cathode follower question
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 24th August 2012, 08:00 PM #11 DF96   diyAudio Member   Join Date: May 2007 See posts 3 and 5, which in different ways describe how the CF works. Just keep reflecting on it, and read descriptions of how the emitter (BJT) and source follower (FET) work too. Eventually it should 'click' and then you will wonder what seemed so baffling at first.
 25th August 2012, 03:27 AM #12 jjman   diyAudio Member   Join Date: Jan 2009 As the current pulses thru the plate resistor in stage 2 the voltage on the plate pulses from ohms law. You show 180v which would be the voltage with no signal. Let's say it goes up and down from 170v to 190v with signal applied. This is a 20volt peak-to-peak swing. That plate is connected to the 3rd stage's grid with the question mark. So the voltage on the "?" is also pulsing and therefore modulating/pulsing the current thru that 3rd stage. Swinging the voltage on a grid makes the current flowing from the cathode to the plate pulse up and down in unison. That 3rd cathode resistor's current is therefore pulsing so the voltage on that cathode is pulsing, again due to ohm's law. The pulsing on that 3rd cathode is where the cathode follower's output is taken. The output is "following" the voltage on the cathode. The output of the 2nd stage is following the voltage on it's plate. The diagram shows 180v on the plate of stage 2 (grid of stage 3) and 180v on the cathode of stage 3, which is not correct. The 3rd cathode would be a little higher in (DC) voltage than it's grid at idle. Perhaps 185-190 or so. The difference in voltage between a grid and its cathode (with no signal) is called "bias."
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Join Date: Mar 2008
Quote:
 Originally Posted by johnny audio 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!
Let me try - step by step - in your terms:

Lets have a look at the 1st stage in your circuit V1a:
You are correct, there is a flow of electrons from the cathode to the plate.
This flow is influenced by the charge (amount of electrons) on the control grid.
Any change of the charge on the grid causes a change in the flow.
More electrons on the grid (more negative charge) reduces the current between cathode and plate.
Less electrons on the grid (less negative charge) increases the current between cathode and plate.
Hence the tube is in fact a voltage (grid) to current (plate/cathode) converter.
One of the main tube characteristics is therefore transconductance gm or S in the datasheets which is measured in mA/V.
That is milliamps (plate/cathode) per volt (grid).
So changes of grid voltage (input signal) are converted to changes of plate/cathode current.
The change of plate/cathode current in turn causes a change of voltage drop in the plate resistor (output signal).
Signal amplification happens when both gm and plate resistor values are big enough such that the change of the grid voltage causes a bigger change of the voltage drop in the plate resistor. And the cathode resistor is small enough such that there is no significant voltage drop across it.

Lets have a look at the 2nd stage in your circuit V2a:
Again, there is a flow of electrons from the cathode to the plate.
This flow is influenced by the charge (amount of electrons) on the control grid.
Any change of the charge on the grid causes a change in the flow.
Note, that the grid of V2a is connected to the plate of V1a by means of a capacitor.
This capacitor allows only changes in the plate voltage (AC) of the 1st tube to pass on to the grid of the 2nd tube, not the voltage (DC) itself.
Again, changes of grid voltage (input signal) are converted to changes of plate/cathode current.
The change of plate/cathode current in turn causes a change of voltage drop in the plate resistor (output signal).
Same as with V1a.

Finally lets have a look at the 3nd stage in your circuit V2b:
Again, there is a flow of electrons from the cathode to the plate.
This flow is influenced by the charge (amount of electrons) on the control grid.
But now, the control grid is directly connected to the plate of the previous stage, no capacitor is blocking DC.
And there is no plate resistor but the big resistor is now at the cathode.
The big positive voltage at the grid of V2b would cause an enormous current through the tube if the cathode resistor were the same as in the 1st two stages.
But instead we now have a much bigger cathode resistor.
This causes a voltage drop big enough that the voltage on the cathode is about the same as the voltage on the grid without excessive current through the tube.
This is why the cathode voltage approximately follows the grid voltage.
If the cathode voltage follows the grid voltage, the current through the tube doesn't change.
If the current doesn't change, there is no amplification.
Any changes in the grid voltage (input signal) show up as similar changes in the cathode voltage (output signal).
Hence cathode follower.
With some math it can be demonstrated that the driving impedance (capability) of a triode cathode follower for small signals is approximately 1/gm or in the range of 200 Ohms whereas a triode amplification stage has a driving impedance around 10,000 ohms. ECC85 / 6AQ8 as an example.
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