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
P.S.: I was asking this already in the Solid State forum, but nobody answered yet..... So I decided to ask it "at home" in the Pass Labs forum
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
P.S.: I was asking this already in the Solid State forum, but nobody answered yet..... So I decided to ask it "at home" in the Pass Labs forum
Thank You Master!
I am realy sory, but I cann't see or understand where is this (8Ohm and 4Ohm load) in the text?
Is it mean that both shematic and description are about PP amp biasing?
I thought that "the 8-ohm load can be seen as being the equivalent to two 4-ohm loads in series" is true for Balanced operation.
Tyimo
I mis-read the second post.
If you have a push-pull Class A stage, then the bias for each device is set to
1/2 the peak output current. If those devices are in series, then the total bias
is the same value. If the devices are driving a transformer as shown, the total
bias is the sum of the biases for each device.
If you have a push-pull Class A stage, then the bias for each device is set to
1/2 the peak output current. If those devices are in series, then the total bias
is the same value. If the devices are driving a transformer as shown, the total
bias is the sum of the biases for each device.
Tyimo said:
Difference between push-pull and a balanced amplfier?
Can not be told. A balanced amplifier can use push-pull.
These two terms are words to try to describe circuits.
If you show me two circuits, instead of words, then I can try tell difference of circuits.
Otherwise I can not tell anything else, than a difference between two descriptions.
Each amplifier is unique. In details.
A desciption or label, such as PP or Balanced, is a summary of details.
Just like saying a person is 'carpenter' or plumber' will not give you very much idea of a person.
-------------------------
The more abstract the question, the description of a hypotetic situation,
the more abstract the answer.
The more concrete the situation in specific details, the more detailed and real the answer, the result.
John Broskie wrote in the Tube Cad articles for the same Push-pull MOSFET class-A buffer/amplifier circuit about the biasing two different statement:
1, "...12 of the power supply’s 15 volts will be deliverable into an 8Ohm 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 / 8Ohm), 1.5A per MOSFET
2, "...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, the circuits are the same, but the statements are different.
1, "...12 of the power supply’s 15 volts will be deliverable into an 8Ohm 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 / 8Ohm), 1.5A per MOSFET
2, "...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, the circuits are the same, but the statements are different.
Balanced single-ended is an oxymoron, but it describes two single-
ended Class A circuits run in balanced mode.
I don't think it has IEEE status as a circuit description.
Since the inductive loads are coupled to each other, I would probably
call that balanced push-pull.
If the Mosfets each had a separate inductor not coupled to the
other then it would be balanced single-ended.
ended Class A circuits run in balanced mode.
I don't think it has IEEE status as a circuit description.
Since the inductive loads are coupled to each other, I would probably
call that balanced push-pull.
If the Mosfets each had a separate inductor not coupled to the
other then it would be balanced single-ended.
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