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MagneQuest/Peerless Forum Welcome! Need support, you got it. Or share your ideas and experiences. |
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In Reply to: Mike! Do you have any tranny optimized for pushpull parafeed, especially posted by Tanker on October 31, 2004 at 12:45:36:
Hi Tanker:the beauty or benefit of parallel feed in the case of PP doesn't really require any "optimization" of a particular PP trans. Use a good one of course is the cardinal rule. In a sense the optimization is a byproduct itself of parafeeding.
See if I can explain what I have in mind. Some of the alleged benefits of parallel feed is in separating out the AC signal path and the DC supply path. This applies whether the unit is PP or SE. The benefit is that you get much higher PSRR (power supply rejection ratio) in either case.
another benefit (perhaps subsidiary) is that the primary copper circuit has less "work" to do... i.e., once you remove the dc current from the primary of the OT (either SE or PP) then the copper current density is made much better because you don't have the heating effect (and I squared R losses) that accrue from the DC being added vectorially to the AC signal current. Or, in plainer english, the primary copper circuit (say it were wound with a 31 gauge wire) will appear effectively as though it were a larger guage wire once you remove the heating effects of the dc supply current.
Here's an example I worked out;
say a 10 watt output into a 5K CT transformer... the ac volts will be (at full power) 223.6 vrms and the ac signal current will be 45 ma rms. And let us say that there is 60 mils of dc supply current (power supply in series with OT) through the primary winding.
the I sub tot (total current) which is also called the heating current is the vector sum of the ac current and dc current.
In our example the heating current will be 75 ma.
a #31 wire has an area of 79.21 circular mils.
therefore the copper current density (always expressed as circular mils per amp) will be 1056. which isn't bad at all.
But if you remove the dc current and it's heating effects and divide the 79.21 (area of conductor in circular mils) by just the ac signal current of 45 ma rms then your same wire guage will have a current density of 1760 circular mils per amp. This is an improvement of 166 percent.
In order to get the series fed unit the same copper current density as our parafed example you would need to go to a 29 guage wire which has an area of 127.7 circular mils. Now divide the 127.7 by our total heating current (in the series fed example) of 75 ma and you get a copper current density of 1702 circular mils per amp.
so the series fed unit would (due to the additional burden of carrying the heating current of the dc component) have to be wound with a wire 2 guages thicker to nearly equal the "performance" of the parafed unit.
and, from a practical point of view, this would be impossible. there is not enough room in the transformer to wind the same number of turns with a wire guage two sizes larger...
So.. the improvement in copper current density from excluding the dc current provides for us a more efficient transformer with lower copper losses. The copper circuit doesn't heat up as much and it's resistance doesn't increase as much as it would if it has to carry both the ac and dc currents mentioned above.
this same benefit also accrues to parafeeding single ended output transformers... your copper circuit becomes much more efficient in the absence of the heating effects of the dc current when it is excluded from the priamry winding.
So... after all that... from a practical point of view... use a good quality PP output trans, parafeed it, and then just count up your benefits...
1)improved power supply rejection ratio (keep the power supply out of the music business)
2)your primary's copper circuit will become more efficient and smile at you.
hope this helps....
also... Jeff might be able to add some insights as he has been experimenting with parafeed in his SETH amps....
best of,
msl
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- push pull parafeed benefits - mqracing 21:24:55 11/03/04 (0)
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