTP's gain.

The triode amplifies a signal that is on its grid referenced from its cathode. The gain is set by the ratio of the cathode resistor and the plate resistor along with the amplification factor of the triode. The tail resistor is outside of the grid-cathode loop. The tail resistor also gets the current from the second triode. When one goes up the other goes down. So you get a constant (more or less) current through the tail resistor. So you do not get any AC voltage developed across it as you would a cathode follower or a cathodyne PI.

With no AC signal on V3A's grid there's 34v DC at the cathode generated by the 'electron drag' of the tail and bias resistors. That drag reduces the gain of V3A and V3B.

That is a dc voltage drop not AC. Also it is not within the grid-cathode loop so the triodes do not see it. It is the same as a dropping resistor in your power supply.

Now I have explained this a number of times to you and I am not going to again as I have better things to do. If you have a specific question that needs answering I will do that. As something to think of, a few years before I started hanging out on this forum I have taught a number of engineering course in a college and the intro to electronics course would help you see how the circuit actually works. You need to think of voltages being developed by a source and dropped around a circuit. They all have to add up to zero.

 

[robrob]

printer2, thanks for the detailed breakdown and answers.

But I have one last question, you're saying a normal amplifier valve will not function, not even temporarily if the grid leak resistor is clipped? I see the voltage loop as running from grid to cathode and the only function of the grid leak resistor is to keep the bias from drifting due to grid current. This is the root of my confusion.

 

[printer2]

Never said an amp will not function without a grid leak resistor, it will develop its own voltage on the grid and float to where it is happy. It may not nessisarily do what we want though.

But we still need a continuous circuit to get our guitar signal into the triode so we need a return ground path. In the case of our guitars we have a volume pot and it takes the place of a grid leak resistor. But relying on the volume pot as a grid leak resistor is not a good idea as noise on the cable shield gets introduced onto the grid.