is current in speaker load to be anyway closer to psu current ratings?
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 8th December 2012, 06:17 PM #21 Mooly   diyAudio Moderator     Join Date: Sep 2007 Post #51 here might give you some insight. Put your own numbers in, Power Supply Preference
donpetru
diyAudio Member

Join Date: Jun 2007
Location: Romania
Quote:
 Originally Posted by rhythmsandy i was looking at the nelsons article where he states about for 200watts he recommends using 700w and for stereo to have a 2kva and sounds pretty sufficient.. but i have a practical example for 300 I=sqrt(P/load) sqrt(300/8) = 6.12amps now here the voltage ratings of the trafo is 56-0-56 so considering this when compared to the psu ratings it has to be 2 x 300 = 600W as common mode of selection which will result in Arms in trafo = 600/(56*2) = 5.35 amps... now the question is when there is 6.12 amps in speaker then where does the extra current come from? or 50% has to convert into heat so we need 6.12 amps more to burn into heat... so the ratings of the trafo has to be atleast min of 12.24amps which results as (56*2)x12.24 = 1370VA and with 5-10% losses its roughly 1500VA as standard size... if that is right? when we take a 4 ohm load into 500W the same amplifier which works in 300watt in 8 ohm.. the the current in speaker is 11.18 amps and when we double this its 22.36 so (56*2)x 22.36 = 2504VA now it looks like its 5 times the ratings are the right ones to deliver the continuous rated output.. are these calculations are right?
Let me explain more detailed example taken in the above quotation. It's something you do not understand well.

Let's start with your example. You say have two 300W audio amplifier where each amplifier works on impedance of 4 Ohms. So,

sqrt(300/8) = 6.12amps

two amplifiers ---> 12.24amps

Then say you have a 600W transformer with two windings 56-0-56. Rated current of the transformer is:

600/(56*2) = 5.35 amps

So far everything is OK, now let's see what happens with DC current value (in the picture below see the current note with Ir0 - see diagram from the middle):

Ir0 current value is calculated based on the type of bridge rectifier using the formulas presented in the PDF below. On the other hand, the same Ir0 can be determined knowing the maximum current that can flow amplifier it. In this case we consider that the audio amplifier works with sinusoidal test and thus the current through the load of 4 ohms will be sinus. Follows:

- first, shall calculate continuous supply voltage (Ur0) of the amplifier to have 300W / 4 Ohm:

Urms = P / 6.12 = 300 / 6.12 = 49.01Vrms --->>>>

Ur0= (1.41*49.01)+Usat, where Usat is the saturation voltage of the output transistors. For simplicity of calculations: Usat=5V.

Ur0= 69.11+5= 74.11VDC which symmetric mean at least +/-37VDC.

- now go back to the PDF posted below and see where the formula (see row called "RMS voltage of the transformer secondary"):

Us = 0.8*Ur0 = 0.8*74.11 = 59.28 Vrms ~ 60Vrms or at least two windings 30-0-30 VAC. Basically it is best to choose at least 35Vrms for each winding to avoid getting large distortions nominal power amplifier. This in conjunction with the selection of high-value capacitors for filtering brings us in a position to build a good feeder or PSU.

Now let's see what happens if you choose a voltage greater than 35V.
As we are approaching the nominal power amplifier, then Ur0 will drop dramatically and the amplifier will distort increasingly louder lower frequencies (mostly). If we talk to a real 4 Ohm load consists of inductance and resistance (ie, LR amplifier load), then we have a small phase shift between current and voltage; you have to keep it because otherwise the speakers will not sound out correctly . Previous statement is even more true as we approach the lower frequencies. To keep the phase shift that will have the filtering capacitors have a minimum value (usually, it also says, a higher value is the better).

REMEMBER: Ideally, for an audio amplifier to sound good at all frequencies and all types of power that can flow, maximum current flowing through the speaker must be smaller than the current through the secondary windings of the transformer, regardless of the value of filtering capacitor. In practice, this goal is difficult to meet. The average value of an audio signal, with few exceptions, is less than the rms value of a sinusoidal test signal. For this reason, the PSU can resize without fully complies previous desire. Here, the theory is much larger and more intricate aspects.

Finally, you should remember (way of example quoted above):
*) A smaller current in the secondary windings will cause the reduction of supply voltage of the amplifier and thus will decrease the current through the speaker;
**) Smaller current through the speaker will involve a pressure drop at low frequencies, which can result in poor quality reproduction of the bass.
***) For quality audio reproduction using appropriate voltages to achieve the desired output power.
Attached Files
 Typical phase rectifier diagrams.pdf (55.0 KB, 13 views)
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 8th December 2012, 08:36 PM #23 rhythmsandy diyAudio Member   Join Date: Aug 2012 how do u calculate the Arms in the trafo?
 8th December 2012, 08:41 PM #24 dmills   diyAudio Member   Join Date: Aug 2008 Location: High Wycombe Sorry,I am not quite shure I see it. From P = V^2/R I get Vrms (300W@4R) = ~35V, so Vpeak (for a sine wave drive) = ~35V * sqrt (2) = 49V near enough, so say rails at +-55V or so for the sake of argument. Now if I allow say 10% ripple on the DC rails, that means my transformer has to put out about 60-0-60 at peak at the output of the rectifier bridge. Allow for say 2V across the diode bridge and I need 62-0-62V at the tranformer secondary, allow for say 10% low mains voltage and I now need something like 68-0-68 peak to ensure I will meet spec at low input voltage. Convert to RMS valtage for transformer secondary I get 68/sqrt(2) = 48-0-48 as the desired transformer. Now, as to current rating, we really have two considerations here, one is what average power level we expect the thing to work at and the other is what transformer regulation we are prepared to accept. Note that in the general case the transformer does NOT supply the speaker current directly (The diode bridge only conducts when transformer voltage exceeds cap voltage after all), so the issue is strictly one of heat, conduction angle and regulation, also note that transformers are mostly rated for long term thermal limited power, they will (within the limits of their regulation) supply large amount of power for short intervals if required (but see my comments on the conduction angle). While it is undoubtedly true that the average current in the transformer windings must equal the average current in the speaker (Ignoring minor supply currents in the amp, and note average, NOT RMS), that does NOT mean that the transformer must be sized as it if had to supply those currents directly to the speaker. Regards, Dan.
 8th December 2012, 10:08 PM #25 stevensoosai2527 diyAudio Member   Join Date: May 2012 Audio amplifiers are not dc to ac power inverters, thus contious output of 15 amp will not occur unless the material being played is extremely compressed audio. Important to be noted, continous 15A current flowing into a 4 ohm load will fry up voice coil in 2 minutes unless your woofers are water cooled type. Audio signal of most recorded material have intermittent high transients but at most of the time they tend to be at much lower level, so there is no way any amplifier need to work at peak power continously. So trafos can be moderate, filter caps can be hugh, good enough for most cases.
 8th December 2012, 10:15 PM #26 dmills   diyAudio Member   Join Date: Aug 2008 Location: High Wycombe Yep, a common assumption is 9dB crest factor (average power is 1/8th peak), which is in practise low for anything except some hypercompressed dance music in a multiway rig. One place where misunderstandings here can bite is if you do a simple minded swap of a SMPSU in place of a conventional transformer only to have it go into hickup mode as soon as the volume is turned up (Hey, I was 13 years old at the time). Regards, Dan.
 9th December 2012, 02:25 AM #27 rhythmsandy diyAudio Member   Join Date: Aug 2012 i got an R-core Transformer from a Aerospace / Defence electric companies who actually cater to them. The grade of construction if you see it you dont believe it how much it weighs as well. We have requested for 1KVa but its buit to 1.32KVa and overall the regulation is made to 2% and im about to go with 2.5KVa by next month and 6KVa in near future... the designer said its built in such a way even after drawing complete current rated at 1kva there will be no voltage drop at that ratings and especially when you use for audio application he even said the music is dynamic so the ratings can be minimal then I just said one thing before he built it I said.. im using this transformer with a unique amp design which runs in classA and 100% duty cycle and will be used in theaters running ruggedly so do consider this factor or I dont want to pick it... he said fine if you want to get it that grade its fine... Imageshack - photo0294o.jpg here consider that im using it super ruggedly even for 21 inch subwoofer applications in theater....
 9th December 2012, 04:01 AM #28 JMFahey   diyAudio Member     Join Date: Mar 2009 Location: Buenos Aires - Argentina Short (and correct) answers: 1) if split rail power amp (direct coupled to speaker, +/- V rails) and sinewave *just* clipping, each rail "sees" a resistive load = Pi x Rl being Pi=3.14 and Rl =speaker impedance 2) if single rail power amp (capacitive coupled to speaker, only +V rail) and sinewave *just* clipping, the single rail "sees" a resistive load = 2 x Pi x Rl being Pi=3.14 and Rl =speaker impedance. Very useful equation to design resistive loads to *test* PSUs under real world conditions. Straight from the Motorola Solid State Amp Design Handbook , around 1971. They will ripple, drop and overheat just the same as if loaded with the amp they are intended to feed.
AndrewT
diyAudio Member

Join Date: Jul 2004
Location: Scottish Borders
Quote:
 Originally Posted by rhythmsandy ........................the designer said its built in such a way even after drawing complete current rated at 1kva there will be no voltage drop at that ratings.................
you have got this wrong.
There is no way a transformer designer ever told you that:
Quote:
 there will be no voltage drop at that ratings
Every transformer has a voltage drop when running at it's rated output.
That is how the transformer manufacturer specifies the transformer regulation.
regulation = [output voltage on no load / output voltage feeding rated load *100]-100
eg.
reg = [36.3/35.1*100]-100 = 3.42%

This (3% to 4%) would be a typical value for a 1kVA toroid transformer of retail quality, Not super quality, just standard retail quality.
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regards Andrew T.

 9th December 2012, 11:21 AM #30 JMFahey   diyAudio Member     Join Date: Mar 2009 Location: Buenos Aires - Argentina Agree and add: transformer makers always specify regulation on resistive loads, and can prove it in their bench if you ask them to. BUT a capacitive filtered PSU is anything but resistive. Capacitors first charge to peak voltage, pulling a lot of current (the turn-on thump), then supply current to load on demand, and re-charge to peak voltage again on short, narrow bursts. Which can, for example, be 10X the average value, go figure, causing, as you should imagine by now, 10X the resistive loss you previously imagined. I learnt it the hard way, in long and heated discussions with my custom transformer supplier, when my PSU voltage dropped around 20% under load, and he proved, in his bench, that regulation was better than the 3% he had promised. The hard realities I learnt my lesson and now wind my own (for the last 35 years or so) , care less about magnetic loss and more about resistive one. My amps need fuses 50% larger than usual to survive turn-on thump, but that's a very small price to pay.

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