Interpreting Static Anode Characteristics Load Line

[Gary_S]

Ok i understand the basic idea of this through reading Merlin's opening chapter, however i'd like some clarification please of this point. It's all to do with AC versus DC again 

On the load line on the horizontal axis the Va; is that DC volts or AC? I thought initially that was DC  because you get the bit where you reach cut off and the voltage is the same as the HT which is DC volts, or B+ you guys call it that are from the US.

I just find this little thing a bit confusing. I in my simplistic thinking thought that the DC originally was just the HT supply voltage and the audio signal was always the AC and they didn't combine so to speak.

This is the bit that i find the most confusing; how the two things AC and DC are combining

[six el six]

You don't have "simplistic thinking". These are difficult concepts to grasp when beginning to learn about electronic circuits.

PRR, HPB, and Sluckey are good at explaining these things.

There were many tube circuit primer books published in the 1940s and 50s that had illustrations that help to understand these "basics" of electron tube operation. You might find them helpful.

[Gary_S]

Hi six el, yeah the interactions of the AC with the DC is the thing i'm finding most complicated. I always assumed that the DC was the power supply voltage feeding the plate and so on in the tube, and the other power that the circuit needs. And the AC was the audio signal coming in from the guitar and going onto the grid. Merlin's book is brilliant and exposes you to all this stuff, but i read a bit and then think about it to try to make sure i understand the concept fully before i go on. And you're right; it's not easy, it's complicated and quite confusing as you sometimes forget and say "what voltage am i  dealing with here ? is it AC or DC?" Probably pretty simple once you have it down, but initially there's a lot to think about.

But i console myself with the fact that it must take years and years of hard study to totally understand all this. You're not gonna get your head round it in a few weeks or a few months, i don't think. 

I always think this is the difference between people who totally understand amps and how they work inside out, and someone like myself who replaces caps and resistors and does biasing but is not fluent with the theory of how everything works and flows.

This is why i'm studying it though, to hopefully end up understanding more. I might not get as professional as some of the guys on this site but a large improvement would be good.

[HotBluePlates]

Hi six el, yeah the interactions of the AC with the DC is the thing i'm finding most complicated. I always assumed that the DC was the power supply voltage feeding the plate and so on in the tube, and the other power that the circuit needs. And the AC was the audio signal coming in from the guitar and going onto the grid.

Well, it is. All of what you said there is correct.

You need to think about changing your frame of reference. Specifically, time.

I'm assuming somewhere in his book, Merlin touches on the basic idea that we talk about sound in terms of sine waves to make things easy, and how the wave repeats over time. Specifically, the thing that is repeating is the size variation, or amplitude, in both a positive and negative direction over time. The frequency of repetition is described in "cycles per second" or "Hertz" (Hz). If you think about it more, if you keep slowing the frequency of repetition, you eventually run into a limit of 0Hz, or d.c.

So d.c. is no variation or change in the amount of voltage (or current), and a.c. is some fast or slow change of voltage (or current) over time.

So, if you also keep making the narrow window of time over which you're thinking about the a.c. smaller and smaller, you can make it so small an instant that it's not changing. In other words, in that snapshot of time, the a.c. is one set amount of voltage.

Why do you care? Read on...

[HotBluePlates]

So you've found out you draw a loadline from the value of the d.c. supply voltage at the bottom right to the value of current that would exist if the full supply voltage were across the plate load resistor at the top left.

Look at the gridlines that were on the graph before you drew your line. Where these lines cross your loadline, you can find what value of plate current and plate voltage would exist if you apply that amount of voltage between the grid and cathode. So if you had -4v measured from the grid to cathode, you could look where the -4v line crosses the loadline and see what the resulting plate current and plate voltage would be.

Let's pretend that your tube is biased such that there is -2v from grid to cathode (that might be a case where your grid is grounded through a reference resistor and the cathode is at 2v due to the cathode resistor), and that your applied a.c. voltage will be 1v peak. If your incoming a.c. was at the 0v starting/ending point of the cycle, the net voltage from grid to cathode is -2v.

The signal voltage increases until it reaches the 1v peak; at that snapshot in time, the net voltage is -2v + 1v = -1v. To see what the resulting plate voltage/current will be at that moment, look at where the -1v gridline crosses your loadline.

You can do the same thing with the negative peak of the applied signal. When that signal reaches its -1v peak, the net voltage from grid to cathode is -2v + -1v = -3v. Again, look at where the -3v gridline crosses the loadline to see what the plate voltage and current will be at that instant.

To the tube, there is almost no difference if the tube bias is being changed from its idle value by a.c. (like the signal we applied above), or if you actually caused the tube bias to be changed to settle on -3v or -1v all the time. If you change your time frame of reference to be impossibly small, there's no difference between d.c. and the net voltage resulting from the combination of a.c. and d.c.

[Gary_S]

Cool stuff HBP and this is augmenting what i've learnt from the book so far.

I couldn't get my head around the voltage swing on the load line!  I said "but that's the bias voltage that's sent to the grid! and it's DC! it can't be swinging up and down or the tube bias will go all over the place!"

So i went back and read the opening chapter a bit more and thought about it. It's the AC signal voltage on the grid that's fluctuating around the steady DC bias voltage. At least i think it is!!!

How long you been studying this HBP? a long time i bet!  you don't get that kind of knowledge overnight, no way!

All this and i haven't even asked the questions i've got stored up about equivalent circuits

Thanks HBP you help me out a lot with looking at the thing from the right angle.

[HotBluePlates]

Been reading about guitar amps for going on 20 years.

What I found as I learned more is I was making the basics way too hard! Imagine that... But once I understood the stuff, I realized I was making it more complicated by trying to imagine some complex way everything fit together. In the end, it's simpler than you think, if you can approach it with a blank slate and accept some concepts on faith.

The wealth of poorly-phrased information, wrong info or things with an overly-complex presentation really mess things up for a lot of folks (me included).

[HotBluePlates]

All this and i haven't even asked the questions i've got stored up about equivalent circuits

Just remember this: equivalent circuits are an engineer's/student's way of breaking down a complex circuit into something simple that can be understood. The only hard thing about equivalent circuits is the process of going from a complex circuit to its simple equivalent.

And, most equivalent circuits (and even my earlier explanation to you) make assumptions that simplify the situation. If you understand the basic overall situation, it's easier to tell you how there are exceptions that go beyond the basic model.

[tubeswell]

I couldn't get my head around the voltage swing on the load line!  I said "but that's the bias voltage that's sent to the grid! and it's DC! it can't be swinging up and down or the tube bias will go all over the place!"

In fixed bias, a -ve bias voltage is fed to the grid. In cathode bias a +ve bias is applied to the cathode via a cathode resistor (and the grid voltage is therefore '-ve with respect to the cathode'). Either way, the signal on the grid swings +ve and -ve around whatever the grid idle voltage is.

...It's the AC signal voltage on the grid that's fluctuating around the steady DC bias voltage. At least i think it is!!

Indeed. And if the signal swing gets too big on the '+ve' side of the swing, it will be prevented from going more positive than the grid-to-cathode voltage (and you will get 'grid-current limiting' causing clipping). Whereas, on the other hand if the signal swing gets too big on the '-ve' side of the swing, it will force the tube to stop conducting current altogether (i.e.: the tube will be in 'cutoff').

[Gary_S]

Ok i''m on to looking at the cathode bypass capacitor and how it stops any AC on the cathode getting on to the cathode bias resistor which would affect the bias of the tube.

This is where i am at the moment in my understanding of the preamp flow, this would be valve 1 first valve in the amp: AC signal from the guitar go's to the grid of the tube, this causes current to flow from the space charge that's formed near the cathode and it get's attracted to the grid because it's slightly more negative but the voltage on the anode or plate causes the electrons to bypass the grid and go to the plate.

What's getting me a bit jammed up is though that conventional current flows the opposite way. As well as the electrons heading to the plate from the cathode, there is also a flow of current from the anode to the cathode that then makes the cathode want to increase it's voltage to follow the grid. This is where the bypass capacitor comes in to stop that flow of AC ruining the bias of the stage.

My big question here is; what current is flowing from anode to cathode? is it AC, DC or a mix of both?

Sorry if this is wordy and complicated sounding, i maybe could have phrased it better but it's the best i could do at the moment!!! I'm studying this so hard trying to get 100% fluent with it in my mind.

[HotBluePlates]

What's getting me a bit jammed up is though that conventional current flows the opposite way.

Don't do that. 

Either pick conventional current flow or electron current flow, and use it without thinking about the other. Which gives the easiest concept to understand depends on the circumstance. When picturing what happens inside a vacuum tube, electron current flow is easier to grasp, so use that exclusively.

Basically, the concept of current flow was developed before there was a solid understand of subatomic particles and the fact that electrons are the particle that is relatively mobile and free to move from atom to atom in a conductor. So the model developed was that of positive charges being mobile; they had a 50/50 chance to make the right choice and they picked wrong.

That said, it's easier when analyzing circuits in general to use conventional current flow and think in terms of positive charges moving through a circuit. It's also easier in some cases to think of positive charge carriers in some solid-state devices.

So, use the notion of conventional current when looking at a guitar amp in a macro sense, and electron current flow when thinking in the micro sense of what happens inside the tube.

[Gary_S]

Don't do that.  

Either pick conventional current flow or electron current flow, and use it without thinking about the other. Which gives the easiest concept to understand depends on the circumstance. When picturing what happens inside a vacuum tube, electron current flow is easier to grasp, so use that exclusively.

Basically, the concept of current flow was developed before there was a solid understand of subatomic particles and the fact that electrons are the particle that is relatively mobile and free to move from atom to atom in a conductor. So the model developed was that of positive charges being mobile; they had a 50/50 chance to make the right choice and they picked wrong.

That said, it's easier when analyzing circuits in general to use conventional current flow and think in terms of positive charges moving through a circuit. It's also easier in some cases to think of positive charge carriers in some solid-state devices.

So, use the notion of conventional current when looking at a guitar amp in a macro sense, and electron current flow when thinking in the micro sense of what happens inside the tube.

Cheers HBP, i am getting a handle on things slowly then i read about the direction of the conventional current flow and i'm like WHAT!

What about the electron flow from cathode to anode is that pure AC?

[HotBluePlates]

If you're inside the tube, the only thing you'll see is electrons emitted from the cathode and flowing to the plate if the control grid allows that to happen.

But you're asking how does the current "go negative" during the negative half of the a.c. signal. Don't you remember what I'd said before about waves on the ocean? The troughs of those waves don't suck in underneath the sea floor...

The tube can only pass electrons out of the cathode and into the plate. That means the 0mA horizontal line of plate curves is the bottom; the tube can't pass -1mA. At 0mA, the tube is cut off.

That's why if you are operating the tube single-ended, you turn the tube on to some middle value of current with the d.c. bias. Then the a.c. has room to both increase and decrease the plate current without turning the tube off (unless you want the distortion caused by cutoff).

Saying another way: the tube is turned on by the bias. The positive half of the a.c. input turns the tube more-on compared to the idle current (more cathode current). The negative half of the a.c. input turns the tube less-on compared to the idle current (less cathode current). If the tube is not driven to cutoff, it's a continuous flow in one direction, just more or less of it.

Saying a 3rd way: Go to your kitchen faucet. Turn it on halfway up. now wiggle the faucet above and below that halfway setting, but not far enough to turn off the water or to turn the tap wide open. Water always flow down (like electrons only going from cathode to plate), but has a varying rate of flow that changes with you wiggling the handle. If you could mentally or physically isolate just the amount of variation, that's the a.c. component, and your starting middle value of water flow is the d.c. component.

At the plate, you can imagine there are both the d.c. and a.c. components of plate current. That the current must flow through the plate load resistor causes a voltage drop across the resistor, and varying current causes a varying voltage drop. The B+ voltage minus the varying voltage drop leaves some remaining variable voltage at the plate itself. The coupling cap blocks the d.c. voltage, but allows the varying voltage component to pass through. The following tube grid just sees the varying voltage component.

[Gary_S]

Great post HBP and clears a lot of little niggles i had in my mind about what's going on. 

The current is flowing, or the electrons are flowing, to the plate from the cathode. The voltage that is imposed on the cathode where is that coming from? is it the anode? through the load resistor? because in Merlin's opening chapter that's the way i interpreted it. Then the cathode bypass cap takes any AC signal out so that no signal go's through the cathode bias resistor to cause problems with the steady DC bias.

I'm indebted to you for your knowledge. The book is great but sometimes you need someone to explain the concept further before you get a handle on it!!

[sluckey]

The voltage that is imposed on the cathode where is that coming from?

Pure Ohm's law. Current flowing thru the tube (as a result of tube operation) must also flow thru the cathode resistor. This current thru the resistor causes a voltage to be dropped across the resistor. Ohm's law states E = I x R. So, for example, take a 12AX7 that has a 1500Ω cathode resistor and a 100,000Ω plate resistor. Tube conditions are such that 1mA flows thru the tube. This 1mA current also flows thru the 1500Ω cathode resistor and causes a voltage drop that is equal to I x R = .001 x 1500 = 1.5V. Likewise, since this same current flows thru the tube and the plate resistor, you can calculate the voltage dropped across the plate resistor. That voltage would be = .001 x 100,000 = 100V. So, if your B+ supply was 300V and the plate resistor dropped 100V, the voltage measured from plate to ground would be 200V.

Then the cathode bypass cap takes any AC signal out so that no signal go's through the cathode bias resistor to cause problems with the steady DC bias.

That's a good way of looking at it.

[Gary_S]

Thanks Sluckey, very good information and makes me look at it from a different perspective. When you read a post like yours there you realize how important and useful Ohm's law is!

The current that is flowing through the tube; is it strictly AC or DC or both? That's one of the things that's foxing me. I know at the plate output we've got DC for instance, but we've also got the AC signal voltage going to and from the coupling cap of the next stage.

Is it the same within the tube? we have both types of current flowing?

[HotBluePlates]

The current that is flowing through the tube; is it strictly AC or DC or both?

Start with simple building blocks and work your way up.

If a cap is two conductive plates separated by an insulator, you can understand how this looks like an open circuit, or the same as two insulated wires next each other with no electrical connection. For now, accept on faith that the cap looks like a short-circuit to a.c. (the factual micro-level explanation of why this is the case will be confusing right now).

So with no signal, but with d.c. applied from the power supply, the tube initially feels a large voltage at the plate, and has a hot cathode emitting electrons. The electrons near the cathode feel the pull of the voltage at the plate, which at the first moment equals the supply voltage, and they all try to rush to the plate.

Ground is the source of the electrons to replenish the ones sucked away from the cathode toward the plate, but there is a resistor between the cathode and ground. If the electrons look to the parallel path of the cathode bypass cap, they just see an open circuit they cannot cross. As the replenishing electrons flow from ground through the resistor to the cathode, a voltage is developed across the resistor. That raises the cathode above ground potential, and it is no more-positive than the grid. This establishes the grid-to-cathode voltage which biases the tube.

At the other end of the path, the power supply voltage tries to suck electrons out of the plate after they arrive from the cathode. But there's a big plate load resistor between the plate and the power supply. As the electrons flow out of the plate, they set up a voltage across the plate load resistor. Because the voltage from power supply to ground does not change, the relative voltage at the plate itself drops from being equal to the supply voltage down to the value determined by ohm's law and the amount of current flowing through the plate load resistor.

If the electrons look to the right at the coupling cap, they again see an open circuit and instead flow through the resistor.

When a positive-going a.c. input signal is applied, the voltage of the grid rises making the difference of voltage from grid-to-cathode smaller (because the cathode was already positive of the grid). This smaller grid-to-cathode voltage allows more electrons to pass through from cathode to plate and increases the tube current.

That bypass cap looks like a short-circuit to changing current which is what a.c. is. So the amount of current change from the steady idle value is bypassed around the cathode resistor, and there is no change of voltage across the resistor. The positive change of current travels to the plate and is pulled through the plate resistor to the power supply. It then travels through the filter cap and back to the ground end of the cathode resistor to complete the circuit.

Because the changing current was pulled through the plate load resistor, it also creates a changing voltage drop in accordance with ohm's law, which results in a bigger voltage drop across the plate load due to bigger current, and leaves less of the supply voltage at the plate itself. The coupling cap allows the following tube stage to sense this drop in voltage because it is passing a changing voltage as though it were a short circuit and the output signal to the next grid appears to be a negative-going voltage. This is why the plate output inverts the signal.

The following stage's grid resistor (perhaps 1MΩ or similar) is large enough that the changing current takes the path of least resistance through the plate load rather than passing through the coupling cap. Therefore, the changing voltage output is what is sensed by the following stage's grid.

The whole process is exactly the same for the negative-going portion of the tube's input signal, except the net change of grid-to-cathode voltage is such to increase the voltage difference, which reduces tube current and reduces voltage drop across the plate load, and increases the resulting voltage at the plate. So a negative-going input voltage signal results in a positive-going voltage output at the plate, again explaining the inversion of the signal output compared to the input.

Direction of current does not change inside the tube, but the change of more- and less-current exists simultaneously, so there is both d.c. and a.c. inside the tube. The external circuit components split the paths that the d.c. and a.c. components travel, and control the tube while taking a voltage input at the grid and presenting a resultant voltage output at the plate for use by the following tube stage.

[Gary_S]

Immense stuff HBP Really top class, and clarifies what i've been picking up from the book. It slowly unfolds this, at it's own rate and it takes a lot of work and studying it.

I work with the book (Merlin's) every day. What i do is take a small part and read it, then put the book away and think about the concept in my mind to see if i can totally understand it. It's not like a novel that you can just bash through page after page    you really have to think about what you're reading and try to understand the concepts.

I read the section on equivalent circuits and the equations for working out various things. What do designers use these circuits for. Those are a bit confusing because; yes the EC does look simpler than what you started with, but i'm not understanding the deep reason behind why you'd use them?

[HotBluePlates]

In the scenario I outlined, the coupling cap is a special case. It does not look like a short-circuit to all a.c. all the time. A designer might use two or more equivalent circuits to study how the circuit behaves at different frequency ranges.

Again for the cathode bypass cap. At some low frequency, it looks like a large impedance and not a short circuit anymore. If an a.c. signal momentarily changes the effective bias of the tube, that might have an impact on tube voltage gain. Again, a designer might use an equivalent circuit to study the effects under different situations.

Mostly equivalent circuits break down a complex circuit to more easily teach them, or to describe the circuit action in general terms that apply in special cases or all the time, without relying on specific component values or voltages or currents.

Equivalent circuits help researchers develop mathematical models that help predict circuit performance.

[Gary_S]

As far as the plate load resistor: the power supply sends the voltage to the plate, which puts a charge on the plate. Then the current flows from cathode to the plate. Then that output current splits in two, the AC being able to travel through the coupling cap to the grid of the next stage, while the DC current go's back the other way through the load resistor?

How do you control the P to P voltage swing of the input signal to the grid? I mean your guitar output, be it a strat or a Les Paul, has a specific output and you can't vary that. Reason i ask that question is because i read in Merlin's book that the greater the range of the voltage swing on the load line the more gain you can work into the stage.

HBP Just to clarify; when you use the term short circuit i take that to mean almost zero resistance in the context we're using it here?

[HotBluePlates]

... the power supply sends the voltage to the plate ...

Voltage is a measure of electrostatic charge, and you can mentally equate it to water pressure. It's also called electromotive force (EMF), and is the force which causes electrons (or other charge-carriers) to move in a directed manner.

So once power is applied, the supply voltage is present at the plate until current results in a voltage drop measurable across the plate load resistor; the remaining voltage which is the difference between the supply voltage and the plate load voltage drop is present at the plate.

To further the water analogy, current is like gallons-per-minute; current is the rate of flow of electrons. The higher the voltage (water pressure), the more current there is (more gallons-per-minute). The exact definition of the unit for current specifies a particular quantity of electrons moving past a point per second.

Resistance is then defined as an opposition to the flow of current; it is defined in terms of the other two units in that 1Ω is the opposition to current such that [a pressure of] 1v results in a current of 1 ampere.

You can go the other way in that if you know you have 100kΩ and measure 100v from leg to leg of the resistor, then 100v/100kΩ = 0.001A or 1mA is flowing through the resistor. So 1mA of current through a resistance of 100kΩ results in a voltage drop of 100v (or "100v dropped across the resistor").

... output current splits in two ...

Not exactly; for your level of understanding, assume current has only 1 path to flow in this case.

An electrical circuit has to be a continuous conductive circle for current to travel around. That circle runs from ground through cathode resistor to cathode to plate to plate load resistor to power supply to ground.

Pretend for the moment that the coupling cap only allows a changing voltage at the plate to be sensed and its force/pressure/influence to be passed along to the next stage.

More accurately, the bulk of alternating current flows through the plate load resistor. But your tube doesn't operate on a current input, it operates on a voltage input. So the changing voltage drop is what gets passed by the coupling cap. Additionally, there is some smallish a.c. current which does flow through the cap. But this doesn't contribute anything useful to the operation of the tube, and a good design usually makes this 2nd current a very small portion of the total a.c. current.

How do you control the P to P voltage swing of the input signal to the grid? I mean your guitar output ... has a specific output and you can't vary that.

Uhh... pick harder/softer or turn the guitar's volume control up/down. Use a pedal whose electronics boost the signal level or turn it down.

... i read in Merlin's book that the greater the range of the voltage swing on the load line the more gain you can work into the stage. ...

I haven't read Merlin's book, but I'll bet on a closer reading, you'll find he's saying that if the load and operating point are such that there is a large swing in plate voltage, that might mean the stage exhibits greater gain. Or depending on the scenario, it could be less gain (amplification), but more headroom (meaning larger maximum signal output before significant distortion).

And in a technical text, "gain" just means amplification or "how many times bigger" the signal is, and is not the same as distortion.

Just to clarify; when you use the term short circuit i take that to mean almost zero resistance in the context we're using it here?

Yes; when I say a cap looks like a short-circuit, I mean to some a.c. frequency it looks like a piece of wire. To 0Hz, or d.c., a cap is the opposite and looks like an open circuit or infinity ohms.

[Gary_S]

Thanks HBP, all your help is invaluable to me. Thing is; you can have a text book and read it and understand it, but there are always concepts that don't quite make sense 100% and it takes listening to someone who is more knowledgeable about the subject to explain it so that it sinks in. This is why this forum is top quality!!

I decided a while back that i wasn't happy with just being a parts changer ie. switching caps and resistors etc. That i wanted to understand how the amp worked from the input to the output and why it reacted the way it did. This is why i bought Merlin's preamp book and i'm asking all these questions. I'm like a sponge soaking up all this knowledge 

It gets frustrating when you're not moving ahead fast with it but it's so complicated that it's gonna take years and years to get this stuff completely understood and fluent.

[Gary_S]

So the electrons are coming up from ground and going through the cathode resistor and on in the manner you described in your post HBP.

I had a problem picturing that because on my Marshall 4010 PCB i can't see where the cathode resistor and bypass cap make the connection to circuit ground. The resistor and the bypass cap connect to the cathode and then their other ends to the PCB.

The connection is obviously being made somewhere but i don't see where  

This is what i don't like about PCB's; the connections are harder to see, unless you've got everything out and can access the other side of the board.

[JPK]

This is an absolutely wonderful explanation of how a tube amp works. Thanks for asking all the questions Gary. And thanks for taking the time and having the patience to answer all the questions HBP. That was a way of explaining that I haven't seen before. I learned alot. Especially about the current not changing direction through the tube (only varies in amplitude). Not sure why, but I just thought ac,...it must be reversing current through the tube. Then after reading all this I realized how silly that was. Also I finally realized why the signal inverts via the tube conducting more or less current through the plate load resistor, which then either gives close to full B+ when the tube is off, to less than B+ when the tube is conducting. Is this basically a pull-up resistor? It sure reminds me of how a transistor works with a pull up resistor but it's been a while.

When I first saw this thread yesterday, I told myself "man you gotta read this later". Well I just did and wow was it worth the effort to read (slowly!).

[Gary_S]

Great stuff JPK! I'm delighted that someone else has learnt something from the great explanations put up by HBP

It's a complicated subject so i expect to be asking for advice for a long time 

[JPK]

Thanks! I just wanted HBP (and you) to know how many other people like me are reading his words and really appreciate the knowledge. It's quite a service to us green horns. You can't get this in school.

HBP, I have a question. I think I finally realized why the B+ is attached to the center tap of the primary on my P-P OT. Is half the transformer like the plate load resistor for 1 power tube, the other half is for the other power tube? What creates actual final speaker volume is the voltage dropped through the OT primary halves created by the current out of the power tubes? Or is that not right?

[HotBluePlates]

... the plate load resistor ... Is this basically a pull-up resistor? It sure reminds me of how a transistor works with a pull up resistor but it's been a while.

Similar. The problem is transistors are current-operated devices, where tubes and FETs are voltage-operated devices. So not all the same logic or visualization will work when talking about transistor stuff.

Not sure why, but I just thought ac,...it must be reversing current through the tube. Then after reading all this I realized how silly that was.

Not silly, because that is how a.c. is described when it is by itself. The mid-point of the a.c. cycle is generally assumed to be 0v (or zero current if you're thinking in terms of a sine wave of current), so the positive-going portion is current in one direction, while the negative-going portion is current in the other direction. This is all true if the mid-point of the cycle is 0v.

But with a.c. superimposed on d.c., the midpoint of the cycle is not 0v (or zero current). Instead, it is the value of the d.c. voltage or current. So the positive-going half of the cycle adds to the existing voltage or current, and when the a.c. changes direction (if it was considered on its own), it is now subtracting from the existing d.c. voltage or current.

Ground is not a magic hole to dump electrons. It's really just a reference point we choose to call 0v. If you measure from ground to another point at ground potential, you see 0v on your meter because the two points are at the same voltage.

Pretend for a moment you could measure the tube current, but you set your reference equal to the idle current or d.c. through the tube. Now when an a.c. signal is applied to the tube, the resulting a.c. would be measured as being + and - your starting reference. If would look like you were expecting (current changing direction in the tube), because you changed the reference point.

The problem with making an experiment out of this is that your meter will automatically interpret the negative half of the current fluctuation in the same way as it does the positive half, so you wouldn't see a meter reading changing from + to -.

You could see the effect on a scope, but what would happen is you see the exact same waveform, but the whole wave is shifting higher or lower on the screen. You would come to the conclusion that the shift up/down was due to the d.c. component, and all about what you choose as a reference.

I was gonna go on about how you could build an amp with a -400vdc power supply instead of a +400vdc power supply, but I didn't want to confuse anyone. The whole amp looks essentially the same, but "ground" is now the line you drew at first as being your B+ rail. You'd have to move the dropping resistors of the power supply down to what is the ground line in a regular amp. As far as the tubes are concerned, nothing has changed; they operate the same way they always did, we just chose to call a different point in the circuit 0v, and now everything is negative compared to it.

[JPK]

That's good stuff. 
  
Re-posted from above:
HBP, I have a question. I think I finally realized why the B+ is attached to the center tap of the primary on my P-P OT. Is half the transformer like the plate load resistor for 1 power tube, the other half is for the other power tube? What creates actual final speaker volume is the voltage dropped through the OT primary halves created by the current out of the power tubes? Or is that not right?

[HotBluePlates]

In a  nutshell, yes.

But a coil is the opposite of a cap; looks like a wire to d.c., but a large impedance to increasing frequency a.c.

Rather than use a resistor in the plate circuit of the power tubes, you use a coil (part of the transformer).

Primary side is isolated from secondary side so they can be at different d.c. voltages. Use of a coil reduces waste heat (voltage drop*tube current) that would be created if the tube's current had to move through a plate load resistor.

What creates actual final speaker volume is the voltage dropped through the OT primary halves created by the current out of the power tubes?

You might look at it that way, as in voltage variation at the plate caused when tube current interacts with OT primary impedance at the freqeuncy range of interest. Then the voltage variation developed at the plate is passed from primary to secondary by inductive coupling, and transformed to a new value by the turns ratio of the transformer.

You might also look at it as the OT primary presenting a load for the tube to work against, and current variation in the primary being coupled to the secondary while current is stepped up (at the same time voltage is stepped down) for use in the speaker.

Two sides of the same coin, existing at the same time.

[JPK]

Thanks, I have a much better understanding of what's going on due to this thread.

[Gary_S]

I have a question about resistors and their ratings and how they function: Say for instance the plate load resistor is a 100k one now if i increase that to say 200k will that result in more overall gain from the stage? I thought that the higher rating a resistor was the more it would attenuate the signal and the gain would be less? But i'm probably viewing it back to front or something.

I know Merlin says in the book that increasing the size of the load resistor will usually increase gain whereas reducing the size of it will increase current. Is that right?

[HotBluePlates]

It depends...

Do you know how voltage dividers and volume controls work?

[Gary_S]

Not really HBP i know that if you have two resistors in parallel then the voltage has another route to go through instead of just the one like if you had two resistors in series. Is that what you mean?

[HotBluePlates]

No, they will be series resistors. See the diagram below.




I will keep this very simple and specific, to help you see how gain of a preamp triode is impacted by the plate load.

A voltage divider is two (or more) resistors in series, with a voltage input and voltage output. The end of the divider opposite the input is ground, which is common to both the input and output circuits.

The input voltage causes a current to flow through the resistors. For a series circuit, the current through each resistor is the same, because there's only one path for it to follow.

Because ohm's law (voltage = current * resistance) informs us that there will be a voltage drop across a resistor in proportion to the current passing through it, and because in our divider the current in each resistor is the same, then the input voltage is divided between the two resistors in proportion to their resistance.

Example:
-  Say the input voltage is 10v, the upper resistor R1 is 9kΩ and the lower resistor R2 is 1kΩ. Divide the input voltage by the total series resistance to find the current through the divider; 10v/(9kΩ+1kΩ) = 10v/10kΩ = 1mA.
-  Find the voltage across each resistor using ohm's law (1mA = 0.001A); 9kΩ * 1mA = 9v and 1kΩ * 1mA = 1v.
-  So we have 9v across R1 and 1v across R2, so with these resistors, Vout is 1v.

You should see that the divider took an input voltage and created an output voltage in proportion to the ratio of the resistances of R2 and R1.

There is a shortcut method to calculate the output voltage without figuring the divider current (it uses only the values of the resistances in the divider), but we'll wait for now until you absorb the basic concept below.

Any volume control in your amp is also a voltage divider. Imagine Vout is the wiper of the volume control, and Vin is the input from the previous tube.

Let's say you have a 1MΩ volume pot. If you turn the volume all the way up, the wiper (Vout) moves all the way up to Vin, R1 is 0Ω and R2 is 1MΩ. Because the upper resistor R1 is 0Ω, there is no voltage dropped across it, and all of the input gets passed to the output.

If you turn the volume all the way down, the wiper (Vout) moves all the way down to ground, R1 is 1MΩ and R2 is 0Ω. Because the lower resistor R2 is 0Ω, there is no voltage dropped across it, and there is no voltage difference between Vout and ground. Voltage output is zero; the volume is "off."

You could reach the conclusion that voltage output is proportional to R2 in the diagram. When it was equal to the total value of the divider (volume pot all the way up), there is no voltage division and output equals input. When R2 is extremely small or zero (volume pot all the way down), all voltage is dropped across R1 and there is zero output.

[Gary_S]

Thanks HBP it makes sense now reading it all!

Whoever dreamed up the idea of how conventional current flows as regards real electron flows should be put on trial for confusing electronics novices 

Also reading an electronics book today they were talking about batteries and they said the negative pole in the battery is the anode and the positive pole is the cathode!  this kinda throws you out when you're used to reading up on tubes 

Thanks for your patience in answering our questions 

[HotBluePlates]

Once you have a good handle on the information above, continue on below.

Every triode has an amplification factor based on its internal geometry. In a perfect world, your 12AX7 would exhibit a voltage gain of 100 because its mu (amplification factor) is 100.

The problem is, we need a plate load resistor to convert tube current variation into a voltage variation that we can couple to the next stage with a cap. The plate load resistor works against the internal plate resistance of the tube to form a voltage divider, and reduces the exhibited gain of the circuit below that which the tube's mu would suggest.

The plate load resistor is R2 of our previous diagram, the internal plate resistance is R1, and "ground" is the filter cap positive terminal feeding the stage.

A helpful fiction to visualize what's happening is to pretend the tube amplifies an input voltage by the amount of its mu. Then this resulting voltage becomes the input to a voltage divider of internal plate resistance (R1) and plate load resistor (R2).

If the plate load resistor is smallish compared to the internal resistance of the tube, voltage output is very much reduced, and the tube exhibits an output much lower than you'd predict by looking at the tube's mu. If you could make the plate load resistance infinitely large, there would be no voltage division of our pretend output signal, and the tube would exhibit a gain equal to its mu.

The problem is how do you get tube current to pass through an infinite resistance, which is like an open circuit? Let's say you solve this problem by picking some resistance that is very large but not infinite. Let's say you also determined for some reason that you want your tube to idle at 1mA, and you figure a 5MΩ resistor ought to be big enough to make the tube's internal plate resistance look "like nothing" (maximizing R2).

5MΩ * 0.001A = 5,000v (!)

To use a very, very large plate load to maximize "R2" of our voltage divider and get a tube voltage gain equal to mu, we would have to use absurdly high supply voltages to compensate for the plate load.

I said above that "for some reason" you had determined you wanted a tube current of 1mA. You may get clever and look at our last statement and say, "Well, what if I use a very big plate load and choose not to stick to 1mA of plate current by using the lower supply voltage I already have?"

Simply put, the tube's internal resistance is lower at high plate current, and higher at low plate current. This makes sense if you think of the tube itself as a variable resistor. If something occurs to cause the tube to pass more current, this is the same as the tube looking like a lower resistance.

Combining these ideas:
-  You have an amp stage with a given supply voltage, and want to get "more gain" out of it, so you increase the plate load resistor value skewing the "voltage divider" for more output.
-  The bigger plate load with the existing supply voltage reduces tube current which increases the internal tube resistance, skewing the "voltage divider" for less output.

It's not hopeless though, because you have a cathode resistor to set tube bias and idle current, so when you increase the plate load, you could tinker the cathode resistor to restore the tube current to the previous value.

That said, relatively small changes of plate load resistance do follow the general rule you asked about. However, there are the competing factors I mentioned, so designing a tube stage for a desired amount of gain is all about balancing trade-offs. If you carry the idea of bigger plate load = more gain too far, you get diminishing returns and runs into the competing factors.

We haven't even talked about the fact that voltage gain is not the same as distortion, or that you might go to a lot of trouble to ensure more gain only to find out you run into some other limit of the tube (plate current cutoff or current saturation due to driving the grid positive of the cathode). The net result is that we usually accept the competing factors and settle for a voltage gain of 50-70 from a 12AX7.

I'll acknowledge here there are advanced methods to wring every last bit of gain from a preamp tube, pushing a 12AX7's exhibited gain closer to 100, but that either these are beyond the scope of this discussion of the basics or that the means of implementing the advanced method is wasteful of tubes (or that even more gain can be had from typical tube stages in cascade).

[Gary_S]

Thanks HBP this is all really cool stuff, and my fervent hope here is that there are other people just learning the technical details who are studying up on this and learning alongside me. This material is invaluable for someone who has just been a parts changer in their amp experience, but wants to go a step up and really understand things more fluently as regards tube amplification and how it works.

One thing that has me still scratching my head a bit is the original example with the two resistors as the voltage divider: input voltage is 10V we're getting 9v dropped on the first resistor, then 1v on the second resistor going to ground. So that's the 10v input accounted for. In my beginners ignorance i thought that would have meant there was no Vout because the whole 10 volts was already eaten up by the two resistors. Although i'm also thinking that after the current go's through the first resistor then it always likes to take the path of least resistance so i can see how it would prefer to head to the Vout.

[JPK]

Thanks HBP this is all really cool stuff, and my fervent hope here is that there are other people just learning the technical details who are studying up on this and learning alongside me. This material is invaluable for someone who has just been a parts changer in their amp experience, but wants to go a step up and really understand things more fluently as regards tube amplification and how it works.

Funny, I just got through reading HBP's last post and was about to post exactly what you said above and saw that there was a new post...so I went back and did a refresh. Yes keep asking questions because I'm here reading everything. I was going to say thanks for asking all these questions. I'm learning a lot here, and I'm sure there are other "lurkers" too.

[HotBluePlates]

One thing that has me still scratching my head a bit is the original example with the two resistors as the voltage divider: input voltage is 10V we're getting 9v dropped on the first resistor, then 1v on the second resistor going to ground. So that's the 10v input accounted for. In my beginners ignorance i thought that would have meant there was no Vout because the whole 10 volts was already eaten up by the two resistors.

Assume the Vin is d.c. for a moment. Assume the Vout terminal is not connected to anything else outside the circuit fragment as-drawn.

You put your meter's black lead on the ground terminal, then have 3 choices of where to place your red lead: ground, Vout or Vin.

If you put your red lead on Vout, you measure 1v. If you put your red lead on Vin, you measure 10v, same as the input voltage.

Now move your black lead to Vin. You have 3 choices of where to place your red lead: Vin, Vout or ground.

Now when you put your red lead on Vin, you measure 0v. When you put your red lead on Vout, you measure -9v. When you put your red lead to ground, you measure -10v.

The two cases are not contradictions; they are looking at the same phenomenon from different perspectives. When we say "voltage drop" we don't mean voltage was somehow thrown away. We mean the resistor (and ohm's law) caused a portion of the voltage to be dropped across the resistor.

So in our second case, where you measured from Vin to Vout, you can see that 9v was dropped across R1, as indicated by the -9v reading. There still exists 10v across the entire circuit, 9v across R1 and 1v across R2.

At this point, I think it would be very beneficial for you to read NEETS Modules 1 and 2 (you can skip the stuff on atoms in the beginning of Module 1 if you like). What I think will help you best is to get basic electrical concepts solidified, as even the most thorough writer assumes the reader has some fundamental knowledge... otherwise the book becomes extremely enormous.

[Gary_S]

Thanks for the link HBP much appreciated i'll be sure to explore that as well as the books i'm looking at just now.

[Gary_S]

Thanks HBP this is all really cool stuff, and my fervent hope here is that there are other people just learning the technical details who are studying up on this and learning alongside me. This material is invaluable for someone who has just been a parts changer in their amp experience, but wants to go a step up and really understand things more fluently as regards tube amplification and how it works.

Funny, I just got through reading HBP's last post and was about to post exactly what you said above and saw that there was a new post...so I went back and did a refresh. Yes keep asking questions because I'm here reading everything. I was going to say thanks for asking all these questions. I'm learning a lot here, and I'm sure there are other "lurkers" too.

Thanks JPK. I meant to put your name in that post and say; i hope that as well as myself and JPK  others are studying this and learning .

But after i'd typed it out i hit the button and posted it in!

Yes; before, i was a parts changer guy only. What i mean by that is that i could do cap jobs, replacing filter caps, resistors, retubing, biasing and stuff like that, but i really knew nothing about how the circuit worked technically. It's ok being like that for folks that are ok with that, but i get curious about everything i enjoy and want to know the innards of how it all works. I enjoy learning it and learning by my mistakes as well  

Keep on tuning in here we have much we can learn  

[PRR]

> they said the negative pole in the battery is the anode and the positive pole is the cathode!

The battery delivers power.

The tube absorbs power.

Battery and tube are in a *loop*.

Just like in audio, we connect outputs to inputs, we connect cathodes to anodes.

"In a device which consumes power, the cathode is negative, and in a device which provides power, the cathode is positive"
http://en.wikipedia.org/wiki/Cathode

[Gary_S]

Thanks PRR, yeah it makes sense, but i'm so used to thinking of the cathode biasing of the preamp tube and the cathode being negative in relation to the plate.