Plate load for push-pull - diyAudio
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 9th April 2008, 06:40 PM #1 diyAudio Member     Join Date: Nov 2005 Location: SoCal Plate load for push-pull There are many websites and threads that discuss plotting push-pull loadlines, but they all say something different regarding 1 thing. Which is true as an aproximation of what each tube will see in a push pull config. 1) Ra-a/4 for both Class-A and Class-B 2) Ra-a/2 for both Class-A and Class- B 3) Class-A, Ra-a/2 ; Class-B Ra-a/4 4) Class-A, Ra-a/4 ; Class-B Ra-a/2 Ra-a refering to transformer primary impedance anode-to-anode.
 9th April 2008, 08:12 PM #3 diyAudio Member     Join Date: Nov 2005 Location: SoCal cerrem, Thanks for the explenations. I don't take load lines as an exact science. Just looking to get in the ballpark. The class-A logic you described makes sense and is what I have been going by.There has been some people that seemed to know what they are doing claim otherwise. Which made me second guess. For class B however, it would seem that since one end of the primary is essentially open, that the inactive half of the primary would electrically disappear from the circuit. Leaving 1/2 of the primary with the full load. But that would mean the tube would see Rp-p/2 (similar to cutting the primary in half and using it single ended). How is that thinking flawed?
 9th April 2008, 09:10 PM #4 diyAudio Member     Join Date: Jan 2005 Location: Hartford I didn't follow that explanation at all. Could you maybe give a quick example? Assume you have a tube with a rp of X, for which, single ended, one would use a transformer with a primary Z of Y (I usually use Y >= 5X, but I'm sure others disagree). For push pull, class A, primary Z should be >Y, =Y,
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Join Date: Jun 2005
Location: San Diego, CA
Quote:
 Originally posted by Jeb-D. cerrem, Thanks for the explenations. I don't take load lines as an exact science. Just looking to get in the ballpark. The class-A logic you described makes sense and is what I have been going by.There has been some people that seemed to know what they are doing claim otherwise. Which made me second guess. For class B however, it would seem that since one end of the primary is essentially open, that the inactive half of the primary would electrically disappear from the circuit. Leaving 1/2 of the primary with the full load. But that would mean the tube would see Rp-p/2 (similar to cutting the primary in half and using it single ended). How is that thinking flawed?
You are correct for Class B, in that one half would electrically disappear... Here is what is going on... 1/2 the primary windings is 1/4 the impedance....
So if you have 4000 ohm plate-to-plate load and there is 1000 turns of wire on the primary.... That means there is 500 turns on each side or half of the primary with respect to Center-Tap...
Well, looking into 500 turns would be 1000 ohms load, not 2000 ohm load... This is because of the plate load is the "SQUARE" of the turns ratio... You need to account for the square law when halving the number of turns....thats were the 1/4 comes from...

Chris

 9th April 2008, 09:53 PM #6 diyAudio Member     Join Date: Nov 2005 Location: SoCal cerrem, That makes more sense, halving the primary divides the VOLTAGE by 2, which means the impedance is divided by 4. Thanks again. That's really quite simple, but it tricked me anyways.

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