[hiker101]

Quote:

Originally Posted by Koyaan I. Sqatsi /img/forum/go_quote.gif Ok, here's a little Electronics 101. Below is an illustration showing the two basic configurations. A is the passive ground configuration, B is the "active ground" configuration. The transistors represent a typical complimentary push-pull output stage. The red arrows represent conventional current flow (as opposed to electron current flow which would be in the opposite direction) for a positive going waveform. For the passive ground configuration, we can begin at the positive plate of the reservoir cap for the positive supply. From there it flows through the positive supply rail to the NPN output device. From there into the positive side of the driver. Then, from the negative side of the driver straight to and through the ground node and on to the negative plate of the positive supply rail's reservoir capacitor. In the active ground configuration, everything is the same up to the negative side of the driver. From there, instead of going straight to the ground node as it does in the passive ground case, it goes into the PNP output device of the ground amplifier. From there to the negative supply rail, where it doesn't stop, nor dribble onto the floor, nor evaporate into the air, but instead continues on to the negative plate of the negative supply rail's reservoir capacitor, then the positive plate of the same capacitor, and just as in the passive ground configuration, through the ground node and on to the positive supply rail's reservoir capacitor. So the active ground does not in any way divert load currents away from ground. Load currents go to ground with the active ground just as they do with the passive ground. The only difference is that the load currents take a more circuitous route to ground in the active ground configuration. Remember, current only flows through a closed loop. You can't have current start out someplace without it ultimately returning to that point. k

Umm....

This diagram violates conservation of energy.

A positive current cannot flow from -V to +V without the assistance of a battery or power supply- a source, in other words.

The power supplies in these amps span the rails, so the return current entirely skips the ground node in between.

 

[tomb]

Quote:

Originally Posted by Fitz /img/forum/go_quote.gif Heh, now I see what you mean, but it is somewhat entertaining though. Since Koyaan is incapable of making the necessary measurements himself, I was considering going and knocking together a simple little test amp to measure active vs passive ground, so he could tell me something was wrong how I grounded the passive one as the only explanation . If only my workbench wasn't down in the basement, which is completely flooded right now.

Maybe you could invite the Prune-like troll to go fishing in your basement.

I would, but my crawlspace is still dry, thank goodness. I had to drive 2 hours out of the way to get home tonight, though - all the other roads were closed.

Sorry to hear about the basement ... those recent pics of your equipment looked very nice. I hope that stuff didn't get damaged.

 

[Steve Eddy]

Quote:

Originally Posted by hiker101 /img/forum/go_quote.gif Umm.... This diagram violates conservation of energy.

No, it doesn't.

Quote:

A positive current cannot flow from -V to +V without the assistance of a battery or power supply- a source, in other words.

The "battery" in this case is the energy stored in the capacitors. I didn't bother including the rest of the power supply as it was irrelevant to the point being made.

Quote:

The power supplies in these amps span the rails, so the return current entirely skips the ground node in between.

No, it doesn't.

It can't.

The only reason you have current delivered to the load is due to the charge entering and leaving the capacitors. You can't draw charge off one plate of the capacitor without an equal amount of charge being delivered to the other plate.

k

 

[hiker101]

Quote:

Originally Posted by Koyaan I. Sqatsi /img/forum/go_quote.gif No, it doesn't. The "battery" in this case is the energy stored in the capacitors. I didn't bother including the rest of the power supply as it was irrelevant to the point being made. No, it doesn't. It can't. The only reason you have current delivered to the load is due to the charge entering and leaving the capacitors. You can't draw charge off one plate of the capacitor without an equal amount of charge being delivered to the other plate. k

This is very basic physics.

Your capacitors are acting like batteries. It's a one way trip. Once the charge gets from the + plate to the - plate it needs to be given external energy to get back.

The charge leaves the top + plate and cancels an equivalent - charge when it gets to the bottom -plate. The voltage difference across the capacitors decreases.

Without an external supply, the capacitors just discharge. That's why your power led glows for a second after you pull the plug. The capacitors release their stored energy to the circuit and then the energy is gone.

A battery contains a chemical pump that does electrical work when it moves the charge from low potential to higher potential, it converts chemical energy to electrical energy. There is no such process going on inside a capacitor.

 

[Steve Eddy]

Quote:

Originally Posted by hiker101 /img/forum/go_quote.gif This is very basic physics.

Yes, it is.

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Your capacitors are acting like batteries.

Yes, they are.

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It's a one way trip. Once the charge gets from the + plate to the - plate it needs to be given external energy to get back.

That's correct.

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The charge leaves the top + plate and cancels an equivalent - charge when it gets to the bottom -plate. The voltage difference across the capacitors decreases.

Again, correct.

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Without an external supply, the capacitors just discharge. That's why your power led glows for a second after you pull the plug. The capacitors release their stored energy to the circuit and then the energy is gone.

Yup.

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A battery contains a chemical pump that does electrical work when it moves the charge from low potential to higher potential, it converts chemical energy to electrical energy. There is no such process going on inside a capacitor.

No, there isn't. Nor did I claim there was.

I only used the circuit to illustrate the path taken by the load current and to show that the load current does indeed go to ground with the active ground just as it does with the passive ground. It just takes a more circuitous route in the former, that's all.

I can replace the caps with batteries if you'd like, but it's ultimately irrelevant for the purposes of illustration.

Just imagine the circuit shown as being a snapshot of the majority of the time that the rest of the power supply isn't part of the circuit, such as is the case with typical capacitor input power supplies.

The load current still follows the same path, and as I said in a previous post, it doesn't sit around waiting for the next refresh cycle.

k

 

[hiker101]

In circuit "A" the current flows from +V to the ground node. There is no return current from ground to +V, that is the segment that violates energy conservation.

In circuit "B" the current flows from +V to -V. There is no return current from -V to +V, that is the segment that violates conservation of energy.

Current flows from high potential to lower potential (except in the case of emf induced by changing magnetic flux.) Demonstrating something different would win a Nobel Prize.

 

[Fitz]

Quote:

Originally Posted by tomb /img/forum/go_quote.gif Maybe you could invite the Prune troll to go fishing in your basement. I would, but my crawlspace is still dry, thank goodness. I had to drive 2 hours out of the way to get home tonight, though - all the other roads were closed. Sorry to hear about the basement ... those recent pics of your equipment looked very nice. I hope that stuff didn't get damaged.

Good to hear everything's nice and dry over there; I know there were some really bad spots over towards Lilburn and Lawrenceville, so I was hoping nothing happened to your place. Fortunately nothing of significant value was damaged here... it's more of an annoyance than anything else, nothing that could compare to those who've lost their homes or loved ones.

 

[Steve Eddy]

Quote:

Originally Posted by hiker101 /img/forum/go_quote.gif In circuit "A" the current flows from +V to the ground node. There is no return current from ground to +V, that is the segment that violates energy conservation.

Where in the illustration is current shown flowing from ground to +V?

Quote:

In circuit "B" the current flows from +V to -V. There is no return current from -V to +V, that is the segment that violates conservation of energy.

Where in the illustration is current shown flowing from -V to +V?

k

 

[tomb]

Quote:

Originally Posted by Fitz /img/forum/go_quote.gif Good to hear everything's nice and dry over there; I know there were some really bad spots over towards Lilburn and Lawrenceville, so I was hoping nothing happened to your place. Fortunately nothing of significant value was damaged here... it's more of an annoyance than anything else, nothing that could compare to those who've lost their homes or loved ones.

Yes, several people died this morning from what I heard. They got out before the sun was up and got stuck in the water before they even saw it.

Unfortunately for my neighbors, the Yellow River borders the back of my neighborhood. There's an entire cul-de-sac with 5 houses under water. Next to that, the back street parallels the river - there are at least 4 houses under water - we couldn't even see them all. All you could see of one driveway was a basketball backboard above the surface of the water. Luckily, my house is on a hill at the front of the neighborhood and we're dry - so far.

 

[Beefy]

Quote:

Originally Posted by hiker101 /img/forum/go_quote.gif In circuit "A" the current flows from +V to the ground node. There is no return current from ground to +V, that is the segment that violates energy conservation. In circuit "B" the current flows from +V to -V. There is no return current from -V to +V, that is the segment that violates conservation of energy. Current flows from high potential to lower potential (except in the case of emf induced by changing magnetic flux.) Demonstrating something different would win a Nobel Prize.

This is what you are saying, yes?

So while power supply ground references to signal supply ground, no current flows through the ground. All current flow is from positive to negative rails and vice versa?

 

[Steve Eddy]

Quote:

Originally Posted by Beefy /img/forum/go_quote.gif This is what you are saying, yes? So while power supply ground references to signal supply ground, no current flows through the ground. All current flow is from positive to negative rails and vice versa?

The power supply is across the reservoir caps. And the current flow is just as I'd drawn it in the original illustration. But for the sake of completeness, here it is with the power supply added:

But as I keep saying, the power supply isn't even in the circuit the majority of the time and the current delivered to the circuit for that majority of the time is coming from the reservoir caps. So for the purposes of illustration, there's no need to include it.

k

 

[Beefy]

Quote:

Originally Posted by Koyaan I. Sqatsi /img/forum/go_quote.gif But as I keep saying, the power supply isn't even in the circuit the majority of the time and the current delivered to the circuit for that majority of the time is coming from the reservoir caps. So for the purposes of illustration, there's no need to include it.

I'm no expert, but I know enough to know that you can't just ignore the power supply. Re-draw your diagram assuming there are no more reservoir caps. An amp like this still works; an amp with just reservoir caps and no power supply does not. Then look at the current flow through the whole circuit from its absolute source to its absolute termination in both cases.

In the standard two channel the positive rail will dump current to ground, negative will source current from ground.

In the active ground the positive rail will dump current to the negative rail, negative will source current from the positive. The ground just sits in the middle of the two, providing a reference.

 

[Pars]

Quote:

Originally Posted by Beefy /img/forum/go_quote.gif >snip> In the active ground the positive rail will dump current to the negative rail, negative will source current from the positive. The ground just sits in the middle of the two, providing a reference.

So a simple TLE2426 should suffice, since the ground is not sinking any current?

Sorry, had to interject that...

 

[Steve Eddy]

Quote:

Originally Posted by Beefy /img/forum/go_quote.gif I'm no expert, but I know enough to know that you can't just ignore the power supply.

For our purposes here, yes, you can ignore the power supply.

To understand why that's the case, you have to understand just what the purpose of the power supply is to begin with.

And the only purpose of the power supply in a typical capacitor input supply such as this is to top off the reservoir caps. That's it. And for all the amplifier knows or cares, it's being powered by the reservoir caps.

Think of the reservoir caps as rechargeable batteries and the power supply as a battery charger.

And for just a brief moment every 1/120th of a second (or every 1/100th of a second if you're AC is 50 Hz), the battery charger gets plugged into the batteries to recharge them, and then it gets unplugged, such that for the majority of the time, the battery charger is not plugged into the batteries.

So for purposes of illustrating the path of current flow vis a vis passive ground versus active ground, there's no need to include the battery charger in the illustration. The path the current takes is the same whether the charger is plugged in or not. The only difference is that the red arrows associated with the capacitors will flip direction during the refresh cycle.

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Re-draw your diagram assuming there are no more reservoir caps. An amp like this still works; an amp with just reservoir caps and no power supply does not.

It'll work. Just not for very long.

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Then look at the current flow through the whole circuit from its absolute source to its absolute termination in both cases.

The path is the same in both cases.

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In the standard two channel the positive rail will dump current to ground, negative will source current from ground. In the active ground the positive rail will dump current to the negative rail, negative will source current from the positive. The ground just sits in the middle of the two, providing a reference.

In both cases, all of the current flows through the ground node. It has to.

You can't draw current from one plate of a capacitor without an equal amount of current flowing to its other plate. And the only way for that to happen, given that both reservoir capacitors have one of their plates tied to the ground node is for that current to pass through the ground node.

k

 

[Beefy]

Quote:

Originally Posted by Pars /img/forum/go_quote.gif So a simple TLE2426 should suffice, since the ground is not sinking any current? Sorry, had to interject that...

As AMB pointed out, in the MMM and PPA, yes, a simple TLE2426 does suffice. Because the current return from the headphone dumps to power rails via the active ground channel, rather than dumping directly into the signal ground.

And in this case the reservoir caps are across V+/V-, rather than V+ to GND and V- to GND. The whole output stage is isolated from signal ground. The OPAMP stage is still connected to signal ground obviously; for that stage we aren't talking about large currents that can pollute the ground either.