"In theory, there is no difference between theory and practice. But, in practice, there is." --Jan L. A. van de Snepscheutgpapag said:
gpapag said:
Thank you very much Mr George.
The all issue appears very complex indeed.
Too complex for me.
By the way I will follow this extremely interesting 3D.
The issue is very fundamental to me.
Kind regards,
beppe
By the way I talked with a friend who has upgradaded various units, the last one being the Behringer SRC 2496 dac, just replacing supply caps with better quality and bigger ones.
The improvements has been very remarkable.
Anyway I do not know which kind of improvement will be obtained replacing the transformer instead.
Uhm, nice idea .....
Thank you very much indeed.
Kind regards,
beppe
The improvements has been very remarkable.
Anyway I do not know which kind of improvement will be obtained replacing the transformer instead.
Uhm, nice idea .....
Thank you very much indeed.
Kind regards,
beppe
gpapag said:
Dear Mr George,
actually I found the term "3D" as abbreviation of "thread".
Kind regards,
beppe
plain English would suit me better. I have no intention of learning or translating phone text jargon.
Phone text jargon is severely abbreviated stuff. The l33t sp34k I mentioned is something lame nerds and h4x0rz came up with to sound c001; it clearly doesn't save any typing. AndrewT, you're usually a smart guy, so I'm surprised you'd fail to realize that something that has as many characters as regular English spelling would be used for the lenght-conscious phone texting.
Beppe,
mi piace per leggerli le vostre parole, ogni volta transportano la firma di araldica tradizionale.
Semplicemente, perché sono al corrente di quello dalla mia infanzia.
Cordialemente, il mi pui alto respetto,
jacco.
Fawlty Towers translation:
My name is Manuel. I'm from Barcelona, i know nottin.
mi piace per leggerli le vostre parole, ogni volta transportano la firma di araldica tradizionale.
Semplicemente, perché sono al corrente di quello dalla mia infanzia.
Cordialemente, il mi pui alto respetto,
jacco.
Fawlty Towers translation:
My name is Manuel. I'm from Barcelona, i know nottin.
yes, keep it for phone texting and let those users choose which they use.
Don't bring it here.
BTW,
how many languages have fewer characters in their alphabet than English?
I know of a few that have more.
Eric Juaneda said:
Eric,
Sorry to come in so late, and with the risk that somebody already said this, but your supply DOES have a cap. The cap supplies the amp for most of the time when the cap voltage is higher than the mains voltage (the sharing diode is conducting). Only at the small intervals that the mains is higher than the cap voltage does the sharing diode block and does the xzformer directly supply the amp. The result is not much different than 'normal' supplies except that the ripple will look funny.
Jan Didden
How big does the reservoir capacitor have to be? Here are some back-of-the-envelope musings that may help shed some light on it.
Consider a really good 100 Watt power amplifier at 8 ohms. If it has a strong power supply, it should be able to deliver 200 watts into 4 ohms. This corresponds to 40V peak into 4 ohms, or 10 Amps peak. It should be able to deliver a square wave at 20 Hz at this same peak voltage into 4 ohms, right? Well, it turns out that putting a 40 V peak square wave into 4 ohms is a bit of a torture test for the power supply of this amplifier.
The time between re-charge of the reservoir capacitor corresponds to the period of 120 Hz, or 8.3 ms. Let us see how far an example 10,000 uF reservoir capacitor would discharge in this amount of time, as a worst case. For simplicity, assume a perfect transformer and perfect rectifier diodes.
V = IT/C = 10*8.33-3/1e-2 = 83.3e-1 = 8.33 Volts
This is a huge amount of ripple, considering that the total rail voltage is only on the order of 45-50 Volts. Of course, in this simplified view, we assume an impulse re-charge after 8.3 ms, rather than the normal onset of re-charge due to the half-sinusoid waveform that is there in practice. Nevertheless, this worst-case look is sobering.
If you wanted to build a really good amplifier and wanted only 1 V p-p ripple under these conditions, you would need 80,000 uF per rail.
But we never drive square waves at 20 Hz into 4 ohm loads, you say? Well, a full-power 20 Hz sinusoid is reasonable. But a 20 Hz sinusoid stays up there near its peak current of 10 Amps for quite some time in comparison to the 8.3 ms re-charge interval. The period of a 20 Hz sinusoid is 50 ms. The recharge time of 8.3 ms corresponds to 59.8 degrees of the 20 Hz cycle, or about +/- 30 degrees about the peak. At +/- 30 degrees from the peak, the current is down to 87% of the peak. Over the full range of +/- 30 degrees about the peak, the average current is 94% of the peak, or 9.4 Amps. This is not much of a reduction. So now we have 8 V p-p ripple (10,000 uF) amplitude modulated by the 20 Hz waveform. Not very nice.
Testing a 100-watt/8-ohm amplifier into 4 ohms at full power at 20 Hz is certainly not unreasonable. This simple analysis is just one reason why a goodly amount of reservoir capacitance is desirable, even in a modest 100 watt amplifier.
Suppose we wish to have the ripple be less than 5% of the rail voltage, corresponding to 2.5V p-p on a 50 Volt rail. Assume also that we want to take into account approximately the half-sinusoid nature of the driving signal coming from the rectifier. This will have the effect of slightly lessening the time to discharge between charge refreshment, improving the situation somewhat.
If the sinusoid reaches its peak at 90 degrees, it is 5% below its peak at 72 degrees. The 18 degree difference at 60 Hz corresponds to (18/360) * 16.7 ms = 0.835 ms. From the original 8.3 ms re-charge interval we subtract 0.835 ms at the beginning of the time interval (how long it takes for the half-sinusoid to decrease from its peak to 95%) and another 0.835 ms from the end of the interval, since the half-sinusoid gets back up to 95% 0.835 ms before the end of the interval. The net re-charge interval is thus 8.3-1.7 = 6.6 ms. We thus only need about 80% of the reservoir capacitance that we estimated using the simpler impulsive recharge model. Things have not changed very much.
We use a 20 Hz sinusoid, so the average current over the 6.6 ms recharge interval is about 9.4 amps with the worst case phase of the 20 Hz sinusoid.
C = IT/V = 9.4 * 6.6e-3/2.5 = 24.8e-3 = 24,800 uF per rail.
Cheers,
Bob
Consider a really good 100 Watt power amplifier at 8 ohms. If it has a strong power supply, it should be able to deliver 200 watts into 4 ohms. This corresponds to 40V peak into 4 ohms, or 10 Amps peak. It should be able to deliver a square wave at 20 Hz at this same peak voltage into 4 ohms, right? Well, it turns out that putting a 40 V peak square wave into 4 ohms is a bit of a torture test for the power supply of this amplifier.
The time between re-charge of the reservoir capacitor corresponds to the period of 120 Hz, or 8.3 ms. Let us see how far an example 10,000 uF reservoir capacitor would discharge in this amount of time, as a worst case. For simplicity, assume a perfect transformer and perfect rectifier diodes.
V = IT/C = 10*8.33-3/1e-2 = 83.3e-1 = 8.33 Volts
This is a huge amount of ripple, considering that the total rail voltage is only on the order of 45-50 Volts. Of course, in this simplified view, we assume an impulse re-charge after 8.3 ms, rather than the normal onset of re-charge due to the half-sinusoid waveform that is there in practice. Nevertheless, this worst-case look is sobering.
If you wanted to build a really good amplifier and wanted only 1 V p-p ripple under these conditions, you would need 80,000 uF per rail.
But we never drive square waves at 20 Hz into 4 ohm loads, you say? Well, a full-power 20 Hz sinusoid is reasonable. But a 20 Hz sinusoid stays up there near its peak current of 10 Amps for quite some time in comparison to the 8.3 ms re-charge interval. The period of a 20 Hz sinusoid is 50 ms. The recharge time of 8.3 ms corresponds to 59.8 degrees of the 20 Hz cycle, or about +/- 30 degrees about the peak. At +/- 30 degrees from the peak, the current is down to 87% of the peak. Over the full range of +/- 30 degrees about the peak, the average current is 94% of the peak, or 9.4 Amps. This is not much of a reduction. So now we have 8 V p-p ripple (10,000 uF) amplitude modulated by the 20 Hz waveform. Not very nice.
Testing a 100-watt/8-ohm amplifier into 4 ohms at full power at 20 Hz is certainly not unreasonable. This simple analysis is just one reason why a goodly amount of reservoir capacitance is desirable, even in a modest 100 watt amplifier.
Suppose we wish to have the ripple be less than 5% of the rail voltage, corresponding to 2.5V p-p on a 50 Volt rail. Assume also that we want to take into account approximately the half-sinusoid nature of the driving signal coming from the rectifier. This will have the effect of slightly lessening the time to discharge between charge refreshment, improving the situation somewhat.
If the sinusoid reaches its peak at 90 degrees, it is 5% below its peak at 72 degrees. The 18 degree difference at 60 Hz corresponds to (18/360) * 16.7 ms = 0.835 ms. From the original 8.3 ms re-charge interval we subtract 0.835 ms at the beginning of the time interval (how long it takes for the half-sinusoid to decrease from its peak to 95%) and another 0.835 ms from the end of the interval, since the half-sinusoid gets back up to 95% 0.835 ms before the end of the interval. The net re-charge interval is thus 8.3-1.7 = 6.6 ms. We thus only need about 80% of the reservoir capacitance that we estimated using the simpler impulsive recharge model. Things have not changed very much.
We use a 20 Hz sinusoid, so the average current over the 6.6 ms recharge interval is about 9.4 amps with the worst case phase of the 20 Hz sinusoid.
C = IT/V = 9.4 * 6.6e-3/2.5 = 24.8e-3 = 24,800 uF per rail.
Cheers,
Bob
jacco vermeulen said:
Very good indeed Sir.
Just some little adjustements and your italian is very nice.
My sincere congratulations.
Best wishes,
beppe
janneman said:
Jan,
I read this thread with great interest. You clarify this point at post #38
janneman said:
So, transformer spend most of the time to do...nothing. The SCPS (Shared Current Power Supply) try to reverse this.
...The transformer supplies the amp for most of the time...
Of course a power supply does have a caps. But what kind of caps ? We could not compare effects introduce by 10uF polypropylene capacitor and effects by 1000uF chemical capacitor. When I try to reduce capacitors, I think reducing negative effects of big chemicals capacitors. [Using big or small capacitors was already invoked at Post #4 of this topic. (ouf!)]
Charging (big) capacitors is critical. High energy on short time. Do you think that short time as no effect on sounding quality ?
Another impossible question : what is better, current coming from chemical source (capacitors) or current coming from polluted transformer ?
With SCPS you experiment this question.
There is another important point. Discussion often speak about critical phase where amplifier output maximum power. I don't think people always listen at maximum power ! Most of time amplifier need smaller current. Supply ondulation is very low and transformer deliver current on very short time. I don't think that amplifier must have a good sounding only at full power output. Sound quality on small power output is important too.
janneman said:
Yes, I don't give applicative solution, just comprehensive and funny schema. I don't think with my 'funny ripple' you obtain reasonable sounding. With complexe schematic diagram, with snubbers and capacitors everywhere, I don't think that someone could understand anything.
Eric Juaneda
I followed this discussion for quite a while and would like to thank both Bob Cordell and the many contributors here. Indeed, a very interesting and important topic!
So, large reservoir capacitance is a must if you want to put out full power at low frequencies. If the rail ripple becomes too large, I would speculate that this would not only reduce the maximum power capability (compared to midband frequencies), but could also induce distortion. A significantly increased ripple will also influence the input- and driver stages and probably create further distortion there. This all depends of course on the CMRR and the amount of rail filtering for these previous stages. Using a separate supply, possibly even from a separate transformer could be beneficial.
On distortion plots vs frequency for various power outputs, my recollection is, that they were bathtub shaped at higher power outputs, i.e. showing rising distortion at both high and low frequencies.
In an earlier post, it was said that, for best sound, reservoir capacitance should neither be too small nor too large, i.e. an optimum exists for best sound. This seems to conflict with a design goal for good LF power handling with low distortion (where a really large cap is best).
On the other hand, one could question if there is any real reason to worry about potential distortion at low frequencies near maximum power, since speakers driven under such conditions exhibit significant distortions of their own.
Kurt
So, large reservoir capacitance is a must if you want to put out full power at low frequencies. If the rail ripple becomes too large, I would speculate that this would not only reduce the maximum power capability (compared to midband frequencies), but could also induce distortion. A significantly increased ripple will also influence the input- and driver stages and probably create further distortion there. This all depends of course on the CMRR and the amount of rail filtering for these previous stages. Using a separate supply, possibly even from a separate transformer could be beneficial.
On distortion plots vs frequency for various power outputs, my recollection is, that they were bathtub shaped at higher power outputs, i.e. showing rising distortion at both high and low frequencies.
In an earlier post, it was said that, for best sound, reservoir capacitance should neither be too small nor too large, i.e. an optimum exists for best sound. This seems to conflict with a design goal for good LF power handling with low distortion (where a really large cap is best).
On the other hand, one could question if there is any real reason to worry about potential distortion at low frequencies near maximum power, since speakers driven under such conditions exhibit significant distortions of their own.
Kurt
Here's a power supply "effect" -- this supply has 39,000 uF/80V caps on each rail and usually runs at +/- 50 - 55VDC on the rails -- the output devices are lateral MOSFETs -- the output was 50 Watts into 8 ohm at 1kHz -- the respective supply voltages were 51V for 0.01% THD, 34.0 V for 0.10 % THD. 32.2 V for 1.0% THD and 31.5V for 3.0% THD. The plots on the spectrum analyzer are up to 1 MHz and represent the distortion residuals with no low pass filtering -- this should demonstrate that as the supply runs out of gas the distortion rises tremendously:
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Javin5 said:
A reason to make the hardship even worse ?
This stuff reminds me of the French Hiraga/Kaneda amps.
Those really suckked, this fool even went to Paris to pay big money for the advised semi's.
Only thing that made them sound any good were the giant cap, CRC, and battery powersupplies.
And the relic single voltage rails still continues to be the common thing to do on a power amp.
Raw Data, I guess? Just exactly what was that jackinnj? Were you maintaining 1kHz, at 28.2volts peak, on an 8 ohm load and manually lowering the powersupply voltage? I did not quite understand?
I would be interested in seeing the 20Hz signal, and the supply rail waveform, from a transformer fed, 39,000uF supply, as Bob C was suggesting 😀
I would be interested in seeing the 20Hz signal, and the supply rail waveform, from a transformer fed, 39,000uF supply, as Bob C was suggesting 😀
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