Help understanding cathode follower....

[hesamadman]

Hi all,


I have recently been trying to understand the cathode follower as I am installing my first fx loop in my next build. I have done a great deal of reading but I am still having a hard time understanding.


I see two parts of a triode in use. The first is setup normal with a plate resistor but the plate gets coupled directly to the grid of the next stage without a coupling cap. So this puts the B+ on the grid of the next stage. The next stage does not have a plate resistor but still has the B+ connected directly to the plate. It has a rather large cathode resistor in comparison to a normal stage. In the case of a 2204 tone stack cathode follower there is a 100k.


1st question. Is the first half triode in this configuration called the "volt amp"? I have heard this term a few times and not exactly understood what they meant.


2nd question. My assumption with a normal gain stage is that current flows from cathode to plate and through the plate resistor. Is current flowing from the cathode of the first half of the triode to the grid of the next half of triode and out of the cathode?

[shooter]

Read this,
http://www.valvewizard.co.uk/dccf.html

I had the same ?'s before I read it

[kagliostro]

And if you want, also this

http://www.valvewizard.co.uk/accf.html


Franco

[tubeswell]

I see two parts of a triode in use. The first is setup normal with a plate resistor but the plate gets coupled directly to the grid of the next stage without a coupling cap. So this puts the B+ on the grid of the next stage. The next stage does not have a plate resistor but still has the B+ connected directly to the plate. It has a rather large cathode resistor in comparison to a normal stage. In the case of a 2204 tone stack cathode follower there is a 100k.

You are referring to a DC-coupled pair. The first triode is a driver (normal inverting)  gain stage. The second triode is the cathode follower stage. The grid of the CF is indeed DC-coupled to the plate of the driver stage. This means the CF grid and driver plate are at the same potential. The cathode of the CF stage is virtually (also) at this same potential (as the grid), so the CF stage is biased really hot and there is a lot of grid current. The CF ends up 'stealing' a lot of plate current from the driver stage (and this produces more asymmetry in the output signal under heavy signal conditions). The signal at the plate of the driver stage will be virtually the same amplitude as the signal at the cathode of the CF stage. The output impedance of the CF cathode is really low, so the signal there will have good bandwidth and be able to drive a heavy AC load really well without losing bandwidth.

1st question. Is the first half triode in this configuration called the "volt amp"? I have heard this term a few times and not exactly understood what they meant.

Never heard it called that before. A Volt-amp is a Watt. The 1st triode of a DC coupled pair is called a 'driver' or inverting gain stage.

2nd question. My assumption with a normal gain stage is that current flows from cathode to plate and through the plate resistor. Is current flowing from the cathode of the first half of the triode to the grid of the next half of triode and out of the cathode?

Electrons flow from cathode to plate. Conventional current 'goes' in the opposite direction, from plate to cathode. (This follows the 'positive-to-negative' convention that Benjamin Franklin came up with). So in this DC-coupled pair, current is going from the HT (B+) rail through the CF plate to the CF cathode and through the (100k) CF cathode resistor to Ground, and also going from the HT rail through the driver (100k) plate resistor and through the driver plate to the driver cathode and the driver cathode resistor to Ground. Some of the plate current for the driver is diverted at the CF grid and ends up going through the CF cathode and CF cathode resistor to Ground. (Hence my above reference to the CF 'stealing' plate current from the driver stage). This pulls the driver stage's plate voltage down slightly.

[sluckey]

Never heard it called that before. A Volt-amp is a Watt.

In this context "volt amp" refers to voltage amplifier, not literally voltage ampere.

[PRR]

> puts the B+ on the grid of the next stage.

"B+" is the battery voltage, or HT Rail, often 300V.

The voltage amplifier plate sits at some lower voltage like 150V.

> the CF stage is biased really hot and there is a lot of grid current.

The classical direct coupled cathode follower takes no grid current.

A common mistake on Fendery plans, using 1.5K instead of Leo's 820r, sets the voltage amplifier plate higher then the cathode follower can follow. This does lead to large and very-varying grid current. It may be "wrong" but some people like it.
________________
> current flows from cathode to plate

Terminology.

*Electrons* flow cathode to plate.

Electrons are NEGATIVE current. You can equivalently say that positive current flows plate to cathode. All old tube-guys draw current flowing plate to cathode. They know the physics inside the glass are electrons moving cathode to plate, but who cares? we can't get inside the glass. The math is simpler reading current as the negative of negative electrons. FWIW, some decades ago many of the basic EE textbooks were published in two editions, electron flow and current flow.

[jjasilli]

Also:  for gory detail "The Fender Bassman 5F6-A", by Richard Kuehnel, Ch's 3 (Cathode Folder & 4 (Tone stack); and Merlin's Preamp Book with a long chapter on the CF with interesting comments on the Bassman 5F6-A.  Radiotron 3 is silent. Haven't checked 4.

[shooter]

*Electrons* flow cathode to plate.

That's how the Navy taught me, then I used my GI bill to get a degree, and they "tried" to teach me that holes flow from + to -, the civilian explanation was when an electron moved it left a hole

[tubeswell]

Never heard it called that before. A Volt-amp is a Watt.

In this context "volt amp" refers to voltage amplifier, not literally voltage ampere.

Indeed. I should have seen this straight away.

[tubeswell]

> the CF stage is biased really hot and there is a lot of grid current.

The classical direct coupled cathode follower takes no grid current.

A common mistake on Fendery plans, using 1.5K instead of Leo's 820r, sets the voltage amplifier plate higher then the cathode follower can follow. This does lead to large and very-varying grid current. It may be "wrong" but some people like it.

I would never question your knowledge. I am curious about which classic DC-coupled pair you refer to?

[kagliostro]

Fender 5F6 Bassman ?



Franco

[jjasilli]

Yes, and let's give credit to Merlin for the drawing.  He states in his Preamp book that the Bassman CF makes a tonal difference by causing composition, which results in smoother overdrive tone. 

[PRR]

> composition

?? Compression?

Merlin's sketch of Fender 5F6 Bassman may be taken out of context? If the cathode follower runs at 180V grid and 190V cathode, the grid is NEGative of cathode and draws negligible grid current.

However those voltages do not look right for 12AX7. 135V plate-cathode and 10V grid cathode do not accord well with Mu=100. We would expect grid-cathode to be more similar to the first tube, 1-something volts.

Ah, Fender's schem shows 180V both places.

[jjasilli]

You'll have to take that up with Merlin. 

[HotBluePlates]

2nd question. My assumption with a normal gain stage is that current flows from cathode to plate and through the plate resistor. Is current flowing from the cathode of the first half of the triode to the grid of the next half of triode and out of the cathode?

The above discussion aside, simplify and assume (electron) current in the cathode follower flows cathode-to-plate.

The cathodes of both triodes suck free electrons from ground (through a resistor), which are then pulled to the plate (and through the resistor in the "normal" gain stage), then on to the + end of the filter cap.

Wherever current flows through a resistor, a voltage drop is created/left-behind.  True for the cathode bias resistor and plate load resistor of the first stage, as well as the cathode load resistor of the follower.

You could/should just see a cathode follower as a regular plate-loaded stage with the load transferred from the plate to the cathode.  And you should look at a split-load/cathodyne/concertina phase inverter as a stage where half the plate load has been transferred to the cathode.

[Ed_Chambley]

*Electrons* flow cathode to plate.

That's how the Navy taught me, then I used my GI bill to get a degree, and they "tried" to teach me that holes flow from + to -, the civilian explanation was when an electron moved it left a hole

Ahhh yes, the old direction.  Which way does the electron flow.  Shooter, "they" taught me when current was established in a wire electrons are attracted to positive.  When 1 moves another takes its place.  I believe this is the hole you speak of. When 1 moves completely around that is 1 hertz.

So I asked what happens to all the ones that get lost?  Well I got a bunch of blank stares from classmates and the teacher.  Resistance in wire has to knock some backwards and even yet if we could see them I believe some would not move.  Of course, this course was about power generation grids and AC distribution, so I was quickly told we did not need to get off track talking about DC which was a ridiculous comment IMO.

One of the thing I took away from this course in power generation, and that is if you really want to understand how things work, ask someone who knows.  We have left practical application to be replaced by theory.

I think reading about "free electron theory" opened up a couple of places in the mind.
All the above gibberish is simply an example of our current (no pun) educational system.  The course I took was last summer at Clayton State University.

Back on topic:
So may I ask, why would anyone not use a MOSFET for a CF?  When using CF wouldn't a mosfet be preferred driving a tone stack.  Is it a better alternative in terms of linear response or does the tube add desirable coloration?

I have made 2 design changes to an amp I have been working on now for a while.  The amp has 2 channels, a plate driven stack and a CF which I recently changed to MOSFET.  It seems I get better response from the MOSFET, so I tried it in a FX loop and by golly liked that too.

Just asking because I really do not know why a tube is preferred.  It is not like i am abandoning tubes.

[shooter]

So I asked what happens to all the ones that get lost?

I went the practical, fix-it n go home model
In MRI you whack a bunch of atoms with enough RF to tip them 90degrees, then listen while they come back up-right.  I asked the same question, "what happens to the ones that don't", the Instructor went and got a physicist to explain, I quit asking dumb questions

[jjasilli]

So I asked what happens to all the ones that get lost?  Well I got a bunch of blank stares from classmates and the teacher. 
Metal is a common electrical conductor.  It has an interesting chemical property.  Given a lump of metal, the electrons in the outer shell of each atom spontaneously and randomly move about to adjacent atoms.  If we apply a proper force we can cause a quantity of those electrons to move in the same direction.  Then we can get them to do useful work for us. 

When 1 moves completely around that is 1 hertz.
Hz refers to the wiggle of AC; not to the time it takes the electron to flow around a circuit.  It does that at near the speed of light; not at, say a mere 60 times per second.  I think an AC voltage wiggles, like at a wall outlet, even if there is no complete circuit -- i.e., no current.  In a DC circuit, the electrons also complete a circuit.  But there is no wiggle, hence no Hz.

Mosfets do have followers (sorry couldn't resist).  There have been threads on this Forum.  I suspect they aren't used more because: many tube guys refuse to put "sand in the signal path"; and it complicates the power supply.

[HotBluePlates]

When 1 moves completely around that is 1 hertz.
Hz refers to the wiggle of AC; not to the time it takes the electron to flow around a circuit.  It does that at near the speed of light ...

I made the same mistake saying this.

Imagine a tube just big enough for a golf ball to pass through.  Let's say this tube is 15 golf-balls-long.  You fill the tube with 15 golf balls, then try to push one more in.  One pops out the other side "at the speed of light".

Pretend the tube is 100 golf-balls-long, and the last ball to fit in is colored red.  You add another ball, one pops out the other side "at the speed of light" but your red ball has only moved 1-ball in space.  Even if you're putting golf balls in twice a second, it still takes 50 seconds before that red ball comes out the other side.

Current is a "directed drift of electrons" because a force (voltage) pushes them in a direction.  The directed drift itself can be slow (if you're peeping a single electron), but the effect (ball out the other side) seems instantaneous.

So I asked what happens to all the ones that get lost?  ...  Resistance in wire has to knock some backwards ...

The class was on electrical generation & distribution?  All the electrons are on the wire (or inside the insulation)... There's nowhere for them to go.

"Current" is defined as a quantity of electrons moving past a point per second.  You could equate it to "gallons per minute" of water.  So it's a "rate of change/flow".

"Voltage" is the "push to get the water to flow"... water pressure.

Resistance doesn't knock the electrons back, it just slows the rate of flow.  5v / 1Ω = 5A (big flow), but 5v / 5kΩ = 1mA (tiny flow).  But the electrons go the same way they always did given the direction of pressure.

You turn on your garden hose (with no spigot on the end), water just comes rolling out.  Let's say it's big current, 5A.  You mostly-cover the end with your thumb... Let's say the gallons-per-minute stays the same.  You just impeded the water 500Ω to the 1Ω it used to see.  But the volume of water per second is still the same: 5A * 500Ω = 2500v.  The water seems to shoot out faster/harder.

But that water isn't flowing backwards through the hose.  And I mostly care about the stream of drops that come out the end, rather then the few drops that stay clinging to the inside walls of the hose.

[Ed_Chambley]

Thanks for the clarifications and sorry I got off topic.  Yes, DC has a cycle, not hertz and I should not post without reading.  The backwards comment was meant tongue and cheek.  I will explain it was only losses I was concerned with and was not amplifier related.  The class was much less expensive than the electrician proposal.  I had to either buy a new press and 2 additional operators to keep my needs under 400 amps, or figure out a way to install a UV dryer on a 40" 6 Color Printing Press which drys the sheet so you can turn it over and run back through and can cut and finish without drying time.

The problem.  The UV dryer coming on at the same time as the press along with put me over 400 amp service.  The press inrush surge current is 190 (varies some) amps.  Adding the dryer would overload unless I could "find" some lost amps.  Some things were simple, like rewiring motors to 240 for 120.  Heat shrink tunnel to 208....etc.

Reading Shooters reply simply reminded me of the electricians proposal which was based on theory without any change to the application and remembering the no talk of DC by the instructor and I am thinking of all the DC which is used on service.

Well the press has the dryer and the backwards electrons is a comment simply as a joke.  When the unit switches on the conduit moves from the electrical surge and when asked why by the pressman I said the electrons were running away from jumping into the fiery UV dryer.  When the press gets hot it also uses more current, so chillers for ink and water were needed, but needed anyway.

The solution was very simple after regaining lost current.  We just made the UV dryer switch activate by a Keyance sensor in the delivery end of the press because once the press is up to speed its current demand drops to about 40 amps.

Again, I am sorry I got so far off topic and am glad you 2 jumped in and made some clarifications to my inconsistent thought plane.

I have heard the water analogy and think it makes for good understanding. 

Also, the MOSFET Followers comment was great.  I have never really encountered any power supply problems using them, but I do not doubt you in saying it happens.

[jjasilli]

I have never really encountered any power supply problems using them, but I do not doubt you in saying it happens.
That's my lack of knowledge of SS.  See this thread Reply #15 and following:  http://el34world.com/Forum/index.php?topic=19805.msg228760#new

" I'm such a tube snob I didn't know SS components could run on "tube" voltages -- off B+ supplies in the 200V - 300V range.  That makes SS reverb & trem easier to implement in a tube amp.  E.g., geofex.com has a SS trem circuit that needs a string of zeners to knock down the supply voltage."

[shooter]

I didn't know SS components could run on "tube" voltages

It's quite the eye-opener when you get your fingers on the "wrong" side of a nixie-driver IC chip!!

[hesamadman]

This was helpful. Even as much as it branched away. I found all info to have answered some questions and cleared up a little. Im no cathode follower expert yet but I have a much better picture. I post questions here rather than reading (well I had been reading actually) because the explanations are usually much better. I have ALL the books everyone usually suggests but the problem is a mild form of dyslexia that makes me seriously have to read things multiple times before I can absorb any of the info....which makes being self taught extremely difficult 


Thanks everyone.

[PRR]

> All the electrons are on the wire (or inside the insulation)... There's nowhere for them to go.

A tidbit. In isolation, we don't need insulation to "keep the electrons in". The same forces which let the electrons slip around inside the metal prevent them from leaving the metal. If you pluck an electron and pull it away, the place where that electron is now missing acts like a positive charge and sucks it back.

We can run naked wires everywhere, except where they might get touched by things we don't want to touch (people, dirt, or other wires). Then insulation is wanted.