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20th July 2012, 09:49 PM  #101 
diyAudio Member
Join Date: Oct 2011

transformer specs
according to #98, the transformer will saturate at startup, however what we need to be concerned about is the current drawn during normal operations, if you get the capacitance right then the transformer will only need to be capable of supplying the loads current plus a little more to keep up the charge.
Thus too little capacitance makes the transformer work harder during normal operation and rails vary alot Too much capacitance, blows your mains, the transformer does nothing after that 
20th July 2012, 10:42 PM  #102 
diyAudio Member

just a thought: a bank of smaller caps is better than two XBOX HUGE caps, because when they fail it will be cheaper to replace one 10 mF instead of one 100 mF
when i said there can't be too much capacitance it is implied that the rest of the circuit is built to withstand the current they will demand. now i have to think what happens when you start up a psu (trannyrectcaps). the transformer will provide as much current as the capacitors demand. which is, how much? is it limited? i think not. So it will peak. Perhaps then a circuit should delay turning on the psu until the AC crosses zero so at least that way it won't be BAM max current to the caps but isn't the transformer also a set of coils? doesn't it have inductance? perhaps it is too small to delay rising of the current... or it already does that so it doesn't matter hm. it's time i put the oscilloscope to my amp's psu. now, where did i stash that fire extinguisher... 
21st July 2012, 02:52 AM  #103  
diyAudio Member

Quote:


21st July 2012, 04:25 AM  #104  
diyAudio Member

Quote:
You might find that the phase angle at the transformer, when the AC was last switched off, if it leaves the core magnetized in a particular way, can affect how bad the next startup transient can be. I've forgoten the rest of the details but it's on diyaudio somewhere. In any case, although the capacitor currents will have very large peaks, they are only that large for a relatively short time. So certain types of components' limits that are exceeded will not necessarily result in automatic immediate destruction of anything, basically because the duty cycle is so low. But the bottom line here is that you should check the specs for the affected components (especially the rectifier bridge), to see how they are rated for such currents. Some types of components might fry at the very first pulse while others could operate fine with them, for decades. I remember designing a small (22 Volts, 4 Amp) regulated power supply, once, where I had to use an inrushcurrent limiter for the caps, because otherwise the adjustable regulator couldn't have a largeenough cap to bypass its adjust pin, because that would make the regulator's inputoutput differential voltage possibly go beyond the maximum spec for the regulator, during startup. Anyway, for the first 170 ms or so, a oneOhm resistor had to carry the full inrush current, which peaked at about 20 Amps (down from 38 Amps without the limiter). At the peaks, the resistor had to dissipate 400 Watts. After looking at resistor datasheets, it turned out that I could use a 5Watt resistor (of a particular type) and still have so much safety margin that the resistor never got warm at all. I attached a couple of screen shots from LTSpice simulations of the startup, with and without the inrush limiter. These might give people an idea of what happens at startup. This is for a 22Volt regulated supply, with 4 amps being drawn by the load after the output voltage stabilizes. (Note that one potential advantage of using multiple parallel caps is illustrated: The total ripple current is divided among the caps, making each cap's ripple current requirement much lower.) The power transformer was also modeled, with more than the usual amount of detail. It was a 120 VA Hammond toroidal type, with paralleled dual secondaries, which made the rating 25V@4.8A. Upstream from the transformers, I even added a model for a portion of the AC Mains wiring, that someone had given to me. (But it looks a little wrong, now.) The schematic is actually still on line, at Spice Component and Circuit Modeling and Simulation . The transformer modeling information (and downloadable spice model and transformer measurement instructions) is also there, at Spice Component and Circuit Modeling and Simulation . (Note that nothing is for sale, on any of my webpages. Some of them still look like it but there should be something saying nothing is for sale, any more. I left most of them on line, at first, because people liked having them for references. It's been longenough now that I could take some of them down but I haven't taken the time for that. Anyway, nothing is for sale there.)  Tom 

21st July 2012, 08:18 AM  #105 
diyAudio Member
Join Date: Nov 2005
Location: East Coast of South Africa

Harrison, I don't even grasp the question.
However regarding 101, firstly one must remember that the charge voltage is not a DC pulse, but a sinusoid that rises from zero to max and that the capacitor will store this max voltage it reached. The next charge pulse would rise from the previous max to a new max in time RC. Also the voltage lags the current by 90 degrees. During the first charge impulse the voltage across the capacitor would rise to some value determined by the duration of the charge pulse, the envelope of the charge pulse, the value of the capacitor, its internal equivalent series resistance and the series resistance of the transformer secondary (and of course the inductance of the windings and magnetic properties of the core). When you say the transformer would saturate while attempting to charge the capacitors, what exactly do you understand by the term saturating a transformer  does the secondary become zero volts and zero amps due to the magnetic field flat lining?
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21st July 2012, 08:26 AM  #106  
diyAudio Member

Quote:
Unregulated Power Supply Design Their equation for the peaktopeak amplitude of the ripple voltage is equivalent to the equation I gave for calculating capacitance based on Δi, Δt, and a desired Δv: (1) Δv = i / (2fC) where f is the supply frequency (60 Hz, in our case). Solving for C, we get: (2) C = i / (2fΔv) which is equivalent to the equation I gave in post # 61: (3) C ≥ (Δi Δt) / Δv if we realize they are using 1 / 2f in place of Δt, because the rectified AC mains (and ripple) frequency is twice the mains supply frequency, f, and the time for one cycle at frequency f is 1/f, AND we realize that in (3), the Δi is usually PEAK current, not RMS, since it's usually applied for transients or short time periods. First we can note that the "3300 uF per Amp" rule would give, using equation (1), with C = 3300uF x 3.54A rms = 11682uF, a pp ripple amplitude of Δv = 2.52 Volts. (Or was that rule for Amps PEAK? That would be 16500 uF, and 1.79 Volts pp ripple.) But the ripple voltage is not the only rail disturbance. And the ripple calculation in (1) only took into account a constant load current. Our load will pull current from the caps that will not be constant. It will be sinusoidal, probably a halfcycle sine at a time, in the simplest case of one pure tone. It might be worse than that but that means that the capacitance requirement for that can be a lower bound, i.e. a minimum. So I started out thinking that the lowest frequency would need the longest current draw and would be the worst case. I looked at 15 Hz as an example. The period for one cycle of 15 Hz can be found using (4) T (sec) = 1 / f (Hz) as 1 / 15 = .0667 sec. So we know that just to supply the current for the rising half of the positive part of a 15 Hz sine wave, we would have to go from 0 Amps to 5 Amps Peak (worst case) in onefourth of a cycle, or 0.0167 sec. Using equation (4), we can get C in terms of the Δv we decide we can tolerate: C = (5A)(0.0167 sec) / Δv, or C = .083500 / Δv , for 15 Hz, which can also be rearranged to give Δv = .083500 / C for 15 Hz. BUT THEN I realized that since the charging pulses are faster than 15 Hz, and would actually recharge the caps up to four times during a halfcycle of 15 Hz, that the equations in the previous paragraph were not valid because of that, and that 60 Hz would probably be a better "worst case". 60 Hz is 0.00417 sec per 1/4th cycle. So, C ≥ (5A)(0.004170 sec) / Δv, or (5) C ≥ 0.020850 / Δv , for 60 Hz (note that it's 20850 uF for 1V Δv), and (6) Δv = 0.020850 / C , for 60 Hz. Note that the Δv in (5) and (6) is in addition to the mains ripple. If we used the 11682 uF given by the "3300 uF pr Amp" rule, that would give Δv = 1.78 Volt of ADDITIONAL railvoltage disturbance, which is 71% more (1/√2 more) than what we thought the 3300 uF rule would give us. (Or if that rule was supposed to be for PEAK amps, it would have given 16500 uF and 1.79 V pp 120 Hz ripple, which would give an additional 1.26 Volts of rail disturbance due to the positive half of our 60 Hz tone, or 42% more.) That might be especially inconvenient if we were also using a linear regulator, since the current humps pulled by the load would cause voltage sags that would essentially pull the troughs of the ripple voltage downward farther than planned, by up to 71% of its originallycalculated amplitude, which might be likely to cause the regulator's input minus output voltage difference to become less than it's dropout voltage spec, during parts of the troughs, which would get really ugly. It would only possibly be 71% [or 42%, if the rule was for peak amps] worse when we were driving the amp all the way to the rail. But the bottom of the ripple waveform would come down lower than we thought, by some percentage, at every volume setting. If we were happy with the mains ripple that the "3300 uF per Amp" rule was going to give us, then it looks like we should ADD 71% (or 42% if the rule used peak current instead of RMS) to the capacitance value it gave us, AND to the rule, in order to keep the worstcase rail disturbance the same. In any case, we might as well come up with a general equation that will take into account BOTH the worstcase mains ripple AND the transient voltage sags that our signals might induce. There might be some "overlap" between those causes and effects but this should give a safe minimum capacitance: For the transient component, from (3) we can substitute 1/4f for Δt, assuming that 1/4 cycle at the AC Mains frequency is the worst case, and get (7) C = ipeak / (4fΔv) and in order to take into account both the DC and transient components of Δv we can combine (7) with (2) to get the total C needed: (8) C = [ iRMS / (2fΔv) ] + [ ipeak / (4fΔv) ] Then we can solve for Δv: (9) Δv = [ iRMS / (2fC) ] + [ ipeak / (4fC) ] If we convert iRMS to ipeak, pull out the common denominator factor, then clean up the remaining fractions, we get: (10) Δv = (0.6036)(ipeak)/fC or (11) C = (0.6036)(ipeak) / (f Δv) where f is the AC Mains frequency, ipeak is the power supply's peak (rail) voltage (not RMS), i.e. ipeak = Vrail/Rload, and Δv is your choice of an acceptable worstcase voltagerail variation. (So it looks like the new rule should be "≥ 5600 uF per Amp (peak)".) So, assuming that the "3300 uF per amp" rule meant PEAK amps, and without considering whether or not the 1.79 Volts worstcase peaktopeak ripple amplitude that it gave is "good enough", the minimum total reservoir capacitance needed in order to REALLY get a worstcase of less than or equal to 1.79 Volts Δv should be, using (11): C = [(0.6036)(5)] / [(60)(1.79)] = 28100 uF We might also want to add 15% or more to that, to account for variations in mains voltage and transformer regulation. So maybe 33000 uF would be a better number. (I hope that someone will check my math and logic.) Last edited by gootee; 21st July 2012 at 08:52 AM. 

21st July 2012, 08:32 AM  #107 
diyAudio Member
Join Date: Nov 2005
Location: East Coast of South Africa

Tom, short and sweet... What capacitor value would you arrive at for the reservoir having to dump a worse case square wave at 3.54A into an 8 ohm load while the transformer secondary is moving through a minimum, i.e. the diodes are turned off.
Thus this cap will supplying the energy to dissipate 100 watt into an 8 ohm load for probably a period of 20mS. In other words the Vc cannot be less than 28V and Ic cannot be less than 3.6A. This should be a straight forward calculation.
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Kindest regards Nico 
21st July 2012, 09:04 AM  #108 
diyAudio Member
Join Date: Sep 2001
Location: Melbourne, Australia

Tom,
Interesting work, thank you. OK, we put in 33mF and in a second amp, identical trafo and rectification, just 10mF. How different do they sound subjectively at normal (60dBA) to loud (90dBA) listening levels? Can you give us a few words on what to expect here? Cheers, Hugh 
21st July 2012, 10:41 AM  #109  
diyAudio Member

Quote:
Hugh, Thanks. Interesting? Yes. Accurate? Not quite sure yet. It might simply be about twice as much as is needed. I'm not really an expert at this stuff. I'm just sortof making it up as I go along. I havn't even simulated this one, yet. I'm not completely comfortable with using the 1.79 V ripple figure, since it came from the 3300uF/amp rule, which tends to imply that 3300uF/amp is already good enough. But in one sense, I am comfortable with it, because it might mean that 3300 uF/amp is USUALLY enough. Maybe one good way to think better about it would be to look at the two contributions to the capacitance, separately. In this case, the "standard" equation for reservoir capacitance, based on a max load current of 5 Amps and maintaining less than a 1.79 V pp mains ripple amplitude gives 16480 uF and the one for suddenly providing the current for driving the beginning of a 60 Hz sine from 0 to the 40V rail (or almost to the rail) into 8 Ohms contributes 11638 uF. Could the 16.5 mF do that job, instead, making the 11.6mF unneeded? It should at least be able to supply a CONSTANT 5 Amps with the ripple not getting bigger than 1.79 V. And there would be nothing for the 11.6 mF to even do, in terms of transient currents, if the current was already at max. On the other hand, if nothing is happening and then the 11.6 mF is suddenly required to blow ALL of its charge out, to change the current from 0 to 5 amps in order to make the load go from 0 to 40 V, then it can't help at all with the ripple, and its voltage is going to have to come down by 1.79 Volts, just to make that amount of current come out of it and change it at that speed. Of course there's more than 11.6mF around, since the other 16.5mF is in parallel with it. So the rail should come down less than 1.79 V and there should still be the equivalent of 16.5mF that's not blowing its whole wad in order to perform the worstcase transient event, and it should be able to keep the ripple in check as usual, since it's able to do that even when the current goes to max. Thinking again about the 16.5 mF being there alone, if an 11.6mF portion of it needed to blow ALL of its charge, to change the load current from 0 to 5 amps, then there would only be 4.9 mF left with any charge, to do the job we think would require the whole 16.5 mF. I guess maybe I'm now convinced that there could be cases where the 16.5 mF by itself would not be enough. And I think I'm also convincd that there could be cases where the equivalent of 11.6 mF is just finishing a worstcae transient, and the current is at almost at 5A, and the whole remaining 16.6 mF would need to also be available to keep the ripple within spec. But I'm not sure how rare or common such cases might be, or even how significant the effects might be if there wasn't enough capacitance available, much less how it might affect the sound. But it would either not be noticeable or it would be bad to some degree. I guess that a 60 Hz bass tone at max output would create one of the situations I just mentioned. Any very large transient could also do it, or at least partially create those conditions. The main question that's left, then, seems to be whether it would require an extreme event, near the rail, to show any difference, or how much difference it might make under moreaverage output conditions. Because of parasitics like rail inductance, it might be a little more beneficial than the calculations imply, too. Regards, Tom Last edited by gootee; 21st July 2012 at 10:44 AM. 

21st July 2012, 11:43 AM  #110 
diyAudio Member
Join Date: Nov 2005
Location: East Coast of South Africa

Nicely put Tom.
Hugh maybe music does not represent even close to a worst case situation, however if one is aware that a certain value of cap would be required for a worst case, then one can decide whether you want to go that far or not. Think of it, how often here are we are adamant in placing X number of output devices in parallel to cater for the occasional low impedance worst case current, but do not have the energy available from the power supply to actually perform the function of driving such load. Then what is the point. If one insist on the best of the best then obviously one must consider all the factors  only then would you have the "best "sounding amplifier, regardless of topology or does topology make up for other basic shortcomings. I cannot say that I have encountered a fully designed amp in any thread on DIY so our colleagues presenting us with a schematic insinuating the best of the best while little consideration was given to what the power supply requirements are would be misleading. This probably applies equally to the commercial and most highend amplifiers out there.
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Kindest regards Nico Last edited by Nico Ras; 21st July 2012 at 11:53 AM. 
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