Re: Kathode folower question

Ian Iveson wrote:
Ian Bell wrote:

How do I calculate the cathode follower cathode resistor in case of DC coupling between the 1st and the 2nd stage, the 2nd tube grid is directly
connected to the 1st tube anode .

Depends on the tube type and what sort of load you want to drive and the its stage tube and its anode voltage. But in simple terms - you know the anode voltage of the preceding stage so you know the CF grid voltage. Then CF needs to be a few volts above this - how many we don't know yet, but it is only a few volts so it is usually safe to say that the operating point of the CF is with a plate voltage of the CF plate supply minus its grid voltage. On the CF tube plate characteristics draw a vertical line at this voltage.

Now comes the hard part. You need to decide what static current the CF is to draw. To find this you need to know the load impedance the CF will drive and the voltage you want to drive in it - this will give you the load current. As a rough guide you should set the static current to four times this value. Now is the time to check you have chosen a tube that is meaty enough by calculating its dissipation as static current times plate to cathode volts.

Assuming that's OK, draw a horizontal line at this current on the plate tube characteristics. Where it intersects the vertical line is your desired operating point and you can see what the grid bias voltage needs to be so add this to the grid voltage and this is the desired cathode voltage. You know the standing current and the cathode voltage so working out the cathode resistor is just a matter of dividing the latter by the former.


In practice it is not quite that easy because it is a good idea for the cathode resistor be be at least twice the tube's plate resistance and this method takes no account of that.
This seems to involve such a common misconception that sometimes I wonder if I've got the wrong end of the stick.

For me, it is the *total load* on a triode that should, as a rule of thumb, be somewhere around 2.5 times the anode resistance. Why should you, and many others I have read, consider only the cathode resistor's contribution to the load?

You have not said what you mean by the total load but I presume you mean the parallel combination of Rk and the AC load.

Could it mean anything else? Didn't think I would need to spell it out. "AC load" is a bit ambiguous anyway, because Rk is also a load for AC. Hopefully we know what we mean.


I did not expect it to be something else but I would rather be certain.

This is taken care off by setting the standing current to 4 times the load current - it effectively means the load is three times Rk so its effect on the total load is small.

Do you mean 4 times the total load current, or 4 times what you call the AC load current?

AC load current. The load current has to come from the supply via the tube so the standing current has to be greater than the load current.

Doesn't the AC load current depend on the signal level?

Yes, you apply ohms law to the desired output voltage and load resistance.

Do you mean at max signal
amplitude?

Since the load current has to be supplied from the standing current then of course yes.

Perhaps it would be clearer if your original rule
were stated in terms of impedance rather than current?


No, what we need to work out is the load current in order to design the stage (to define the standing current). You know the (max) desired output voltage and (min) load.

Anyway, if you mean that the driven load is three times Rk, then that seems a big effect on the total to me. I generally consider a parallel load to be of acceptable insignificance when it is at least 10 times. Three is far too small. Better than most CF applications I've seen though, so at least we agree that it should be bigger rather than smaller...


Check ouit RDH4 which goes into this in more detail.


Now, if the total load met that condition, then the CF wouldn't be able to serve it's common purpose of driving low impedance loads.

That is a common misconception. A CF is designed to produce a low impedance source, not to drive low impedance loads.

Should be so designed, I would say, but most often isn't, due quite possibly to the common misconception.

Agreed.

Further, a
low impedance source is only ever required for low impedance loads, for why else would a low impedance source be necessary?


To drive high capacitance cables without HF losses.

You might say my objection could equally be applied to any situation in which feedback is used to achieve an acceptable impedance ratio. You would probably be right...I am generally suspicious of the technique, and try to meet ratio requirements open loop. Perhaps this objection should be extended to active filters, now I come to think of it.


What people forget is that a CF output impedance is only low (due to NFB) for small signals - the math is only valid for small signals. The weakness for large signals is the CF is single ended. It can only source current from the supply via the tube. It has no active sink, the only sink being Rk, hence the many multi-tube variations on the CF.

Consequently it might be said that the CF
has no place in high fidelity, because it abuses a triode and then tries to cover up with loads of feedback. May as well use SS, which has the advantage of not needing heater current.

Emitter followers have much the same problems and limitations as CFs.

Quite, and vice versa, except SS doesn't need heater current.


|Well, that true of and SS design versus its tube equivalent, not just in the case of the CF. I am not quite sure what point you are trying to make.


My point was the EF like the CF is single ended and its NFB reduces op Z only for small signals.


Anyway, in answer to the OP's question, surely calculation of the cathode resistance needed to establish a particular operating point is the same regardless of the voltage on the grid?
No, because the grid voltage determines the cathode to plate voltage which is one part of the operating point. You can only choose the operating current because the grid voltage in this example is given.

That is, your choice of operating point gives you Vgk

Well, yes. I think we're saying the same things, but in a different order...

and Ia. Simple ohms law returns Rk, bearing in mind that Vgk = Vg - Vk.

Except you have completely ignored the drive requirements.

So far...but I remarked on the problem of calculating signal Vgk from signal Vg later. I realise that an operating point is chosen with a certain maximum signal in mind, and have identified this as one part of the reason for the comparative difficulty in calculation, following paragraph. I try to read a whole post before I reply to part of it.

The main reason a CF is more difficult to design
than an AF is that Vk is a much more significant part of Vak, so you don't know the necessary Va until the end of the calculation.
Sorry, that makes no sense at all to me. You could say the same about the voltage across Rl in an AF.

No, because you wouldn't be using Rl to adjust the bias current. Let me put what I mean another way: when using Rk to adjust the bias current of an AF, the effect on the total standing DC load is negligible because Rk is small compared to the anode load resistance. Not true for a CF, where Rk is the only standing DC load.


Well, first off, that is only (possibly) true because the particular instance of CF we are considering has *fixed* bias and you are comparing it with a *cathode* biassed AF. For a cathode biased CF the situation is identical.

Second the effect of Rk on the standing dc load in a AF is amplified by mu + 1 so in most cases (mu+1)Rk is not small compared to the anode load resistance.

Hence not only does the current
change because the of the adjustment of bias, but also because the total load varies as a result of that adjustment.

If I were to write what I wrote again, I would have made the point more clearly. What I'm really getting at is that, whereas an AF can be set up with arithmetic, more or less, a CF needs algebra, or iterative arithmetic, because of the feedback.


An AF has feedback as far as dc operating point is concerned.

That's why the OP asked, I feel sure. OP needs a good book coz ngs don't make algebra easy to convey.


Agreed!!! I prescribe RDH4 ;-)

Often you will be designing for a given HT
voltage, so you end up with an iterative process of rethought and recalculation. Also, the signal Vk is a much more significant part of the signal Vgk so it's not so easy to choose the best operating point in the first place.

OTOH, perhaps if I were to use a CF, I shouldn't bother too much about choosing an operating point...just choose Va and Ia, mess with Rk until it fits, and let Vgk and Vak be whatever they fancy. Feedback willing, it'll be alright on the night.

Not a design methodology I would recommend.

But probably the most common.


Unfortunately most likely true.

Cheers

Ian
Ian


.

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