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 johnny audio 20th August 2012 07:27 PM

cathode follower question

Hello,

I'm just learning about tube amps and have come to a stumbling block on the issue of the cathode follower.

It's seems my head is stuck on the idea that in a vacuum tube, electrons flow from the negative cathode to the postive anode (plate) 'amplifing' the grid input, but when it comes to a cathode follower (increases current not voltage) it's seem that electrons are flowing from the plate to the cathode (I know this can't be right, is it?) and thus current is being drawn from the grid signal to the cathode.

I've read and I'm trying to understand this explaination from ampbooks.com but still not getting it...

"A basic triode amplifier is great for increasing the voltage of a signal. Its major problem is that it can't drive a very heavy load. Place a low-valued resistance across its output and it easily gets bogged down, which reduces the gain and makes it dependent upon the characteristics of the load. If the next stage is a voltage amp or a phase inverter, this isn't much of a problem. A tone stack, however, typically demands a lot of current from the driving stage, the amount of which depends on the signal characteristics and the tone control settings. A cathode follower mitigates this problem. It has no gain (in fact a slight loss), but it can easily drive the most demanding of tone stacks.
An increasing input signal voltage causes the plate current to increase, which increases the voltage across the cathode resistor, thereby making the cathode more positive with respect to ground. If the input increases the grid-to-ground voltage by 1 volt, then the increased plate current raises the cathode-to-ground voltage by almost 1 volt, leaving the grid-to-cathode voltage almost unchanged. The grid voltage increases but the cathode voltage "follows" it. Thus we get this circuit's name."

I know there must be some simple, laymen explination that I'm just not getting, if someone could enlighten me I would greatly appreciate it!

Cheers
John

 ChrisA 20th August 2012 07:38 PM

You are thinking like a Physicist about electrons. The "correct" way to think about tubes is like an EE. THink about "holes" that cary positive carge flowing from plate to cathode. Engineeers don't give a hoot about electrons, current is always "positive".

The cathode follower is simple when you think about positive holes moving: Ohm's law causes a voltage across the load resister that depends on current. That voltage is the output signal. That's it.

 DF96 20th August 2012 09:03 PM

There are no 'holes' in a valve. Don't think about electrons or holes, just think about AC signal currents and voltages. It's all about using the appropriate level of abstraction. You need electrons (and holes for semiconductors) to understand how an active device works internally, but you can then just treat it as a 'black box' in a circuit. Using appropriate levels of abstraction is really thinking like a physicist!

If a valve grid goes more positive then the cathode-anode current will increase. This will drop more voltage across any resistor in the cathode or anode circuits. If the resistor is in the cathode circuit then the change in voltage opposes the grid change, so it is an example of negative feedback.

 soulmerchant 20th August 2012 09:36 PM

Another way to think of it is in term of impedances. The cathode follower is a buffer stage with very low output impedance. It protects your voltage gain stage from the signal distorting impedances of the external world. For example, you might increase the AC 'swing' with a comon cathode amplifer, but the cable, attenuator and input impedance of the next stage will ultimately distort that beautiful AC 'swing'. The cathode follower will protect (buffer) it.

 HJWeedon 21st August 2012 01:26 AM

Cathode follower.

Hi

Come on guys. please do not confuse a new-comer.
We can compare a cathode follower to the power steering that we all love in our cars. The grid of a cathode follower is like the steering wheel of a car with power steering, the cathode output is like the steering knuckle controlling the front wheels. The wheels do nothing unless the steering wheel is turned, similarly the cathode does nothing unless the grid tells it to.

There are bias levels in a tube that you have to contend with, that is done with capacitors that pass AC voltages, but leaves the DC bias voltages in place.

When properly done, the cathode follower can drive just about as much current into the load as it's standby bias current. in a typical tube that might be 100mA, not much drive for a speaker, but usually enough for a pair of headphones.

Just like the power steering, where the wheels do not do anything more than the steering wheel tells it to do, the output of a cathode follower does not do anything more than the grid input tells it to, but at significantly increased power level. If you have about 2V available for driving a pair of headphones, but not enough power behind it, put a cathode follower in between the signal; source and the load.

Hans J Weedon

:cop: Unprotected email address removed by moderation to prevent spam.

 Miles Prower 21st August 2012 07:54 AM

Quote:
 Originally Posted by johnny audio (http://www.diyaudio.com/forums/tubes-valves/218262-cathode-follower-question-post3133857.html#post3133857) 've read and I'm trying to understand this explaination from ampbooks.com but still not getting it... "A basic triode amplifier is great for increasing the voltage of a signal. Its major problem is that it can't drive a very heavy load. Place a low-valued resistance across its output and it easily gets bogged down, which reduces the gain and makes it dependent upon the characteristics of the load. If the next stage is a voltage amp or a phase inverter, this isn't much of a problem. A tone stack, however, typically demands a lot of current from the driving stage, the amount of which depends on the signal characteristics and the tone control settings. A cathode follower mitigates this problem. It has no gain (in fact a slight loss), but it can easily drive the most demanding of tone stacks. An increasing input signal voltage causes the plate current to increase, which increases the voltage across the cathode resistor, thereby making the cathode more positive with respect to ground. If the input increases the grid-to-ground voltage by 1 volt, then the increased plate current raises the cathode-to-ground voltage by almost 1 volt, leaving the grid-to-cathode voltage almost unchanged. The grid voltage increases but the cathode voltage "follows" it. Thus we get this circuit's name." I know there must be some simple, laymen explination that I'm just not getting, if someone could enlighten me I would greatly appreciate it! Cheers John
It's no wonder you're not "getting it": your source material is pure, absolute, unadulterated bull. The cathode follower is just another VT circuit that can not perform the magic of making a high voltage, low current, device into a high current, low voltage device. It's the same loadline regardless of whether the load's connected between the plate and DC rail, or whether it's connected between the cathode and DC ground. (Or whether it's split between the cathode and plate. It's always the same loadline.)

A stiff load pulls down the gain and linearity of a triode. However, the same applies to a cathode follower: it makes no difference where the load is connected. None. Absolutely none!. The cathode follower is a good device for presenting a Lo-Z source impedance, but only to a Hi-Z load (like the aforementioned tone stack).

The misapplication of cathode followers by running them into stiff, near vertical, loadlines is what's responsible for the negative reputation of the topology.

 DF96 21st August 2012 10:17 AM

Quote:
 Originally Posted by Miles Prower However, the same applies to a cathode follower: it makes no difference where the load is connected. None. Absolutely none!.
No. CF has much more negative feedback than a grounded cathode stage. It can't do magic, that is true, but for a given load a CF will drive it with less distortion until clipping begins. For a given load line a CF will provide about 1/mu as much distortion as the corresponding grounded cathode, as well as about 1/mu as much gain - exactly as the NFB equations predict.

The source material quoted by the OP is woffly, but it is not as bad as you say.

 disco 21st August 2012 04:05 PM

NFB depends on the available amount of amplification...

 DF96 21st August 2012 08:57 PM

Which for a CF is a little below mu, so adequate if an appropriate valve has been used. You could easily have 20-30dB of feedback.

 johnny audio 24th August 2012 05:44 PM

1 Attachment(s)
Thank you all for your responses, I must admit though, I'm still at a loss. Mine you, I'm just learning basic electronics (Chapter 9 of 23 of our schools electronic course) but am burning to understand how guitar amps work.

If we could take a step back for one moment. I think what I'm having difficulty is understanding the actual guitar signal path. I know that in a triode amplifying stage a very small AC guitar signal enters the grid, this signal is 'amplified' by a stream of electrons flowing from the cathode to the plate (anode). This occurs because their is a positive potential at the plate with respect to the cathode, thus their is a one way current flow from cathode to plate. If the plate was made negative with respect to the cathode, electrons would be repealed and the current would stop.

So what is actually happening in a CF stage, between the Plate and the Grid, between the Grid and the Cathode and between the Plate and the Cathode with respect to voltages, current and actual guitar signal?

I've attached a diagram of a Fender 5F6-A Bassman's (normal channel only) that I've been using to try to get my head around this wall.

Thanks

John

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