Quasi-Balanced Method

[Stee]

Hi
my new Hybrid amplifier works with a particular system
preamplifier is a SRPP of 6SN7
amplifier are double on each channel (Lite A-680 DIYGENE) in bridge
but "Cold" amplifier don't have signal in input - Not Connected
sounds GOOD

http://www.esafono.it/quasi_balanced.htm

[GK]

Originally Posted by Stee

...in bridge but "Cold" amplifier don't have signal in input - Not Connected
sounds GOOD]

[janneman]

Originally Posted by Stee

[snip]
amplifier are double on each channel (Lite A-680 DIYGENE) in bridge
but "Cold" amplifier don't have signal in input - Not Connected
sounds GOOD

http://www.esafono.it/quasi_balanced.htm

So why is it there? Why not just ground that side of the load?

jd

[boraomega]

Originally Posted by janneman

So why is it there? Why not just ground that side of the load?

jd

Because he was afraid that it could look even more uninventive (I have to resist to write more appropriate words)!

[janneman]

Originally Posted by boraomega

Because he was afraid that it could look even more uninventive (I have to resist to write more appropriate words)!

Well he must have a reason, maybe it is a good one. I looked at his website but couln't figure it out, my Italian isn't that good.

jd

[Stee]

The compensation of the "character" of bipolar devices in complementary symmetry amplifiers
Normally occurs with the overlap of the curves PNP and NPN entire waveform:
Class A inappropriately use the method of complementary
. Because it does not comply with the direction of the curve.
To obtain an accurate compensation is required
Synchronizing the operating point of the two curves P and N
In the direction of work (increase / decrease) consistent.
This consistency in the pattern of vectors is "Bridge" or Balanced (Class AB)
To obtain an excellent result with the Class AB balanced (or bridge)
It requires, however, a perfect symmetry of the signals between the hot and cold branch.
Quasi-Balanced configuration is not bound by this requirement
As the signal is single (unbalanced) then the amplifier "cold"
It acts passively by providing current as a function of the amplifier "hot"
(to maintain balance in the DC voltage)
Making a component Tonal balance seemingly useless
But can cancel like a mirror with great accuracy the emphasis of the components.
(Google translator)

[Stee]

for better understanding

[cbdb]

To obtain an excellent result with the Class AB balanced (or bridge)
It requires, however, a perfect symmetry of the signals between the hot and cold branch.

This is wrong. The math is simple and probably easy to find in www. land.

[Stee]

http://www.esafono.it/ISY.ZIP

[janneman]

Originally Posted by Stee

http://www.esafono.it/ISY.ZIP

I see what you mean. Interesting thinking, but I am not sure it works that way. Let us suppose that the driven amp has the 'N' curve. This curve is the result of several things, like non-linear amplification and non-zero output impedance.

The not-driven amp only has an error due to it's non-zero output impedance because it is not driven. Why would that error be the exact complement of the non-linear amplification of the N part?

jd