OK, we all know about the post-PI cathode follower for increased drive current.....

[Chuck H]

There is no current drive or very little. Tubes are voltage source devices not current devices. There is no current drive and there is no gain. The voltage that was present at the plate gets transformed to the Cathode and the impedance changes from high to low but current stays the same what very little ua's there is.

Certainly there are current differences in a tube under operating conditions. But thats not the discussion. If as you say a tubes current is constant then the op's consideration of parallel triodes does indeed double the current capabilities of the tube. But it's not current being drawn from the tube that is the issue here because 1) that doesn't happen, and 2) AC coupling limits the tubes current due to bias shifts caused by the time constant of the grid circuit load. Still:

Whether that happens or not. The problem is that when a tube is pushed into conduction it tries to draw current through it's grid as well as it's cathode. When this happens the grid can load the coupling capacitor causing a negative bias shift unless 1) the time constant of the circuit load allows the capacitor to discharge faster than the loading effect, or 2) the grid circuit of the tube is not AC coupled and is of suitably low impedance to allow current required by the grid. This is why amps that run in AB2 (requiring grid current) aren't AC coupled.

This is as much as I know about it so don't ask for any more technical information. But it's enough to know this much most of the time. Tubes certainly are variable current devices as well as variable voltage devices. In fact they can be operated as either or a balance of both.

[redelephant]

Basically a Cascode follower aka a White Cascode and there are some other variations like boot strapping to get a tad more drive out of it but it's still a follower with best unity gain.

Ure right there is less than unity voltage gain, but there is an increase in potential current gain. Driving grids requires current, and CF can do that.

[defaced]

AK, you're looking at the wrong side of the system. The gain is not to be had from the cathode follower, we all agree it's a "unity" stage. What is to be had is the ability to drive the power tube grid positive and transition operation from AB1 (no grid current flows) to AB2 where grid current does flow and THIS is what causes the voltage gain. Since typical circuits are AC coupled here, and the capacitors block the DC current needed for positive grid operation, you need some way to deliver the current: direct coupled buffers, AKA cathode followers are one of many ways to do that. This is the best explanation of what happens that I've been able to find in a short amount of time. I'm sure there's more our there, but I don't' have the time right now to dig: EL34 in class AB2 triode mode push-pull? - diyAudio

Edit: Seems Chuck got a good reply in while I was screwing around with mine.

[Amp Kat]

You guys can keep talking till the cows come home but the CF is not driving crap. It is basically passing the signal on and basically impeading the signal. The whole CF scenario is blown way out of proportion and you can't drive anything without drive. There is no current gain because there was never any to begin with. Stick em in there if you want and knock yourselves silly if you like. I do use them but not for driving tone stacks. I use them to supress noise at the 1st stage and they work well for it.

[Amp Kat]

Ure right there is less than unity voltage gain, but there is an increase in potential current gain. Driving grids requires current, and CF can do that.

Where is there current gain ? I'm not seeing any at all without doing serious bootstrapping techniques that are very complex and out of the ordinary.

[Amp Kat]

AK, you're looking at the wrong side of the system. The gain is not to be had from the cathode follower, we all agree it's a "unity" stage. What is to be had is the ability to drive the power tube grid positive and transition operation from AB1 (no grid current flows) to AB2 where grid current does flow and THIS is what causes the voltage gain. Since typical circuits are AC coupled here, and the capacitors block the DC current needed for positive grid operation, you need some way to deliver the current: direct coupled buffers, AKA cathode followers are one of many ways to do that. This is the best explanation of what happens that I've been able to find in a short amount of time. I'm sure there's more our there, but I don't' have the time right now to dig: EL34 in class AB2 triode mode push-pull? - diyAudio

Edit: Seems Chuck got a good reply in while I was screwing around with mine.

Once again it's just coupling the signal not driving anything. There is nothing to drive with !

[redelephant]

All right. Oh, I gotta take back my apology. Seriously, get urself some EE books.

[defaced]

Once again it's just coupling the signal not driving anything. There is nothing to drive with !

Ok. This can go the other way. Do you have any primary literature or publications to back up your rebuttal?

[Steve Conner]

The purpose of the cathode followers is to drive the power tube grids positive. Drive AC coupled through a capacitor can't do that, the capacitor charges up and prevents it.

It's the driving of the grids positive that produces more power. And it's the DC coupling that allows that, not the presence of a CF as such. If you could figure out how to connect the power tube grids directly to the PI plates, and you used a beefy enough PI tube with low-valued plate resistors, then you could drive them into AB2 just the same.

In a preamp, people sometimes say that the cathode follower in a Marshall gives more gain. This is sort of true. A 12AX7 stage has an output impedance at the plate of about 56k. If you load that with 100k (a reasonable load for a tone stack at some settings) then you lose one-third of the signal voltage. A CF has a much lower output impedance, less than 1k for a 12AX7. So you get that third back, for small signals at least. The large-signal behaviour of CFs is more complicated than that: while one might increase the amplification of small signals by a third, it might actually decrease the available headroom before clipping.

In a really big power amp with lots of tubes in parallel, you need CFs to drive the low-valued bias feed resistors that are mandated to prevent thermal runaway. In this case, a big PI tube or two sets of paralleled triodes would work the same.

[Chuck H]

Three cheers

I alluded to the impedance of the PI circuit being the issue but my post was warped by the subject of current. As in, if a tube is to draw current through it's grid the grid circuit needs to have a low enough impedance to allow it. Otherwise the grid loads up and the bias shifts.

This is where the confusion is I think. You do need a driver with enough current avilable to get into AB2, but you need a lower impedance circuit feeding the grid so the grid can draw current. But what if the current available from the PI isn't enough??? Same as the AC coupled circuit. The bias shifts. So, it doesn't matter whether it's a cathode follower, a MOSFET source follower or an obscene amount of gain coupled to a very small load resistor. The point is to allow the grid to draw current. Since this must be coupled with voltage gain to drive a tube into this condition it becomes complicated as compared to normal guitar amps.

As for a cathode follower feeding a tone stack not driving anything, well, AK, I guess it's your right to interpret things however you like but I think your just being difficult. "driving" a tone stack is common terminology. Acclimate yourself and quit trying to be contrary for any reason you can find. And as far as nothing to be gained, I have to side with Steve on this. The whole point of the cathode follower in the 5F6A Bassman amp was to reduce losses through the tone stack. So there is a gain benefit. Certainly not as much as if Fender had just used that triode as another cascade gain stage but hey, their amp, their rules. And besides that it also sounds different. In a typical coupling circuit the phenomenon of impedance rising with frequency is actually pretty bad. A cathode follower is less prone to this so it does sound different.

How does a cathode follower used at the first stage supress noise?

[octal]

Three cheers

I alluded to the impedance of the PI circuit being the issue but my post was warped by the subject of current. As in, if a tube is to draw current through it's grid the grid circuit needs to have a low enough impedance to allow it. Otherwise the grid loads up and the bias shifts.

This is where the confusion is I think. You do need a driver with enough current avilable to get into AB2, but you need a lower impedance circuit feeding the grid so the grid can draw current. But what if the current available from the PI isn't enough??? Same as the AC coupled circuit. The bias shifts. So, it doesn't matter whether it's a cathode follower, a MOSFET source follower or an obscene amount of gain coupled to a very small load resistor. The point is to allow the grid to draw current.

Chuck,
We've mostly agreed throughout this thread, and perhaps I'm being overly pedantic here, but I think bias shift is strictly a product of AC coupling to the grids of the output tubes. (Assuming a non-cathode biased amp.) If a DC coupled driver circuit "runs out of" current on the positive swing as the output tube grids conduct, you certainly could get a form of non-linearity in the output, but it would look and sound different from bias shift. It wouldn't cause that funny "cored out" effect around the center crossing of a sinewave.
This is a GROSSLY simplified explanation of bias shift in an AC coupled circuit which omits many factors (time constants, pull up current limitations of the PI) Let's say that with no signal, the plates of the PI tube are resting at +175DC, the grid of the output tubes are sitting at -35, so there's 210 DC across the coupling cap. A sinewave comes into the PI and tries to drive the plate voltage up 40V. The plate swings to +215 but the grid of the power tube becomes essentially a short as the grid voltage hits 0. The coupling cap now charges so it has 215 volts across it. As the sinewave descends from its peak, and crosses the zero line the plate of the PI will again be sitting at +175, but the grid of the output tube will be sitting at +175 the cap still has 215 across it so the output tube grid rests at =-40 the bias is shifted 5v more negative and will remain so until the "extra" charge bleeds off the coupling cap though the grid resistors. I'm guessing you knew all this.
If my understanding is correct, in a DC coupled circuit, there's no mechanism for this to happen... there's nothing that will shift the DC potential difference between the CFs and the grids of the output tubes, so therefore no bias shift.

Nathan

[Chuck H]

Never assume I know anything

The way I saw it, though it may be wrong, when a power tube grid is driven the tube will try to conduct correspondingly. If there aren't adequate electrons from the cathode to do this it falls on the grid to provide them since it is the gate (or valve). This is when a grid needs to be able to draw current. If it can't a negative charge can build up and stay until enough electrons are provided or the drive voltage reduces and the need for those electrons is gone. Similar to the way the voltage on a cathode bias rises more the harder the tube conducts. If this is the case then any grid circuit that doesn't have a low enough impedance or otherwise fails to provide the needed current would allow a bias shift whether it's AC coupled or not. Could be hogwash.

It's certainly easy enough to see your explaination in practise to some degree on many (if not most) guitar amps.

EDIT: Space charge has been mentioned. I should have mentioned it myself since it's part of the above theory. When the grid needs to draw current it is trying to release some of it's space charge (which will be replaced by electrons from the cathode). But it can't do this unless available current via a low impedance allows it.

[redelephant]

The hot cathode has a cloud of free electrons hovering around it. Those electrons create what's called the 'space charge' and are not attached to any nucleus, and b/c of electron's miniscule mass, can be accelerated and pulled away in a heartbeat.
As we know, the positive plate attracts electrons, but the much closer grid can control the electron flow by placing a negative charge between the space charge and the plate.
If the grids become positive, and even just a bit, some electrons will be attracted to the grid. As simple as that. The thin wires making up the grid holds no charge at all, simply attracts electrons that are negative in respect to itself (when the grid becomes positive in respect to the space charge).
Since current flow must have a path all around a circuit, the resistance in series with the circuit sets the amount of electrons that the grid can attract. If the resistance allows for 1mA, then the grid will go as much positive as 1mA requires...so forth...

The output power is of course only related to plate current, not grid current. We understand that almost all electrons pass thru the grids and right up to the much more positive as well as pysically larger plate. This electron flow to the plates continues to the plate in about the same ratio as the grids go positive, which is why there is an almost linear increase in power as the grids go positive. Actually some tubes draw relatively large amounts of grid current and in those there is a noticable change in plate current relative to grid voltage. Some tubes that are made for AB2 have almost no change in their curves as they draw grid current. The amount of grid current relative to plate current is still very small. Anyways...

A capacitor charges to a given voltage according to the current going into it. 1amp will charge 1farad to 1volt in 1 second. Man...I cannot remember the formula right now, but basically (Voltage over cap) = (amps/capacitance)*time. So the larger the coupling cap the less it will charge up given the same current and time. But it will also discharge slower.
1mA will charge a 10nF cap to 10000volts in one second! Obviously in our amps, the cap will only charge up to the difference between grid voltage and plate voltage of the preceding stage.
But u can AC couple and drive the grids if the cap is large enough.

Anyways, point being, the grids hold no charge, all the charge is in the coupling cap. So if we want to allow the cap to go positive and thus charge up, we must find some path for that charge when the grid goes negative (and at that point presents an open circuit). The grid resistor. But a grid resistor small enough to do that fast enough will be just as difficult to 'drive'. Using CF or any other type follower is the simplest way. The CF itself wont have its grids going positive (they can, but not in our amps) so the cap to the CF has no charge buildup. U can also to some degree have a powerful triode in a common cathode gain stage as a follower. It's gain will be so low it's called a common cathode follower, instead of common cathode gain stage. To me the best way to drive grids positive is using transformer coupling, say with a interstage transformer with a 3:1 ratio, driven by a medium power triode.

ok...enuff ramblings....sorry for taking off...even boring for me...

[Steve Conner]

I alluded to the impedance of the PI circuit being the issue but my post was warped by the subject of current. As in, if a tube is to draw current through it's grid the grid circuit needs to have a low enough impedance to allow it. Otherwise the grid loads up and the bias shifts.

This is where the confusion is I think. You do need a driver with enough current avilable to get into AB2, but you need a lower impedance circuit feeding the grid so the grid can draw current. But what if the current available from the PI isn't enough??? Same as the AC coupled circuit. The bias shifts.

Again this is an oversimplification. (Everything in electronics is.)

Impedance isn't necessarily a constant. As I suggested above by talking about small and large signals, it can vary with the signal amplitude. But it can also vary with frequency.

If you just want to "drive" a tone stack harder then AC coupled drive is fine. But if you want to drive power tube grids into AB2, you need a driver with low impedance at DC (aka, zero frequency) because the grids rectify the drive and draw a net DC current.

No AC coupled circuit will work in this respect, because no matter how big the capacitor and how beefy the drive, all capacitors have infinite impedance at zero frequency.

This is an oversimplification again: the capacitor might have infinite impedance at DC, but the bias feed resistor is in parallel with it. So for AB2 purposes the driving impedance of an AC coupled PI circuit is that of the bias feed resistor plus bias supply. You could get AB2 with AC coupling by making the drive very powerful and the bias feed resistors really small. The bias supply would also need to be shunt regulated, because this kind of AB2 drive forces current back the wrong way into it: you sometimes see this in tube RF amplifiers for ham radio.

How does a cathode follower used at the first stage supress noise?

It doesn't. But it shouldn't add much extra noise either, because the tube's internal noise is degenerated by the CF's internal feedback. Thermal noise from the grid leak resistor would have been there anyway, but the amp now has one more grid leak resistor, so maybe the noise floor goes up by 3dB.

In this case, maybe the first two stages contribute equally to the noise: a violation of the rule that the first stage determines the noise performance. Again, this is a popular oversimplification: in a guitar amp the noise floor is more likely to be set by the reverb recovery or FX loop return stage.

[Amp Kat]

Three cheers

I alluded to the impedance of the PI circuit being the issue but my post was warped by the subject of current. As in, if a tube is to draw current through it's grid the grid circuit needs to have a low enough impedance to allow it. Otherwise the grid loads up and the bias shifts.

This is where the confusion is I think. You do need a driver with enough current avilable to get into AB2, but you need a lower impedance circuit feeding the grid so the grid can draw current. But what if the current available from the PI isn't enough??? Same as the AC coupled circuit. The bias shifts. So, it doesn't matter whether it's a cathode follower, a MOSFET source follower or an obscene amount of gain coupled to a very small load resistor. The point is to allow the grid to draw current. Since this must be coupled with voltage gain to drive a tube into this condition it becomes complicated as compared to normal guitar amps.

As for a cathode follower feeding a tone stack not driving anything, well, AK, I guess it's your right to interpret things however you like but I think your just being difficult. "driving" a tone stack is common terminology. Acclimate yourself and quit trying to be contrary for any reason you can find. And as far as nothing to be gained, I have to side with Steve on this. The whole point of the cathode follower in the 5F6A Bassman amp was to reduce losses through the tone stack. So there is a gain benefit. Certainly not as much as if Fender had just used that triode as another cascade gain stage but hey, their amp, their rules. And besides that it also sounds different. In a typical coupling circuit the phenomenon of impedance rising with frequency is actually pretty bad. A cathode follower is less prone to this so it does sound different.

How does a cathode follower used at the first stage supress noise?

Because the CF has a low impedance it shunts the grid circuit of the next stage to ground. Since the Cathode load resistor is in parallel with the next stage grid leak resistor the cathode output impedance of the tube is the reciprocol of the tube gm which is around 600 to 700 ohms. Putting these values in parallel with the Rk = 100k & Rg=1M and it will shunt the external noise to ground.

As for the 5F6A Bassman amp yes they did use that circuit but the next version they removed the feedback resistor to make up for the lack of gain to warm it up. Later they got away from it all together because of it just wasn't doing justice not being a gain stage. Marshall however uses it continuosly and probably the best in the Marshall 30th Anniversary where they did a half-ass bootstrap and pulled the gain up close to unity.

Sorry for ranting about not liking the follower and I did make a mistake in saying it wasn't driving anything because in reality it really is just not in the way I see it. Certainly transformers and transistors are a much better approach of getting grid conductuction than the follower and if it weren't for a large negative voltage to the Cathodes which is where the driving current comes from I'm not sure it would be able to drive the grids positive. IMO the follower by itself isn't enough to drive them but it's low output impedance is a must as would a transformer would do much easier.

As for not liking the Cathode follower in normal designs I just don't see where all the hype is for driving capabilities and it's bandwidth is not close to a plate driven load but many still use them and will continue to use them not knowing what the hell is really going on or why they are using them. They do increase tube count though so if you want more tubes for nothing then grind away. Hope John isn't too pissed we diverted his thread as that wasn't my intent and I actually like his design and think if he bootstrapped it and sent say about-150 to the Cathodes of each tube it would work ok but not sure how much extra-current drive it would supply.