[j4cbo]

Quote:

Originally Posted by 00940 /img/forum/go_quote.gif is this better ?

Much

The other thing to remember is that unless your amp is wired with a perfect star-grounding scheme, ground is not a "node" in the EE sense - it's a whole bunch of nodes all connected together by various low, but nonzero, impedances, and loads of extra parasitics. The exact permutation of connections makes a big difference to which currents flow where, as anyone who's ever tried to troubleshoot grounding-related hum will know.

 

[nikongod]

Quote:

Originally Posted by Koyaan I. Sqatsi /img/forum/go_quote.gif Whether class A or class B is irrelevant.

I disagree, and you contradict this statement twice in this same quoted post.
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The purpose of the illustration was to show the current path, and to illustrate that current flows through the ground node in both cases.

You are showing LOAD current in both examples, and IMO mistaking it with idle current. The amplifier in example B will behave VERY differently depending on whether it is in class-A or class-B.
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I'm already well aware that when running in class A, the bridged circuit results in a constant current draw from the power supply and have stated so several times throughout this thread.

This disagrees with your first quoted statement by showing a clear difference running the amplifier in example B class-A or class-B

You say right here that the load currents sum to zero in the active stages of example B running class-A, how can they then get to ground if they have summed to zero?
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But what was at issue here, and was the whole purpose of the illustrations, wasn't whether the current draw was constant or dynamic, but rather the claim that the current does not flow through the ground node. As amb claimed earlier in this thread: In an active ground amp, the [size=small]load current [/size]doesn't go to ground. It's sourced from one amplifier's positive rail and sinked to another amplifier's negative rail (and vice versa).

Its good to know you agree with AMB that the load current dosnt go to ground. Arent we talking about the load current here?
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All of the current goes through the power supply, and does so through the ground node. The only difference is that in A, the current draw is dynamic, and in B, the current draw is constant when running class A, which has some benefit of its own.

in this quoted section you again mention a BENEFIT to running the amplifier in example B in class-A due to the constant load that it represents to the power supply. this contradicts the first quoted bit AND you maintain elsewhere that these topologies are without merit or benefit! come now. pick one.

all of the current goes through the PS, but none of the SIGNAL CURRENT goes to the power supply in the amplifier in example B running class-A. it is not difficult to see a long list of advantages from this.
Quote:

Originally Posted by j4cbo /img/forum/go_quote.gif That image is a nice strawman, but it is not relevant to the discussion at hand, because it does not match the actual schematic of the M³. Please try again with reference to the schematic.

I prefer the simplified schematic although it is a strawman because it illustrates the faults of a class-B amplifier where class-A amplifier works realllllly well.
This discussion long ago stopped being about the M3/B22 designs, but about the design idea in general.

 

[Steve Eddy]

Quote:

Originally Posted by j4cbo /img/forum/go_quote.gif That image is a nice strawman, but it is not relevant to the discussion at hand, because it does not match the actual schematic of the M³. Please try again with reference to the schematic.

Um, if you'll take a moment to read the subject of this thread, the "discussion at hand" is the B22, not the M³. And it was because of this that I made the illustration as it is.

But even if the "discussion at hand" was the M³, it still doesn't change the fact that the load current flows through the ground node.

se

 

[Steve Eddy]

Quote:

Originally Posted by j4cbo /img/forum/go_quote.gif The other thing to remember is that unless your amp is wired with a perfect star-grounding scheme, ground is not a "node" in the EE sense - it's a whole bunch of nodes all connected together by various low, but nonzero, impedances, and loads of extra parasitics. The exact permutation of connections makes a big difference to which currents flow where, as anyone who's ever tried to troubleshoot grounding-related hum will know.

Yes. Which is precisely why it's important to know that an "active ground" does not bypass ground as is often claimed. That all of the current flows through the ground node, instead of spilling out onto the floor, or evaporating into the ether, or as amb claimed, return to the power transformer (which as I stated numerous times isn't even in the circuit the majority of the time).

So thank you for helping to further illustrate my main point here.

se

 

[Steve Eddy]

Quote:

Originally Posted by nikongod /img/forum/go_quote.gif You are showing LOAD current in both examples, and IMO mistaking it with idle current.

No, I'm not.

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The amplifier in example B will behave VERY differently depending on whether it is in class-A or class-B.

They both behave the same in that all of the currents flow through ground.

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This disagrees with your first quoted statement by showing a clear difference running the amplifier in example B class-A or class-B

The only difference in example B in class A versus class B is that in class A, the current will be constant, whereas in class B the current will be variable.

This has absolutely nothing to do with the primary point that all of the current flows through ground. That contrary to claims made, ground is not bypassed by using an "active ground" channel.

Quote:

You say right here that the load currents sum to zero in the active stages of example B running class-A, how can they then get to ground if they have summed to zero?

You're misunderstanding what "summed to zero" means.

All currents in a given node "sum to zero" (see Kirchhoff's Law). But that doesn't mean that there is NO CURRENT through the node. What it means is that the current entering the node must be equal to the current flowing out of the node.

Quote:

Its good to know you agree with AMB that the load current dosnt go to ground. Arent we talking about the load current here?

I don't agree with AMB that load current doesn't go to ground. Because it does go to ground.

Once again, it doesn't just spill out onto the floor or evaporate into the air.

Quote:

in this quoted section you again mention a BENEFIT to running the amplifier in example B in class-A due to the constant load that it represents to the power supply. this contradicts the first quoted bit AND you maintain elsewhere that these topologies are without merit or benefit! come now. pick one.

I'd previously acknowledged a number of times throughout this thread that keeping power supply current draw constant could be of some benefit.

Quote:

all of the current goes through the PS, but none of the SIGNAL CURRENT goes to the power supply in the amplifier in example B running class-A. it is not difficult to see a long list of advantages from this.

ALL of the current goes to the power supply.

Again, it's a matter of whether or not the current is constant, or variable.

But that doesn't change the fact that the claims made that an "active ground" bypasses ground are erroneous. And that's the issue that the illustrations were intended to illustrate.

Quote:

I prefer the simplified schematic although it is a strawman because it illustrates the faults of a class-B amplifier where class-A amplifier works realllllly well.

It's not simply a matter of class A. A class A amplifier does not inherently draw a constant current. A simple class A amplifier with a complimentary push pull output stage will draw varying current from the power supply depending on the signal. To get a constant current, you need to bridge it with another class A amplifier with a complimentary push pull output stage.

And that's really all an amplifier with an "active ground" is, is a pair of bridged amplifiers. The only difference is that one of the amplifier channels isn't driven with a signal.

se

 

[Iniamyen]

My 3-channel active ground B22 sounds really awesome.

 

[Steve Eddy]

Quote:

Originally Posted by Iniamyen /img/forum/go_quote.gif My 3-channel active ground B22 sounds really awesome.

Great. I'm glad you enjoy it. That's what it's all about at the end of the day.

But that doesn't make the claim that an active ground channel bypasses ground any less erroneous.

se

 

[digger945]

Quote:

Originally Posted by Koyaan I. Sqatsi /img/forum/go_quote.gif The only difference is that one of the amplifier channels isn't driven with a signal.

If the ground wire from the headphone drivers is connected to the output of the active ground channel amp directly between the output mosfets, then how does that active ground channel know how and when to source or sink current, and from what rail, without a signal?

 

[Steve Eddy]

Quote:

Originally Posted by digger945 /img/forum/go_quote.gif If the ground wire from the headphone drivers is connected to the output of the active ground channel amp directly between the output mosfets, then how does that active ground channel know how and when to source or sink current, and from what rail, without a signal?

Because the "active ground" channel acts like most every other amp and strives to keep its output voltage at some constant relative to its input voltage. For example, if the amplifier has a gain of 10, and you feed its input 1 volt, it will want to output 10 volts, and will sink/source sufficient current through the load to maintain that 10 volts on its output.

Similarly, if its input is at 0 volts, it will try to maintain 0 volts at its output.

So now consider an "active ground" channel with its input at 0 volts and its output tied to one end of a headphone driver or even a loudspeaker for that matter.

On the other side of the headphone driver or loudspeaker is the output of the amplifier proper.

When the signal gets fed into one end of the driver, the "active ground" channel sees that signal voltage at its output. And in response to that, it will sink/source the appropriate amount of current in order to maintain its output at 0 volts.

So there doesn't need to be a signal at its input.

Do you follow?

se