Phase Inverter - How?

[gearhead]

Alright,

My what-seems-to-be weekly question is: why and/or how do you get phase inversion for the second half of the 12AX7 in the PI? I've read Aiken and quite a number of other sources, but none of them quite address the question.

First stage is obvious. In-phase signal directly feeds the grid and equals out of phase from the 82K plate resistor. Seeing as the second stage has a 100k plate resistor, believe that tells me that the second stage is being fed a lower signal than the first.

Since the ouput of the second stage is in-phase, whatever fed the second stage grid has to be out of phase. Obvious would be the out-of-phase first stage signal, but don't see any connection (plus it's probably too high to begin with?). Either I'm missing a connection, or it's something else.


Appreciate any responses


Keep feeling like I'm asking Tubes and Electronics 101 questions, but hope others are benefiting from many of the very excellent responses. Am I embarrassed? A little bit, but my curiousity is much stronger at this point.

[nickt]

The amount of current through the pair of triodes is constant. So if you increase the flow in one it must decrease in the other and vice versa.

Think of it this way: you're drying your backside with a towel. You pull one side up the other goes down...

Does this help?

[UR12]

Think of it this way. On a long tail phase inverter the first triode and the second triode have their cathodes tied together and like you said the first triode acts like a inverting amp with the signal applied to it’s grid and the output taken from the plate. On the second triode you basically have the grid tied to AC ground through the cap from the grid to the cathode and the top of the bias resistor. This sets up the stage kind of like a grounded grid amplifier. Remember in a cathode follower circuit where the output is taken from the cathode the signal remains in phase with the input signal. The second triode’s signal input is actually the signal being developed across the tail resistor from the first triode and since it is also connected to the second triodes cathode, the signal on the cathode is the input to the second triode and the output is taken off the plate and is in phase with the signal being applied to the grid of the first triode. Grounded grid amplifiers do not change the phase of the signal applied either. Do I have you completely confused yet?

[Fischerman]

To expand on what UR12 wrote...I think the 'lightbulb' is the following fact about vacuum tube triodes (especially the second sentence):

A signal input at the grid is amplified out-of-phase at the plate.
A signal input at the cathode is amplified in-phase at the plate.

If you attempt to look at it like; grid and plate are out-of-phase while grid and cathode are in-phase; then it might lead you to think that the signal at the second plate would still be out-of-phase (just like the first triode) since the signal at both cathodes is the exact same signal...but it's not because of the little factoid above.

Hope that helped.

[gearhead]

Do I have you completely confused yet?

Yes

On the second triode you basically have the grid tied to AC ground through the cap from the grid to the cathode and the top of the bias resistor.

As a common reference, am looking at Dr. Hulsey's schemo numbers. Which cap do you mean? C10 and/or C11? Top of the bias resistor as in the junction of the 4 resistors (long-tail) or as in where it connects to the cathodes?

This sets up the stage kind of like a grounded grid amplifier. Remember in a cathode follower circuit where the output is taken from the cathode the signal remains in phase with the input signal.

You saying that the first triode can also be treated like a cathode follower?

The second triode’s signal input is actually the signal being developed across the tail resistor from the first triode and since it is also connected to the second triodes cathode, the signal on the cathode is the input to the second triode and the output is taken off the plate and is in phase with the signal being applied to the grid of the first triode. Grounded grid amplifiers do not change the phase of the signal applied either.

Let me see if I got this straight. Probably some holes, but here's a shot:

Assume signal input is rising (going positive).

Input signal going positive causes the Triode 1 (T1) grid voltage to increase. This decreases grid bias, which results in increased T1 cathode to T1 plate current. As stated, the output is inverted when measured at the plate resistor (R15).

When T1 cathode to plate current increases, current throught the bias resistor (R13) increases, resulting in an increase of voltage across R13. This increases the T1 grid bias? Normally, this would then result in a decrease in T1 cathode to T1 plate current. I can only assume that this increasing grid bias just retards (a bit) the T1 grid to plate current increase, not completely reversing it? The input signal is great enough to overcome this tendency?

Since this is a shared bias resistor, it directly affects Triode 2 (T2).


As above, the voltage across R13 bias resistor is going increasingly positive. This increases the T2 grid bias, which causes a corresponding decrease in the T2 cathode to T2 plate current (since there is no direct input signal as with T1, there is nothing to keep it from dropping?)

Measuring off the 100k plate resistor results in an increasing signal (at this point, it still inverts.) , which is "in phase" with the PI's input signal.

All these voltages are scaled/phased versions of the input signal.

This in the ballpark? Or up in the stands behind home plate? LOL.

[Jack]

Read Randall Aiken Tech page about the schmitt splitter. I think he even has the original article on his website.

[gearhead]

I have, a number of times. Would not have posted here without doing a lot of research and reading first. Am able to work out 95% or so without posting here; but not this one

[UR12]

Does this help any? It is another way to look at the same circuit. Please excuse my crude drawing

[Fischerman]

gearhead,
I look at it like this:
When you have an unbypassed cathode, the signal 'exists' on the cathode...that's the local NFB people are talking about with respect to unbypassed cathodes in a regular gain stage. Think of the first triode as a regular gain stage (input at grid, inverted output at plate). So when you input signal to the first grid...that signal also exists on the cathode. But the cathodes are tied together and that signal that exists on the cathodes is now the input signal to second triode. And since the input for the second triode is at the cathode...the amplified signal on the second plate is in-phase with that input signal (which also is in-phase with the main input signal).

I think the 82K and 100k plate resistors are backwards in the above schemo...usually it's the main input triode that gets the 82k plate resistor.

[novosibir]

I think the 82K and 100k plate resistors are backwards in the above schemo...usually it's the main input triode that gets the 82k plate resistor.

Yes!

Larry

[UR12]

Dang

Tough crowd. I changed the resistors

[gearhead]

Thanks for the bottom-line conventions on common-cathode, common ground tube circuits. Will go to those immediately in the future.

However, what has got my curiosity are the "mechanisms" that cause the changes.

Went around this a couple times on this thread: https://tubeamparchive.com/viewtopic.ph ... highlight=

For example, there is a reason the "signal" exists on the cathode for an un-bypassed cathode. I -believe- it's because the input signal, to the grid, changes the voltage on the grid, which changes the bias. This change in biase increases (or decreases) the cathode to plate current. That (electron flow) current is sucked thru the cathode pin. With a bypass cap, the increased/decreased current will not flow thru the bias resistor but thru the cap. Hence no change in voltage at the cathode. However, take it away, and there will be increased/decreased current thru the resistor (and therefore increased/decreased voltage). Which happens to be in synch and phase with the input signal.

Or I could be way off base

[PRR]

I'm swamped by Real Work.

As others said:

The second side acts like a "grounded grid" amplifier, driven by the first side functioning as a "cathode follower". You don't see a GG stage anywhere else in audio, so you may have to brush-up on it.

Or look at Differential Pair.

> 82K plate resistor ... 100k plate resistor

Red herring. With a specific (non-infinite) value common cathode resistor, the plate resistors "should be equal". In practice the difference is small even for equal resistors; the 82K/100K fudge is a frill and not exactly "perfect" either. Fortunately it hardly matters if the PI is 25% out of balance. May even sound "better" off-balance.

[UR12]

Thanks for the bottom-line conventions on common-cathode, common ground tube circuits. Will go to those immediately in the future.

However, what has got my curiosity are the "mechanisms" that cause the changes.

Went around this a couple times on this thread: https://tubeamparchive.com/viewtopic.ph ... highlight=

For example, there is a reason the "signal" exists on the cathode for an un-bypassed cathode. I -believe- it's because the input signal, to the grid, changes the voltage on the grid, which changes the bias. This change in biase increases (or decreases) the cathode to plate current. That (electron flow) current is sucked thru the cathode pin. With a bypass cap, the increased/decreased current will not flow thru the bias resistor but thru the cap. Hence no change in voltage at the cathode. However, take it away, and there will be increased/decreased current thru the resistor (and therefore increased/decreased voltage). Which happens to be in synch and phase with the input signal.

Or I could be way off base

Hmmm... The bypas cap does 2 things it acts as a filter just like in a power supply and helps to keep a constant voltage on the cathode. This keeps the cathode at pretty much the same voltage positive in respect to the grid. This in effect makes the grid negative in respect to the cathode or provides a "negative Bias" for the grid. It also provides an ac path to ground so it does have some frequency componet. If you change the 22uf electrolytic to a .68 you will notice a change in the tone of the amp stage.

The bias we have on the grid of the triode with no signal applied set's it's operating point. You usually look at the tubes characteristic curves published in the tube manual for a given plate voltage and pick a linear portion for the tube to operate in. If the negative bias voltage were set to say 2 volts then you could put in a 4 v peak to peak signal before the bottom of the wave form would be clipped. (top of the wave form if you are looking at the signal coming off of the plate.) It's not really a change in bias that you mentioned that changes the signal. We usually only talk about the bias as being the voltage on the grid while no signal is present on the grid. To a point, the more negative bias on the grid the larger the signal we can apply to the grid before clipping occurs.

From your question I think you really want to discuss what happens when the cap isn't there. One thing to remember is that a small signal on the grid controls a large signal through the tube. That's why we get gain out of the triode. We are using a small signal to control a large change in current. You can think of the triode as a variable resistor. Its resistance between cathode and plate changes depending on the signal applied to the grid. In other words as the grid becomes more pos (Less negative) more current flows from cathode to plate. When more current flows from cathode to plate you also get more current flowing through the cathode resistor and plate resistor since they are in series with the cathode and plate.

The more current flowing through the cathode resistor the more positive the cathode becomes in respect to the grid and you loose a little gain. As the signal becomes more negative on the grid the less current flows from cathode to plate and the voltage drop from cathode to plate becomes greater.

[Fischerman]

Red herring. With a specific (non-infinite) value common cathode resistor, the plate resistors "should be equal". In practice the difference is small even for equal resistors; the 82K/100K fudge is a frill and not exactly "perfect" either. Fortunately it hardly matters if the PI is 25% out of balance. May even sound "better" off-balance.

Could you elaborate on this? I've tried building some amps using equal plate resistors in the PI and they never stayed that way for long...almost always back to 82k/100k just because it sounded infinitely better (some amps were 91k/100k...usually ones that have a 22k/24k tail resistor instead of 10k). In a Vox...even if you didn't use both inputs the PI would be way more balanced using both 100k plate resistors than say a Marshall or Fender would (using two 100k plate resistors) because of the 47k tail resistor (larger tail resistors increase balance and IME if the PI has a 10K tail then it really needs the 82k plate resistor on the input triode).

One of these days I'm gonna build an amp with a negative rail and see what the PI sounds like when you don't have to 'lift' the whole thing above ground.