Basic question about electron flow to tube grid

[Planobilly]

Hi,

Consider the circuit from the guitar pickups to the control grid of a typical 12AX7.

I think I understand how this works but I would like some confirmation that I am correct.

The control grid is a single point or node on the loop or circuit connecting the pickups to the control grid.

As the string moves this mechanical force disrupts the magnet field in the pickup magnet causing a voltage to exist and a current to flow to the control grid. I understand there are pickup winding's, resistors, capacitors,impedance resistors, bleeder resistors and so on.

Also is there any reason other than safety that this circuit would be required to to be connected to earth ground?

Do I have a clear and correct understanding of this circuit?


Thanks,

Billy

 

[rayma]

Consider the circuit from the guitar pickups to the control grid of a typical 12AX7.
Also is there any reason other than safety that this circuit would be required
to to be connected to earth ground?

An amplifying tube's grid works by changes in voltage with respect to the cathode, not current flow.
The pickup provides this changing voltage. It isn't necessary for current to flow to have a voltage,
for example in a battery.

True earth grounding is seldom necessary for a circuit's operation, but when HV or the AC line
are involved, it is a safety requirement.

 

[Planobilly]

Hi rayma,

Let me ask this in a different way.

When there is a voltage generated by the pickup more electrons are present on the control grid which makes it more negative or less electrons exist on the grid during the negative part of the sign wave which makes it more positive. IS this correct?

And thanks, I understand how swinging the control grid more positive causes more electron flow from the cathode to the plate and also the effect of screen grids and suppressor grids.

Because the control grid is only one connection I understand no AC current from the pickup exist there. Well..I assume that to be true.

Thanks,

Billy

 

[cbdb]

There is current, the electrons that flow on and of the grid, but very small. Ohms law. A volt or so from the pick ups into a Meg or so input impedance of the amp give about a microamp of current.

 

[rayma]

When there is a voltage generated by the pickup more electrons are present on the control grid
which makes it more negative or less electrons exist on the grid during the negative part
of the sign wave which makes it more positive. IS this correct?

When connected to a source of potential, such as a battery (or ground with cathode biasing),
no excess electrons are present on the grid. An ohmic connection from the grid to the voltage source
causes the grid to assume the same electrical potential as the source, without transfer of charge to the grid.
Parasitic effects will cause deviations from this ideal behavior, though.

 

[indianajo]

More simply, electrons push on each other the way magnets push on each other north to north.
The control grid is close to the cathode which is hot and boils off electrons. So a few electrons deposited on the control grid keep a lot of electrons from flying down to the plate, which is plus and attracts them.
They bias the control grid negative so there are a lot of electrons on there already.
The "capacitance" of the control grid is low, so it only takes a few electrons to do a lot of pushing. So not much current flows into the control grid of a tube. Changing the voltage on the control grid makes a few electrons flow in and out, ie the current is very low. Many more electrons are affected flowing from cathode to plate, so the current change coming out the plate is dozens or hundreds of times bigger (S factor). The plate current is run through a plate resistor, so the current change becomes a big voltage change. S on datasheet is current change out the plate divided by voltage change in the control grid.
Grounding the guitar amp is important because tubes sometimes short plate to control grid. This could put 275 v on the pickup with enough current to kill people. Guitar amps get banged around, and the distance between all these wires and bits of metal in a tube is very small.
So ensure your control grid is separated from the pickup by a nice safe 400v rated capacitor or something when designing or copying a diy guitar amp. Also ground the chassis per instructions. Diy solder joints are particularly prone to fall apart and let wires fly. Mine do.

 

[DF96]

Also is there any reason other than safety that this circuit would be required to to be connected to earth ground?

I am going to answer a question you didn't ask, because it is an issue which often confuses newbies and because it may be that this is the question you meant to ask.

Q: "Since no current flows to the valve grid, why is it necessary to connect both sides of the pickup? Surely it is enough to just connect one side to the grid, as that will put the right voltage there."
A: There are two answers, both of them are correct.
A1: A tiny current does flow so it needs a complete circuit. It doesn't matter whether this circuit loop is connected to ground.
A2: There is no such thing as 'a voltage'. All voltages, with no exceptions, are the difference between two points so therefore you need two wires, one to the grid and one to the cathode. Whenever someone speaks of 'a voltage' there is always a reference point (whether he realises it or not); this reference point may be called 'ground', but don't run away with the idea that it has to be ground, or that there is one unique ground (two different connections to 'ground' will always be at slightly different voltages).

 

[Planobilly]

A simple input circuit.

[/IMG]

 

[Planobilly]

Thanks guys for all the feedback.

I assume that....

In the above circuit the only reason current flows from the pickups is due to the 100k resistor completing the circuit to ground

A change of potential would appear on the control grid regardless if the cathode was connected or not.

My question was not about how a tube works as I think I understand that pretty well. Just the electron path from the guitar to the control grid.

Am I missing anything?

Thanks,

Billy

 

[DF96]

In the above circuit the only reason current flows from the pickups is due to the 100k resistor completing the circuit to ground

Almost true. There is a tiny current into the valve grid and out of the cathode too.

A change of potential would appear on the control grid regardless if the cathode was connected or not.

The grid would have a potential with respect to something else. This would do nothing at all if the only valve electrode connected was the grid.

My question was not about how a tube works as I think I understand that pretty well. Just the electron path from the guitar to the control grid.

"Electron path" implies circuit. Circuits need loops.

Am I missing anything?

I suspect so, but your denial makes it hard to discover exactly what it is. Please carefully read my post 7 again, and ensure that you understand and agree with what I say there about voltages and circuits.

 

[Malcolm Irving]

Not sure if this will help, but here goes:

The fundamental quantity in electricity is ‘charge’. It is measured in coulombs. An electron has a fixed, very small, negative charge. Current is the rate of flow of charge and is measured in amperes (= coulombs per second).

Potential (measured in volts) is a less fundamental unit and arises as follows: There is a force of repulsion between two charges of equal sign (and attraction between charges of opposite sign). If we move a charge ‘against’ this force, work needs to be done, which is measured in Joules. A potential of 1 volt exists between point A and point B, if 1 joule would be required to move 1 coulomb from A to B. (A volt is a joule per coulomb.)

 

[DF96]

Yes. Hence we cannot say that point A 'has a voltage of 1V' - this is completely meaningless. We can say that there is a voltage of 1V between A and B. So we cannot say that the voltage from the pickup goes to the grid. We can say that the voltage across the pickup is applied across the grid-cathode. People often say the former, and actually mean the latter (if they understand electricity); this simplifies speech but confuses newbies.