Hi I was looking around to gauge wattage and speakers when I stubled upon this long article: https://www.parasound.com/reviews/hfc404-hint.pdf
The integrated amplifier have one 1KW transformer and delivers 45A current and 2 x 160 at 8 ohm / 2 x 240 at 4 ohm!
How is this possible?
The integrated amplifier have one 1KW transformer and delivers 45A current and 2 x 160 at 8 ohm / 2 x 240 at 4 ohm!
How is this possible?
So what is worrying you
Those specs look VERY reasonable, nothing to write home about.
1kW capable transformer for a 2 x 240W RMS rated amplifier?
Check.
If you were to nitpick, the transformer should be actually rated as 1000VA.
Not sure you know the difference or even the word, but, hey, Google is your friend.
2 x 160W RMS into 8 ohm integrated amplifier?
Available since at least late 70´s, early 80´s ; way back then from the great Japanese Hi Fi equipment makers and certainly from American ones.
That said amplifier can put ot 240W RMS per channel if loaded with 4 ohms?
That´s what´s usually expected , I would be surprised if it is not
45A **peak** current capability?
Again what´s to be expected.
Math says that for 240W RMS into 4 ohms you need Sqrt(240W*4r)=31A RMS.
Even more Math
says that the peak current needed is 31A*1.4142=44AAre you surprised Designer chose to make it capable to put out (at least) 45A peaks?
FWIW he might have designed it to put out, say, 60A peaks, just to have some extra margin.
And in fact I am quite certain he must have done something like that, he´s VERY experienced and KNOWS speakers very often wander below rated impedance, thus needing even higher current peaks.
Speaking of that, Mr John Curl is a respected Designer, happens to be a Member of this Forum and no doubt you can ask him directly about your doubts.
Just check the Math first so as not to waste (his) time on inane questions.
If it had an "infinite" transformer, you would expect 2X160W@8r to imply 2X320W@4r.
The fact it "only" makes 240W @ 4r says the transformer sags about 15% more with 4r sine-test. Which is a very "reasonable" rating, and infinitely less costly than an infinite transformer. Indeed very likely the PT VA is around twice the Sine Audio Output, and twice 480W is a 1KVA transformer.
The "45 Amps" is really about hFE fall-off past 12A, which it will do in 4 Ohms for short transients. It is very common for the "max" current of a transistor to be several times higher than the "best gain" current. And also this means it will survive short bursts with loads at 1 Ohm, such as some sucky crossovers or ribbon speakers, although somewhat higher THD.
The fact it "only" makes 240W @ 4r says the transformer sags about 15% more with 4r sine-test. Which is a very "reasonable" rating, and infinitely less costly than an infinite transformer. Indeed very likely the PT VA is around twice the Sine Audio Output, and twice 480W is a 1KVA transformer.
The "45 Amps" is really about hFE fall-off past 12A, which it will do in 4 Ohms for short transients. It is very common for the "max" current of a transistor to be several times higher than the "best gain" current. And also this means it will survive short bursts with loads at 1 Ohm, such as some sucky crossovers or ribbon speakers, although somewhat higher THD.
Okay interesting! But how would such a transformer be configured?
I mean 160W / 31A = 5V ?!
JMFahey I am not sure that I read right, but you seems a bit hostile, I think.
If you look at my question, you will find that my question is "how it is possible".
I am new to DIY and are therefor trying to learn and how better to learn then ask here? I do really like your good and detailed answer, thanks a lot for that!
EDIT: The Saint you arrive to the result of 31V and 7.75A where JMFahey gets 31A I am not sure how to understand that?
I mean 160W / 31A = 5V ?!
JMFahey I am not sure that I read right, but you seems a bit hostile, I think.
If you look at my question, you will find that my question is "how it is possible".
I am new to DIY and are therefor trying to learn and how better to learn then ask here?
I do really like your good and detailed answer, thanks a lot for that!EDIT: The Saint you arrive to the result of 31V and 7.75A where JMFahey gets 31A I am not sure how to understand that?
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I don't see the problem. It's a 500w class AB so may need 750w just for that RMS power. But RMS is only 0.707 of peak, so to cover that we need say 1250w and lets face it, some overhead is always nice for transient peaks. And what good is an amp that can't meet the peak demands without sagging. Lets say 1500va.
I see it now... Your saying 1000va isn't enough. From the look of it, it's about a 150mm wide and 90mm tall, so I don't think it is any bigger than 1000va.
All off the top of my head, but it doesn't seem far off. Other posters have done the math more precisely, but I don't feel it's needed. He can build what he wants.
I see it now... Your saying 1000va isn't enough. From the look of it, it's about a 150mm wide and 90mm tall, so I don't think it is any bigger than 1000va.
All off the top of my head, but it doesn't seem far off. Other posters have done the math more precisely, but I don't feel it's needed. He can build what he wants.
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Okay interesting! But how would such a transformer be configured?
I mean 160W / 31A = 5V ?!
JMFahey I am not sure that I read right, but you seems a bit hostile, I think.
If you look at my question, you will find that my question is "how it is possible".
I am new to DIY and are therefor trying to learn and how better to learn then ask here?
EDIT: The Saint you arrive to the result of 31V and 7.75A where JMFahey gets 31A I am not sure how to understand that?
Hi FriedMule
JMFahey has made a mistake in his calculation by misunderstanding the Ohms Law Formula Wheel.
As I explained in my last post to find how much amperage or current into a given load you must use this formula; if all you know is the power output and the load. I = to the Square Root of P/R
We know the power output is 240 watts and the load is 4 Ohms. So to work out the current delivered into the load it is 240 watts divided by 4 ohms and we take the Square Root of that which is 7.75 Amps.
I hope this helps
Cheers Anthony
The trick in understanding the specs is that 45A are not available PERMANETLY.
As to how the transformer is configured:
This is based on the 160W into 8 ohms basic spec. The peak voltage needed for that is 51V, so the power supply for the amplifier output must at least be +-51V worst case, which means at the lowest mains voltage and highest peak current. This usually means it is about 15% higher with no load and nominal mains voltage, plus there is an implied minimum voltage drop between power supply and output at maximum current, which can easily be a few V. So, let's say 60-65V dual power supply, when unloaded.
We know also that the power into 4 ohms does not double, which is what would happen if the power supply voltage remained the same under twice as much current load. THis usually means that the transformer and power supply filter capacitors are dimensioned to let the voltage sag somewhat to keep the current from power supply through amplifier and to load limited to values that the components can withstand. Using much the same maths as above, we get that for 240W into 4 ohms we need 44V peak, so the power supply sags about 7V when the load current goes from 6.375V peak (8 ohm load) to 11A peak (4 ohm load).
The basic power supply running at about 60-65V with no load will have the transformer secondary windings at the unloaded voltage divided by square root of two, plus 1V or so to compensate for the drop in the rectifier. This gets us to around 85-90V center tapped secondary or 2x 42.5-45V secondary windings. These are RMS voltages as that is how the transformer is specified. Given that we can have up to 11A peak current, or ~7.8A RMS, but only from one secondary (or one half in the case of the CT secondary) at a time, we can calculate that te power requirement is 45 x 7.8 = 350W per channel, when the amp drives a 4-ohm load. Of course this is very simplified math and in reality the voltage and current waveforms are complex and do not simply multiply to get the power requirement. Suffice to say, since the transformer is dimensioned in VA, you could only get that same number as W if the load was an ideal resistor, so all waveforms would be sines and with no phase difference, In reality, the transformer has to be over-dimensioned and usually by a factor of around 1.5 or even more, depending on how the transformer is actually wound and how the power supply filtering is done.
But using even this simple rule of thumb, we get very close to the actual spec - assuming we need W x 1.5 to get transformer VA, we are already at 525VA per channel, so 1050VA for the whole amp, and here is where the 1000VA transformer spec comes from.
So, the actual transformer would be wound as 90VCT 11.11A if one wanted a common power supply for both channels, or 2x (90VCT 5.56A) for separate power supplies, or 4x (45V 5.56A) for completely separate rectification of all power rails.
Now back to the 45A peak spec. This is not uncommon, because of several reasons. First, a typical power supply uses capacitive filtering. The power supply plus capacitors can easily deliver hundreds of A on a short term (ms) basis, which is what can happen if the output of the amplifier is shorted or a very low impedance load is connected. Amplifiers generally don't survive this without some form of current limiting, that is more complex than a fuse.
Also, depending on what kind of semiconductor is used in the output stages, there may be limits to current conducted vs voltage drop over the semiconductor, commonly known as SOA (safe operating area) limitations, which happen for real (reactive) loads, unlike idealized testing loads. To make a long story short, the semiconductors are as a rule over-dimensioned compared to ideal conditions so they can operate reliably. Maximum theoretical peak currents can often be 3-5x more than the idealistic calculations into dummy loads. Given that we have calculated 11.11A peak for an ideally resistive 4 ohm load, 45A max theoretical peak is well within these guidelines.
As to how the transformer is configured:
This is based on the 160W into 8 ohms basic spec. The peak voltage needed for that is 51V, so the power supply for the amplifier output must at least be +-51V worst case, which means at the lowest mains voltage and highest peak current. This usually means it is about 15% higher with no load and nominal mains voltage, plus there is an implied minimum voltage drop between power supply and output at maximum current, which can easily be a few V. So, let's say 60-65V dual power supply, when unloaded.
We know also that the power into 4 ohms does not double, which is what would happen if the power supply voltage remained the same under twice as much current load. THis usually means that the transformer and power supply filter capacitors are dimensioned to let the voltage sag somewhat to keep the current from power supply through amplifier and to load limited to values that the components can withstand. Using much the same maths as above, we get that for 240W into 4 ohms we need 44V peak, so the power supply sags about 7V when the load current goes from 6.375V peak (8 ohm load) to 11A peak (4 ohm load).
The basic power supply running at about 60-65V with no load will have the transformer secondary windings at the unloaded voltage divided by square root of two, plus 1V or so to compensate for the drop in the rectifier. This gets us to around 85-90V center tapped secondary or 2x 42.5-45V secondary windings. These are RMS voltages as that is how the transformer is specified. Given that we can have up to 11A peak current, or ~7.8A RMS, but only from one secondary (or one half in the case of the CT secondary) at a time, we can calculate that te power requirement is 45 x 7.8 = 350W per channel, when the amp drives a 4-ohm load. Of course this is very simplified math and in reality the voltage and current waveforms are complex and do not simply multiply to get the power requirement. Suffice to say, since the transformer is dimensioned in VA, you could only get that same number as W if the load was an ideal resistor, so all waveforms would be sines and with no phase difference, In reality, the transformer has to be over-dimensioned and usually by a factor of around 1.5 or even more, depending on how the transformer is actually wound and how the power supply filtering is done.
But using even this simple rule of thumb, we get very close to the actual spec - assuming we need W x 1.5 to get transformer VA, we are already at 525VA per channel, so 1050VA for the whole amp, and here is where the 1000VA transformer spec comes from.
So, the actual transformer would be wound as 90VCT 11.11A if one wanted a common power supply for both channels, or 2x (90VCT 5.56A) for separate power supplies, or 4x (45V 5.56A) for completely separate rectification of all power rails.
Now back to the 45A peak spec. This is not uncommon, because of several reasons. First, a typical power supply uses capacitive filtering. The power supply plus capacitors can easily deliver hundreds of A on a short term (ms) basis, which is what can happen if the output of the amplifier is shorted or a very low impedance load is connected. Amplifiers generally don't survive this without some form of current limiting, that is more complex than a fuse.
Also, depending on what kind of semiconductor is used in the output stages, there may be limits to current conducted vs voltage drop over the semiconductor, commonly known as SOA (safe operating area) limitations, which happen for real (reactive) loads, unlike idealized testing loads. To make a long story short, the semiconductors are as a rule over-dimensioned compared to ideal conditions so they can operate reliably. Maximum theoretical peak currents can often be 3-5x more than the idealistic calculations into dummy loads. Given that we have calculated 11.11A peak for an ideally resistive 4 ohm load, 45A max theoretical peak is well within these guidelines.
Gentlebeings, JM understands all this perfectly well, but made a mistake (V vs I).
The question is unclear.
Are you totally un-clear? "Superman flies; how is that possible?" Then you want to learn about flying, and why Superman can not fly like he does in the movies.
Are you questioning the numbers? "The 2002 Honda 2.2 liter engine makes 148 horsepower, but the new ToyoRia SX7 SUV claims 250HP from their 2.2l engine; how is that possible?" And my 1962 Willys got 72HP from its 2.3l engine. We can talk all day about different ways to build engines.
The Audio Amplifier is NOT a simple resistor. When you want to derive an equivalent resistance, you MUST remember that we feed it DC but conventionally test with SINE tone. Which throws a couple of 1.414 into the math, and we get "Sine Power" about half of what we "could" get with Square Waves.
Sadly this is something most experienced designers worked-out a long time ago and never wrote-up. It is not(?) in Self or Cordell or Sloane. Someone should write the basic math of audio power amps. But not me today.
The question is unclear.
Are you totally un-clear? "Superman flies; how is that possible?" Then you want to learn about flying, and why Superman can not fly like he does in the movies.
Are you questioning the numbers? "The 2002 Honda 2.2 liter engine makes 148 horsepower, but the new ToyoRia SX7 SUV claims 250HP from their 2.2l engine; how is that possible?" And my 1962 Willys got 72HP from its 2.3l engine. We can talk all day about different ways to build engines.
The Audio Amplifier is NOT a simple resistor. When you want to derive an equivalent resistance, you MUST remember that we feed it DC but conventionally test with SINE tone. Which throws a couple of 1.414 into the math, and we get "Sine Power" about half of what we "could" get with Square Waves.
Sadly this is something most experienced designers worked-out a long time ago and never wrote-up. It is not(?) in Self or Cordell or Sloane. Someone should write the basic math of audio power amps. But not me today.
A further thing to consider. Music is not sine wave. Unless you're running a PA amp at the limit for a long time with compressed material, you can get away with a much smaller transformer (except for class A amps).
DIYers almost always oversize their PS but the benefits aren't that significant. To quote Dan Fraser: "a VA of about 2/3 of the amp's RMS power rating is fine for a Class AB amp. With Class D amplifiers, the VA of the transformer can be 55% of the RMS power rating."
For home hifi, Hypex will suggest a single 160VA transformer for a stereo UCD180... think about it.
Un-clear YES as much you can imagine!
My question was so open because I did know about until now.
I could easily have invented 10 question that would prove how little I know.
I.e. your superman example: How does hi fly, does it depend on the suns rays? On his name? The color of his cape? And so on.
And someone had to tell me no no no, not that, we have to start all over.
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