I accept what was said only for reference, was there anyone who, according to the scheme I posted above, could say specifically - at a power of 100 watts, at an 8 ohm load, and +–40 volts, how many amperes is on the shoulder, according to him?
You received the answer several times in the topic, but it seems that they are not good if they do not confirm the idea that you have already formed.
This is what Rod Elliott's website says:
Any transformer having between 300VA and 1000000VA will be able to power a 2x100W class AB amplifier.
But anything higher than 400VA-500VA is nothing but a waste.
You ask how many ampere. If the transformer has 2x40Vac, then for 400VA it result that the transformer will have to have two secondaries of 40Vx5A.
Does this answer satisfy you?
PS: in the post #29 you showed 5A at the output of the capacitors, which is not correct because there you can have very large current pulses.
This is what Rod Elliott's website says:
Any transformer having between 300VA and 1000000VA will be able to power a 2x100W class AB amplifier.
But anything higher than 400VA-500VA is nothing but a waste.
You ask how many ampere. If the transformer has 2x40Vac, then for 400VA it result that the transformer will have to have two secondaries of 40Vx5A.
Does this answer satisfy you?
PS: in the post #29 you showed 5A at the output of the capacitors, which is not correct because there you can have very large current pulses.
That 5A is only at the peaks! You are not amplifying DC, are you?
Draw the sinewave, and note the voltages and currents - the voltages and currents are also half waves, and the currents only half of the time.
The 40V/8ohm is 5A for the pos sine, then 5A for the peak of the neg wave, so 5A peak values only.
Draw it and you will see, continue to speculate will not get you anything except conflicting advise.
Now these values are peak values; the RMS value, which is the equivalent DC value (which is important because the power supply delivers DC)
is 3.5A. And this is with with everything at full power continuously which NEVER happens in music.
Jan
Jan-
You must often feel like you are
😗
I think you'll agree that it is, however, a good idea to not operate components right at their design specifications and to de-rate everything a reasonable amount. Just because "the numbers say it will work" does not mean that there are not other design considerations. A point that is lost on some.
Hal
You must often feel like you are
😗
I think you'll agree that it is, however, a good idea to not operate components right at their design specifications and to de-rate everything a reasonable amount. Just because "the numbers say it will work" does not mean that there are not other design considerations. A point that is lost on some.
Hal
Yes I do.
Yes I agree you need some leeway. But the idea that a 100W amp constantly draws 5A while amplifying music is total BS.
It's easy to see as a 1st approximation: if the amp is continuously outputting 100Wrms in 8 ohms, then according to an Ohms law rephrasing, I^2*R=100 which makes I = 3.5Arms. Note this is with 100W rms continuously, which probably make the amp give up the magic smoke long before the power supply runs out of current.
So if you size the power supply for 3.5A continuously, you'll have about a factor 5 leeway during normal operation, amplifying music.
Of course, transformer manufacturers will have their ownmarketing engineering math.
Jan
Yes I agree you need some leeway. But the idea that a 100W amp constantly draws 5A while amplifying music is total BS.
It's easy to see as a 1st approximation: if the amp is continuously outputting 100Wrms in 8 ohms, then according to an Ohms law rephrasing, I^2*R=100 which makes I = 3.5Arms. Note this is with 100W rms continuously, which probably make the amp give up the magic smoke long before the power supply runs out of current.
So if you size the power supply for 3.5A continuously, you'll have about a factor 5 leeway during normal operation, amplifying music.
Of course, transformer manufacturers will have their own
Jan
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Oh, I completely agree - I was just thinking in general terms and that peak current draw can be an issue to be considered; not necessarily here but more because this discussion seems to have a lot of people wanting to be given an absolute answer when the truth, as it usually does, lies somewhere in the middle.
Hal
Hal
OT anecdote: I've worked in software development, and at one time we had a customer rep who demanded error-free software.
My boss at that point called the customer team leader and asked if that person could be replaced by someone who knew what he was talking about.
We did get the contract ;-)
Jan
My boss at that point called the customer team leader and asked if that person could be replaced by someone who knew what he was talking about.
We did get the contract ;-)
Jan
My day job is for a company providing the development of precision metrology and testing for the military - a LOT of money being spent by a few very clueless people. Sigh. Half the job, sometimes, seems to be convincing people that "wanting" is not the same as "possible."
Hal
Hal
It is a good writeup and explains the math well. One more example of what is taught not being the whole story: most new EEs we get fresh out of school would only consider the RMS current when sizing rectifier diodes and would not have a clue that the peak current must also be factored in. And that the big 'honkin low ESR bulk capacitors they put after them effectively present a short at turn on and for quite a bit thereafter until they are charged. A lot of people who don't know better (and some that should) think that when it comes to bulk capacitance "if a little is good, more is better." Then wonder why all that smoke decided to come out. From what I have seen capacitors and their uses, specifications, parasitics and other qualities are the most misunderstood component, and not just in Audio circles.
Hal
Hal
One thing to note, though omn
Hal
One thing to note, though, is that all calculations - including ripple current - are for a steady state supply. Initial current draws will be much higher until normal operating conditions are met. The primary limitation will be the winding resistance of the transformer and it's degree of saturation.
Hal
And what about Dejan Veselinovic's three part article "Solid State Power Amplifier Supply" - https://www.tnt-audio.com, are the calculations there correct?
Don't know, haven't looked... but it should be easy to check if you want. All the calculations are basic entry level stuff and can be found easily in reference material or just Google.
Hal
His calculations are OK, I mean, simple algebra can hardly be wrong.
But I don't agree to all the assumptions they are based upon.
Example: speaker impedance can drop to 3 ohms. Yes, at a very small freq band where the music comes only once in a while.
Sizing the supply for continuous use into 2 ohms??
Using two bridges instead of one is 'better'? What kind of better? We're not running a train here, do we?
Sure, you can do that, but I would rather spend the money on music.
So it is always - translate anxiety into grossly overrated stuff.
Jan
But I don't agree to all the assumptions they are based upon.
Example: speaker impedance can drop to 3 ohms. Yes, at a very small freq band where the music comes only once in a while.
Sizing the supply for continuous use into 2 ohms??
Using two bridges instead of one is 'better'? What kind of better? We're not running a train here, do we?
Sure, you can do that, but I would rather spend the money on music.
So it is always - translate anxiety into grossly overrated stuff.
Jan
For me was enough to read until market text:
This is a completely wrong statement.
The capacitors do not have speed!
Is a completely silly assertion and it proves that that article is written to fool people who don't understand electronics or, worse, it's written by someone who doesn't understand electronics well.
Whatever the truth, the rest of the article is not even worth reading.
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No. We rate audio amps as "Sine power", which is much less than full steady DC.
The no-BS technique is to forget bipolar (it just confuses you), and assume the full-Sine-roar amplifier acts-like a 50 Ohm load. (This is probably a couple of pi times 8 Ohms, but I won't derive that tonight.) (If you are running a 3-Ohm, or a train, adjust accordingly.)
40+40V is 80V. 80V on 50 Ohms is 1.6Amps. This is the "average"; the 5 Amp peak comes from the main cap(s) half the time.
This is 128 Watts DC. The 40V 5A peak implies 100 Watts Sine output power. 100W out for 128W in is 78% efficiency, and 78% is the well-known efficiency of a class B audio amplifier. https://en.wikipedia.org/wiki/Power_amplifier_classes#Class_B
The DC Watt to AC VA conversion factor is 1.6 to 1.8. So your transformer needs to be 205VA to 230VA. Depending a lot on how much sag you can tolerate. That's for steady Sine output. In domestic hi-fi you can get away with very much less; or build very much bigger and impress your friends.
NOW re-think bipolar, if desired. The transformer stays the same except needs a CT. Instead of one >1000uF >80V cap you need two >2000uFd >40V caps. (These are real minimal; we often use 10X the C and should use 2X the voltage.)
To the speed of caps: I've seen that before. I don't know where it comes from. it may have to do with the ESR.
Ultimately, if you suddenly put a voltage across a cap, it is the ESR (and ESL) that determines how fast the cap can charge.
But it is pretty irrelevant in this context as the charging current is limited by the source impedance (xfrmr, rectifiers, wiring etc).
Jan
Ultimately, if you suddenly put a voltage across a cap, it is the ESR (and ESL) that determines how fast the cap can charge.
But it is pretty irrelevant in this context as the charging current is limited by the source impedance (xfrmr, rectifiers, wiring etc).
Jan
I understand what you mean and even my friends and I have thought about it, but here the way the information is structured makes you think that it refers to the slew rate of the amplifier, which is completely wrong.
Anyway, what does the charging/discharging speed of the filter capacitors have to do with the slew rate of the amplifier, slew rate which is determined by the scheme of the amplifier and how the compensation is done. Anyway, a 10000µF capacitor has a resonance frequency below 20Khz and above the resonance frequency the behavior is inductive. In order to improve the filtering above the resonant frequency of the electrolytic, other capacitors (multilayer film or ceramics) placed in parallel are used.
But we were already talking about something else.
The discussion on this topic is a clear example of acute intoxication with extreme audiophile literature.
Anyway, what does the charging/discharging speed of the filter capacitors have to do with the slew rate of the amplifier, slew rate which is determined by the scheme of the amplifier and how the compensation is done. Anyway, a 10000µF capacitor has a resonance frequency below 20Khz and above the resonance frequency the behavior is inductive. In order to improve the filtering above the resonant frequency of the electrolytic, other capacitors (multilayer film or ceramics) placed in parallel are used.
But we were already talking about something else.
The discussion on this topic is a clear example of acute intoxication with extreme audiophile literature.
Friends, this is not about whether someone is an audiophile or not. It's about getting the right way to calculate the power supply of an amplifier, so that the amplifier will deliver the power of 100 watts into 8 ohms at its output for a long time. In this regard, I ask for the following clarification - we assume that by dividing the voltage on a rail, 40 volts is divided by the resistance of the load, which is 5 amperes. Then in reverse order multiplying current times voltage 40x5 from the power formula, does that give us a peak power of 200 watts per rail, or 400 watts per channel? If this is not correct, please explain simply why it is not correct, I think this will be useful for a lot of people.
Your math is correct: 40V x 5A = 200W. And obviously, 2 x 200 = 400. But that's not the issue.
What does that number mean in this context?
It only tells us that when you have 40V across an 8 ohms load, it dissipates 200W.
What is missing is to translate this into a power supply requirement.
That is what several people here have been trying to explain.
Did you actually read the posts about music power versus continuous DC 100W output?
If you draw the waveform and annotate voltages and currents and power you would immediately see the nonsense of thinking you need 5V continuously at 40V or even twice that, and we all could move on to other things.
Jan
What does that number mean in this context?
It only tells us that when you have 40V across an 8 ohms load, it dissipates 200W.
What is missing is to translate this into a power supply requirement.
That is what several people here have been trying to explain.
Did you actually read the posts about music power versus continuous DC 100W output?
If you draw the waveform and annotate voltages and currents and power you would immediately see the nonsense of thinking you need 5V continuously at 40V or even twice that, and we all could move on to other things.
Jan