Looking back at post 106, at http://www.diyaudio.com/forums/solid-state/216409-power-supply-resevoir-size-11.html#post3099489 , I probably should have left the signal portion out of it, since that should mostly be handled by the decoupling caps (and I also didn't take into account the effects of having the rail inductances between the two types of caps), and so the "safe minimum" capacitance that was arrived at is probably "much too safe".
I still like the part about using the supply frequency as the worst case so I might look at that some more, later, in a better context.
--------------
So here's the "standard" version:
The equation for the peak-to-peak amplitude of the ripple voltage is:
(1) Δv = i / (2fC)
where f is the supply frequency in Hz, Δv is the peak-to-peak ripple voltage, and i is the load current. C is capacitance in Farads.
Solving for C, we get:
(2) C = i / (2fΔv)
Choose your desired Δv and i and out pops the approximate capacitance needed.
-------------------------------------------------
What about Clipping?
So you could use the usual rule, something like 2200uF or 3300 uF per max output Amp.
(Of course, you won't KNOW the max output Amps until AFTER you calculate the ripple and find out what the max output voltage can be without clipping. But that's one reason that those equations are just approximations.)
In our example, we were using i = 3.54 A RMS, max. But we can't even get there, without clipping (running into the ripple voltage), unless the capacitance is infinite!
-----
So maybe we should first find out where the maximum output would run into the ripple voltage, as a function of the reservoir capacitance. (But keep in mind that everything below will still be just an approximation.)
Let's see... I guess we're assuming that the peak of the ripple voltage is at Vsupp = 40V and its p-p voltage subtracts from the supply voltage to give us a smaller usable voltage envelope, like usual.
The ripple voltage, Δv, given by (1), is peak-to-peak. So the bottom of the ripple voltage waveform would reach down to:
(3) Vmax_peak = Vsupp - Δv
So that would be our maximum peak output voltage, just at the point where an output signal would start clipping (i.e. running into the bottom of the ripple voltage waveform).
In RMS volts, that would be
(4) Vmax_rms = ( Vsupp - Δv ) / √2
So the maximum RMS current through the load, just befoe clipping, would be
(5) imax_rms = ( Vsupp - Δv ) / ( (√2)(Rload) )
Substituting the right side of equation (1) for Δv, in (5), and solving for imax_rms gives
(6) imax_rms = Vsupp / ( Rload∙√2∙(1 + 1 / ( 2∙Rload∙f∙C∙√2 ) ) )
where everything is a known constant except for C,
so we can directly calculate the maximum output current just before clipping (or voltage or power, since we know Rload), that we could get with different values of C.
For our example, with Vsupp = 40, f = 60, and Rload = 8, equation 6 becomes:
(7) imax_rms = 3.54 / (1 + ( 1 / ( 1358∙C ) ) )
Examples:
With C = 10000uF, we would get imax_rms = 3.297 Amps, and max rms power = 86.9 Watts, at clipping onset. The 3.297 A rms gives 26.38 Vrms across 8 Ohms, which is 37.3 Vpk, leaving 2.7 volts as the maximum Δv peak-to-peak ripple voltage.
With C = 4700uF, we would get imax_rms = 3.06 Amps, and max power = 74.9 Watts, at clipping onset. The 3.06 A rms gives 24.48 Vrms across 8 Ohms, which is 34.62 Vpk, leaving 5.38 volts as the maximum Δv peak-to-peak ripple voltage.
With only 4700 uF, the ripple is already a little more than 10% of the supply voltage. The equations above are approximations and should only be trusted when the ripple is less than 10% of the supply voltage. (Also, which way the error goes depends on the characteristics of the diodes and transformers upstream, and could go either way.)
-----------
So what capacitance should we pick, and why?
How should I know? That's what we're here to discuss. <grin> If we are concerned about the performance at the maximum output power, and we use the rule-of-thumb that says ripple should be 5% to 10% of the supply voltage, then that would put us between about 14000 uF and about 6630 uF, which give about 2V and 4V of ripple, respectively, and max output currents of about 3.35 A rms (4.72 A peak) and 3.18 A rms (4.5 A peak), and max output powers of about 90 Watts and 81 Watts. According to those, the "rule" should be "3000 to 1500 uF per max peak output Amp".
---
But some have said that using less capacitance sounds better.
Viewing the filter capacitance and the load resistance as a high-pass filter has also been mentioned (by Onaudio).
Onaudio's example was: Down to 20Hz is desired, so aim for a 2 Hz RC filter cutoff, so 8 Ohms gives C = 1 / ( 2 Pi R f ) = 9947 uF, or about 10000 uF.
That "filter" point of view seems valid if we look at the reservoir capacitance as a current source with a capacitance in series with it.
In that case, looking at it as a high-pass filter with 8 Ohms as the shunt resistance is valid only when the output device has about zero resistance, i.e. it's wide open and the amp is at max output power.
But how much power is typically used, when listening? It's a very low number, if I recall correctly, in the single-digit Watts.
If we use 10% of the maximum power, maybe we can see what might be different at that level.
In that case, the output device has some resistance, which will add with the 8 Ohms, and that will lower the capacitance needed to still get the 2 Hz cutoff frequency.
---
Let's just try it with 10 Watts into 8 Ohms, to make it easier.
Power = i²R = v²/R = v∙i
So the output RMS voltage and current would have to be 8.944 Volts and 1.118 Amps.
So the voltage across the output device would have to be Vsupp - 8.944 - Vripple, or, more generally,
(8) Vdevice = Vsupp - √(Rload∙Pout) - Vripple
The current through the output device will be the same as through the load, which would be
(9) idevice = √(Pout/Rload)
The resistance of the output device must be vdevice/idevice, which is
(10) Rdevice = vdevice/idevice = ( Vsupp - √(Rload∙Pout) - Vripple ) / √(Pout/Rload)
In terms of a high-pass filter, the shunt resistance would be the sum of the device and load resistance:
(11) Rshunt = Rload + Rdevice = Rload + ( Vsupp - √(Rload∙Pout) - Vripple ) / √(Pout/Rload)
Such a filter's cutoff frequency is
(12) f = 1 / ( 2 Pi Rshunt C )
But we know we want f = 2, so solving for C gives:
(13) C = 1 / (4 Pi Rshunt)
Substituting (11) and all of the known values into (13), it becomes
(14) C = 1 / ( 449.6 - (11.24 ∙ Vripple) )
I know that everyone would have just loved it if I had kept all of the quantities as variables and had gone on for pages to derive a general closed-form solution. Sorry to disappoint you. <grin>
In (14), there are two unknowns, C and Vripple. We will need two independent equations to solve for two unknowns. We can use equation (2), which must be equal to (14) since they a both expressions for C in terms of Vripple:
(15) 1 / ( 449.6 - (11.24 Vripple) ) = i / (2 f Vripple)
Substituting all of the known values and solving (15) for Vripple gives
Vripple = 3.792 Volts
and then using (14) or (2) gives
C = 2457 uF
The ripple seems relatively small for such a low capacitance until we remember that we are at 10 Watts and the load current is only 1.118 Amps rms.
Anyway, having that value for Vripple, we can now calculate a value for the Rshunt of the high-pass filter, which is:
Rshunt = 32.37 Ohms
Using the values for C and Rshunt, and (12), we get
f = 2.001 Hz
Bingo!
We could also plug that C value into (7), even though the results will be less than accurate because the ripple will be too high, just to get some idea of what the max output power conditions would be: We would get imax_rms = 2.724 Amps, and max rms power = 59.3 Watts, at clipping onset. The 2.724 A rms gives 21.79 Vrms across 8 Ohms, which is 30.82 Vpk, leaving 9.18 volts as the maximum Δv peak-to-peak ripple voltage.
We "could" say that the "rule" from this configuration would be (from 2457uF/2.724 Amps=) "900 uF per max peak output Amp".
But since it was not designed for that output level, maybe we should say (from 2457uF/(1.118 A rms x 1.414)=) "1500 uF per average peak output Amp", assuming that the average "peak output current" occurs at the average listening level.
Interestingly, it then still falls within the "3000uF to 1500 uF per amp" rule, from between (7) and (8) above.
-----
But would the amplifier sound better (or somehow perform better), at the lower output power level (10 Watts), than it would if there was more capacitance?
If it does, I should patent an amplifier and power supply design that switches in more reservoir capacitors as the volume knob is turned up, and also one that uses a gyrator circuit (instead of, or in addition to, caps) that automatically sets its equivalent capacitance in proportion to the output current level.
---
I have no idea, yet, which way sounds better.
The problem is that the various reports of "it sounds better with less capacitance" AND "it sounds better with more capacitance" are not comparing the same amplifier or even the same design and layout techniques (nor the same listener).
And even if one real amplifier did sound better with more, and worse with less, or the other way around, that fact would not prove that all amplifiers would produce the same result.
My guess is that a lot of people don't correctly use decoupling capacitors at each output device (i.e. with a few hundred uF extremely close and made up of paralleled caps for lowest inductance, and then also some much, much larger uF as close as possible, also made up of smaller caps in parallel, or however it fits best), and many people might not have good power and ground layouts, and might not worry enough about their inductance, and who knows what else.
So the people who don't get the large decoupling caps done right should probably hear an improvement when adding reservoir capacitance, especially at lower frequencies.
But I don't yet know what could cause someone to hear a degradation in sound quality when adding reservoir capacitance, unless maybe it has something to do with the impedances UPSTREAM from the reservoir caps. Or maybe they don't use proper decoupling caps at each device, or they use only 0.1 uF (which is not for decoupling but for high-frequency BYPASSING, for stability in the face of the hidden HF positive feedback loop through the power rails that most transistor amps have, inherently). What it's CALLED is not important. We just need to think more about the capacitors that should be near the output devices.
Inductance is significant. Decoupling caps are like little power supplies at each transistor. They keep the fast large currents in a very small loop so other voltages are not upset. They are CLOSER, so their response is much faster and more accurate. You cannot get low-enough impedance by going through more than a few centimeters of copper.
Cheers,
Tom
I still like the part about using the supply frequency as the worst case so I might look at that some more, later, in a better context.
--------------
So here's the "standard" version:
The equation for the peak-to-peak amplitude of the ripple voltage is:
(1) Δv = i / (2fC)
where f is the supply frequency in Hz, Δv is the peak-to-peak ripple voltage, and i is the load current. C is capacitance in Farads.
Solving for C, we get:
(2) C = i / (2fΔv)
Choose your desired Δv and i and out pops the approximate capacitance needed.
-------------------------------------------------
What about Clipping?
So you could use the usual rule, something like 2200uF or 3300 uF per max output Amp.
(Of course, you won't KNOW the max output Amps until AFTER you calculate the ripple and find out what the max output voltage can be without clipping. But that's one reason that those equations are just approximations.)
In our example, we were using i = 3.54 A RMS, max. But we can't even get there, without clipping (running into the ripple voltage), unless the capacitance is infinite!
-----
So maybe we should first find out where the maximum output would run into the ripple voltage, as a function of the reservoir capacitance. (But keep in mind that everything below will still be just an approximation.)
Let's see... I guess we're assuming that the peak of the ripple voltage is at Vsupp = 40V and its p-p voltage subtracts from the supply voltage to give us a smaller usable voltage envelope, like usual.
The ripple voltage, Δv, given by (1), is peak-to-peak. So the bottom of the ripple voltage waveform would reach down to:
(3) Vmax_peak = Vsupp - Δv
So that would be our maximum peak output voltage, just at the point where an output signal would start clipping (i.e. running into the bottom of the ripple voltage waveform).
In RMS volts, that would be
(4) Vmax_rms = ( Vsupp - Δv ) / √2
So the maximum RMS current through the load, just befoe clipping, would be
(5) imax_rms = ( Vsupp - Δv ) / ( (√2)(Rload) )
Substituting the right side of equation (1) for Δv, in (5), and solving for imax_rms gives
(6) imax_rms = Vsupp / ( Rload∙√2∙(1 + 1 / ( 2∙Rload∙f∙C∙√2 ) ) )
where everything is a known constant except for C,
so we can directly calculate the maximum output current just before clipping (or voltage or power, since we know Rload), that we could get with different values of C.
For our example, with Vsupp = 40, f = 60, and Rload = 8, equation 6 becomes:
(7) imax_rms = 3.54 / (1 + ( 1 / ( 1358∙C ) ) )
Examples:
With C = 10000uF, we would get imax_rms = 3.297 Amps, and max rms power = 86.9 Watts, at clipping onset. The 3.297 A rms gives 26.38 Vrms across 8 Ohms, which is 37.3 Vpk, leaving 2.7 volts as the maximum Δv peak-to-peak ripple voltage.
With C = 4700uF, we would get imax_rms = 3.06 Amps, and max power = 74.9 Watts, at clipping onset. The 3.06 A rms gives 24.48 Vrms across 8 Ohms, which is 34.62 Vpk, leaving 5.38 volts as the maximum Δv peak-to-peak ripple voltage.
With only 4700 uF, the ripple is already a little more than 10% of the supply voltage. The equations above are approximations and should only be trusted when the ripple is less than 10% of the supply voltage. (Also, which way the error goes depends on the characteristics of the diodes and transformers upstream, and could go either way.)
-----------
So what capacitance should we pick, and why?
How should I know? That's what we're here to discuss. <grin> If we are concerned about the performance at the maximum output power, and we use the rule-of-thumb that says ripple should be 5% to 10% of the supply voltage, then that would put us between about 14000 uF and about 6630 uF, which give about 2V and 4V of ripple, respectively, and max output currents of about 3.35 A rms (4.72 A peak) and 3.18 A rms (4.5 A peak), and max output powers of about 90 Watts and 81 Watts. According to those, the "rule" should be "3000 to 1500 uF per max peak output Amp".
---
But some have said that using less capacitance sounds better.
Viewing the filter capacitance and the load resistance as a high-pass filter has also been mentioned (by Onaudio).
Onaudio's example was: Down to 20Hz is desired, so aim for a 2 Hz RC filter cutoff, so 8 Ohms gives C = 1 / ( 2 Pi R f ) = 9947 uF, or about 10000 uF.
That "filter" point of view seems valid if we look at the reservoir capacitance as a current source with a capacitance in series with it.
In that case, looking at it as a high-pass filter with 8 Ohms as the shunt resistance is valid only when the output device has about zero resistance, i.e. it's wide open and the amp is at max output power.
But how much power is typically used, when listening? It's a very low number, if I recall correctly, in the single-digit Watts.
If we use 10% of the maximum power, maybe we can see what might be different at that level.
In that case, the output device has some resistance, which will add with the 8 Ohms, and that will lower the capacitance needed to still get the 2 Hz cutoff frequency.
---
Let's just try it with 10 Watts into 8 Ohms, to make it easier.
Power = i²R = v²/R = v∙i
So the output RMS voltage and current would have to be 8.944 Volts and 1.118 Amps.
So the voltage across the output device would have to be Vsupp - 8.944 - Vripple, or, more generally,
(8) Vdevice = Vsupp - √(Rload∙Pout) - Vripple
The current through the output device will be the same as through the load, which would be
(9) idevice = √(Pout/Rload)
The resistance of the output device must be vdevice/idevice, which is
(10) Rdevice = vdevice/idevice = ( Vsupp - √(Rload∙Pout) - Vripple ) / √(Pout/Rload)
In terms of a high-pass filter, the shunt resistance would be the sum of the device and load resistance:
(11) Rshunt = Rload + Rdevice = Rload + ( Vsupp - √(Rload∙Pout) - Vripple ) / √(Pout/Rload)
Such a filter's cutoff frequency is
(12) f = 1 / ( 2 Pi Rshunt C )
But we know we want f = 2, so solving for C gives:
(13) C = 1 / (4 Pi Rshunt)
Substituting (11) and all of the known values into (13), it becomes
(14) C = 1 / ( 449.6 - (11.24 ∙ Vripple) )
I know that everyone would have just loved it if I had kept all of the quantities as variables and had gone on for pages to derive a general closed-form solution. Sorry to disappoint you. <grin>
In (14), there are two unknowns, C and Vripple. We will need two independent equations to solve for two unknowns. We can use equation (2), which must be equal to (14) since they a both expressions for C in terms of Vripple:
(15) 1 / ( 449.6 - (11.24 Vripple) ) = i / (2 f Vripple)
Substituting all of the known values and solving (15) for Vripple gives
Vripple = 3.792 Volts
and then using (14) or (2) gives
C = 2457 uF
The ripple seems relatively small for such a low capacitance until we remember that we are at 10 Watts and the load current is only 1.118 Amps rms.
Anyway, having that value for Vripple, we can now calculate a value for the Rshunt of the high-pass filter, which is:
Rshunt = 32.37 Ohms
Using the values for C and Rshunt, and (12), we get
f = 2.001 Hz
Bingo!
We could also plug that C value into (7), even though the results will be less than accurate because the ripple will be too high, just to get some idea of what the max output power conditions would be: We would get imax_rms = 2.724 Amps, and max rms power = 59.3 Watts, at clipping onset. The 2.724 A rms gives 21.79 Vrms across 8 Ohms, which is 30.82 Vpk, leaving 9.18 volts as the maximum Δv peak-to-peak ripple voltage.
We "could" say that the "rule" from this configuration would be (from 2457uF/2.724 Amps=) "900 uF per max peak output Amp".
But since it was not designed for that output level, maybe we should say (from 2457uF/(1.118 A rms x 1.414)=) "1500 uF per average peak output Amp", assuming that the average "peak output current" occurs at the average listening level.
Interestingly, it then still falls within the "3000uF to 1500 uF per amp" rule, from between (7) and (8) above.
-----
But would the amplifier sound better (or somehow perform better), at the lower output power level (10 Watts), than it would if there was more capacitance?
If it does, I should patent an amplifier and power supply design that switches in more reservoir capacitors as the volume knob is turned up, and also one that uses a gyrator circuit (instead of, or in addition to, caps) that automatically sets its equivalent capacitance in proportion to the output current level.
---
I have no idea, yet, which way sounds better.
The problem is that the various reports of "it sounds better with less capacitance" AND "it sounds better with more capacitance" are not comparing the same amplifier or even the same design and layout techniques (nor the same listener).
And even if one real amplifier did sound better with more, and worse with less, or the other way around, that fact would not prove that all amplifiers would produce the same result.
My guess is that a lot of people don't correctly use decoupling capacitors at each output device (i.e. with a few hundred uF extremely close and made up of paralleled caps for lowest inductance, and then also some much, much larger uF as close as possible, also made up of smaller caps in parallel, or however it fits best), and many people might not have good power and ground layouts, and might not worry enough about their inductance, and who knows what else.
So the people who don't get the large decoupling caps done right should probably hear an improvement when adding reservoir capacitance, especially at lower frequencies.
But I don't yet know what could cause someone to hear a degradation in sound quality when adding reservoir capacitance, unless maybe it has something to do with the impedances UPSTREAM from the reservoir caps. Or maybe they don't use proper decoupling caps at each device, or they use only 0.1 uF (which is not for decoupling but for high-frequency BYPASSING, for stability in the face of the hidden HF positive feedback loop through the power rails that most transistor amps have, inherently). What it's CALLED is not important. We just need to think more about the capacitors that should be near the output devices.
Inductance is significant. Decoupling caps are like little power supplies at each transistor. They keep the fast large currents in a very small loop so other voltages are not upset. They are CLOSER, so their response is much faster and more accurate. You cannot get low-enough impedance by going through more than a few centimeters of copper.
Cheers,
Tom
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So I went hunting for possible candidates
Checkout the specs
PLV1J470MDL1TD Nichicon | 493-3874-1-ND | DigiKey
PLV1J270MDL1TD Nichicon | 493-3873-1-ND | DigiKey
B41691A8107Q7 EPCOS Inc | 495-3715-ND | DigiKey
Nice!
Hi,
Here is a symetrical power supply 2*10000 µF.
Each rail is loaded by a 39 Ohm resistor connected to ground.
Mains was provided through a Variac transformer with a slow increase
in voltage at start-up to avoid current peaks.
A power supply using 2*10000 µF,
voltage mains adjuted to get 2*29V DC voltage.
Voltage and waveform on each rail.
Details on the ripple waveforms.
Addition of 44000µF in parallel with the positive rail 10000 µF cap.
Details of the ripple waveform with 54000 µF.
As expected, the ripple is reduced, however its shape is not
the familiar almost triangular waveform and should content
more high frequencies components than with 10000 µF.
More or less current does not change much this shape.
I wonder what is the cause of this "distorsion".
I have done these experiences with two power supplies from amp
manufacturers, B&O (3300 µF for a 2*20W) and Marantz
(2*10000µF for 2*100W shown here). With these values,
they certainly did not want oversize the size of their reservoir.
Could the preservation of the neat triangular waveform of ripple
be a criterion to choose the optimal value of caps of a power supply ?
Here is a symetrical power supply 2*10000 µF.
Each rail is loaded by a 39 Ohm resistor connected to ground.
Mains was provided through a Variac transformer with a slow increase
in voltage at start-up to avoid current peaks.
An externally hosted image should be here but it was not working when we last tested it.
A power supply using 2*10000 µF,
voltage mains adjuted to get 2*29V DC voltage.
An externally hosted image should be here but it was not working when we last tested it.
Voltage and waveform on each rail.
An externally hosted image should be here but it was not working when we last tested it.
Details on the ripple waveforms.
An externally hosted image should be here but it was not working when we last tested it.
Addition of 44000µF in parallel with the positive rail 10000 µF cap.
An externally hosted image should be here but it was not working when we last tested it.
Details of the ripple waveform with 54000 µF.
As expected, the ripple is reduced, however its shape is not
the familiar almost triangular waveform and should content
more high frequencies components than with 10000 µF.
More or less current does not change much this shape.
I wonder what is the cause of this "distorsion".
I have done these experiences with two power supplies from amp
manufacturers, B&O (3300 µF for a 2*20W) and Marantz
(2*10000µF for 2*100W shown here). With these values,
they certainly did not want oversize the size of their reservoir.
Could the preservation of the neat triangular waveform of ripple
be a criterion to choose the optimal value of caps of a power supply ?
Just using electrolytes is not sufficient, they are too slow to supply the initial current, but a fraction after T=0 they do start to show their significance. (avoiding the L of the supply wire). No, these electrolyte buffers always need an MKT film cap of about 100nF and if you want to go for the best parallel it with a ceramic of around 100pF.
You people must not forgot that it is the audio signal that is pulling this 'instant' current through the output devices. No; it is the closed-loop HF regulation frequency that must be supplied instant current through local caps.
so you mean that while electrolytics (not electrolytes) are a good way to stabilize the voltage, they are not the best source of power when the amplifier stages need it? and that for that purpose you need smaller but faster responding capacitors?
frankly i doubt electrolytics are "slower" to respond. they are pretty much as fast as electrons can respond.
frankly i doubt electrolytics are "slower" to respond. they are pretty much as fast as electrons can respond.
I think what you see on the scope shots is the charging and de-charging slopes. if you cap bank was faster or bigger the rising slope would be steeper, the de-charging follows the current pull through the load resistor and would as such resemble at DC current pull through the amplifier. For me this looks like the two things that the caps needs to do, one: to get charged and hold the charge, two. to supply the current to the music load. I still see a separation of these two events as a positive. So the supply must could/must then be divided in two...One crude and big for the charging and one smaller faster for the signal.. the two banks is separated by a resistor or a coil to ensure a smallish voltage drop to the Signal-bank
I think that your waveform distortion on the positive supply is voltage drop caused by the rectifier current.
It's either due to where the Oscilloscope probe is connected when the extra voltage drop is due to the resistance of the wire and connections (What does the waveform look like when the Oscilloscope probe tip and ground are connected directly to the capacitor terminals).
Or the electrolytic capacitor ESR is rather high. The capacitors suggested in post #54 look good.
I miss the larger main caps with screw terminals (e. g. Sprague etc.)
as usual to find by older models like this:
AudiogoN Forums: Review: Classe Audio 70 Amplifier
thank you for this detailled report.
actually you need an additional test with an amp and tone bursts at various level and various amplifier loads similar like this about
http://www.marklev.com/IMF/TLS80/tls80tonburstbass.jpg. Then more differences there occurs.
There are to find several amplifier tests in this kind in the German test magazine "Production Partner" - go to
http://www.quoka.de/unterricht-lite...oduction-partner-jahrgang-1996-neuwertig.html
As load will be used impedances from 16, 8, 4 and 2 ohms.
the increase in output power nearly inversely proportional to the reduce of the load impedance down to 2 ohms, one can be sure, a strong power supply was installed.
Unfortunately you will not find such investigations in typical audio magazines (i. e. for home audio and car hifi)
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