Power Supply Resevoir Size

[AP2]

Originally Posted by liching1952

The only thing needed is a BIG cap(around 1000uF for each Watt power), not a Capacitance multiplier and some smaller condensors paralel because a big cap gives large inductance that must be avoided for high frequencies. There is no issue about this matter.

Honestly, I do not see the relationship with my post.
anyway thanks for the advice.

[alayn91]

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.

[liching1952]

The capacitor must big enough to deliver current as long as the peak power last. It has nothink to do with a filter. In principle the capacitor is NEVER big enough. The best way is the use of a car battery and charge it with the average needed current.

[Nico Ras]

Originally Posted by liching1952

The capacitor must big enough to deliver current as long as the peak power last. It has nothink to do with a filter. In principle the capacitor is NEVER big enough. The best way is the use of a car battery and charge it with the average needed current.

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.

[Nico Ras]

Guys could the solution to audio power supply lie in an adequately rated SMPS simply diode decoupled from calculated value of distributed capacitance around the output devices purely to supply a sharp rising edge of a transient? It appears to make some logical sense.

[tommy1000]

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?

[liching1952]

Originally Posted by Nico Ras

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.

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.

[MiiB]

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.

[liching1952]

there is no such as minimum power supply. If the power supplv comes down, the power output comes down as well. Thats all.

[DF96]

Originally Posted by tommy1000

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.

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.

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!!