Going further with the thread derailing, this is what WW2 soldiers in their foxholes and even concentration camp inmates used when needing an "audio diode" and Mouser was still 40 years into the future: enter the mighty razorblade_and_pencil diode
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3) Multi-grid tubes have much higher dynamic plate load values than triodes and thus present LESS loading to subsequent circuits.
That should be the other way around - they are more (not less) heavily loaded by subsequent circuits (because, as you pointed out, they have higher internal anode resistance, and behave more like current sources).
Parallel a 100KΩ input load with a 2KΩ triode output and you get something less than 2KΩ upon the input load.
Parallel that same 100KΩ input load with a 33KΩ pentode output and you get about 25KΩ...more than 10-times LESS loading upon the input load.
It's easier to generate a voltage GAIN across a larger resistance/impedance than it is across a lower resistance/impedance, because it requires less current to create that GAIN.
...and the Devil said: "...yes, but it's a DRY heat!"
Parallel a 100KΩ input load with a 2KΩ triode output and you get something less than 2KΩ upon the input load.
We may have a little terminology confusion going on. Loading is about the output impedance of a driving stage acting as a voltage divider with the input resistance of a subsequent stage. If the voltage is substantially lowered, that's heavy loading. If the voltage is barely affected, that's light loading.
Parallel a 100k input load with a 2k triode output, and you get 98% of the signal voltage making it through the voltage divider - this is called very little loading (only 2% of the voltage is lost). The load (the subsequent stage) has very little effect on the output voltage of the preceding (triode) stage.
Parallel that same 100KΩ input load with a 33KΩ pentode output and you get about 25KΩ...more than 10-times LESS loading upon the input load.
Load a 33k pentode output with a 100k input load (subsequent stage input resistance), and only 75% of the output voltage makes it to the subsequent stage. About 25% of the voltage is lost. This is much heavier loading.
It's easier to generate a voltage GAIN across a larger resistance/impedance than it is across a lower resistance/impedance, because it requires less current to create that GAIN.
Quite true, but how is that relevant to the question of loading caused by input and output impedances?
"Following" circuits are not always amplifying tubes, quite often they're RL/RC-tone shaping circuits, which attenuate passing signals QUITE a bit, so more loss means less output to subsequent circuitry, which has to MAKE UP (and usually exceed) those losses. A good designer (employed) uses as few components as necessary because of cost.
...and the Devil said: "...yes, but it's a DRY heat!"
Sorry, still not the best description of what´s going on.
Parallel a 100KΩ input load with a 2KΩ triode output and you get something less than 2KΩ upon the input load.
You mean the triode (generator) has 2k internal impedance.
It is driving the next stage, which has a 100k grid resistor.
Also applies to passive elements, next stage might have no tube whatsoever but consist of just a 100k resistor (passive load) to ground and nothing else.
Or a capacitor or inductor or any combination which, at least at that frequency, shows 100k impedance.
Any of these is loading the triode (generator) which has 2k internal impedance so in principle it´s lightly loaded.
The 2k is not "loading" anything, it´s the generator internal impedance, and definitely is not loading itself.
Parallel that same 100KΩ input load with a 33KΩ pentode output and you get about 25KΩ...
Again, the 25k pentode internal impedance is not loading itself and can not be called "a load".
To be more precise:
* "a load" goes from generator output to ground.
* "internal impedance" is in series with generator output.
Clearly these two concepts are very different.
more than 10-times LESS loading upon the input load.
how can you load a load?
A load loads a generator.
If you put 2 loads in parallel (such as 2 speakers in parallel) would you say one is loading the other?
The control-grid of a triode is 'normally' a HIGH impedance, so it presents a HIGH-Z load to whatever is sourcing/driving it. But, as soon as that grid goes positive enough to draw current (even slightly) it promptly becomes a LOW-Z load back to what is driving it...it has loaded its source beyond what it was originally intended to do.
As investigators like to say: "...Follow the current (instead of money)..."
Last edited by Old Tele man; 10-14-2017, 09:25 PM.
...and the Devil said: "...yes, but it's a DRY heat!"
The control-grid of a triode is 'normally' a HIGH impedance, so it presents a HIGH-Z load to whatever is sourcing/driving it. But, as soon as that grid goes positive enough to draw current (even slightly) it promptly becomes a LOW-Z load back to what is driving it...it has loaded its source beyond what it was originally intended to do.
As investigators like to say: "...Follow the current (instead of money)..."
Maybe, maybe not; that’s a clipping circuit. Either it’s an intentional clipper or it’s a linear circuit being pushed beyond its linear range.
Whatever, I don’t understand how that relates to your hypothesis of ‘a source impedance loading the following input’?
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