Tube drive amplifier question

I understand that avoiding curtailing HF response means the effects of reactance are avoided, but how and why on Earth does low impedance "swamp" stray capacitance?

It does so by effectively shorting it out (the stray capacitance is effectively shorted out or nulled by the low impedance), is the simple answer, Iam sure the more mathematically astute on here will be able to expand on this.

There are quite a few ways of looking at this. Here's one:

Let's think of the drive stage as a perfect source of oscillating voltage in series with an impedance. This impedance is the output impedance. Let's imagine that it's mostly resistive. If the drive stage is feeding into significant stray capacitance then current will flow into and out of that stray capacitance as the drive stage tries alternately to charge and discharge it. That current will cause voltage to be dropped across the output impedance. This voltage drop means that the voltage at the input of the next stage will be reduced relative to what it would be if the drive stage was feeding an open circuit. This loss of voltage is generally a bad thing.

There are two ways to fix it. First we could minimise the stray capacitance. If it was small it would charge up really quickly and the current which is causing the voltage drop would thus be kept small. Second we could reduce the output impedance. That would mean we wouldn't drop many volts when the charging and discharging current flowed.

The reason that the problem gets worse as the frequency goes up is because at high frequencies we don't have so much time to charge the stray capacitance before the volts reverse and we have to start discharging it again. So we never reach the point where the stray capacitance is fully charged which is when the voltage lost across the output impedance would be zero.

I think their use of the word 'swamp' is a bit colloquial. But basically they are saying that a low impedance stage can deliver a lot of current without losing much voltage. If you like, the large current 'floods' into (and out of) the stray capacitance and quickly charges (and discharges) it.

Hope this helps,

VB

I addition to my post above,
The stray capacitance will not have simply vanished, it will still be there, just by providing a suitable low impedance driver the effects of it are effectively made negligable and are as your paragraph relates are 'swamped' or overridden by the much lower impedance of the driver impedance.

Iam trying my best to answer this as a simple 'mender', rather than as a design engineer (which Iam not!), and it is somewhat a simplistic view, one of the 'electronic design' posters on here may well be able to provide a better and much more comprehensive explanation.

Edit: cross posted with Valvebloke

Edit II: the effects of stray capacitance on the driver are manifested on screen by a smearing or streaking of the picture, or of the affected gun in a CTV.
I have seen this first hand on a certain Panasonic CTV Chassis where one of the driver 'peaking coils/inductor' has gone O/C.

Another way of looking at what is going on can be found in simple audio amplifiers.
On the tone control there is a "domesticated" capacitor connected in series with the POT that cuts the treble when adjusted to low resistance. The driver has a high output impedance in order to make this work correctly.
Cathode follower set ups are a good choice for dealing with the issue being discussed in the book.

A bifilar wound choke is invisible to the heater current (as they cancel) and isolates the capacitance of the heater transformer. But I have no idea if people repairing TVs knew this in 1950s to 1970s. It was known to designers, as "common grid" direct filamentary cathodes use the scheme at RF. It's also used to feed HT or Bias to push pull RF amplifiers.

The capacitance of the CRT circuit and the resistance of the driver valve is a simple RC low pass filter. You can actually measure both and predict the effect.

A cathode follower has no voltage gain, but maybe 100 to 1000 times less output impedance, so the frequency of the low pass filter created by driving resistance and load capacitance is 100 to 1000 times higher.

Try looking at it another way.

Say that you have a Video Output valve with a 10K load resistor. The total capacitance from all sources, including tube and valve inter-electrode capacitance, is a mere 50pF.

This is effectively in parallel with the load resistor. So what effect does 50pF have?

At DC, none. At 1kHz its reactance is 3.2M - negligible. But, as frequency increases, the reactance decreases.

At 1MHz it is only 3.2k and at 5.5MHz (the upper limit for UK 625 line transmissions) it is a mere 580 ohms in parallel with that 10,000 ohm load!

If, for convenience, we assume that the voltage output to the CRT is directly proportional to the effective load resistance - the load resistor in parallel with the reactance of the load capacitance at any specific frequency - we find that it has dropped to to a mere 24% at 1MHz and plummets to only 6% at 5MHz!

If we now add a cathode follower with an effective output impedance of 100 ohms, the situation is entirely different, dropping by only 3% at 1MHz to 97% and by just 14% at 5MHz to a healthy 86%

I think the cathode follower can indeed be said to have swamped the capacitive effects!

For more detail see the tables below: