Signal bleeding through

[jeff]

I'm trying to hunt down a problem in one of my amps and I have a general question first. In these two circuits would turning the pot down kill the signal, or would both still operate weither or not the pot was turned down?

Thanks
 Jeff

[stratele52]

Like this ,this circuit won't work.

[Willabe]

What are you trying to do?

Do you want to put the 2 triodes in parallel with a single input and control them both from the 2'nd triodes grid?


                         Brad      

[jeff]

No. I'm not actually trying to build or use these circuits. I'm trying to understand better how tubes work.
 Let's take the first circuit:

 It's my understanding that what makes a tube operate is that the grid controls the current through the tube. When a singal is applied to the grid it allow more or less current to flow through the circuit which allows more or less current to flow through the plate resistor. This changing current through the fixed plate resistor causes a change in the voltage at the plate. So the relationship of the grid to the cathode controls the current which changes the voltage at the plate.

 If instead of changing the grid voltage, the first schematic changes the cathode voltage. Because of the shared resistor, when the first section is operated the current through the shared resistor changes the voltage at that resistor which in turn changes the votage on the other tubes cathode. Because of this the second tubes grid to cathode voltage is changed operating the second tube.

Right so far?

What I'm wondering is weither the grid has 1M to ground or is grounded(pot up or down) isn't the grid basically at ground potential with respect to the cathode? So as far as the grid to cathode voltage goes is it the same weither the pot at 1M or ground?

So my two questions are:
-In the first schematic would putting a signal on the first tube operate the second?
-Would grounding the grid by turning the pot turn the second tube off?

[Willabe]

OK, but I have a question, how does the input signal from V1a plate get to V1b grid in both of your drawings?

Some of the others guys can answer your questions and explain it better than I can. I will say any time you ground the grid in a common cathode stage, you turn off the tube.

In the mean time this might help you;

http://www.freewebs.com/valvewizard1/gainstage.html


                                Brad      

[HotBluePlates]

It's my understanding that what makes a tube operate is that the grid controls the current through the tube. ... So the relationship of the grid to the cathode controls the current which changes the voltage at the plate.

You're almost there.

So you're right, except that you should add the condition that the "grid-to-cathode voltage controls the plate current, which can result in a plate voltage change."

So your line of thinking is the pot will change the grid-to-cathode voltage and make the (d.c.) plate voltage change.

Problem: You're changing a resistance to make the grid voltage change. Since the two different properties, there has to be something to link them.

Answer: The "something" is Ohm's Law. Voltage = Current*Resistance.

If you can change the resistance, you'd think you can change the apparent voltage at the grid. But there is no current in the grid circuit of a typical tube stage operating properly. Notice the pot is between the grid and ground, and has no other connection to a source of voltage other than 0v. As a result, for any setting of the pot, the grid still appears to be at ground potential, so there is no change in the grid-to-cathode voltage. This is why volume controls work well to adjust volume but don't impact the d.c. conditions of the tube.

So why does a volume control do anything at all? It helps to think in terms of a constant d.c. voltage with a varying a.c. component superimposed on it. Because the volume control/grid circuit is isolated by a coupling cap in most tube stages, there is no available d.c. other than ground in the grid circuit. However a.c. signal passes through the cap, adding to and subtracting from the standing d.c. grid-to-cathode voltage. The amount of a.c. that makes it to the grid (and so the size of the variation/signal/volume) can be altered by the volume control.

So if you want to be able to vary the plate current and therefore plate voltage of a tube stage, you would need a source of d.c. voltage other than  ground, which you might then vary with some form of control. This seems like more trouble than it's worth for a preamp tube stage.

[jeff]

It's my understanding that what makes a tube operate is that the grid controls the current through the tube. ... So the relationship of the grid to the cathode controls the current which changes the voltage at the plate.

You're almost there.

So you're right, except that you should add the condition that the "grid-to-cathode voltage controls the plate current, which can result in a plate voltage change."

Isn't that what I'm saying in the missing ... part of my quote? The change in current across a plate resistor(typical 100k) causes a change in voltage?

V=IR    so change in I X 100K=change in V?

The plate voltage change is a result of the current change through the plate resistor.
Or am I misunderstanding you?

[jeff]

Here's the real problem I'm having. I've removed parts trying to hunt down the problem so this schemnatic is my amp as it is now. There's no wire actually connecting one stage to the next, ant the second stages cathode resistor is removed,... BUT...when I play through it I can hear it through the speakers. Not full blast but pretty loud(like on 2, loud enough). Somehow the first tube is "talking" to the 3rd(layout,capacitance,whatever).

What I want to know before I rip anything else out is: 
Seeing as if I turn the pot down it kills the sound, somehow my problem the grid is picking up signal from the 1st tube. I'm just trying to eliminate the possiblity the the plate or cathode is picking up signal.
If turning that pot down kills the sound it must be the grid picking up signal, right?
I just don't want to be barking up the wrong tree if somehow the cathode or plate is picking up signal.

[HotBluePlates]

Isn't that what I'm saying in the missing ... part of my quote? The change in current across a plate resistor(typical 100k) causes a change in voltage?

Yes, but where's the signal? You have one triode drawn with an apparent signal source. A second triode is shown with no signal source, only a pot doing nothing connected to its grid.

So for that second triode, where is the change of grid-to-cathode voltage? If there is no current through the resistance between grid and ground, there will also be no voltage developed across the resistor, so the voltage at the grid equals the voltage at ground.

So I think we're pointing at two different parts of the circuit.

[HotBluePlates]

There's no wire actually connecting one stage to the next, ant the second stages cathode resistor is removed,... BUT...when I play through it I can hear it through the speakers. Somehow the first tube is "talking" to the 3rd(layout,capacitance,whatever).

What I want to know before I rip anything else out is: 
Seeing as if I turn the pot down it kills the sound, somehow my problem the grid is picking up signal from the 1st tube. I'm just trying to eliminate the possiblity the the plate or cathode is picking up signal.
If turning that pot down kills the sound it must be the grid picking up signal, right?

Good chance that you're on the right track. A picture might be worth 10,000 words in this case, because you might not know "what right looks like." (to borrow a military phrase)

So here's what's not likely:
Coupling through a shared cathode resistor is somewhat unlikely, because the resistance is too small. Think about a long-tail inverter; there is some amount of cathode coupling, because the shared cathode resistance is relatively large. If you have a resistor on the order of 1k instead of 100k, cathode coupling is reduced 100 times.

It is possible to have cross-talk with a tube. It's not clear from what you've described so far if the triodes in question are inside the same envelope.

Also, along the same lines as the cathode-coupling example, but opposite, a high impedance point is susceptible to picking up stray signals. So a 1M resistor will translate a small stray current into a voltage 1,000 times bigger than a 1k resistor. So if you have wiring for a large signal (typically phase inverter output) pass near a 1M resistor or pot (or wires leading to them), you create good conditions for the 1M to develop a stray signal voltage.

A common high-resistance ground could cause some strange problems. The cathode of a triode has a low input impedance, but it also has a larger gain than the grid input. If there is a shared ground path that is poor quality (appreciably bigger than 0Ω), it is conceivable that the input signal develops a small voltage across the poor ground which serves to inject a signal into the cathode of a later stage. However, hum usually results from this cause, as power supply current are likely to be much bigger than signal currents.

A way too small decoupling resistor between filter caps may allow signal to bleed from one supply node to the next through the power supply. However, low-frequency oscillation (0-to-several Hz, or motorboating) is more often the result as higher frequency signals like audio are typically shunted to ground by the filter caps (unless they are toast).

So aside from poor solder joints, or a wire soldered in the wrong place, the most likely cause is the plate output of one stage passing too close to the grid wiring of the later stage. This is especially true if the wires are parallel to each other, as that maximizes inductive coupling.

[jeff]

It's not between two triodes in the same tube

"A common high-resistance ground could cause some strange problems. The cathode of a triode has a low input impedance, but it also has a larger gain than the grid input. If there is a shared ground path that is poor quality (appreciably bigger than 0Ω), it is conceivable that the input signal develops a small voltage across the poor ground which serves to inject a signal into the cathode of a later stage. However, hum usually results from this cause, as power supply current are likely to be much bigger than signal currents."

 So would this set up a situation like example A? And seeing that when the pot is grounded no sound comes out can I rule this out?(I probally asked this whole post backwards but that's what I was getting at with circuit A)

And If it was something in the power supply or with the cap, would that setup a situation like ex.B? And again since grounging the grid kills the sound can I rule that out too?

"plate output of one stage passing too close to the grid wiring of the later stage"
This is probally the likely cause. Now that I look at it, the coupling cap of the first stage(first stages plate) is next to the coupling cap of the second stage(feeding 3rd stages grid). If those two are "talking" then what comes out of the first stage is being feed to the 3rd.

Sounds right. The problem I was having with the amp was when the gain was on 0 I heard sound(guitar). As I turned it up it it got quieter till 2 then louder again.
So I think I was hearing the first stage, with the second stage grounded at 0(what I'm hearing now with all the parts out).
As I turned it up the 2nd stage was getting louder BUT it's output was getting cancled out because it was out of phase with the signal bleeding from the 1st.

So at 0 Stage 3 was picking up stage 1 - so I heard guitar
     at 2 stage 3 was picking up stage 1 plus getting stage 2 and those were out of phase - so it was getting quieter
    past 2 stage 2 was getting louder that what bled through from stage 1 - so it was getting louder

I'll move one of those caps.
Thanks
  Jeff

[HotBluePlates]

I'll move one of those caps.

Sounds like a plan!

[jeff]

OK, It's just that I used terminal strips, the kind that look like a T not with the hole in the middle. I wound the leads around them pretty good so it's gonna be a pain getting the caps unsoldered and off to move 'em around. I just wanted to make sure this is the most likely cause of the problem before I removed them.

Sounds like it is. I gotta do what I gotta do.

Thanks for all the help, and the in depth explainations.

  Jeff

[jeff]

Interesting side note. I found this:
http://www.webphix.com/schematic%20heaven/www.schematicheaven.com/ampegamps/reverb_rocket_12r.pdf

V2As plate feeds the reverb cicruit which feeds V2Bs grid and V2As cathode feeds V2Bs cathode.
Neet. This is the real world application of my orignal question: "If the cathode is picking up signal will grounding the grid kill the signal?"
The ironic part is I'm planing to build this and had the schematic because I have a lot of 6SL7s/6SN7s. Should have given it a closer look before posting.

Thanks again, time to move some parts.
  Jeff

[Willabe]

V2As plate feeds the reverb circuit which feeds V2Bs grid and V2As cathode feeds V2Bs cathode.

I don't think that's the same thing as you've been trying to say? Wild guess, 22K connecting V2a (pin 3)/V2b (pin 6) cathodes is feed back? Those 2 triodes are at the opposite ends of the reverb loop with 2 triodes and the verb tank inbetween them.  


                        Brad      

[HotBluePlates]

V2As plate feeds the reverb circuit which feeds V2Bs grid and V2As cathode feeds V2Bs cathode.

I don't think that's the same thing as you've been trying to say? Wild guess, 22K connecting V2a (pin 3)/V2b (pin 6) cathodes is feed back?

What Brad said.

Without saying the crazy idea I'm thinking, I'd want to measure the gain of those two stages (by applying a signal) with/without that resistor in place to confirm its function.

[sluckey]

"If the cathode is picking up signal will grounding the grid kill the signal?"

no

Google "grounded grid amplifier".

[HotBluePlates]

"If the cathode is picking up signal will grounding the grid kill the signal?"

no

Google "grounded grid amplifier".

And remember, your original question stated that the "input voltage" is really a voltage difference from grid to cathode.

If the grid is grounded (and therefore stays at a steady 0v), but the cathode voltage is moving higher and lower, then the grid-to-cathode voltage is changing and there is an input signal to the tube.

[PRR]

> Think about a long-tail inverter; there is some amount of cathode coupling, ... ... 1k instead of 100k, cathode coupling is reduced 100 times.

Not the shared resistor alone; that against the internal cathode impedance(s).

Say the cathode impedance is 1K. If the tail is 100K, 99% gets to the other cathode; if shared resistor is 1K, 50% gets to the other cathode.

I've over-simplified this, but the difference is nothing like 100:1. Usually from 9+/10ths to 4/10ths. Half-ish, not one-hundredths.

So sure, you can couple through a common cathode resistor, with a loss-of-gain similar to one stage, which on a Audio-taper VOLume pot may be a lot like the difference from "10" to "2".

There's other reasons to NOT share cathode-stuff from one stage to the *next*. With the normal plate-grid path too, it's positive feedback.

Even without shared-cathode paths, it is "normal" for an "unconnected" grid to have audio on it. Capacitance is everywhere. Grids are very high impedance. Very small capacitance and large grid resistor will leak a little bass and some to a lot of treble.

[jeff]

V2As plate feeds the reverb circuit which feeds V2Bs grid and V2As cathode feeds V2Bs cathode.

I don't think that's the same thing as you've been trying to say? Wild guess, 22K connecting V2a (pin 3)/V2b (pin 6) cathodes is feed back?

What Brad said.

Without saying the crazy idea I'm thinking, I'd want to measure the gain of those two stages (by applying a signal) with/without that resistor in place to confirm its function.

Isn't it's function the dry signal?

 Am I seeing this right? Remove the 22K resistor and all the signal passes through the echo unit(spring tank).
Without that resistor wouldn't you have no dry?
It looks to me "dimension" is their name for reverb. Picture "dimension" turned down. Now the grid of the tube feeding the PI(V2B) is grounded.

So the only way signal gets from V2A to V2B is off the plate through the reverb circuit/tank to V2Bs grid(wet) or off V2As cathode through the 22K resistor to V2Bs cathode(dry).

Right?
I think I'm having trouble expressing what I mean.

http://www.webphix.com/schematic%20heaven/www.schematicheaven.com/ampegamps/reverb_rocket_12r.pdf

[sluckey]

Am I seeing this right? Remove the 22K resistor and all the signal passes through the echo unit(spring tank).
Without that resistor wouldn't you have no dry?

That's absolutely correct.

So the only way signal gets from V2A to V2B is off the plate through the reverb circuit/tank to V2Bs grid(wet) or off V2As cathode through the 22K resistor to V2Bs cathode(dry).

You understand very well and your explanation is clear (to me).

That's just a good method of mixing dry and wet signals together. Magnatone used that same basic circuit in some of their reverb amps too. Look on page 4 for another example using the cathode as the input for dry signal and the grid as input for the wet signal.

http://home.comcast.net/~seluckey/amps/magnatone/Magnatone_M10A.pdf

[Willabe]

Ahh, I never noticed or thought about the dry signal path.       

Jeff now I see what you were getting at. 

                      Brad         

[jeff]

I think the 22K resistor is used to reduce feedback from the reverbs output to it's input. If the reverb was off it could be a straight wire and still work.

 V2As grid controls V2B through the cathodes. So with the dimension off the dry gets through. But.... the opposite is also true. V2Bs grid controls V2A. If the cathode resistor was shared and there was no 22K, the dry would work but I think there would be promlems when reverb is turned up.

It's hard for me to think of/describe a feedback loop were it starts, what controls what, but here's what I'm trying to say.

 The pre on grid of V2A causing a signal at the plate that runs the reverb circuit feed V2Bs grid. Feeding V2Bs grid causes V2A to operate through the cathode, this feeds the reverb circuit which feeds V2Bs grid which.........Etc.Etc. So the shaking springs would cause the springs to shake more, that would cause the springs to shake More, that would cause the springs to shake MORE.....
 
I think the 22K resistor is used to reduce feedback from the reverbs output to it's input. In most reverb there is a path from output to input, Weither it be Fenders 3M3 or the Marshally 100K~220K but it's kept under control. I think the 22K keeps this managable.   (Maybe someone can put what I'm trying to say more elequentally???? Or tell me if I'm misunderstanding this???)