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 croccodillo 7th August 2006 09:03 AM

Parafeed doubt

Hi,

I was thinking about CCS loaded parafeed configuration, and there's something I cannot understand.

As you know, the biggest advantage of the CCS loaded parafeed topology is that the lower tube is feed with a constant current (obviously), making it work in a horizontal working curve (I hope I'm clear).
THIS is the schematic I'm talking about.

Actually, this assumption is true only if you are in an idle condition; it becomes false if you have an AC signal driving the output tube grid.

I'll try to explain:

1. Idle condition
Imagine you are in idle condition (no input signal): let's say the output tube (lower tube) plate voltage is of about 300V.
There's no AC signal, so the current flowing through the output capacitor is zero, and all of the CCS current flows through the lower tube.
In this case the current in the output tube is exactly the same of the CCS current.

2. Positive pulse on Input
If you apply a positive pulse at the input, the lower tube will change the working point, and the plate current will drop from 300V to a lower value (let's say, 200V).
In this case we'll have a current flowing through the output capacitor into the output tube: the current will ADD to the CCS current, and then the total current flowing through the output tube is the SUM of the two currents: CCS current + Output capacitor discharge current.

3. Negative pulse on Input
The same of the previous point: the output tube will change its plate voltage to somewhat higher than 300V (let's say 400V), and then you'll have a current flowing through the output capacitor, but from the CCS to the ground.
In this case the total current through the output tube will be the DIFFERENCE between the CS current and the output capacitor current.

If I'm right, that means that the current in the lower tube is not constant, but continously changes with the input level.
Moreover, the output tube is not working on a straight line (constant current working curve), but on a different curve (or better, on a different series of curves).

The circuit in the schematic uses the R5 resistor to sense the current, and then CHANGES the current through the upper tube in manner to always keep the current in the lower tube constant.

In fact the VCS will change its current as needed.
In case of a positive pulse on input, the VCS will decrease its current in manner the sum of the two currents (VCS current + output capacitor current) is always the same.
Viceversa, in case of a negative pulse, the VCS will increase its current in manner to feed the output capacitor and, in the same time, feed the output tube with the correct current.

With this topology the output tube is ALWAYS fed with the right current, increasing the linearity.

I hope everything is clear.

Please tell me if I'm right or if I've made some stupid mistake analizyng the topology.

Ciao,
Giovanni

 croccodillo 7th August 2006 09:16 AM

Another question:

I want to build the amplifier in the schematic, but I need a triode for the VCS that can handle up to 600V, and can go down to 100V plate voltage, with a limited grid voltage (high mu?).
With such a triode I could drive the upper tube (VCS tube) grid from 600V to 100V, and grant in this way a reasonable output voltage swing.
I was considering a 6SN7 tube, but the maximum voltage is not enough.

Can you help me? what kind of triode should I try?

I think I will try also a mosfet, connected in this way.

Thanks again,
Giovanni

 EC8010 7th August 2006 10:07 AM

Let's assume that your circuit works and that it allows a perfectly constant current in the lower valve despite AC signal changes. If so, you are extracting no power from that lower valve, so what is the point in it dissipating any power? Might as well have a small-signal valve. If that's so, then any power extracted from the circuit must come from the upper valve. In effect, what you would have is a power cathode follower. But think a little further. If it's a power cathode follower, than why would you want the device below it to pass a large constant current? That would mean that at the instant that the AC load requires a peak positive current, the upper device has to supply that plus the DC constant current for the lower device. That means you've reduced the efficiency of the upper valve and increased its dissipation unnecessarily. Sorry, but I don't see any advantage.

 croccodillo 7th August 2006 10:13 AM

As usual, you're right.
Thanks for the explanation, now it's clear.

Ciao,
Giovanni

 EC8010 7th August 2006 10:20 AM

You're welcome. Bear in mind, though, that any parafeed circuit has the effciency problem I mentioned, but it's tolerated because avoiding passing DC through (SE) output transformer is a worthwhile advantage.

 croccodillo 9th August 2006 07:35 AM

The previous schematic, slightly modified: the output transformer is connected not anymore to ground but above the R5 current sensing resistor.
In this way the circuit has been transformed in a standard parafeed stage, the upper tube becomes a CCS, and the output tube makes all the dirt job.
The big advantage of this stage is that implements an all-tube CCS.

If it can work, I need a suitable device for the triode V3A.
It should work up to 600V, have an high mu and a low bias voltage.

Does it worth to connect hte upper tube as a pentode, and how?

Thanks,
Giovanni

 EC8010 9th August 2006 08:08 AM

Looking at your circuit, it strikes me that any change in the relative gains of V2 and V3A will affect the impedance of the CCS, possibly changing it from positive to negative (which would cause oscillation). I suggest you test the principle on the bench (perhaps with smaller valves) to see what happens.

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