Basic Theory Question

[silverfox]

Is there anything wrong with these statemenst?

In a vacume tube:

The source of electrons is ground. The vacume poses an obstruction to the flow, (launching) of the electrons. The heating element provides the energy needed to escape the surface of the cathode and begin the trip towards the anode. The grid offers varying degrees of obstruction to the passage of the electrons.

Different conclusion here: The flow of electrons in the preamp tubes does not change from positive to negative, (it is not an AC signal), but rather varys between states of more or less positive? At some point these variations are split into positive and negative going signals- The phase inverter??

Thanks,

Silverfox.

[Chuck H]

The vacuum isn't there to pose an obstruction to the electron flow. It's there to keep the tube element from quickly oxidizing with operation.

Yes, the grid poses a variable obstruction that is controlled like a "valve" with the input signal. This the name they use for tubes on the other side of the pond.

I may be only part right on my assessment of this next question, but here goes... The high plate volts are indeed NOT AC but DC fluctuated from a floating point. The fluctuated portion becomes useful AC +/- only after the coupling cap or a transformer unless the DC level is used to derive a bias operating point.

The phase inverter is exactly that. Phase is a separate issue, though analogous to voltage. Having to do with when the cycles happen rather than how vigorously or frequently they happen.

[Enzo]

The vacuum takes air molecules out of the way of the electrons. Vacuum obstructs nothing therefore, in fact exactly the opposite.

Under normal conditions, the flow of current is only one direction through the tube. As current is varied, then the voltage across a series resistor will change by Ohm's Law. That voltage will rise and fall, but will remain positive.

Yes, a cap can isolate the change (AC signal) from the DC.

A phase inverter does not make anything negative, it just splits the signal into two versions that have opposite polarities. When one side goes more positive, the other side goes more negative (actually less positive). But like above, the current is always positive.

[loudthud]

In a push pull power amplifier the signal is split into two phases and then reassembled by the output transformer. When the current in one of the output device increases, the reassembled output goes positive and when current in the other output device increases, the reassembled output swings negative. Thus an amplified AC signal is recreated.

In a solid state power amp, the situation is somewhat different because there are "positive" (NPN) and "negative" (PNP) transistors. The phase inversion and reassembly is more or less automatic. There are no tubes that are constructed so that electrons move from the plate to the cathode or some kind of positively charged particles move from cathode to plate. Electrons have a negative charge so if the plate was a negative voltage, it would repell the electrons. Protrons would be attracted to a negative plate, but they are too big to be very mobile and it would take a nuclear reaction to set them free from the cathode.

[silverfox]

Making Progress

Okay I think I'm getting somewhere finally with all this.

The guitar signal is an AC signal. When it is transfered to the grid of the input tube, only half of the signal is amplified by the tube, (valve)? The signal must then exit as a pulse I presume. In order to recreate in whole the input wave form the tube would need to be connected to a negative supply as opposed to ground.

ultimately the phase inverter generates a corrosponding negative going portion of the signal? I think that was stated but just need to make it concrete.

Hope you can understand up to this point I was under the assumption that an AC signal, in it's entirety, was being amplified by the preamp valves in a tube based guitar amp. The theory I've read so far never mentions the fact that only half the input signal is amplifed at the input stage. Maybe the books did mention it but I never picked that up and things just never made sense at a subconscious level.


Thanks again,

Fox.

[Jazz P Bass]

Why would an 'amplifier' only amplify half of the sine wave?

The preamp tube is biased in such a way that it accepts & passes on the full AC signal.
(Class A)
In a clean (ie: non distorted) preamp circuit, what leaves the plate, via a plate load resistor, is a duplicate of the sine wave at the grid, only it is bigger (amplified)

Please try to limit your understanding to the preamp for now.

Once you fully grasp that you can move on to the output section.
For now, forget the phase inverter.

The very first tube was a a heated cathode & a plate.
Boring!
Lee DeForest managed to put a grid between the two & realized that there now was a way to control the amplification, by the grid itself.

[g1]

Agree with JPB, ignore the phase splitter and push pull type power amp for now. Try to stick with some simple single ended amp like a Champ, where all the tubes are amplifying the entire signal waveform.

[silverfox]

Well I'm off to the chassis with an O-scope to figure this out.

Thanks and I'll let you know what I find.. Not disputing but I've got to understand this.

Silverfox.

[Mick Bailey]

The guitar signal is an AC signal. When it is transfered to the grid of the input tube, only half of the signal is amplified by the tube, (valve)?

Taking just the preamp;

I can see why you'd think this, but it comes down to biasing, which means that current is flowing through the tube even when no signal is applied. The bias voltage applied to the grid sets this current flow. Think of it like controlling the flow of water in a pipe with a gate valve; if the valve was set midway to allow water to flow, you could either increase or decrease the flow.

Your AC guitar signal is doing this to the electron stream on the tube. It's not causing current to flow, but is causing an existing DC current flow in the tube to increase or decrease. This is why both halves of the sine wave can influence the current flow - because the tube is not 'off' to begin with.

So, the current flow results in a varying DC voltage at the junction of the plate and its load resistor. Even with no signal on the grid, there is still a voltage present at this point. In a simplified model you could set this with biasing to be 1/2 the stage supply voltage to give a 50% upward swing towards b+, or a 50% swing towards ground. Now if a capacitor is connected to the plate/load resistor junction, only the AC component will be present on the other end of the cap. It blocks the DC. With no signal on the grid there is no change at the plate, so no voltage on the other end of the cap. This becomes the zero axis for your output.

[Chuck H]

Pretty good!

I might even go more layman and say that the tube is operated at a floating DC so that both plus and minus AC at the grid can influence the current flow with "swing" potential dictated by that floating point. This is why the whole AC input can be mirrored instead of cut off on one side. So now some of the DC is waving around, and therefor not DC anymore, because it's alternating. A coupling cap blocks the DC component and only the fluctuations end up at the other end as the amplified AC.

[silverfox]

I forgot all about the biasing

I completely forgot about the biasing. I was going to put a scope on the input circuit but now remember I don't have a preamp section in the power amp I've got.

I'm making progress but will look again at the theory in the books an be back.

Silverfox.

[R.G.]

There are very few truly AC amplification devices in electronics. Magnetic amplifiers are the only ones that come to mind at the moment, and even they work more on a "bias" principle than true AC.

Vacuum tubes, bipolars, JFETs, MOSFETs, all must have some DC bias/idle condition to amplify. What they do is to set up some middle-of-the-road DC condition that signal then wobbles around. The wobble is the amplified version of the signal, and is taken out of the DC "nest" by DC-blocking capacitors. In fact, one of the beginner's lessons in amplifiers is that the idle condition must be right in the middle of the total available DC conditions so as to get the biggest undistorted swing out of the amplifying device.

It gets tricky with depletion mode devices, like vacuum tubes. Vacuum tubes pass DC current to the maximum unless you do something to turn them off. The common "self bias" setup for a triode or small signal pentode puts a resistor from cathode to ground, and another resistor from the grid to ground. When the grid is at the same voltage as the cathode, it does not shut down current flow to any noticeable degree, so current flows from plate to cathode. The current in the cathode resistor raises the cathode above ground. The zero current in the grid lets the grid resistor hold the grid at ground, so the rising cathode makes the grid effectively negative with respect to the cathode. That cuts down the current a little. When the cathode current in the cathode resistor rises to just the right amount to make the grid let that much current pass but no more, the tube reaches a stable bias point and stays there.

What's confusing about this is that it looks like you just poke AC into the grid with no DC offset, not noticing that the cathode has self-biased itself some DC amount away from the zero-voltage grid. So it looks like it's amplifying AC, but it's really not. It's doing the same old wobbling a DC level on the grid - but the DC level happens to be 0Vdc, and the other tube pins are DC offset from that grid voltage.

[silverfox]

Is this correct?

Much thanks to everyone here as I should be about there on understanding this.

I hope everyone can see I'm not disputing what is stated, just trying to comprehend this and it has taken over a year of hear and there study to actually try to grasp this concept. The conclusion if correct is stated in parts thoughout the discussion above.

The flow of electrons in the preamp tubes does not change from positive to negative but varys between states of more or less positive? The passage of time causes this variation to transform into a waveform. The variations in voltage appear on the other side of the coupling cap as the independant waveform separated from the DC source.

The tube in Class A mode?

The remainder of the preamp stages serve to enhance or filter the tone as well as, if desired, increase the gain.

Fox.

Silverfox.

[Mick Bailey]

The flow of electrons in the preamp tubes does not change from positive to negative but varys between states of more or less positive? The passage of time causes this variation to transform into a waveform. The variations in voltage appear on the other side of the coupling cap as the independant waveform separated from the DC source.

What happens is the cathode emits electrons and these are attracted to the charge on the plate. This gives an electron flow through the tube. The control grid itself is made of wire or mesh (depending on the tube) and offers very little resistance to the passage of electrons by itself, but when a negative voltage is applied (bias) an electrostatic field is created around the grid wire/mesh which repels some of the electrons so they can't all reach the plate.

Now, the negative bias is always present on the grid for a given configuration, but when a signal is applied to the grid, it either makes the grid more negative, or less negative depending on the polarity of the signal. This controls how many electrons can reach the plate, and the current flow is determined by how many electrons per second can pass through the tube.

If the control grid charge is continually varying due to an applied signal, then the electron flow, and therefore the current through the tube is continually varying.