# Wrapping my head around cathode followers. Does this explanation make sense?

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#### Sodacose

##### Member
I'm trying to write some tube descriptions for a layman because I think they're useful and because describing them to others often helps me understand them better. So is the following simplified description of a cathode follower accurate? I think I'm getting it, but I'm not totally sure.

Edit: this isn't meant to be a recipe for a cathode follower or an application for a specific tube/circumstance, it's just meant to explain the most basic concepts of what happens when a cathode follower is doing its thing.

"
[for a grounded cathode] We know that the grid will have a more negative voltage potential than the cathode because otherwise it starts attracting the cathode’s [emitted] electrons instead of allowing them to zip to the anode. When the grid is very negative, it doesn’t allow many electrons (current) through and when the grid is not quite so negative, it lets more electrons through (remember that same charges repel each other). The grid’s negative relation to the cathode is achieved by referencing it to 0V. [replace 0V with "ground or a voltage divider in the cathode"?]

Consider the numbers in this example. What’s the voltage dropped across a 10,000 ohm cathode resistor when the circuit draws 15mA?

V = I * R
.015 amps * 10,000 ohms = 150 volts

Now if a positive change in the grid (making it less negative) allows another 5mA through the tube, what is the new voltage drop across the cathode resistor?

(.005 amps + .015 amps) * 10,000 ohms = 200 volts

So the cathode voltage goes up, but as this happens, the grid’s voltage in relation to it is actually becoming more negative. This lets fewer electrons through to the anode. Fewer electrons means lower current and lower current actually means less voltage across the cathode resistor.

This all happens simultaneously, like a tug of war, and results in an increase in the current of the signal, but no increase in the voltage (it’s actually a slight decrease). It doesn’t amplify the voltage, but it does add power (because watts = voltage * current).
"

Probably a description of impedance would follow this. Am I on the right track here? Please keep in mind that I'm trying to make this as conceptual (and non technical) as possible without being totally inaccurate.

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#### jcx

##### Member
try reading up on how others describe negative feedback - because that's what it is

#### Sodacose

##### Member
Thanks, jcx! I've read about cathode followers as a form of negative feedback, too. Do you happen to know of any especially good readings on it?

#### ultrafi

##### Disabled Account
Sodacose,

I really think you need to understand bias better. Bias is the difference in voltage potential between the grid and the cathode, i.e., Vg-Vk. It is this difference in voltage potential that controls the flow of current from the plate to the cathode. This gives rise to the notion of transconductance based on Ohm's Law where gm=DeltaI/DeltaV. Now, for a tube mu=(gm)(rp).

Putting it in terms of a "stage" as you have done works; but, it is not universal in application. Misapplied, one get's the wrong answer.

I'd read the Radiation Designer's Handbook and pay particular attention to the the distinction between the DC analysis (bias) of a stage and the small signal AC analysis.

True appreciation for the small signal model will show the flaws in your analysis.

#### Sodacose

##### Member

I'm not quite sure I understand your point about bias. I am familiar with what bias is, but not how I'm misrepresenting it above. I will absolutely grant you that the explanation leaves a lot out (intentionally), but again I'm not trying to write a recipe for how to construct a cathode follower, just explain the basic concept of what is happening to a newbie using little more than Ohm's Law and some simple algebra.

I'm definitely not trying to write a replacement for any of the excellent technical texts that already exist!

#### GoatGuy

##### Member
Sodacose, there are several things you need to propose before a relatively simple explanation can be both simple and more-or-less correct.

× 1. tube has to be high gain
× 2. tube has to 'run' with pretty high potential across it nominally.
× 3. input can't exceed about 50% of the nominal grid bias voltage without topology tweaks

Each of these has its own purpose. (1) ensures that the nonlinear transfer characteristics of the chosen tube is reduced (through built-in negative feedback) to the greatest degree. Its important. (2) ensures that the cathode follower will be in its higher gain regime. Even though a lower voltage would suffice, it results in greater non-linearities on output. (3) is a cheeky one: The cathode potential cannot usually drop below ground voltage level. If it is nominally +5 volts (say), then it can only “swing down” to perhaps 50% of 5 volts, or 2.5 volts before intrinsic un-corrected distortion starts creeping in.

AFTER THAT (assuming the above), then you can get on with describing the mechanism.

I prefer to think of it as “a high gain tube always likes its grid to be close to N volts below its cathode”, and will modulate current flow in lockstep with grid fluctuations. These fluctuations in turn pass thru both the cathode load resistor and any load impedance attached. The E = IR Ω's rule (or E(t) = I(t) Z for A/C) ensures that with every upward wiggle in current, there's an upward wiggle in the voltage across the load. This provides the negative feedback which ensures that the changes are almost exactly the same on both.

The actual 'gain' of the “perfect valve” would be μ / ( 1 + μ ) … [ a simplification, but a good one ]. Thus it becomes obvious that low μ tubes would not get anywhere as close to 1.000 pass-thru gain. The bigger the μ, the closer to 1.000 one gets.

And it can be argued that nonlinear distortion (to whatever degree the valve is nonlinear) in turn varies as 1 / (1 + μ) or so. Bigger the μ, lower the distortion (for this circuit), if every other aspect were the same between a set of valves.

. . . . . . . .

As UltraFi sez… it is important to keep in mind that the large-signal or quiescent operating point math be kept separate from the small signal model. one can actually work out decent algebraic forms that govern the whole thing, but its a pain in the âss, and rarely in real-life works as well as the more sophisticated SPICE tube models.

Just saying. I've been there. Done that. Bought the tee-shirt. Had a goat eat it.

GoatGuy

#### Merlinb

##### Member
Start by explaining how an unbypassed cathode resistor causes gain to fall. This will lead you on to the pure cathode follower. Remember, the valve only amplifies what appears between grid and cathode. If you don't hold the cathode voltage constant (e.g. with a cathode bypass cap) then the cathode voltage will tend to follow the grid, as you showed with your sums. If the cathode is following the grid then the difference between grid and cathode shinks, so there is less for the valve to amplify.

#### Sodacose

##### Member
Thanks everyone for your explanations. I am trying to avoid anything technical for the purpose of this exercise, but I appreciate all the input. It sounds like I'm in the ballpark conceptually and that's about where I'm trying to be.

Think about this like a tube amp picture book for kids. A kid's picture book might talk about a car having an engine and a transmission, but it isn't going to go into the chemical reactions required for internal combustion.

Really appreciate it though. Please keep the explanations coming, too. I think I'm wrapping my head around the cathode follower, but explaining something simply requires a really thorough understanding and I don't claim to have that.

#### DF96

##### R.I.P.
Sodacose said:
The grid’s negative relation to the cathode is achieved by referencing it to 0V.
No. The grid is (usually) negative with respect to the cathode. What it is with respect to the point in the circuit which you have decided to call 0V is completely irrelevant.

Am I on the right track here?
Not enough to be useful.

Please keep in mind that I'm trying to make this as conceptual (and non technical) as possible without being totally inaccurate.
You can't explain a technical concept without using a technical explanation. You can use a simple explanation or use analogies, but the end result has to be technical in some sense.

In order to explain something simply you need to really understand it. That is why Feynman's books on physics were so good.

Do a lot of reading of good books. Get to fully understand how a cathode follower works. Learn to do the maths yourself. Do simulations. Then you might be able to write a simple explanation which might be sufficiently accurate to be useful.

#### Sodacose

##### Member
Hey DF96, thanks for the response.

Can you point out specifically where I'm going wrong in my understanding? Just so I have a direction to start searching in for more information.

Edit: Ultrafi, are you referring to this book? http://www.tubebooks.org/Books/RDH4.pdf I love you, Pete Millet.

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#### Sodacose

##### Member
Ok, from Radiotron Designer's Handbook, this is the best simple explanation I've found so far:

The stage gain, however, is necessarily less than 1 since the input voltage [applied between grid and earth] is equal to the grid to cathode voltage plus the output voltage [taken between cathode and ground]...Since the stage gain is always less than unity, the input voltage is always greater than the output voltage. p 317

I'll definitely have to go back and introduce a few more concepts before trying to explain a cathode follower simply. This explanation makes a lot of sense to me, though.

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#### JMFahey

##### Member
Sorry but you won't learn Electronics (or anything else) by asking some random questions in a Forum.

Get a book (or many) , and start reading, step by step, at your own pace.

Do not run before you walk and do not walk before you crawl.

Grasp a good concept (besides repeating the textbook definition) of such concepts (in no particular order) as: volt - current - resistance - capacitance - supply (battery) - frequency - power - dissipation - conductor - insulator - dielectric - switch - electric field - magnetic field - induction - etc.

Often used analogies, such as "current in a wire is like water in a pipe" , "a potentiometer (or a transistor) is like a water faucet", etc. are just analogies, to "explain the unknown based on what's already known" ... better than nothing but often causing new confusion when taken too literally, so it's best to "waste" some time learning things the classic way.

Your explanation on post #1 is both more complex than the proper one and to boot, inaccurate, and "correcting" it would practically mean erasing it and writing a shorter, correct one, so it's better you study from the beginning and write your own.

In fact, in due time, you won't even have to ask us

#### Sodacose

##### Member
Sorry but you won't learn Electronics (or anything else) by asking some random questions in a Forum.

Obviously this was my first mistake, but it's also the motivation behind coming up with a simplified explanation. This forum is great and I really respect all the knowledge in it, but it isn't for beginners or amateurs.

I'll go read some books and keep working on my project. Thank you for the responses.

#### DF96

##### R.I.P.
Sodacose said:
This forum is great and I really respect all the knowledge in it, but it isn't for beginners or amateurs.
Almost everyone on this forum is an amateur at audio electronics i.e we don't get paid for it. All of us started as beginners; some of us are still beginners, although not quite everyone realises this. This forum is good for beginners, as long as they come asking questions and thinking about the answers.

Your mistake was to try to explain something to others which you don't properly understand yourself. You are not the first one to attempt this. Posting a confused explanation and then asking others to, in effect, proof-read it is not a good way to learn.

There are three good ways to think about the cathode follower:
1. as an amplifier which sits on top of its own output
2. as an amplifier which has 100% negative feedback
3. as a modified grounded cathode amplifier which has a large amount of cathode degeneration
Each can lead to a correct understanding, and the correct mathematics. However, before you tackle the CF you should first gain a full understanding of the grounded cathode amplifier, including the use of load lines to choose a bias point.

#### Sodacose

##### Member
There are three good ways to think about the cathode follower:
1. as an amplifier which sits on top of its own output
2. as an amplifier which has 100% negative feedback
3. as a modified grounded cathode amplifier which has a large amount of cathode degeneration

Thank you, DF96. These are the kind of points I hoped to get to further my conceptual understanding.

I'm aware that my understanding is less than complete and I'm sure that further reading of the authoritative texts would help (and I intend to do just that). I'm also sure that forums aren't a substitute for said texts. But I find that attitudes here can be very dismissive at times (your own initial response included, if I'm being honest) when they could instead point the less knowledgable down a certain path of research (like your second post, which I appreciate very much). It was this frustration that lead me to say that it isn't friendly to beginners.

No one owes me anything to be sure but I guess I was just hoping for a little more encouragement for a beginner tube enthusiast. As you pointed out, everyone had to start somewhere.

#### SY

##### Ex-Moderator
Tim, a hint that may help: a tube doesn't know what "ground" is, it changes current with a changing difference in the voltage between grid and cathode. So in a sense, it looks like an op-amp with an inverting (grid) and non-inverting (cathode) input.

#### Sodacose

##### Member
Thanks, Stuart I was in a crappy mood on Saturday and I apologize to everyone for being so easily put-off.

I'm doing some more reading this morning and came across this on Tube Cad:

An intrinsic distortion correcting mechanism known as degenerative feedback keeps the Cathode Follower's distortion low. It works this way: any departure from the desired output voltage will subtract from the input voltage and result in a countervailing change in current flow. In other words, if the cathode is forced more positive, the grid will effectively become more negative relative to it and thus less current will flow through it, which works to decrease the output voltage; on the other hand, if the cathode is forced less positive, the grid will effectively become more positive relative to it and thus more current will flow through it, which works to increase the output voltage.

Tube CAD Journal: Tube balanced amplifiers July 1999

Underline my own. This is the gist of what I had in my head in the original post (though a much better executed explanation). As I tried to think through what happens in a cathode follower, this seemed to be the effect that grid signals would have on current flow through the tube (and in turn voltage output taken from the cathode).

Can anyone comment on Broskie's explanation?

I picked up a text on DC fundamentals as well as one on Impedance (which seems to be mostly dealing with AC signals). I started reading about AC superimposed on DC and I see part of where I was going wrong in terms of bias and signal in my description.

#### SY

##### Ex-Moderator
It's sorta right. Problem is that the CF works perfectly with a CCS load (and hence no change in current). So you have to go a bit further.

So... you have a CCS load in the cathode. Now put a positive-going signal on the grid. If current is constant, then the voltage on the cathode has to rise as well, and to an amount roughly equal to how much the grid rises (remember, for small signals, the current through the tube is constant if and only if the grid-to-cathode voltage is constant).

#### DF96

##### R.I.P.
. . . and the anode-cathode voltage is constant too.

It is always difficult to know when to complicate a simple explanation, but I think if you have a perfect CCS tail then you can't really avoid the O(1/mu) correction to CF gain.

#### SY

##### Ex-Moderator
. . . and the anode-cathode voltage is constant too.

Thus my "small signal" qualifier: it simplifies the problem. That's always a good starting point for understanding.

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