The Very Best Amplifier I Have Ever Heard!!!! - Page 221 - diyAudio
 The Very Best Amplifier I Have Ever Heard!!!!
 User Name Stay logged in? Password
 Home Forums Rules Articles diyAudio Store Gallery Wiki Blogs Register Donations FAQ Calendar Search Today's Posts Mark Forums Read Search

 Solid State Talk all about solid state amplification.

 Please consider donating to help us continue to serve you. Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving
diyAudio Member

Join Date: Jan 2011
Location: Fosser, Aurskog-Holand, Akershus, Norway.
Quote:
 Originally Posted by wahab Actualy the maths is right providing we are talking of a single channel... The upper side of the push pull will suplly the 5.6A half of the time , during positive going of the signal while the bottom side will conduct only on negative cycles of the signal , hence the power supply rails will be sollicited alternatively, the average current drained by the amp on each rail will be 2,3A. On a class AB/B biaised push pull each side conduct only 50% of the time , the non active side conduction being negligible.
Well, this is really creative thinking.

Also the supply from the transformer acts in excactly similar way.
In the positive part of the cycle the transformer is supplying the cirquit theese 5,9Amps, and in the negative part is supply nothing?
Anyhow: The result is the same: The average power is the same: There is 5,9Amps available to supply the positive half, and 5,9Amps to supply the negative half. 44VAC in 8 Ohms is 5,6Amps average power going in both the positive and the negative halves of the full cycle.

But You should have for the creativity.

I could add this: In the seventies TANDBERG built an amp wich in no way was presented as any High End alternative, and probably will not today either. The total power output of this HiFi appliance is 2X75W (150W)
The maksimum powerconsumption is about 360VA.
Please compare theese figures with the 625VA transformers and what to expect from it.
The 625VA transformer is 1,76 times larger than the Tandberg Transformer.
2X75WX1,75 equals 260W.
Got the picture?
__________________
Sooner or later you end up with TANDBERG

Last edited by TANDBERGEREN; 19th February 2013 at 03:43 PM.

 19th February 2013, 03:42 PM #2202 diyAudio Member     Join Date: Oct 2004 Location: At the sea front, Rotterdam or Curaçao But what has the current rating on the transformer label to do with the (0.0083/0.01s) charging spikes ? __________________ It doesn't count how one deals with winning, but how to handle a loss (© DjT) https://www.youtube.com/watch?v=FJEzEDMqXQQ
diyAudio Member

Join Date: Jan 2011
Location: Fosser, Aurskog-Holand, Akershus, Norway.
Quote:
 Originally Posted by jacco vermeulen But what has the current rating on the transformer label to do with the (0.0083/0.01s) charging spikes ?
Who is mentioning that? And for what purpose?
__________________
Sooner or later you end up with TANDBERG

 19th February 2013, 05:27 PM #2204 diyAudio Member   Join Date: Mar 2008 This is the ALFET application note. Read on from page 7 about output power. http://products.semelab-tt.com/pdf/A...nNoteAlfet.pdf
diyAudio Member

Join Date: Nov 2009
Location: algeria/france
Quote:
 Originally Posted by TANDBERGEREN But what about taking this in Your calculations too? Also the supply from the transformer acts in excactly similar way. In the positive part of the cycle the transformer is supplying the cirquit theese 5,9Amps, and in the negative part is supply nothing?
Both rails caps are charged at each alternance , at 100HZ
rate since we are supposed to use a four diodes bridge
but the charging current is function of how much the cap
has been discharged during 10ms.

Now , the amp will drain current alternatively from the two rails
and generaly at higher speed than the one at wich charging
occur unless the output signal is less than 100Hz ,
but even then the amp will NEVER drain some current
from both rails at any given moment.

Quote:
 Originally Posted by TANDBERGEREN Anyhow: The result is the same: The average power is the same: There is 5,9Amps available to supply the positive half, and 5,9Amps to supply the negative half. 44VAC in 8 Ohms is 5,6Amps average power going in both the positive and the negative halves of the full cycle.
What is available is not what is absorbed , as said above the transformer
will only provide what is necessary to compensate for the previous 10ms.

Quote:
 Originally Posted by TANDBERGEREN But You should have for the creativity. I could add this: In the seventies TANDBERG built an amp wich in no way was presented as any High End alternative, and probably will not today either. The total power output of this HiFi appliance is 2X75W (150W) The maksimum powerconsumption is about 360VA. Please compare theese figures with the 625VA transformers and what to expect from it. I will help You a bit of the way: The 625VA transformer is 1,76 times larger than the Tandberg Transformer. 2X75WX1,75 equals 260W. Got the picture?

If the current was 5.9A with the two 44V AC rails simultanously loaded
with this current the total drained power would be at least 88 X 5.6 = 519W
and in fact quite more given that the rectified voltage would stand at about
+-50V with currents of this magnitude , increasing the absorbed power
to an unlikely 100 X 5.9 = 590W , and that s a RMS value....

A better way to make an estimation is to assume that the amp has
70% efficency at full power , hence at 250W RMS the absorbed power
will be 250/0.7 = 357W , the output stage will dissipate 107W of course.

These 357W will be what the transformer will provide for
250W RMS assuming a class AB/B amp , that s the laws
of physics , there s really no creativity involved....

diyAudio Member

Join Date: Nov 2009
Location: algeria/france
Quote:
 Originally Posted by TANDBERGEREN Look at the precise moment the voltage out of the amplifier is 46V At that presise moment (provided there is a clean resistive load) there will run theese 5,75A through the load. This equals 264W to the load.
At full voltage the output will be ALTERNATIVELY at +46V and at -46V
but it cant be at the same time at thoses voltages , this is where you
are confused as you think that the amp is draining the same current
from both rails SIMULTANEOUSLY , wich is not the case.

When the output reach +46V the upper side of the push pull will provide
264W and 5,75A while the bottom side will provide 0 W and 0 A,
when the output is at -46V the bottom side of the push pull will provide the 264W and the 5,75A while the upper side will be switched off , so each side will provide the 264W half the time , hence one side average supplied power (to the load) is 132W , that is , 264W half of the time.

Push?pull output - Wikipedia, the free encyclopedia

Amplifier - Wikipedia, the free encyclopedia

diyAudio Member

Join Date: Jan 2011
Location: Fosser, Aurskog-Holand, Akershus, Norway.
Quote:
 Originally Posted by wahab At full voltage the output will be ALTERNATIVELY at +46V and at -46V
Aha, a P-P-value of only 92V? Or a RMS-value of 32,5V = 132W?

As the power from the transformer sends it power to the amplifier ALTERNATIVELY. But then with p-p-value of approx 160V. Resulting in nice railvoltages of +/- 80V
And by no means, I clearlty states that the power is drawn from the rails ALTERNATIVELY. As from the transformer.

As with the links to the push-pull wiki.
Please look for DISTORTION when trying to get such power out of the amp.

THEORETICALLY efficiency of 70% is not the same as achieved efficiency of 70%. Also, the theoretical efficiency takes no concern to distortion, wich may be 100% at the point where the efficiency is meassured.
Please look at what to do to make a close to distortionfree amplifier.
No way You get the supposed 450W out of amplifier with this transformer.

But it seems as if You really don't understand the basic facts anyway.

For me this part of the discussion is over.
__________________
Sooner or later you end up with TANDBERG

diyAudio Member

Join Date: Nov 2009
Location: algeria/france
Quote:
 Originally Posted by TANDBERGEREN Aha, a P-P-value of only 92V? Or a RMS-value of 32,5V = 132W? But then with p-p-value of approx 160V. Resulting in nice railvoltages of +/- 80V
With a 2 X 44V AC transformer you ll have +-62 V DC at iddle.

Assuming the supply voltage doesnt sag and that the amp
has only 2V losses per rail the output will be at most either
at +60V or at --60V in respect to the ground , because the
load is connected to the ground , isnt it ?

Hence there will be no more than 60V at the output , be it
positive or negative and this will be the peak voltage ,
the RMS voltage will be about 60/sqrt2 = 42.4 V and
the peak current will be on 8R load about 60/8 = 7.5A ,
the RMS current being 7.5/sqrt2 = 5.3A.

RMS power is 5.3A X 42.4V = 224W.

The maximum voltage that will be provided to the load is 60V ,
do not confuse this value with the peak to peak voltage wich
is only an excursion that move +-60V around the ground wich
is the reference.

Quote:
 Originally Posted by TANDBERGEREN As the power from the transformer sends it power to the amplifier ALTERNATIVELY.
See rather the amp as a load that drain current alternatively
from the two rails in function of the output signal polarity
to better understand the power and current requirements
of the PSU....

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are Off Pingbacks are Off Refbacks are Off Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post pkgum Solid State 17 14th November 2013 02:24 PM beauistheman Multi-Way 12 22nd September 2008 06:49 AM Junior Solid State 0 1st August 2007 12:56 AM Staticjeep Music 0 9th March 2006 02:15 AM tschanrm Class D 1 17th November 2005 01:10 PM

 New To Site? Need Help?

All times are GMT. The time now is 12:22 PM.