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Old 9th April 2008, 05:40 PM   #1
Jeb-D. is offline Jeb-D.  United States
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Default 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.
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Old 9th April 2008, 06:27 PM   #2
cerrem is offline cerrem  United States
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Look at it this way...... You have a Center-Taped P-P output transformer... The Center-Tap is essentially AC Ground....
Lets say you use just one tube on only one side of the OPT and not put a tube on the other side of the ouput transformer....neglecting DC core saturation of course...
You have a 8 ohm load termination on secondary, with a 4000 ohm Plate-to-Plate reflected impedance... SInce impedance is turn ratio squared and your only using just half the primary windings...this means that this only tube will see a plate load of 1000 ohms.... which is Raa/4 ....when "looking into" one half of the primary, with respect to ground (Center-Tap)....
Now if you have a true Class B circuit, 2 tubes...that being a tube on each side of the transformer....you still have the same scenario, since each tube acts independent and niether tube is ON at the same time...these 2 tubes alternate, as far as the transformer is concerned, it only sees one tube at a time working, or conducting.... so Class B tubes also see Raa/4 .......
Keep in mind that the two halves of the primary winding, with respect to Center-Tap (ground) are in PARALLEL.....
So in Push-Pull, each tube's plate resistance is coupled in parallel with the other tube's plate resistance as 1:1 ...... In Class B, one tube is ON when the other is OFF, therfore one tube has a plate resistance of infinity when it is OFF, thus not really in parallel..
Class-A push-pull has both tubes, ie.. each side of the primary conducting at all times... In a perfect ideal Class-A P-P situation, there will be no AC current flowing in the Center-Tap, since the net AC current will be 0 , two equal and 180 degree opposite currents will cancel...in real life things are bit different...
The source impedance, will now be considered as the two push-pull tubes in SERIES.... now you don't look at it as 2 tubes...but instead a single signal source of plate resistance x 2 working into the entire Plate-to-Plate load impedance... If you really want to examine what one tube sees in Class P-P, then it's the plate load/4 and the 1:1 coupling of the other plate resistance in parallel, or look at it as rp/2 looking into 1/4 load... That's where composite load-line derivation comes in handy...but after many years of doing this Raa/4 works out close enough.....
Class AB is somewhere between all this, also use Raa/4 .....
You may need to derive these to prove it for yourself...
My advice, don't make an exact science out of load-lines...I did many years ago and wrote many computer programs to do this analysis long before this stuff was popular.... The problem is the tubes you get are SH*T now adays and are all over the map.... and the real life reactive loads vary all over as well....
Find me a .1% tube and a .1% linear load and I will show you a .1% analysis
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Old 9th April 2008, 07:12 PM   #3
Jeb-D. is offline Jeb-D.  United States
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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?
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Old 9th April 2008, 08:10 PM   #4
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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, <Y? I always assumed Z = 2Y, but I don't know where I got that.
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Old 9th April 2008, 08:39 PM   #5
cerrem is offline cerrem  United States
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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
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Old 9th April 2008, 08:53 PM   #6
Jeb-D. is offline Jeb-D.  United States
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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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