Honestly, I do not see the relationship with my post.
anyway thanks for the advice.
Hello,
In the LF domain, can't we think of the reservoir capacitor as a high pass filter ??
As "Gootee" said:
"You're close. But the PSU voltage is not the music signal.
The PSU current is the music signal. "
So, this current will run in the reservoir capacitors, the output BJT/MOS and the loud-speaker.
In we want to have some current in a 8R loud-speaker at 20 Hz, for ex, we will need the impedance of the reservoir capacitors to be less or equal to 0.6R.
So, Cr = > 13,000µF.
Am I wrong or right or something else ?
Regards.
Alain.
In the LF domain, can't we think of the reservoir capacitor as a high pass filter ??
As "Gootee" said:
"You're close. But the PSU voltage is not the music signal.
The PSU current is the music signal. "So, this current will run in the reservoir capacitors, the output BJT/MOS and the loud-speaker.
In we want to have some current in a 8R loud-speaker at 20 Hz, for ex, we will need the impedance of the reservoir capacitors to be less or equal to 0.6R.
So, Cr = > 13,000µF.
Am I wrong or right or something else ?
Regards.
Alain.
Problem with your theory is that you are charging the reservoir capacitor with an half sinusoid. As discussed by other members, if the capacitor is too large then you have a problem with charging it back up to the required rail voltage before another current demand is made.
There may be some synergy in your proposal in using two separate capacitors, the reservoir that is very large decoupled from another critically calculated capacitor by some switch that would replenish this last cap by an exact measure of energy to maintain the rail voltage.
Are we saying that this is a SMPS with Pulse Width modulated output.
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According to Profusions application note:
Power supply considerations.
The power supply should be capable of supplying the module with D.C. equivalent to 150% of the RMS output power. The power supply capacitors should be at least 2,000mF per amp drawn from the supply rail. This will yield approx 1 volt p-p of ripple, which is a good compromise between cost and performance. For applications where the sound quality is of prime importance, the capacitance should be increased to 2 or 3 times this value. The mains transformer should be sized according to expected usage. Hi-fi amplifiers use a transformer rated at approx 1.5 x the total output power (in VA). For a stage or P.A. amplifier the factor should be 2 - 2.5 times the output power. Oversize power transformers will also enhance the sound quality.
If I understand the discussion correctly this will still be good advice then?
Power supply considerations.
The power supply should be capable of supplying the module with D.C. equivalent to 150% of the RMS output power. The power supply capacitors should be at least 2,000mF per amp drawn from the supply rail. This will yield approx 1 volt p-p of ripple, which is a good compromise between cost and performance. For applications where the sound quality is of prime importance, the capacitance should be increased to 2 or 3 times this value. The mains transformer should be sized according to expected usage. Hi-fi amplifiers use a transformer rated at approx 1.5 x the total output power (in VA). For a stage or P.A. amplifier the factor should be 2 - 2.5 times the output power. Oversize power transformers will also enhance the sound quality.
If I understand the discussion correctly this will still be good advice then?
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Are you sure you understand what you wrote? Because I dun understand and you dun understand a bit itself also.
To be clear, there is no problem designing a power supply at all. It is done with a big cap only for a long time and will keep that way. Thats all.
Only here it seems a problem and with solutions that results in poorer performance.
This is as result of poor knowledge of what really happens.
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what if.....the allowable rail-sving was held well below the supply rail... for an example 4V.. if the drivers are then held by their own supply,
We can then calculate or simulate the needed capacitance.. eg to supply a railed-4 Volt 20 HZ tone into a 4 ohm load...in such a manner that the rail doesn't sink below the needed voltage.. this would then be the minimum supply.
We can then calculate or simulate the needed capacitance.. eg to supply a railed-4 Volt 20 HZ tone into a 4 ohm load...in such a manner that the rail doesn't sink below the needed voltage.. this would then be the minimum supply.
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Given the ratio between DC and RMS current in a cap input PSU you would need transformer VA to be something like 5-6 times W before you begin to approach 'oversize'. 1.5 - 2.5 times simply recognises that real music is mostly quiet so current demand is mostly well below peak, so the transformer has time to cool down between peaks.tommy1000 said:
The size of your caps depends on how much droop you can stand, which is essentially the same as ripple: droop = ripple pk-pk (roughly). How much droop you can stand depends on how far above peak signal voltage you started from, among other things.
Assume Ipk=sqrt (2P/R), then Vdroop=0.01 Ipk /C; to avoid clipping you need Vdc-Vdroop>= R Ipk. So Vdc>= (R + 0.01/C) Ipk = sqrt (2PR) + 0.01/C sqrt (2P/R)
Stored energy E = 1/2 C Vdc^2, so E>1/2 C (2PR + 0.0001/C^2 2P/R + 0.02/C sqrt (4P^2) )
E >= PRC + 0.0001 P/RC + 0.02 P/C
to keep the numbers simple I will assume R=5 so
E >= 5PC + 0.02002 P/C
so stored energy E must be at least as big as this, but it is a function of C with a broad minmum. We can approximate it as E >= 5PC + 0.02 P/C.
The minimum occurs in this type of problem when the two terms are equal (you can prove this with calculus or trial and error).
So 5C = 0.02/C so C = sqrt (0.004) = 63 246 uF, and then E = 0.63 P.
What does all this mean? I think I have shown that the minimum stored energy you need is 0.63 times output power, but this requires the optimum Vdc. Most people will use more than this, to give some slack, and then you can get away with a smaller cap (and more droop) but more stored energy because of the higher voltage. Conversely you could use a much bigger cap to get less droop and slightly smaller Vdc, but this would also require more stored energy.
Note that this calculation ignores the voltage drop through the output stage so it is a bit idealised. I often make mistakes in algebra, so use this result at your own risk!!
Are you sure you understand what you wrote? Am sure you are not because there so many flaws in your posting.
alayn91,
I am not one of the experts here that is for sure, but I think that the answer to your question is that no the power capacitors are not considered a high pass or low pass filter. This portion of the power supply if I am following this correctly is not a part of the audio filter. This is before that portion of the audio circuit that would be considered an active filter. Someone correct me if you like. A Duh, dummy here.
It would be nice if the people who really have the answers and know what they are doing would answer the questions in an intelligent manner instead of being facetious with their answers. For those of us who are here trying to learn it doesn't much help to read a bunch of back and forth innuendo about who understands the question.... Please give us the facts and when something is wrong point it out and tell us why or where the error is being made. As a famous American said in the not to distant past, "Can't we all just get along', (Rodney King).
I am not one of the experts here that is for sure, but I think that the answer to your question is that no the power capacitors are not considered a high pass or low pass filter. This portion of the power supply if I am following this correctly is not a part of the audio filter. This is before that portion of the audio circuit that would be considered an active filter. Someone correct me if you like. A Duh, dummy here.
It would be nice if the people who really have the answers and know what they are doing would answer the questions in an intelligent manner instead of being facetious with their answers. For those of us who are here trying to learn it doesn't much help to read a bunch of back and forth innuendo about who understands the question.... Please give us the facts and when something is wrong point it out and tell us why or where the error is being made. As a famous American said in the not to distant past, "Can't we all just get along', (Rodney King).
The power capacitors are in fact a low pass filter, and the better they are at being such the better the real behaviour of the device they're feeding. Ideally they only pass DC, but theoretically they also pass 50 or 60Hz, and some harmonics, the normal ripple people think of. Being low pass means they pass the low frequencies, and block the high frequencies, and the better they are at the latter then the better everything is. The simulations I and Tom did showed that an actual audio signal forcing the amplifier to work hard makes the voltage sag firstly, but also injects audio noise onto the supply voltage, that high frequency stuff up to 20kHz, and beyond.
And that's where the troubles begin. Circuits are notoriously bad at stopping that higher frequency noise from interfering with their operation, which is what this PSRR is all about: the ability of the circuit to reject noise on the voltage rails.
Take an extreme example: a stereo amp with power supply shared between left and right where only one channel is driven hard by a 20kHz signal. The power supply will ripple like crazy at this frequency, and the other channel, of the many power amps that will have very poor rejection at this frequency, then proceeds to output a very, very healthy high frequency tone, and if the feedback operation suffers because of this poor behaviour, lots of harmonics as well. Really terrible, distorted crosstalk, in fact.
Frank
fas42, Okay I follow that and should have remembered about the noise from the ac source.
Next question then. If what we are attempting to do is keep the voltage on the +/- rails from dropping or sagging as you say, is there an ultimate value for this. If we just increase the power capacitor value to some excess value then the slew rate becomes important and we have a problem with charging the capacitor bank fast enough.? If to small a capacitor rating the voltage drops. What are we matching the expected output of the amp plus and loss to heat in the output devices?
Next question then. If what we are attempting to do is keep the voltage on the +/- rails from dropping or sagging as you say, is there an ultimate value for this. If we just increase the power capacitor value to some excess value then the slew rate becomes important and we have a problem with charging the capacitor bank fast enough.? If to small a capacitor rating the voltage drops. What are we matching the expected output of the amp plus and loss to heat in the output devices?
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