Quote:
Originally Posted by digger945 /img/forum/go_quote.gif
I really want to try and wrap my head around what your saying here so bear with me please.
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No problem. As I said previously, I'm nothing if not patient.
Quote:
I really think we could all learn something here as we move from a place where one tries to prove one's self correct or another wrong to a place of thinking about what's actually going on here. |
Exactly!
It's NOT a personal issue. It's not about
who's right or
who's wrong. It's about
what's going on and what's
not going on.
Quote:
Your saying that the current from the rails in a bipolar power supply actually passes directly through the caps tieing them to ground. I really think this needs to be established here if I(we) are to gain an understanding of how the current is traveling full circle here. |
It depends on what you mean by passing "directly through" the caps.
I didn't mean, or intend to imply in anything I've said that current was passing through the dielectric of the capacitors. It's not. So the current can never travel completely full circle. It will always come up short by the thickness of the dielectric.
But let me elaborate a bit here on what I've been trying to say.
When a capacitor is fully discharged, and there's no voltage across it, each plate has a more or less equal amount of positive and negative charges on it which cancel each other out and the charge on each plate is "neutral."
When we charge a capacitor, we do so by pulling some electrons (which have a negative charge) off one plate, and putting them on the other.
Now one plate has a positive charge (the one which the electrons were removed from), and the other plate has a negative charge (the one which the electrons were moved to). And there is now a potential difference, or voltage if you will, across the capacitor. And in the context of what we're talking about here, that voltage will be one or the other rail voltages.
Of course it required some energy to displace those electrons. But that energy didn't go to waste. It's now stored in the charged capacitor.
In order for that energy to be used, such as the energy stored in the reservoir capacitors of a power supply, it must discharge to one degree or another.
As it discharges, as in the case of powering an amplifier circuit, electrons are drawn from the negatively charged plate. But those electrons can't be drawn off the negatively charged plate without an equal number of electrons flowing into the positively charged plate. And that means current flow.
Now let me relate this to what I've been trying to say here.
Here's the original illustration, however I've flipped the direction of the red arrows to reflect electron current flow rather than conventional current flow.
Consider the reservoir cap on the positive side of the power supply in
A, the passive ground example.
Load current flows from the capacitor's negative plate,
through the ground node, and then to the driver's negative terminal, and then on around the loop, where there is an equal amount of load current flowing into the capacitor's positive plate.
Now let's consider
B, the active ground example.
Again, let's start with the reservoir cap on the positive side of the power supply.
Load current flows from the capacitor's negative plate,
through the ground node, just as it does in the passive ground example. The only difference here is that instead of flowing next into the driver's negative terminal, it flows into the positive plate of the reservoir cap on the negative side of the power supply.
Of course that current can't flow into that cap's positive plate without an equal amount of current flowing out of it's negative plate. And this current flows through the negative rail, the output stage of the ground amp, finally to the load itself, and ultimately on back to the positive plate of the positive rail's reservoir cap.
And this illustrates that a ground channel
does not divert load current from ground as has been claimed. Instead, load current flows through ground just as it does in the passive ground scenario. The only difference is that the load current takes a more circuitous route in the active ground scenario.
Does this help?
k