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**Ray Ivers** · 06-23-2006, 02:59 AM

Paul,

Think of B+ as 'audio ground', since it's connected to ground through the filter & decoupling caps - so the plate and following grid resistor have their 'tops' connected together via the coupling cap, and their 'bottoms' each connected to a different AC ground point; in one case it's ground-ground, and in the other it's the AC ground of the B+ supply.

As the upstream side of the coupling cap charges, it pulls electrons to the downstream side by electrostatic attraction - so the net potential difference ("charge") across the cap stays pretty much the same. Note that in guitar amps, the coupling caps often are not "large enough" from the hi-fi (i.e., 20Hz - 20kHz) viewpoint of these articles, so at low frequencies a smaller coupling cap will not charge quickly enough to keep both sides at the same potential, leading to a reduction in low-frequency signal level.

Ray

**Enzo** · 06-23-2006, 03:31 AM

What Ray said.

You are right the resistors appear to be in series between B+ and ground, but at AC freqs, you have to consider what the impedance between those two points is. At audio, B+ and ground are essentially shorted together. Which makes the plate load resistor and the following grid resistor effectively in parallel.

For further science, look into the impedances of power supplies.

Also, if the coupling capacitor doesn't allow the charge across it to
change, how does it work ?

At audio freqs, the impedance of the cap is low. We want that. We want the signal to pass through the cap unimpeded, while we block the DC. When signal is absent, the DC voltages are stable. Let's say 200V on the plate and 0v on the grid. There is 200vDC across the cap. The DC impedance of a cap is very large.

Now the plate voltage starts to wiggle up and down as we apply signal. The wiggle is the signal at the plate. We want that same signal to appear at the following grid, without the DC of course, and the cap allows this. When the plate signal voltage goes up 5 volts, the grid voltage also goes up 5v. SO the plate at this instant is 205 volts, and the grid is +5v at this instant. There is still the same 200vDC across the cap. WHen the signal draws that plate down to 195 volts, then the following grid will likewise go to -5v. At that instant.

The circuit resistances are high, and the cap is large enough that it can maintain that 200v charge over the course of a cycle of the audio signal.

A typical couplign cap might be a 0.01uf cap. That is 10,000pf. The grid resistor keeps the lower end of the cap at ground level, and the plate load charges the top of the cap to whatever the plate is - our 200v here. If we instantly jacked the plate up to 250v and left it, there would be a momentary increase at the grid, but the grid resistor would slowly (relatively) allow the cap to charge up to the new 250v level. There would be 250v across the cap then.

But in audio, we don't leave the plate at 50v higher than we started. The signal goes up to a new peak and right back down. If the cap is large enough, this complete cycle is completed before the actual charge across the cap can change much.

If the cap were tiny, like say 100pf, the same thing happens, but much faster. That same large positive peak can charge the tiny cap much faster, so fast in fact that the cap settles to the new charge before the cycle of the signal can complete. So at the grid end, we never see the whole signal, just the initial leading edge of the signal spike. The net effect of this is that short time transients - higher frequency transients - come through OK, but transients that take longer - lower frequency transients - don't. That is why a smaller cap won't pass as much lower frequency. The lower frequency signal would take longer to pass through the cap than it takes for the cap to change its charge.

**dai h.** · 06-23-2006, 05:20 AM

(look at the first stage of a 5F6-A Bassman)
I look at it sort of like this:

1)The first voltage loop goes through the hot (grid) wire, through the 1Meg then returns to source (towards pickup) through ground (and this positive and negative voltage--since it's AC) is sensed by the grid (which is a coil of wire along and around the cathode).

2)The output current (AC signal) loop goes from the plate of the triode up through the plate resistor, through the positive side of the 8uF filter/de-coupling cap, then the negative side, then through the 820ohm cathode R, and this AC current goes through the 250uF, then the cathode. The 8uF filter puts the B+ side of the plate R at AC ground. The plate resistance of the tube is in parallel with the 100k plate R as well as the 1Meg volume on the other side of the .02uF coupling cap (from an AC perspective). The output signal current flows through a resistance/impedance (the plate resistance and plate R in parallel) and creates a voltage drop. 100k plate R is at AC ground, 1Meg pot at DC ground. This signal is opposite polarity because (with AC you have a positive and negative half), and when the grid side goes positive, the plate side goes negative (and vice a versa). (This is not actual numbers but just to get the general idea), but say the grid side goes to +1V, then the plate side goes to -15V, then through zero, grid side to -1V then plate side to +15V (opposite polarity, and an amplification of the grid signal).

3) The AC voltage for the next stage is tapped off the .02uF output coupling cap, and is a facsimile of the above output current (AC signal) loop which goes from plate to .02uF cap through 1Meg volume back to ground then previous cathode via the 250uF cathode bypass cap. This voltage is sensed by the next stage's grid--same as 1).

so sort of like voltage loop, current loop, then voltage loop again. Small power (pickup signal) controls a bigger power, gets bigger voltage (amplification). It makes more sense if you read a bit how a tube basically works IMO.

**Paul P** · 06-23-2006, 04:47 PM

Thanks for the clarifications. I see that I was missing a large part of the
puzzle. For some reason I imagined the DC part remaining stable while
the AC signal wound it's merry way from grid to plate to following grid and
so on, keeping within the bounds of that line. I now see that the AC is
radiating all over the place

It's going to take me a while to digest the details of what you guys have
just provided. While things are now clearer, they're also a lot more
complicated !

Back to the drawing board...

**MBSetzer** · 06-24-2006, 12:48 AM

there's good replies already here which if I had seen I probably wouldn't have written as much, but here's my 2cents,

Well, I'm going to assume a couple of things and see if my explanation makes sense.

First assume you are talking about a 470K grid resistor which is a proper *grid load* resistor, one which appears between the grid of the following tube, and ground. Not a *grid stopper* which exists between the coupling cap and its following grid.

Also assume Crowhurst would not have used the same terminology or example if it appeared much later in the text, after more technical background becomes developed more accurate verbiage could be assumed to be readily grasped.

Now when considering the *loading down* effect due to different levels of impedance which appear at various nodes along the signal path, you have to figuratively take the point of view of a sine wave sitting right there at that node. At any particular crossroads in the schematic, there sometimes exists a path or more than one path to ground through which your signal will flow whenever it appears at that node. When too much signal flows to ground, like through impedances that are lower than they should be, that could be considered signal waste (although sometimes it is quite intentional to develop more signal amplitude than is needed just so you can dump some of it to ground or attenuate it some other way).

So naturally if the only thing in your circuit was an input jack wired straight to a grid, and the grid has some resistance to ground soldered to it (like a megohm or something), whatever signal appears at the grid will also be flowing straight through the 1Meg to ground. Well, a little of the signal is lost but not a significant amount, a million ohms is a lot so not much signal is *wasted* in this typical example.
At the same time whatever signal is presently remaining there is of course available to the grid and drives the tube like its supposed to.
Now inside the tube, the grid is a piece of metal which is not connected to any other conductors, so there is no easy path for signal at this node to get to ground (or anywhere else) through the tube itself, (unless you drive into grid current, not going there now) but some of the signal will be doing it anyway. Typical grid input impedances are many megohms so that's orders of magnitude less significant signal being consumed by the grid connection compared to the (essential for stability) 1meg grid load.

But we have to draw the line somewhere, and traditionally you just consider the grid impedance of the tube to be infinite, especially since it was beyond the range of most routine ohm-meters for most of the decades when tubes were most popular. So it doesn't come into the calculations.

Now at our initial simplified input node we have not plugged in a guitar, no we are theoretically driving this grid with an imaginary signal source having zero (output) impedance its own self. Or in reality a laboratory signal generator with very low impedance but not quite zero.
What would ideally happen then if you (like at 1KHz) set the sig gen for 1V P-P output by itself, then after you connect it to the grid, the sig gen will still be putting out the full 1V P-P even though some of that signal is being lost through the 1meg grid load. Since the sig gen has such low output impedance, that means you could easily choose lots lower ohm values for the grid load and the sig gen would not be loaded down, still putting out 1V P-P. Its a big lab instrument and as you lower the grid load resistance you demand more current from the lab supply to keep the voltage at 1V P-P on the grid, but it does so automatically up to its rated specifications.

But there is so much signal impedance in a guitar pickup (way more than the measured DC ohms) that it is just the opposite of a low impedance signal source. So then when you reduce the ohms of a grid load, it noticably reduces the amplitude of the signal appearing at the grid node.

So now you see two different ways to look at the same node, but in the guitar vs. lab source situation you really are changing a component which appears at the node, so a dependent difference is not unexpected.
And it should be clear why sometimes you can neglect an *escape path* for signal in case it is truly insignificant or nearly enough so not to affect the math at the required number of significant figures. Also we will be neglecting an escape path (almost by assuming there is an assumed lab source - ha ha) through the source sometimes for theoretical instructional purposes.
In my example input triode wiring, there is the typical 100K plate load resistor which will be considered in conjunction with the following gain stage.

So . . .

Now move on to the second stage, another 12AX7 triode with nominal 470K grid load this time, this would be the stage you are concerned with following the above input stage.
The size of the incoming coupling cap is specified precisely as *large enough*.
How large was it?
*large enough that it does not allow the charge across it to change
during the audio fluctuation, and accordingly the audio signal voltage
at the end of the grid resistor swings with the voltage at the plate
of the stage we are considering.*
Yikes, that's not clear as mud. Must be because it appears so early in the discussion.

What this means to me is that the cap has enough uF to not roll off any of the bass freqencies. That's it.

So the cap imparts none of its own signal impedance at any frequency of interest (which it would be doing if it was rolling off anything in the audio band), and you can then simply neglect its action for this discussion even though it is a component in series with the signal path, it will not introduce any series or parallel impedance when looking for escape paths from different nodes. So the cap acts like a plain piece of wire rather than a frequency-dependent resistor when it comes to passing AC signal.

Just another step to make the math easier, like neglecting any escape of signal through the grid itself.

OK, lets put it all together.
We remove the input jack, grid load resistor, and triode only of my above example, resulting in the lab sig gen in place of the triode plate, putting its sine wave directly into the 100K plate resistor that was there (same one mentioned by Crowhurst) along with the above coupling cap leading to the remaining triode in question, with its attached 470K grid load.

Theoreticaly now, don't really hook up just any sig gen to any plate pin carrying B+ without a *large enough* or *high enough voltage rating* blocking cap if needed to protect it from the DC! Consult your instrument manual or ask here about things like this, these are tube amps.

Anyway, now let's just sit there at that node, at the output of the sig gen itself. And see where you, as an audio signal, could go from here. Well, there's only two ways to go, right through the 100K or right through the 470K. The coupling cap counts as a zero ohm piece of wire for AC signals, so it adds nothing to the 470K it is in series with. Any significant escape through the grid itself is *impossible* since we have branded it a dead-end. And the lab supply itself is assumed to be theroetically ideal so no losses through it.

So with the only two choices being the 100K and the 470K, the signal chooses both, with proportional amounts of AC signal current passing through each resistor depending on their obvious ohm ratio. Since both resistances are effectively both connected to this same node at this same point, they perform in parallel to each other when it comes to signal loss (or preservation) FROM THIS NODE'S POINT OF VIEW. So that's the math you use when you figure values at this node.

Whew. That would be it, but . . .

Reality is not as simple as this elementary example from Crowhurst would indicate, I think he is trying to emphasize only the parallelness of the obvious resistors and I have tried to concentrate on that up unitl now, in a round about way which may be useful hopefully.

But when we have a real 12AX7 triode input stage back in place of the sig gen, then we no longer have a theoretically ideal signal source to drive the second triode in question. So there is the third escape path this time right through the vacuum of the first triode when it is in place normally connected to the node, we have been talking about the plate node of this triode the whole time of course.
And the plate is different than the grid, there is still no *easy* way for signal to be lost here since this electrode also juts into space without connecting to any other conductors itself. But it is a lot easier for signal to be lost compared to the almost infinite control grid impedance, the plate impedance of a 12AX7 runs from 50K to 100K under typical operating conditions, lots of people use the 62.5K value from the data sheet when they are doing calculations. That's why when they are talking about the property of the tube itself they call this value *plate resistance*.
So the plate resistance actually also goes in parallel with the 100K and 470K which all lead away from this same node.
Well, for any signal that appears at the node, but under normal conditions we will not be putting an outside signal into that node and experiencing losses in all these 3 directions. Instead, under normal operation the signal at this node comes from the plate of the first triode itself and has already passed *through* the plate resistance of that triode before showing up on the node. You can think of the signal from this triode as emanating mysteriously from deep space within the tube, and having to pass through 62.5Kohms before it shows up at the plate. Now lets figure the parallel resistance of the 100K and the 470K since this signal will flow in parallel from the node through both resistors on its path away from the node, this works out to be 82.5K in parallel.

So whatever AC signal voltage you measure at the plate of the first triode, it is already the result of all these forces at work, and (with a big enough coupling cap) you get the same reading at the grid of the second triode, which drives that triode. There is basically the ultimate signal source in space inside the first triode, it has passed through 62.5K and beyond that is now powering a 470K and a 100K as it passes toward DC ground (Earth) through the 470K and toward AC ground (massive low-AC-impedance filter caps are the only thing separating a power supply node from earth, and these caps are intentionally transparent to AC, so an adequately filtered B+ node is also an AC ground) through the 100K.

If you change the 100K it changes the gain so lets leave it. With the 470K in parallel as an additional exit path from the node, its a little easier than 100K worth of ohms for signal to get away through, and be reduced in amplitude by the time it reaches the following grid, namely 82.5K calculated. That's in the same ball park as the plate resistance which was estimated at 62.5K to begin with, it would not be unusual for this 82.5K & 62.5K to actually be more equal lots of times.
This means that as a rule of thumb for a 12AX7, you have already lost about half your ultimate source amplitude before it reaches the following grid under these conditions, and this is considered good.
Regardless, if you reduce the 470K to 220K, then the former 82.5K becomes 68.75K so that then results in 83percent of the impedance you were getting with the 470K there.
You really need to include the parallel 100K in calculations like this otherwise you could end up thinking the 220K would reduce the impedance by over 50percent or something.

The plate resistance comes into play when you figure that the grid node is about the mid-point of a signal voltage divider occurring between the mysterious vacuum inside the triode vs. ground as signal passes through the plate resistance and the grid load resistor on its way to ground. For instance if the plate resistance was 80K and there was effectively also about 80K of parallel resistances considering both the 100K & 470K, then a 100V P-P mysterious source signal will only be 50V P-P at the grid of the next triode. Or at the plate of the generating triode which is the first place you could measure it.
You wouldn't be able to directly measure the internal 100V P-P but you could still tell its there for instance by reducing the ohms of the 80K parallel resistance and measuring the resulting drop in amplitude of the former 50V P-P signal at the node.

With something like a 12AU7 having much lower ohms of internal plate resistance than a 12AX7, when you reduce the ohms of a following grid resistor it makes much less relative reduction in grid signal to that tube until you drop to about equal or below the plate resistance of the plate driving it, considering of course the total calculated isolation of that node from ground in ohms, including the paralel ohms of the plate resistor on the 12AU7.

Mike

**Paul P** · 07-05-2006, 04:54 PM

Great book

I just printed out (all 122 pages of) Norman Crowhurst's book Basic Audio
Vol.2 which can be found at :

http://www.pmillett.com/technical_books_online.htm

This book has all the details missing from the magazine articles. Lots of
pictures and diagrams on every page, very easy to read.

I was unable to find a paper copy of volume 2 anywhere on the net. I
did locate copies of volumes 1 and 3.

The three volumes are at :

http://www.pmillett.com/Books/crowhurst_basic_1.pdf
http://www.pmillett.com/Books/crowhurst_basic_2.pdf
http://www.pmillett.com/Books/crowhurst_basic_3.pdf

There are many other books at the above site.

**Ray Ivers** · 07-05-2006, 08:07 PM

Paul,

Those are truly outstanding links - thanks!

Ray

**dai h.** · 07-07-2006, 03:56 PM

Ray, I'm surprised you haven't seen that one. There are a bunch of cool ones on the right hand side links on the old ampage site like this one (starting to read the navy electronics course--looks like a good one to start learning about tubes) :

http://www.tech-systems-labs.com/books.htm

http://www.firebottle.com/fireforum/...sub=garm&mode=

**Ray Ivers** · 07-07-2006, 04:19 PM

Dai,

Yeah, I went to enter the links in my address book and found they were already there.

I hadn't really downloaded anything yet though, so I went ahead and grabbed a bunch of stuff I'll probably never get to.

I had about a third of the titles in book form already, all of which I managed to plow through at various times in the past. I find it so much easier to actually sit down and read a book than computer files off a screen - I really should take these to a printer and have bound copies made if I plan on actually getting them read, or maybe try and find them on eBay (a great source for books! - I've found a bunch of Crowhurst plus various other authors there).

Ray

**dai h.** · 07-07-2006, 06:12 PM

yes, I got one of those big chunky staplers for that purpose. Advanced toilet studies...

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