I know i promised to ignore this thread but i see you need help and are asking for it in a verry polite manner, it would be foolish for me to refuse it, so what you need to understand is that power from a PSU is the product of voltage and current, not the current alone, and that as the voltage increases so does the power. Now let us immagine a constant current source givin a constant 1A, if for example at 12V you supply a resistive load of 12 Ohms that load will need exactly 1A, thus we say we supply the load with 12W of power, now if the load is powered 1 minute or 1 hour it makes no difference, as long as the PSU is capable of that 1A needed by the load it does not matters one bit how long the load is powered, the only thing that would be influenced is the temperature of the power devices on the constant current source.
It is like that with an audio power amplifier as well, as long as you supply the current needed on a given resitive load, as long as the power supply unit is capable of that current and is designed to cope with long term usage 1 minute or 100 munutes of 1,2Kw of power it is the same, the current does not drop in time, or at least it should not cus that is why you take a loong time in designing the power supply.
An AC voltage supplied to a speaker MUST result in a varying current through that load. You can substitute a resistor to make the circuit analysis simpler, but you will still find that the current in the resistor load VARIES with time and in time to the varying voltage fed to the load.
An AC voltages reduces to zero twice every half cycle. It does not stay constant.
The AC current through the load does not stay constant. It too varies.
Don't listen to analogies that suggest anything different.
@AndrewT ofcourse the current varry on a speaker for audio signall, and that is precisely what makes things a bit easyer for the sypply cus it does not need to supply that max current constantly but in pulses. Now when designin the PSU you take in account that max current at max power and max audio signall. y do not know you so i cannot make any claim about you, but i can say that it is dissapointing to see this much missunderstanding.
@jlind54 if you trully want my help than you need to accept my imput and not debate it, this way i will be glad to help any way a can, but if i have to fight your misstrust at every turn then i am affraid i have to say no, and good luck.
Best wishes to all.
Marian.
@jlind54 if you trully want my help than you need to accept my imput and not debate it, this way i will be glad to help any way a can, but if i have to fight your misstrust at every turn then i am affraid i have to say no, and good luck.
Best wishes to all.
Marian.
benb the desired sound is to be as accurate as possible and not exhibit a 'sag' sound similar to a guitar amp. Would having a supply with in sufficent current capacity cause that? My ultimate goal is to have something robust that can take loads of punishment. But I guess there is just something with the voltage and current to a load that I'm not understanding. Typically looking at an impedence curve of a speaker as frequency decrease the impedence decreases. I my mind a subwoofer spends most of its time in these lower octaves therefor would need a fair amount of current...
Marian I'd like to start back at the basics and perhaps you could enlighten me on your thoughts for the current going into the speaker from a mathematical formula standpoint. To make things simple we can base the figures on a 100 watt continuos output into a 4 ohm speaker. This would establish a Vref of +-20 with Vpk +-28.2. Rounding that up a bit to account for losses one could estimate a designed rail voltage of around +-35 possibly a bit more. A transformer with a 26-0-26 winding configuration would be suitable as it would achieve near 35 volts after diode rectification. figure an amp with 50% efficiency is used would require a minimum power supply rating of 200watts. My confusion comes with rating the amp capacity... What's the best rule of thumb? Use the 200watts/70volts = 2.85 amps (power supply only), 200/40volts =5 amps (power supply/calc rms audio Vref), 200/35=5.71 amps (power supply amp draw of one winding side ), or some other combination? Apologies for so many questions to you and again thanks for the information you have provided.
When dealing with a 50/60Hz linear power transformer i have always kept in mind theyr winding voltage for current calculation, and i go like this:
-In your case 100W in 4 ohm resistive load a continous max stable signall voltage ( witch is the most demanding on the power supply ) the signal would be 20Vac indeed.
-Now you have to establish the power rails but first you need to know what power devices does the amp has, cus fets eats a few more volts that BJT's, so if i have a bjt power stage than i would reserve about 2v for them ( it would be more than enough ) so a 30Vcc power rails would be required.
-We know that, now we have to establish the power drawn from the power transformer, and if we are building a stereo amp than we would have 200W audio power needed, and i take in account an efficiency of about 65-70% ( for class B/AB it is a good figure ) and so the power needed from the supply would be 200/0,65=307W minumum power needed.
-We know the dc voltage rails and power needed, now we have to establish the power transformer windings, and for 30Vcc the Vef would be 21,3Vac, at that we add the 1,4V drop on the bridge rectifier and we get 22,7Vac, we round it up to 23 or 24Vac if possible, cus in full load the voltage will drop some.
-Knowing the voltage for the windings is the last step needed to establish the current for the power transformer, and at 300W and 2x24Vac it is 307/48=6,39A, we round it up to 7A.
Those would be the calculations i go trough when desighnin the power supply for a given amplifier, power needed and load. That is for a linear psu witch a trully recommend to most people as long as we talk about ofline PSU, they are much more easyer to bult and ofcourse much more safe for one's life. Hope that helps
-In your case 100W in 4 ohm resistive load a continous max stable signall voltage ( witch is the most demanding on the power supply ) the signal would be 20Vac indeed.
-Now you have to establish the power rails but first you need to know what power devices does the amp has, cus fets eats a few more volts that BJT's, so if i have a bjt power stage than i would reserve about 2v for them ( it would be more than enough ) so a 30Vcc power rails would be required.
-We know that, now we have to establish the power drawn from the power transformer, and if we are building a stereo amp than we would have 200W audio power needed, and i take in account an efficiency of about 65-70% ( for class B/AB it is a good figure ) and so the power needed from the supply would be 200/0,65=307W minumum power needed.
-We know the dc voltage rails and power needed, now we have to establish the power transformer windings, and for 30Vcc the Vef would be 21,3Vac, at that we add the 1,4V drop on the bridge rectifier and we get 22,7Vac, we round it up to 23 or 24Vac if possible, cus in full load the voltage will drop some.
-Knowing the voltage for the windings is the last step needed to establish the current for the power transformer, and at 300W and 2x24Vac it is 307/48=6,39A, we round it up to 7A.
Those would be the calculations i go trough when desighnin the power supply for a given amplifier, power needed and load. That is for a linear psu witch a trully recommend to most people as long as we talk about ofline PSU, they are much more easyer to bult and ofcourse much more safe for one's life. Hope that helps
I'm curious to see how this works out and its good to go over the theory once in a while. And while we all agree music and speaker impedance are anything but constant we have to have some "standard" method of arriving at realistic figures. So here's what I came up with 
100 watts into 4 ohms. That means the required amplifier output voltage is V= √W*Ω that calculates as 20 Volts RMS.
What rail voltage do we need. 20 volts RMS is 28.28 volts peak and 56.56 volts peak to peak. Knowing this enables us to "guestimate" the likely supply voltage. No amplifier can put its supply "across the load", there are losses. These occur in the output devices themselves, in the emitter resistors, in the wiring, in fuses etc etc so we have to make assumptions now. The figure could be anywhere in the 5 to 15 volts range. Lateral FET's in particular have a high Rds value and are consequently "lossy". Lets say 8 volts is a "reasonable" overhead. That implies we need our 28.28 volts peak PLUS another 8 volts.
So we get 36.28 volts as the supply voltage (that's plus and minus 36.28 volts don't forget) because the amplifier can only "use" one rail at a time.
So moving on to the transformer we get 36.28/√2 which is 25 volts. Thats the AC voltage needed to give 36.28 volts after rectification and smoothing. Or is it ? The bridge rectifier has losses and these are higher at high currents. Higher than you might think. We probably need add another two or three volts or more to overcome these losses.
So now we are at around lets say 28 volts AC. For the dual rail supply that means a 28-0-28 (or higher) Vac transformer for our 100 watts into 4 ohm amplifier.
What about current though ? This is where it all gets more complicated and I had to refresh my knowledge on some of the more obscure formulas (college was a long time ago)
The peak current in the load is easy though. Ohms law. 28.28 volts peak across 4 ohms is 7 amps. The average current is Ipk/pi giving 2.2 amps. (Thats for one channel remember and can be used as a guide for transformer current rating. For 2 channel we are doubling the current rating)
What size reservoir cap. That depends what ripple voltage you consider acceptable. Also remember that huge capacitor banks draw hugh peak currents when supplying a steady steady state current into a load. Why ? Because the large cap discharges "little" in between mains cycles yet that energy taken out has to be put back in. This means the bridge and wiring and transformer supply huge peak currents but over ever shorter time intervals. Why so. Because the capacitor peak voltage drops little between cycles so that means that no charge is put back until the transformer voltage reaches at least this value (plus losses) and so the charging occurs only near the "tips" or peaks of the AC waveform.
Ripple voltage across the caps is the average current drawn * discharge time. I found conflicting info for this discharge time value with some formulas just quoting half the mains cycle time. i.e. 10ms for 50 Hz. That doesn't sound logical. Another reference quoted around 7ms on the basis that this is the average discharge time with around 3ms as the charge time. That sounds reasonable so we'll go with that. So ripple voltage is (Iaverage/Tdischarge)/capacitance.
Putting some numbers in and using our 2.2 amps average current and selecting say an 8200uf reservoir cap and we get around 1.9 volts pk/pk of ripple. That ripple voltage "reduces" our steady state DC rail value because the minimum rail voltage is now the lower part of that ripple waveform. So thats another thing to factor in the guestimation. That minimum dip has to be equal or greater than the steady state DC rail voltage needed.
And again remember. The calculations were for one channel not two.
100 watts into 4 ohms. That means the required amplifier output voltage is V= √W*Ω that calculates as 20 Volts RMS.
What rail voltage do we need. 20 volts RMS is 28.28 volts peak and 56.56 volts peak to peak. Knowing this enables us to "guestimate" the likely supply voltage. No amplifier can put its supply "across the load", there are losses. These occur in the output devices themselves, in the emitter resistors, in the wiring, in fuses etc etc so we have to make assumptions now. The figure could be anywhere in the 5 to 15 volts range. Lateral FET's in particular have a high Rds value and are consequently "lossy". Lets say 8 volts is a "reasonable" overhead. That implies we need our 28.28 volts peak PLUS another 8 volts.
So we get 36.28 volts as the supply voltage (that's plus and minus 36.28 volts don't forget) because the amplifier can only "use" one rail at a time.
So moving on to the transformer we get 36.28/√2 which is 25 volts. Thats the AC voltage needed to give 36.28 volts after rectification and smoothing. Or is it ? The bridge rectifier has losses and these are higher at high currents. Higher than you might think. We probably need add another two or three volts or more to overcome these losses.
So now we are at around lets say 28 volts AC. For the dual rail supply that means a 28-0-28 (or higher) Vac transformer for our 100 watts into 4 ohm amplifier.
What about current though ? This is where it all gets more complicated and I had to refresh my knowledge on some of the more obscure formulas (college was a long time ago
)The peak current in the load is easy though. Ohms law. 28.28 volts peak across 4 ohms is 7 amps. The average current is Ipk/pi giving 2.2 amps. (Thats for one channel remember and can be used as a guide for transformer current rating. For 2 channel we are doubling the current rating)
What size reservoir cap. That depends what ripple voltage you consider acceptable. Also remember that huge capacitor banks draw hugh peak currents when supplying a steady steady state current into a load. Why ? Because the large cap discharges "little" in between mains cycles yet that energy taken out has to be put back in. This means the bridge and wiring and transformer supply huge peak currents but over ever shorter time intervals. Why so. Because the capacitor peak voltage drops little between cycles so that means that no charge is put back until the transformer voltage reaches at least this value (plus losses) and so the charging occurs only near the "tips" or peaks of the AC waveform.
Ripple voltage across the caps is the average current drawn * discharge time. I found conflicting info for this discharge time value with some formulas just quoting half the mains cycle time. i.e. 10ms for 50 Hz. That doesn't sound logical. Another reference quoted around 7ms on the basis that this is the average discharge time with around 3ms as the charge time. That sounds reasonable so we'll go with that. So ripple voltage is (Iaverage/Tdischarge)/capacitance.
Putting some numbers in and using our 2.2 amps average current and selecting say an 8200uf reservoir cap and we get around 1.9 volts pk/pk of ripple. That ripple voltage "reduces" our steady state DC rail value because the minimum rail voltage is now the lower part of that ripple waveform. So thats another thing to factor in the guestimation. That minimum dip has to be equal or greater than the steady state DC rail voltage needed.
And again remember. The calculations were for one channel not two.
Nice marathon
All is verry usefull info, and i hope everyone will take it into account. Still i would think that 8V drop for BJT output devices seams like a bit of a stretch, 3-4V would seam more likely, FET's would eat about 8 or even more so, that depends on theyr Rds-on at stated current and also on Vth for the gate witch is usually arround 4V.
Regards,
Marian.
All is verry usefull info, and i hope everyone will take it into account. Still i would think that 8V drop for BJT output devices seams like a bit of a stretch, 3-4V would seam more likely, FET's would eat about 8 or even more so, that depends on theyr Rds-on at stated current and also on Vth for the gate witch is usually arround 4V.
Regards,
Marian.
Thanks for getting me on the right track and explaining things so well. Now that I have a bit better understanding I would guess that a standard mains transformer capable of the output described above would have the same sound as a smps capable of outputing the same voltage/current. Ideally there wouldn't be a corralation between the amps psu and the sound produced correct?
Reality check:
And the fets will never be full on either so rds-on has quite a little to do in the reality with the losses and power consumption of the amplifier.
In reality the amplifier biasing will require the most of the losses, then losses to the heat and some to the rectifiers and very little to the cabling (hopefully) and so on...
You guys are also using quite complex methods to arrive to your conclusions, when you can make it really really simple. Im a simple guy and liking simple things.
To have 100w of music power (mono), you need a 100w powersupply! It is that simple - in reality...
First off all - music is not a constant sinusoidal signal and the most power consuming parts in the music signal are the basses. Everything else is recorded with 'lower volume'.
So how much there is bass in the signal, will in reality determine your power supply requirement.
And now to the fun part!
If you want 100w power, you will need the 100w to the BASSES!
Remember everything else is recorded with lower volume...
Because 100W 2kHz 90dB/W will make you deaf or at least give you permanent hearing disability.
How much bass is there?
Depends on your liking. Lets assume you like some ridiculous tecno beat - is there 50% bass?
So are we already down to 50% of our powersupply power requirement - in reality?
Voltage calculation has been precented.
For current: I=P/U
That is the average.
And for Andrews transients we need a capasitor bank - a big one.
Many of you will now disagree and i accept that, because this is only my opinion based on my limited knowledge of the basic electro physics.
And the fets will never be full on either so rds-on has quite a little to do in the reality with the losses and power consumption of the amplifier.
In reality the amplifier biasing will require the most of the losses, then losses to the heat and some to the rectifiers and very little to the cabling (hopefully) and so on...
You guys are also using quite complex methods to arrive to your conclusions, when you can make it really really simple. Im a simple guy and liking simple things.
To have 100w of music power (mono), you need a 100w powersupply! It is that simple - in reality...
First off all - music is not a constant sinusoidal signal and the most power consuming parts in the music signal are the basses. Everything else is recorded with 'lower volume'.
So how much there is bass in the signal, will in reality determine your power supply requirement.
And now to the fun part!
If you want 100w power, you will need the 100w to the BASSES!
Remember everything else is recorded with lower volume...
Because 100W 2kHz 90dB/W will make you deaf or at least give you permanent hearing disability.
How much bass is there?
Depends on your liking. Lets assume you like some ridiculous tecno beat - is there 50% bass?
So are we already down to 50% of our powersupply power requirement - in reality?
Voltage calculation has been precented.
For current: I=P/U
That is the average.
And for Andrews transients we need a capasitor bank - a big one.
Many of you will now disagree and i accept that, because this is only my opinion based on my limited knowledge of the basic electro physics.
I like simple things too but sometimes it's just not that simple, you cannot espect to get 100W of audio power from an 100W PSU, i think not even Class D power amp does not have anywhere near 100% eff. So you really need to take into account the losses when calculating the power of the supply, and @Mooly presented us with a verry usefull way to go ( cind of more detaliled/complete version of what i have presented ), it is not that complicated at all if you think about it, but it is the correct way to go.
You really need to take in to account the losses when you calculate the VOLTAGE of the supply - this was my point.
Therefore 100w music = 100w power supply.
Well it is not like that ( and you do not seam to understand why ), maybe you are confusing switching with linear, well with linear power amp ( class B for example ) it is not enough to get the right voltage level, you need the correct amount of current as well, and it has already been presented a detailed way to do it, you cannot drop the current just because you have enough volts to get that 100W, cus the less curent you supply, the more volts will drop on full load, and you do have to think about tha too if going for the correct way to design the supply, look at @Mooly way, +/-36Vcc @2,2A ( witch ofcourse i would round it up to 3A ) gives about 150W of power from the supply, for 100W of power to the load, so again you are wrong and i hope u will understand why, sooner rather than later, it will save you poor designes.
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Yes. But:
P=U*I and real music in real life is not constant sinusoidal signal at full power CONSTANTLY, that is why 100w power supply for 100w music is more than you will ever need - in real life situations.
I.e. in this case 100w/(2*36v)=1.4A is what is needed...
You can measure the real life power consumption of the amplifier really easy...
You will be surprised how little it needs when the volume is fully cranked - it needs little... You should know this - because you are the man!
I find it complete waste of money to design for double the max power for a constant sinusoidal (test)signal - complete waste of money...
But this is diy - we can do what we each want. And I do find Andrews transients an important aspect in the design procedure - therefore money saved in the transformer goes to the bank - capasitor bank that is - many many many many of them..........
The one on my drawing board right now the bank holds 2500W of capasitor power!!!!
Amplifier is only 2*100W.
P=U*I and real music in real life is not constant sinusoidal signal at full power CONSTANTLY, that is why 100w power supply for 100w music is more than you will ever need - in real life situations.
I.e. in this case 100w/(2*36v)=1.4A is what is needed...
You can measure the real life power consumption of the amplifier really easy...
You will be surprised how little it needs when the volume is fully cranked - it needs little... You should know this - because you are the man!
I find it complete waste of money to design for double the max power for a constant sinusoidal (test)signal - complete waste of money...
But this is diy - we can do what we each want. And I do find Andrews transients an important aspect in the design procedure - therefore money saved in the transformer goes to the bank - capasitor bank that is - many many many many of them..........
The one on my drawing board right now the bank holds 2500W of capasitor power!!!!
Amplifier is only 2*100W.
Whatever!... i do not have either the time nor the will to get on the right track every misguided, self convinced technician, who ever listens guys like you may verry well pay the price on weak designs, i cannot care less. on the other hand whoever really wants my help, and accepts my imput i will be more than glad to give it any time i can.
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