OK, so none of our Classic Circuits work, because they are swamped by +Positive+ Feedback, and we never noticed for all these decades.
Brimar listed 750femtoFarad. There will be some variation with construction details, but not 1000:1. 12AX7 plates are all about so big, there's room in the bottle, they are all about so far apart.
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Yes. Be careful that you don't overthink this issue.tapehead ted said:
This is one effect, not two. Just two different ways of looking at the same phenomenon.
Valve circuits work. That means that the frightening effects you keep considering are either mistakes or just not big enough to have a major effect. They do have an effect, but this is well known and described in all the textbooks and taken into account by all competent designers.
Thanks all. Being a competent designer myself is nothing I can take for granted.
I keep reading how important it is to keep ground connections low impedance and have not found much about what happens when they are not so low impedance. For instance. I'm just trying to figure out what the consequences are of imperfections we try to minimize, assumedly because there are consequences if something like ground impedance becomes significant.
A lot of classic designs as built commercially oscillate and motorboat and everything else, and I still enjoy them very much- I also enjoy 5% THD, and yet there's a lot of effort to reduce harmonic distortion.
I will try to keep it in proportion and try not to overthink it so much- it's definitely my nature to overthink things- and thank you for that admonition.
I just am aware in a semi-ignorant kind of way that some design details weather the anomalies better than others, and that some defects really bug my ear while others don't disturb me. So I look forward to being a bit less ignorant, and with all your help and patience I am getting a little less so.
Thank you!
I keep reading how important it is to keep ground connections low impedance and have not found much about what happens when they are not so low impedance. For instance. I'm just trying to figure out what the consequences are of imperfections we try to minimize, assumedly because there are consequences if something like ground impedance becomes significant.
A lot of classic designs as built commercially oscillate and motorboat and everything else, and I still enjoy them very much- I also enjoy 5% THD, and yet there's a lot of effort to reduce harmonic distortion.
I will try to keep it in proportion and try not to overthink it so much- it's definitely my nature to overthink things- and thank you for that admonition.
I just am aware in a semi-ignorant kind of way that some design details weather the anomalies better than others, and that some defects really bug my ear while others don't disturb me. So I look forward to being a bit less ignorant, and with all your help and patience I am getting a little less so.
Thank you!
There are two methods to minimise ground impedance problems:
1. minimise ground impedance
2. ensure that currents flow where you want them to, and voltages are referenced to where they should be referenced
The classic designs mostly took account of this, provided they are built according to the original layout. Modern copies often overlook this and hit problems. There is more to a circuit then the circuit diagram, especially where grounding is concerned - which many diagrams conceal (or leave ambiguous) rather than exhibit.
1. minimise ground impedance
2. ensure that currents flow where you want them to, and voltages are referenced to where they should be referenced
The classic designs mostly took account of this, provided they are built according to the original layout. Modern copies often overlook this and hit problems. There is more to a circuit then the circuit diagram, especially where grounding is concerned - which many diagrams conceal (or leave ambiguous) rather than exhibit.
Thank you.
So here's an example: I'd like to slug the dominant pole by lowering the corner frequency of a particular common cathode stage. Let's say I want to roughly halve the corner frequency. I could add series resistance on the way to grid. Or I could add shunt capacitance. I could place a capacitor with a value equal to the miller capacitance between grid and ground, thus doubling the input capacitance.
However, it looks to me like ground noise and worse, signal on the ground from elsewhere in the amplifier now would have a new path to the grid, albeit a frequency filtered one. Or, I could place a small cap roughly equal to Ca-g between the anode and the grid, and let the miller effect multiply the value of all the capacitance, inherent in the tube (Ca-g) and my additional cap, accomplishing the same desired aim without inviting frequency filtered noise and signal from the ground.
I like this last option. Am I missing something again? Without overthinking it, it seems perfectly straightforward to me- then again, I do fool myself fairly often. So I ask here. FWIW, personally I especially dislike the idea of frequency filtered signal mixing with my intended signal, whether adding to it or subtracting from it. So this avenue bothers me more than signal crawling up the grid leak.
My interest in these potential anomalies has been aroused my an amp a very experienced and knowledgeable friend built, that I now own. Quite a few strange phenomena take place with his amp. Controls affect the sound in unexpected ways, and signals mutated in various ways appear in unexpected places. This has led me to try to account for some of the variables that seem insignificant- they certainly had significant effects in this case. I can see how his layout would certainly invite some weirdness.
Thanks again.
So here's an example: I'd like to slug the dominant pole by lowering the corner frequency of a particular common cathode stage. Let's say I want to roughly halve the corner frequency. I could add series resistance on the way to grid. Or I could add shunt capacitance. I could place a capacitor with a value equal to the miller capacitance between grid and ground, thus doubling the input capacitance.
However, it looks to me like ground noise and worse, signal on the ground from elsewhere in the amplifier now would have a new path to the grid, albeit a frequency filtered one. Or, I could place a small cap roughly equal to Ca-g between the anode and the grid, and let the miller effect multiply the value of all the capacitance, inherent in the tube (Ca-g) and my additional cap, accomplishing the same desired aim without inviting frequency filtered noise and signal from the ground.
I like this last option. Am I missing something again? Without overthinking it, it seems perfectly straightforward to me- then again, I do fool myself fairly often. So I ask here. FWIW, personally I especially dislike the idea of frequency filtered signal mixing with my intended signal, whether adding to it or subtracting from it. So this avenue bothers me more than signal crawling up the grid leak.
My interest in these potential anomalies has been aroused my an amp a very experienced and knowledgeable friend built, that I now own. Quite a few strange phenomena take place with his amp. Controls affect the sound in unexpected ways, and signals mutated in various ways appear in unexpected places. This has led me to try to account for some of the variables that seem insignificant- they certainly had significant effects in this case. I can see how his layout would certainly invite some weirdness.
Thanks again.
You need to think carefully about what you mean by this. "The ground" suggests just one ground. If so, by definition it cannot have a signal because it is the reference point for all signals. If you mean 'the ground for that part of that stage' then you need to think about how it might acquire a signal from elsewhere and how you might prevent this.tapehead ted said:
It is relatively easy to make an amp where the controls do 'strange' things, such as the volume control varying the frequency response by a noticeable amount. Some 'high end' items even make a feature of this by talking about 'focus' etc. in order to hide incompetent design.
Probably not if you ground properly. Since a tube amplifies the AC voltage it sees between grid and cathode, if you always ensure that the grid ground reference ties to exactly the same point as its cathode ground reference, there will be very little chance of having a voltage potential difference between those two points. Thus if the two ground points are truly at the same potential then current (ground loop) can never ever flow between them, and thus no signal developed between them.
"Signal on the ground" is really an oxymoron, since if there's signal on the ground, meaning current flowing through the wire that you have mentally labeled "ground," then it's not really at ground potential.
The valve amplifies the voltage appearing between grid and cathode. Adding such a capacitor would approximate to shorting the grid to cathode at sufficiently high freqeuncies (where the cap looks like a small impedance), so it would in fact reduce the possibility of 'ground noise' (ambiguous term) from being amplified.
Thank you, that's very clear.
I've been assuming that the ground is at a potential of 1 ohm for some of these thought experiments, and then considering that it might be more or it might be less. It can't possibly be zero if there's a wire involved.
I'm picturing the ground rod through a hole in the floor directly beneath my friend's test bench, with the enormous wire. I'm thinking that's got to come far closer to approximating zero impedance than my 14 gauge household wiring winding it's way to the ground rod through various circuits- at least there is a ground rod.
I can easily imagine an amp acting one way on his test bench and then revealing some weirdness when plugged into my system.
Probably the ground is the least of it- the power supply has obvious limitations and the layout is cramped, but if a signal can travel on a 1 ohm HT supply then why not on a 1 ohm ground wire?
I do hear you about ensuring there is no potential difference between where the cathode connects and where the grid leak connects. I'm starting to get it.
Thank you all very much for helping to provide this invaluable (to me anyway) resource. Your help is very much appreciated.
This is one of those things I've always wondered about- what happens to resistances in this situation? Very gratifying to learn about this.
After some thought, I can see that if the ground was connected from the same point to grid and cathode and nothing else were connected, there would be no contribution from the grid to the amplified signal. But I'm still seeing a potential divider at the grid from ground perspective where the series resistance is the grid leak and the shunt resistance (to HT and ground) is the anode resistance of the previous stage.
So what arrives at the grid must be different from what arrives at the cathode. So, there's a potential difference after all, and something from ground ends up mixed with our amplified signal. If ground is not in fact at zero impedance. I'd like to be able to accommodate a non-zero ground impedance in a design, as best I can- it seems inevitable unless we regard the amplifier as a fixed installation in a known and controlled location.
So what arrives at the grid must be different from what arrives at the cathode. So, there's a potential difference after all, and something from ground ends up mixed with our amplified signal. If ground is not in fact at zero impedance. I'd like to be able to accommodate a non-zero ground impedance in a design, as best I can- it seems inevitable unless we regard the amplifier as a fixed installation in a known and controlled location.
I assume you don't mean "potential"?tapehead ted said:
Don't make the mistake of thinking of 'ground' as some sort of magic dump for all unwanted signals.
If a potential divider has the same potential at each end (what you call 'ground') than the output will be at the same potential. The only source of extra signal at the grid is the HT/B+, not the ground. As I keep saying, if 'ground' is the reference then by definition it cannot have a signal. Problems occur when the grid and cathode are connected to different 'grounds', because you can have a signal between different 'grounds'.
I really do think you are over analyzing this, and you appear to be conflating impedance with potential. The principles are straight forward:
1. Tube amplifies whatever signal it sees between grid and cathode.
2. Current flows only between two points when there is a potential (voltage) difference between those two points (and of course when there is a conduction path between those two points). It does not matter if this conduction path is labeled "ground" or otherwise.
3. To minimize the possibility of potential difference between the tube's grid ground reference and its cathode ground reference, connect those ground references to the same point, thus no potential difference between those two points can exist, thus no current can flow between those two points, thus no extraneous signal can be generated between those two points.
Thank you again and sorry- I did mean impedance, not potential. Here's what I'm still seeing, possibly it's a total chimera, but none of the replies are addressing it directly so I don't know if it's real or again I'm just confused.
Let's say there is 1 mV of noise on the ground. If cathode and grid were both connected directly to ground, both would see the same 1 mV and it would cancel out completely- there's no potential difference there.
Now say we hook up a grid leak, let's say 1M, and it's an input stage and the input source resistance is 1K. So now our same 1mV arrives at the grid, attenuated approximately 1000 times by the potential divider formed by the series element of the grid leak resistor and the shunt element of the source resistance- from whatever is plugged into the input- and now the grid sees 1 microvolt and the cathode is still seeing 1 millivolt. So barely anything is canceled. Right here at the very input of the amp is where I'm most concerned with noise. Thank you.
Let's say there is 1 mV of noise on the ground. If cathode and grid were both connected directly to ground, both would see the same 1 mV and it would cancel out completely- there's no potential difference there.
Now say we hook up a grid leak, let's say 1M, and it's an input stage and the input source resistance is 1K. So now our same 1mV arrives at the grid, attenuated approximately 1000 times by the potential divider formed by the series element of the grid leak resistor and the shunt element of the source resistance- from whatever is plugged into the input- and now the grid sees 1 microvolt and the cathode is still seeing 1 millivolt. So barely anything is canceled. Right here at the very input of the amp is where I'm most concerned with noise. Thank you.
With respect to what? Where is the supposed noise current flowing to/from?
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Is it too simple just to say suppose it's there? I'm concerned about losing track of my question in the previous post. All the return current in the amplifier is flowing in that ground wire, eventually to earth. I'm not sure if you mean to be making a point that the valve, atop it's resistors to ground, is at a higher potential and thus "uphill" from this flow of current through the ground wire to ground?
I think I see your confusion now. To help explain this, start by assuming a fictitious ground plane sits at 1V DC with respect to some external reference point. You connect the ground references for grid and cathode both to this 1V ground plane. What is the difference between grid and cathode ground reference points now? It is exactly zero.
Now, with respect to this same external reference point, let's say instead of 1V DC sitting on this fictitious ground plane, there is 1V AC residing there. But with respect each other, what is the voltage difference between grid and cathode ground reference points now? It is exactly as before...zero. Both are fluctuating at 1V AC with respect to the external reference point, but with respect to each other there is no fluctuation.
AC noise, then, rides on top of DC, and modulates the DC voltage at the rate of AC fluctuation. So if both grid and cathode ground references are connected to a ground plane with some small amount of AC noise on it (with respect to some external reference point), the voltages at grid and cathode will be fluctuating by the same amount (with respect to external reference point) but with respect to each other, they are NOT fluctuating.
Mathematically, let:
Vg = voltage at grid
Vk = voltage at cathode.
Vd = voltage difference between grid and cathode
N = noise
with respect to external reference point: voltage at grid = Vg + N.
with respect to external reference point: voltage at cathode = Vk + N.
with respect to each other, voltage difference Vd = (Vg + N) - (Vk + N) = Vg - Vk. (see that the noise cancels).
Does this help?
Now, with respect to this same external reference point, let's say instead of 1V DC sitting on this fictitious ground plane, there is 1V AC residing there. But with respect each other, what is the voltage difference between grid and cathode ground reference points now? It is exactly as before...zero. Both are fluctuating at 1V AC with respect to the external reference point, but with respect to each other there is no fluctuation.
AC noise, then, rides on top of DC, and modulates the DC voltage at the rate of AC fluctuation. So if both grid and cathode ground references are connected to a ground plane with some small amount of AC noise on it (with respect to some external reference point), the voltages at grid and cathode will be fluctuating by the same amount (with respect to external reference point) but with respect to each other, they are NOT fluctuating.
Mathematically, let:
Vg = voltage at grid
Vk = voltage at cathode.
Vd = voltage difference between grid and cathode
N = noise
with respect to external reference point: voltage at grid = Vg + N.
with respect to external reference point: voltage at cathode = Vk + N.
with respect to each other, voltage difference Vd = (Vg + N) - (Vk + N) = Vg - Vk. (see that the noise cancels).
Does this help?
Thank you kward. That is very clear and I do follow it. I appreciate your patience with me and the effort to spell things out systematically for me.
I've been struggling with the idea that somehow the potential divider between this ground plane and the grid (formed by the series resistance of the grid leak and the output resistance of the previous stage) has no influence on what voltage appears at the grid. I don't see how that's possible. To the extent that there is any voltage there it must be attenuated by the potential divider, right? Only if that voltage was zero would what arrives at the grid still be zero. If I am following this, you are saying that in effect it is zero for the reasons you describe so clearly.
Thanks again.
I note that for a directly coupled pair of stages the second stage has no grid leak and there's nothing to cancel any noise from ground. Unless the grid leak of the previous stage can serve that way.
I've been struggling with the idea that somehow the potential divider between this ground plane and the grid (formed by the series resistance of the grid leak and the output resistance of the previous stage) has no influence on what voltage appears at the grid. I don't see how that's possible. To the extent that there is any voltage there it must be attenuated by the potential divider, right? Only if that voltage was zero would what arrives at the grid still be zero. If I am following this, you are saying that in effect it is zero for the reasons you describe so clearly.
Thanks again.
I note that for a directly coupled pair of stages the second stage has no grid leak and there's nothing to cancel any noise from ground. Unless the grid leak of the previous stage can serve that way.
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