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Old 19th February 2013, 03:33 PM   #2201
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Quote:
Originally Posted by wahab View Post
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.

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?
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.
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?
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Last edited by TANDBERGEREN; 19th February 2013 at 03:43 PM.
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Old 19th February 2013, 03:42 PM   #2202
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But what has the current rating on the transformer label to do with the (0.0083/0.01s) charging spikes ?
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Old 19th February 2013, 05:02 PM   #2203
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Quote:
Originally Posted by jacco vermeulen View Post
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?
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Old 19th February 2013, 05:27 PM   #2204
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This is the ALFET application note. Read on from page 7 about output power.
http://products.semelab-tt.com/pdf/A...nNoteAlfet.pdf
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Old 19th February 2013, 08:26 PM   #2205
wahab is offline wahab  Algeria
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Quote:
Originally Posted by TANDBERGEREN View Post

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 View Post

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 View Post
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....
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Old 19th February 2013, 09:12 PM   #2206
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I am sorry Wahab.
You got it completely wrong.
There is parameters you completely ignore here.
You should have read anb old article that asket the right question:
Whats WATT in an amplifier.



Your definition on theese watts are close to SONYs definitions, when selling "home theatre systems" back in the nineties. They managed make a whole kilowatt out of a tiny 50VA transformer.

You are not quite there, but close.

"My watts" include the full power output, over the whole audible area, with a distortion less than 0,02%
You will not be able to drain the powersupply with more than there is to get anyhow.
All the "makebelieveWatts" will be far out of reach when thinking of using them for HiFi listening. To make coffe or toast Your sandwich it probably is fine ;-)


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. The rest of the voltage between theese 46V and supplyrail will produce a dissipated wattage of 228W. The same will occour when the voltage is -46V Add theese up and You will see there is consumed 492W, in the one rail only. Add up all the losses elsewhere in the amp and the supply you will soon find the "missing 133W" ( or VA, simplified). In the rectifier for instance.
There You will find nicely 20-30Watts alone.
Run it all through any simulation you find, and you will find that the supply of power from the transformer restricts the output power nicely.
In the ideal world one would find a SMPS who clinged the rail to 80VDC, and just let the amps run away. But to produce the "spike-power" of that nice, undistorted sinewave, running up to around 75V, the SMPS would have to prduce nearly 10Amps Thats 800VA.
But the output of the amplifier would "only" be some 351Watts.
Please look at the bare numbers and spin them through Your calculator, and You will find the thruth.
Not taking any concern about the output transistors SOA in any way, 350W would be the absolute maximum power to get out of this amp. Providing I had a transformer being able to spit out theese 10Amps to make the sinewave nice and round as it is at the input.
Thats at least a 1,1KVA transformer. (OR if I used the absolutely correct voltage: 1,2KVA)
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Old 19th February 2013, 10:18 PM   #2207
wahab is offline wahab  Algeria
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Quote:
Originally Posted by TANDBERGEREN View Post
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
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Old 19th February 2013, 10:37 PM   #2208
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Quote:
Originally Posted by wahab View Post
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.
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Old 19th February 2013, 11:42 PM   #2209
wahab is offline wahab  Algeria
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Quote:
Originally Posted by TANDBERGEREN View Post
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 View Post
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....
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Old 5th March 2013, 04:21 AM   #2210
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Default I AM LATE TO THIS PROJECT AND NEED TO KNOW PLEASE

Hello friends, about an year ago I bought 2 PCB of the Goldmund original project from NagysAudio. I received the original PCB's + schematics in time, with a bad draft drawing of the protection circuit. Nagys gave me some advice via email on how to get the parts and he was a kind person to me.

The PCB's seem to be OK but there is no silkscreen on it. The components layout came in a separate sheet. I bought 99.9% of the original parts from Digikey, Newark, Mouser, etc. I have the original Panasonic relay. My MOSFET's are BZ90X parts.

All this time I couldn't dedicate myself to the project that was kept in a box. Today I returned to the forum ( that I like so much ) to see if somebody could get this project up and running. Honestly, I am sad to know that there was a lot of trouble between some members and NagysAudio. I will keep my opinion about this particular because I didn't participate in the project in any way and also because I don't like to hurt anybody. I am simply sorry for the way the original intention ended.

Before writing a line, I tried to read as much as I could to understand where the project really ended. Please correct me if I'm wrong, so much time passed that I don't feel entitled to give opinions here, specially because I notice a lot of work from a group of members.

For some reason it seems to me that there is a new PCB version,
The Very Best Amplifier I Have Ever Heard!!!! #1087

After reading some posts #1297 ; #1317 ; #1325 ; #1349 ; #1444 ; #1449 ; #1453 ; #1458 ; #1462 ; #1467 ; #1484 ; #1504 ; #1530 ; #1534 ; #1566 ; #1575 ; #1577 ; #1630 ; # 1633 and # 1651 I can see that the original circuit has been modified, parts discussed, added trimpots for bias adjustments, fixed DC-problems, etc.

I don't want to bother at this point but before I attempt to build my actual PCB version I would like to kindly ask if there is a complete final circuit resuming all these changes made, specially concerning the protection circuit that is included in my board. I found this one
http://www.diyaudio.com/forums/attac...-schematic.pdf

I can't see the protection circuit related to this amplifier and I can't tell if it has been modified. I don't really know if these original boards really work

Also , I have read most of the Liliya testing and results and I guess if this circuit reflects all the modifications done regarding parts and circuit.

Should I buy a new pair of corrected PCB's from bigpanda ? Or the originals may still work ?

I would very much appreciate I you could tell me what the best option is.

Thank you in advance.

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