Trying to learn more about C4S used as source (VA) or sink (CF)

xjb123 · 962

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Offline xjb123

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I have an old Anticipation C4S kit (circa 1998) that I am finally going to install in my Paramour III. I've been reading the C4S handbook written by John Camille trying to understand exactly how the C4S circuit works. I think I got the fundamentals down pretty good when the C4S is used as a current source (VA), but it is confusing to me when it is used as a current sink (CF). Are they essentially equivalent, just operating on different sides of a tube?

I have 2 specific questions:

1) Is the R1 value that sets the current limit level the same for both use cases (VA or CF)?

2) If I install two C4S loads in a circuit that uses both halves of a tube - one stage as a current source followed by a second stage as a current sink (e.g. the Paramour III) - would I use the same R1 value for both? Seems like I should because they are running on the same tube - therefore, the same plate characteristics apply, right?

Any explanation would be much appreciated!



Offline Paul Birkeland

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1) Is the R1 value that sets the current limit level the same for both use cases (VA or CF)?
Yes.
2) If I install two C4S loads in a circuit that uses both halves of a tube - one stage as a current source followed by a second stage as a current sink (e.g. the Paramour III) - would I use the same R1 value for both? Seems like I should because they are running on the same tube - therefore, the same plate characteristics apply, right?
Do you mean the Foreplay III? 

The R1 value is determined based on the operating current you desire.  For example, in the Anticipation kit for the older Foreplay kits, I think the current source runs a bit more current than the current sink.  Or in a kit like the Crack, the 6080 runs 10x the current of the 12AU7.

Paul "PB" Birkeland

Bottlehead Grunt & The Repro Man


Offline xjb123

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OOOPs - good catch!

I was listening on my 2A3 Paramours when I typed the question.... Yes - I was actually referring to my Foreplay III with C4S.   ;)

OK - so then I guess that creates another question for me... Why would the current source run more current than the current sink? Both are operating in the same tube so I would assume there is an "ideal" operating point and therefore they should be the same. But I also heard that the operating point of a current sink can be affected by the load it is connected to (as a parallel resistance) - so is this the reason the selected current values might be different?

Has anyone ever published guidelines how to determine the best operating values for C4S circuits with various tubes? I've seen many for selecting resistance loads using tube curve graphs but nobody seems to spell out how it applies when a constant current source is applied - other than the slope of the load line just becomes completely horizontal.

Selecting optimal values seems like black magic to me.

Cheers,
Anthony




Offline Paul Birkeland

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The power transformer in the Foreplay I and Foreplay II has limited current.  The designer of the old Foreplay has passed away, so we can't ask him.  The power transformer did limit how much HV current could be sucked down.

The operating point of a current sink under a cathode follower doesn't vary based on the load, but the loadline will.  The current sink just sets the quiescent DC current and provides a very high AC impedance. 

Since the load line for a voltage amp is approximately horizontal, then you can draw horizontal load lines till you find one with nice equal spacings.  It is important to remember that the grid leak resistor on the following stage is in parallel with the impedance of the current source, so the 249K grid leak resistor in your Paramour amps would become the load line impedance you would want to use.  In something like a Foreplay, the grid of the cathode follower is a negligible load, so the horizontal line is an adequate approximation.

Paul "PB" Birkeland

Bottlehead Grunt & The Repro Man


Offline xjb123

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Got it. Makes sense - thanks for the feedback!

I finally realized that I was over-thinking the situation and that is probably what got me confused in the first place. Since I'm not trying to design the circuit from scratch, just find appropriate operating current levels - all I have to do is look at the original circuit and draw out the load line for the given tube. Assuming the designer did a good job (which I'm sure he did) - the optimal operating point will probably be near the center of the load line. From there all I need to do is draw a horizontal line across the graph and then look at the corresponding current. Now selecting R1 is not so mysterious after all and I feel a lot better about experimenting with different values knowing that I have a safe starting point.

There is always something new to learn with this hobby!



Offline Paul Birkeland

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the optimal operating point will probably be near the center of the load line.
This gets a bit more complicated since the load line with a current source hits the Y axis, then goes on infinitely, so there ends up being no center really.  Therefore you can look for regions that look good on the line, or you can consider how much voltage swing you're going to need and work backwards from there.  In a preamp, for instance, we don't need to position the load line to swing +/-200V. 

-PB

Paul "PB" Birkeland

Bottlehead Grunt & The Repro Man