Post #51 here might give you some insight. Put your own numbers in,
http://www.diyaudio.com/forums/power-supplies/224767-power-supply-preference-3.html#post3274569
http://www.diyaudio.com/forums/power-supplies/224767-power-supply-preference-3.html#post3274569
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.
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.
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.
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.
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.
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.
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....
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....
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.
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.
you have got this wrong.
There is no way a transformer designer ever told you that:
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.
no load voltage = 36.3Vac
Rated load voltage = 35.1Vac
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.
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.
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.
No.
10times the current gives 100times the power loss. That's why we call it IsquaredR losses (P=I^2*R)
Use a soft start (with the correctly sized resistor for start up current limiting). That will allow close rated fusing.
it means that current still exists after the load has drawn entire 1kva there is still enough current till 1.3kva to get the voltage drop on load.
I agree that there will be the voltage drop once we put the load but the loads are considered for this so the ratings are actually more than what stated.
I agree that there will be the voltage drop once we put the load but the loads are considered for this so the ratings are actually more than what stated.
Acvtually the whole is quite simple.
One only have to have this simple rule in mind:
The current drawn out of a cirquit is equal to the current that is put in to it PLUS losses.
No matter how big capasitorbank one have in an amp, a continious power drawn via the output transistors to the speakers have to be substituted by the transfrormer, and the transfomrer alone. The capasitor bank is there to make the DC-voltage smooth and as free from ripple as possible.
This means if ine draws 6,25A through the speakers, then the transformer must deliver 6,25Amps to the power amplifier.
But since music is very dynamic we have no need to make the PSU as big as this would indicate in every aspect.
To demonstrate:
If the amp delivers 25V into 4Ohm, the power output is 156W.
This represents 6,5A to the speakers.
An amp to deliver this power continously shouls have at least a transformer wich can deliver theese 6,5A continously. As for voltage, 2X35V is good enough.
This would be +/- 50V DC supply to the amp.
With two channels we have a total of 12,5A, or 875VA or 2,8 times the power output.
The power here wich does not go to the speakers remains in the power amplifier and is dissipated as heat. A good 563W
No other losses taken in mind that is.
One only have to have this simple rule in mind:
The current drawn out of a cirquit is equal to the current that is put in to it PLUS losses.
No matter how big capasitorbank one have in an amp, a continious power drawn via the output transistors to the speakers have to be substituted by the transfrormer, and the transfomrer alone. The capasitor bank is there to make the DC-voltage smooth and as free from ripple as possible.
This means if ine draws 6,25A through the speakers, then the transformer must deliver 6,25Amps to the power amplifier.
But since music is very dynamic we have no need to make the PSU as big as this would indicate in every aspect.
To demonstrate:
If the amp delivers 25V into 4Ohm, the power output is 156W.
This represents 6,5A to the speakers.
An amp to deliver this power continously shouls have at least a transformer wich can deliver theese 6,5A continously. As for voltage, 2X35V is good enough.
This would be +/- 50V DC supply to the amp.
With two channels we have a total of 12,5A, or 875VA or 2,8 times the power output.
The power here wich does not go to the speakers remains in the power amplifier and is dissipated as heat. A good 563W
No other losses taken in mind that is.
I agree "Sooner or later you end up with TANDBERG"
like in the above post we cant deny the laws of physics and btw its better to overengineer the product as the classic one than building a japanese compact product
i agree Andrew has given good information but what he said is about precision but here we are here to build an over engineered product
better to build like krells than a entry nad...
my question to Tandberg is :
when we draw 100W of power form the psu then 50w is delivered at the output and 50w will be generated as heat right? not considering the losses..
like in the above post we cant deny the laws of physics and btw its better to overengineer the product as the classic one than building a japanese compact product
i agree Andrew has given good information but what he said is about precision but here we are here to build an over engineered product
better to build like krells than a entry nad...
my question to Tandberg is :
when we draw 100W of power form the psu then 50w is delivered at the output and 50w will be generated as heat right? not considering the losses..
Eh, NO.
There will be voltage over both speakers AND output transistors.
When current is drawn through them, they generate some other form of energy.
Remember the basics from school when talking about energy: The sum is always a constant. In the speakers the energy is converted to movement in the diagraphm, and some heat. In the output transistors the electric energy (I*V) is transformed to HEAT.
So there is no summation of the amps here. They flow through it all.
The voltage from the PSU is there as a charge over it all, and is divided between the output transistors and the speaker.
Simple answer to the transformer question.
The transformer VA should be roughly between one times and two times the total amplifier maximum output power.
A single channel 100W amplifier will work well with any 100VA to 200VA transformer.
It really is no more complicated than that.
If you can multiply by 1 (one) and multiply by 2 (two), then you can determine your own transformer requirement.
The transformer VA should be roughly between one times and two times the total amplifier maximum output power.
A single channel 100W amplifier will work well with any 100VA to 200VA transformer.
It really is no more complicated than that.
If you can multiply by 1 (one) and multiply by 2 (two), then you can determine your own transformer requirement.
Yes, that's correct
... for *power* loss ... but we were talking *voltage* loss ... we were talking regulation.Maybe instead of using the words "resistive loss" (which I thought was clear enough, given the context) I should have written "voltage loss due to wire resistance"
Yes, on hindsight, it sounds klunkier but more precise. Thanks.
Anyway in that case, I was speaking loosely trying to give the idea that the current that must be considered to calculate regulation is not the average power amp draw, nor 1.4 times it or 2x it as some seem to be saying, but a much higher one, because we have a duty cycle problem here.
To put it more precisely: the current charging the filter caps has low duty cycle (which is bad, and worse because power losses are squared)
And not only on the amp transformer, the pulsing energy demands travel all the way back to the Utility Company, if you think about it for a few seconds.
In fact, it must be considered also when rating the diodes themselves, datasheets include an "I squared T" parameter ... even fuses do .
Yes, PSUs *look* deceptively simple, "just a few parts" ... but they are more complex than that.
Back to the original problem, the conclusion that a 3% regulating transformer gives birth to an around 20% regulating PSU (as a whole) is valid.
And why 20% and not 30%?
Because loss is not only resistive but has other components which do not react the same way.
Yes, that's a technically correct suggestion, can't disagree with that, but I'm a minimalist at heart and found the 50% extra fuse rating both enough to survive years of normal functioning and at the same time protect the PSU from load shorts.
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