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In the Forum: Melquiades Amplifier
In the Thread: To drive the 6C33C...
Post Subject: The expressed perfect operation point is a fiction in a way.Posted by Romy the Cat on: 11/3/2010
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 N-set wrote:
Risking that the topic has already been discussed and that I post
in a wrong category (Romy feel free to (re)move this post:
Is there any sense in trying 6C33 at higher volatges and lower currents?
More precisely, I've been wondering if  1/2 6C33 @ 300-320V, 30-40mA
is worth a try or the tube is bad non-linear there?
The amp I'm contemplating is a Romy-style-but-Push-Pull
amp for electrostatic headphones: something for the input - 6E5P - DC -  1/2 6C33 Push-Pull
stacked PSU's, the driver and the output PP choke loaded.
The load in a rough approximation is a 100pF cap (stator-stator).
 
N-set, I moved it into this thread; I hope you do not mind. I think the question is a bit irrelevant as there is absolutely no concept of linearity projected to operation point of a tube. Let me to explain.

You have an output tube, any abstract tube with any own characteristics. You have a load – a loudspeaker with all-imaginable parameters. Then you have single ended output transformer. Warn you right the way that I know nothing about PP and know nothing about OTLs. You pick transformer a ratio that would assure a proposer harmonic structure of your sound – in this care it will be proposer loading of your tube projected to given type of your load. Then you have a very simple task to assure that under the given load that your amplifies will give you max power. The concept of max power is an absolute key – there is no other considerations that need to be considered as at max power all seating come clean. You power your 6C33C with any voltage and any currant you wish but load it to your actual load and via your actual transformer. Then you begin to increase the input signal. You do it unit you have a first clip. The clip will be at the bottom of sinusoid or at the top of the sinusoid top of the sinusoid represent current, the bottom of sinusoid represent voltage (or opposite - I never remember). Let say the tope is current and the top got clipped. Then you increase current and reduce voltage. Playing like this you need to find a configuration when your tube has absolutely symmetrical clipping at tope and at bottom. In this configuration the tube will give up max power but the most important it will not run at voltage or at current starvation. Whatever voltage and current you will end up to be will be the “perfect operation point” for 6C33C”.

It might be a bit tricky and change of current will drive the plate impedance a bit and it would demand a change of transformer ratio, that would reset all your searches for symmetrical clipping. Any other tube, and particularly 6C33C will behave differently, particularly in beginning and in the end of it’s life. Still, the concept is there.

If the concept is there then why I declared the perfect operation point is a fiction. Because just naming the operation point meaningful only for standard typical load. What if I use my headphones from submarine sonar? They have impedance of 400R and to drive them with my custom transformer I would have symmetrical clipping at 300V and 25mA? Or alternatively: if I have 8 drivers of my woofer towers all connected in parallel and total impedance of 0.8R. I have a crazy transformer for this operation but the symmetrical clipping in this case mish take place at 50V and 1A of current on the plate. They all are hypothetic examples but all you need to understand is that in ANY situation of using ANY SET output stage it will be always the condition for symmetrical clipping and it will happen ONLY in one single dead locking between voltage and current at a given power rating of plate. THAT will be the ONLY ONE non-fictional operation point and this operation point worse only to you and your conditions.

Rgs, Romy the Cat
 

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