Question about loaded impedance of PP OPT

Some people load a single ended 300b into 3k.

So for a Class A PP pair, that would be 6k (each tube sees 1/2 the total primary impedance).

4 tubes would be 3k and 8 tubes would be 1500 ohms.

Now if you are one who gives up a little power for less distortion at all other power levels and would load a single 300b with 5k then a pair in PP would be 10k, a quad 5k and a octet 2500 ohms.

Tre'
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RE: Question about loaded impedance of PP OPT, posted on April 21, 2016 at 10:27:18
Tubenstein Audiophile Posts: 251 Joined: August 5, 2011	Let's say operating each 300B at 80ma for class A, and 8 of them in push pull are load to 5K push pull OPT, how much power can we get?

RE: Question about loaded impedance of PP OPT, posted on April 21, 2016 at 12:09:12
Eli Duttman Audiophile Posts: 10455 Location: Monroe Township, NJ Joined: March 31, 2000	Keep in mind the power formula for this situation is P = V²/R. You need a smaller step down in the O/P trafo to realize more power in the load. 3 into 2 does not go. Eli D.

RE: Question about loaded impedance of PP OPT, posted on April 21, 2016 at 14:11:27
PakProtector Audiophile Posts: 12360 Joined: May 14, 2002	So four-x PPP into 2k5 a-a. That is the same as a single pair into 10k a-a, or the 5k SE load. The grid=0V intersection with a 5k SE loadline will give you the available voltage swing from your OP...and you did not mention what your B+ will be... cheers, Douglas Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.

RE: Question about loaded impedance of PP OPT, posted on April 22, 2016 at 08:30:44
Tubenstein Audiophile Posts: 251 Joined: August 5, 2011	Sorry, the B+ should be 400v. How much power should be for those 8 x 300B in PP if loaded to OPT with primary of 4K?? Thanks!

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Question about loaded impedance of PP OPT

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