Idealized grounding arrangement per Merlin!

[Wilder Amplification]

I think it still demonstrates that there is no reason to object to conventional current.

But...it makes things VERY confusing when trying to grasp the concept of electron tube operation when viewed from the angle of conventional current. Simply because protons cannot move (at least last time I checked...so long as no changes have occured to the laws of atomic physics) so they would as such have no way to travel from the plate to the cathode.

However, electrons CAN move. But within a vacuum tube, they cannot migrate from cathode to plate until the cathode has been heated to the temperature sufficient enough for thermionic emission to occur.

Once we have acheived this, electrons must flow from cathode to the plate in order for the positive charge flow from the plate to "find its way back to the negative end via ground" as Merlin states on page 7 of his grounding article. IMHO (AFAIK, FWIW and all that good stuff)...that leaves much to be desired in terms of demonstrating how things "flow" within a tube because of the fact that he's completely left out electron flow, which is the standpoint from which electron tube operation has been taught from day one. As such, I would think that this would lead to a myriad of confusion since it's been stated for over a century that electrons boil off of the cathode and flow to the positively charged plate, not from the standpoint of conventional current flow of positive charge to the cathode.

[Merlinb]

But...it makes things VERY confusing when trying to grasp the concept of electron tube operation.

I understand the point you are trying to make, but I don't agree that is makes it "VERY difficult". You learn that electrons boil off the cathode, and you must learn that conventional current flows the opposite way, from anode to cathode. But that's it, and that's hardly very taxing on the brain, right?

In any case, you only learn "how a vacuum diode works" on the page 1 of your old textbook. After page 1 there is no need to consider where the electrons might be coming from. There was a time when I had literally no idea how a semiconductor junction worked (and I only have a sketchy idea now), but since I'm not in the business of designing semiconductors or valves, I don't need to know. I just design circuits. When I think of current flow in a circuit, the notion of electrons doesn't even enter my mind; it's just current.

[GainFreak]

Again, haven't read the book, but grounding the chassis to the input jack and isolating it from all other "grounds" except the safety ground is a good thing to do.

I never quite got that. Does it mean that your input jack ground should contact the chassis?
In the amps I built so far I never did that - all my jacks were plastic which but I never had hum, buzz or any noises although I'm mostly into high gain stuff.

[Steve Conner]

I'm not sure what Merlin will say, but I think that's also a reasonable way to do it: use a plastic input jack and ground the preamp bus somewhere else. I've also built amps this way.

Drawbacks: RF and ground currents picked up on the guitar lead's shield can go right into the amp. With Merlin's approach they get diverted to the chassis.

[Russ]

If you believe this is so, than it would be easy to prove. Right !!!!!!
Not that I personally give a flying flip, but if the piece of aluminum does get charged ; which I am highly doubting because I understand the insides of a battery produces a voltage as a result of a chemical reaction inside the cell ; then one should be able to "untouch" from the negative post, and then "touch" it to the positive post, and then measure a current ?

Ya, Not Happening......

But hey, I'm not calling myself "merlin", so . . . . .. .


-g

Yes, I think that is what happens because a battery, by iteself, has an electrostatic field surrounding it.

So when a conductor, such as one end of the bulb string, is connected (or even brought close) to one end of the battery, that E field effects the electrons (charge) in the string, which then redistributes in the string. If no circuit is forms then no current flows, but the charge in the string does redistribute itself forming a brief current which eventually stops as the E field stabilizes and becomes static once again around the battery and the string as well. This disturbance of the E field propogates out as an EM wave which eventually disipates and stops until the E field stabilizes and becomes static again.

Now if the other end of the string is connected, a circuit is formed and charge has a place to go and thus moves forming current. This movement starts at the point of contact where the circuit is formed (at the switch for instance). At this contact point an EM wave propagates out and then disipates and stabilizes and, in this case of DC current from a battery, you're left with current flowing through the string and a static E field around the string.

Russ

[GainFreak]

I'm not sure what Merlin will say, but I think that's also a reasonable way to do it: use a plastic input jack and ground the preamp bus somewhere else. I've also built amps this way.

Drawbacks: RF and ground currents picked up on the guitar lead's shield can go right into the amp. With Merlin's approach they get diverted to the chassis.

Maybe that's why Marshall and Mesa for example are using RF beads at the input.

[Gollum]

I'm sure I'm just the idiot here, but if the +/- charge of a piece of aluminum attached to a single terminal of a battery didn't change +/- value then why the HECK do capacitors work?....

Point being that by definition (I thought at least) a capacitor is two conductive plates separated by a non-conductive medium (air being a darn good one). By touching the aluminum to one battery terminal you've just made a huge capacitor plate haven't you? Or say you put a second piece of aluminum on the other terminal as well... separated by nothing more than air. That's a capacitor isn't it?

Because if I'm right in thinking this way then merlin must be right. The light bulbs will light sequentially from the switch. If a light bulb could truly light at any electron flow then the mere act of hooking up a small enough light to a big enough battery + or - could cause it to glow while it reaches a standing point. The circuit would never NEED to be completed.

Or I've just gotten this whole thing wrong, which is ok by me.

[Steve Conner]

Because if I'm right in thinking this way then merlin must be right. The light bulbs will light sequentially from the switch. If a light bulb could truly light at any electron flow then the mere act of hooking up a small enough light to a big enough battery + or - could cause it to glow while it reaches a standing point. The circuit would never NEED to be completed.

Seems reasonable. The battery might be the source of the energy, but the switch is the source of the initial transient.

Using a Tesla coil, radio transmitter or similar, you can indeed light a bulb with only one terminal connected. The current that lights the bulb is what Maxwell called "displacement current". It was too small to be detected experimentally with the equipment available in Maxwell's day, but he reckoned that it had to exist because his wave equations wouldn't work without it.

[R.G.]

I'm sure I'm just the idiot here, but if the +/- charge of a piece of aluminum attached to a single terminal of a battery didn't change +/- value then why the HECK do capacitors work?....

The charge does change. But like everything else, the devil is in the details.

All conductors have a capacitance to every other conductor in the universe. This capacitance is calculable (to some extent!) and obeys both the inverse square law for capacitor value, and the limitation that any electromagnetic change can only propagate at the speed of light. So the capacitance to things that are modestly far away is truly tiny. The capacitance from a piece of aluminum to everything else in the universe may be viewed as the "self capacitance", and is quite small for small objects. Maybe a couple of pF. Maybe. The charge needed to move 1pF by 12V is pretty small.

Point being that by definition (I thought at least) a capacitor is two conductive plates separated by a non-conductive medium (air being a darn good one). By touching the aluminum to one battery terminal you've just made a huge capacitor plate haven't you? Or say you put a second piece of aluminum on the other terminal as well... separated by nothing more than air. That's a capacitor isn't it?

It is. But there's that separation thing. For two plates of a conductor separated by a small distance, you get this:Parallel Plate Capacitors. For plates which are not largely overlapping and a small distance apart, the math gets ugly. Capacitance falls off very rapidly with increasing distance and non-overlap.

Because if I'm right in thinking this way then merlin must be right. The light bulbs will light sequentially from the switch. If a light bulb could truly light at any electron flow then the mere act of hooking up a small enough light to a big enough battery + or - could cause it to glow while it reaches a standing point. The circuit would never NEED to be completed.

You're correct as far as this goes. Simply connecting a conductor to a battery forces the conductor to go to the battery voltage. Current flows in a transient until it charges up the self-capacitance, perhaps a distributed self capacitance, and brings the conductor to the source voltage. When it gets there, current stops because there is no voltage difference to drive it.

You can approximate this by hooking one end of a light bulb to a BFC (Big Freaking Capacitor) which connects to one end of the battery. A wire connected from the other end of the bulb to the other end of the battery causes the same transient, but much, much bigger by the Big Freaking amplification factor. The charge that moves is much bigger because the BFC is much larger than the self capacitance. But only the magnitude and time scale are larger. There is also a loss of the distributed capacitance and any relativistic effects, which is good for being able to see what's happening.

Using a Tesla coil, radio transmitter or similar, you can indeed light a bulb with only one terminal connected. The current that lights the bulb is what Maxwell called "displacement current". It was too small to be detected experimentally with the equipment available in Maxwell's day, but he reckoned that it had to exist because his wave equations wouldn't work without it.

This is an alternate (and probably more correct!) way of saying "self capacitance effects". The Tesla coil and RF transmitter are forcing EM waves off into the rest of the universe. The bulb picks up some of that and the self capacitance of the bulb to the rest of the universe completes the circuit. It's handy to think of it this way, anyway.