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Old 7th October 2008, 01:14 PM   #1
Tyimo is offline Tyimo  Hungary
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Join Date: Nov 2002
Location: Hungary
Default Push Pull or Balanced?

Hi!
I am a bit confused by John Broskie's Tube Cad articles.

In the article:
http://www.tubecad.com/index_files/page0021.htm
He wrote for the Push-pull MOSFET class-A buffer/amplifier about the biasing, that :

"...This amplifier is effectively the equivalent to a normal source follower amplifier with +/- voltage rails twice that of this amplifier’s single rail; the magic of inductors. In this case, we can assume that 12 of the power supply’s 15 volts will be deliverable into an 8W load, so 24 volts of peak voltage would require 3A of peak current (as one terminal goes up the other goes down). Thus, an idle current of 3A is needed (24V / 8W), 1.5A per MOSFET. ..."

In another blog articles: http://www.tubecad.com/2004/blog0006.htm

He wrote a different text for the same circuit:
"... This amplifier is effectively the equivalent to a normal source follower amplifier with +/- voltage rails equal that of this amplifier’s single rail; the magic of inductors. In this case, we can assume that 12 volts of the power supply’s 15 volts will be deliverable into each side of an 8-ohm load, so a total of 24 volts (peak voltage) would require 3A of peak current (as one terminal goes up the other goes down). Thus, a total idle current of 6A is needed (24V / 8 ohms ), 3A per MOSFET . In other words, the 8-ohm load can be seen as being the equivalent to two 4-ohm loads in series, whose common connection is grounded; thus, 12V/4 ohms = 3A. ..."

Well, which statement is true?
I think the first is for PP and the second is for balanced amp. (As I learnt)
Anyway what is the difference between PP and Balanced????


Greets:

Tyimo
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