
Home  Forums  Rules  Articles  diyAudio Store  Gallery  Wiki  Blogs  Register  Donations  FAQ  Calendar  Search  Today's Posts  Mark Forums Read  Search 

Please consider donating to help us continue to serve you.
Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving 

Thread Tools  Search this Thread 
11th January 2013, 09:14 PM  #1691  
diyAudio Member

Quote:
Yes, I had decoupling caps in mind, partially, but also just the ablity of a cap, in any situation, to be an accurate source for lowfrequency maximumamplitude sine output currents (or any frequency). I also noticed that if we take the equation relating frequency content and rise time that I saw in Henry Ott's EMC Engineering book, trise = 1 / (πf) and substitute it into, for example, equations like 13a, 13b, and even 17, as a Δt in the numerator, those equations suddenly look much more familiar. I just thought that was interesting. By the way, I did verify that LTSpice simulations give _exactly_ the same numbers as the equations do. Last edited by gootee; 11th January 2013 at 09:21 PM. 

17th January 2013, 04:43 AM  #1692  
diyAudio Member

Quote:
How would we set the Δv_max based on the PSRR and the required headroom? Also, after thinking about it again, I don't believe that we can say that the ripple calculation would "dominate". I think that "the ripple calculation" would be dependent on the currentsupplying events required by the music signal. When real music is playing, at significant output power levels, we will almost never see a textbooktype periodic ripple waveform. We WILL see ripple CURRENT that IS, almost exactly, the music signal. And due to the caps' ESRs and the voltage variation due to the music current being discharged, the ripple voltage might then also somewhatresemble the music signal. I showed the caps' currents, almost exactly matching the music signal, before, at: Power Supply Resevoir Size So what actually happens? Without the rectifiers in the circuit, the caps' voltage would be according to equation 11 that I gave before: (11a): v(t) = (1/C) ∫ i(t) dt + ESR∙i(t)  ESR∙i(0) + v(0) where the integral is from 0 to t. In that equation, i(t), the capacitor inflow current, is simply iload(t): (18): i(t) = iload(t) But WITH a rectifier, we would have irect(t) = i(t) + iload(t), so now (19): i(t) = irect(t)  iload(t) i.e. the load current is no longer necessarily just the negative of the cap current, except when the rectifier is not conducting. So we would have the nowfamiliar case where the system switches between a firstorder differential equation, when the rectifier is not conducting, and a secondorder equation (if transformer leakage inductance is included), when the rectifier is conducting. We could use essentially the same numerical solution of the differential equation(s) as I used in the spreadsheet, except using the music current instead of the constant load current, which would probably complicate the calculation of the rectifier turnon and turnoff times. The present spreadsheet doesn't do that, of course, but the constant current that it does assume, with amplitude equal to the maximum peak sine current for a given rated sine output power, should still give the worstcase minimum v(t), including for any music signal, even though with music the ripple voltage waveform would have a completely different shape. We could also write some inequality that would enable us to determine when v(t) would fall below the specified ripple voltage minimum value, but again, we would have to integrate the music current to be able to get a numerical answer. And we could derive an equation that would tell us whether or not the caps' voltage would be able to be sustained, by the charging pulses, within the specified ripple envelope or if they could instead be drained to lower and lower voltages, which would let us determine the required charging pulse capability. But again, we would have to integrate the music current to get a numerical answer. At any rate, the caps are still providing current, but that current is the music, and so the ripple voltage will be irregular, compared to the usual case of an average current or a worstcase constant current. So the "true" ripple calculations, while still mathematically possible, are probably mosteasily done (and are definitely moreeasily generalized) by considering a bounding case, as the spreadsheet does. Sorry if that went offpoint. Regards, Tom Last edited by gootee; 17th January 2013 at 05:13 AM. 

17th January 2013, 09:34 AM  #1693  
diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders

Quote:
We can see that easily on a scope with both sinewave and with music. There is a tertiary case: The signal gets progressively more distorted as clipping is approached. I have seen FFT plots showing the increase of high harmonics as clipping is approached. I suspect this tertiary case is showing the effect of non linearity of the amplifier as Vce of some devices becomes similar to Vbe or even worse becomes less than Vbe. As device Vce approaches zero, then we see the bounding case of clipping.
__________________
regards Andrew T. Sent from my desktop computer using a keyboard 

17th January 2013, 09:49 AM  #1694 
diyAudio Member
Join Date: May 2007

If the ripple calculation assumes the worstcase current draw, as it should, then no music current will exceed this. Music at frequencies higher than the (doubled) supply frequency are guaranteed to draw less than the worst case, as for part of the time between charging pulses the music will be drawing current from the other supply rail. Music at frequencies lower than this can only draw the worst case current, and no more. I think we are agreeing?
The only exception to this would occur if capacitor stray inductance (+wiring) caused extra voltage drop for higher frequencies, but in reality this inductance is just as likely to reduce the voltage drop across much of the audio range as it has the opposite sign to the capacitive reactance. Naive bypassing could screw this up, but people using naive bypassing are unlikely to do ripple calcs anyway. Now if the final PSU cap is well separated (electrically) from the reservoir cap (e.g. by a choke) then the ripple requirements for the reservoir cap and the music current requirements for the final cap get partly decoupled. We can assume the final cap gets fed with a DC current which on average is exactly right to meet music current requirements. The problem now is that the voltage drop in the final cap has to be added on to the average DC droop (i.e. half the maximum droop) in the reservoir cap. We can no longer count on the charging pulse to reset things every 10ms so low frequency requirements are more demanding. So if you have LC smoothing and you want good bass then the final cap may need to be larger than you would need for a single reservoir cap, as it may need to supply current for 25ms (half a cycle at 20Hz) rather than 10ms. 
17th January 2013, 05:28 PM  #1695 
diyAudio Member
Join Date: Nov 2005
Location: East Coast of South Africa

Tom/DF/Frank,
following is a set of practical measurements from our latest commercial development. After the amplifier design was finalized and frozen the power supply was optimized with the objective that started this thread. The amp is essentially 100 watt into 8 Ohm and I will qualify this so that we are all aware of the criteria used to obtain the results. The measurement was intended to detect any artifacts in the output signal compared to the input signal added by the amplifier while delivering the rated rms voltage across an 8.2 Ohm resistive load at 20 Hz sine wave input stimulus(mains =50Hz; 240VAC). The observations were made by comparing the signals between the inverted (NFB) and noninverted inputs of the amplifier under test using a scope with AB and dB meter at rated output nulled at 1kHz as reference output. The amplifier uses four pairs of LMOSFET as output devices and 0.22 ohm noninductive drain resistors. There is no current limit applied, no "zobel" across the output nor any series output inductor. 6.3 mm spade terminals were crimped to the caps so that we could connect several in parallel onto an 8 x 8mm copper, positive and negative rail. A 35 Amp soft recovery fast bridge rectifier type SRDB3500P1A was used between the transformer and reservoir without any capacitors across the diodes. We used multiple Hitano ELP range 4700 uF/100V 30 mm diam caps to arrive at an observed optimum value. We used the same 368VA toroidal transformer wound to supply 36  0  36 to 46V  0  46 VAC at 2 V intervals (taps) for all measurements. Our method consisted to set a transformer voltage by connecting to the lowest tap, then starting with 4700uF, and increase the capacitance until the differential voltage on the scope/meter was as low as possible. If after adding further caps the difference was hard to detect we removed the last cap and returned to the 1 kHz reference signal to make sure this has either improved or remained the same as before. We then stepped to the next transformer voltage and incremented the capacitance by 4700uF until we reached the point that an observed improvement was uncertain and so on. The instruments used for this exercise was HP 54645A scope, HP 33120A function generator and HP 33401A multimeter. The final results was achieved by using four paralleled 4700uF capacitors per rail and 42V setting on the transformer which was higher than one would expect the rail to be for a 100 watt amplifier. In fact the max average power measured with the final settings was 132 watt into 8 ohm (to produce the optimum performance at 100 watt which we find interesting) Tom/DF/Frank can these observations be verified by any of the theoretical scenario?. There is still a lot of measurements and tests before the amp goes to production, thus I will perform other measurements you may require such as transformer inductance or whatever is needed to plug into the formulas. I will not divulge the schematic but it consists of a typical Hitachi topology using double differential amps driving LMOSFETs, you all know the topology.
__________________
Kindest regards Nico Last edited by Nico Ras; 17th January 2013 at 05:31 PM. 
17th January 2013, 06:30 PM  #1696 
diyAudio Member
Join Date: May 2007

Quick backofenvelope calculations at 100W say the peak current will be about 5A, so voltage droop will be 2.7V over 10ms. Allowing 1.5V drop in the bridge, say 4V in the output stage (including 0.22R), gives a requirement of 34V RMS. Add 15% to get 132W means 39V RMS. Not too far off 42V?
It could be that the FET output devices are a bit 'softer' than BJTs, so they drop a bit more voltage and begin to suffer earlier from insufficient headroom. Or maybe the circuit design has poor PSRR? PSRR has two aspects: direct injection of supply rail signals into the output, and supply rail variations modulating the gain and so creating IM. Note that only a small part (33%) of the total difference between peak offload DC and the onload minimum comes from capacitor droop. This means that a small change in any of the other voltage factors will cause a big change in required cap value. The calcs here take no account of transformer resistance and inductance, so must be regarded as best case. 
17th January 2013, 08:24 PM  #1697 
diyAudio Member
Join Date: Nov 2005
Location: East Coast of South Africa

Hi DF, you are correct that the MOSFETS drop more voltage than BJTs as well as having a diode in series with the rails to decouple front end from the power end so the drive swing is slightly lower than if driven directly.
The DC power supply rails drop by about 2 V at 100watt/ch both channels driven into 8 ohms and no visible clipping. I did not measure PSRR because I needed to finalize the power supply first laid out PCBs and all the good stuff in order to commence with CEA490AR2008 compliance testing to see if we make the grade else it is back to the drawing board. Thank you for your input DF. I am also really keen to get some feedback from Tom regarding a tie up between the spreadsheet and the measurements I got. My last comment above might be of interest in that, although the amp was rated and tested for best performance at 100 watt, the transformer voltage had to be adjusted to be able to deliver 130 watt to provide the best performance at 100 w, thus adding three more capacitors had much less benefit than increasing the rail voltage a tad. If you have any clues here it could make for an interesting comment.
__________________
Kindest regards Nico Last edited by Nico Ras; 17th January 2013 at 08:29 PM. Reason: Seems that 1/2CV˛ should be considered and wire cost (transformer winding) to Cap cost should be compared. 
17th January 2013, 11:15 PM  #1698  
Banned

Quote:
But I will say that comment reminded me of results I was getting earlier on: if you're purely worried about voltage drop the easiest solution was to bump up the voltage from the transformer slightly, for a toroid just wrap a couple more turns around and add to the end of the secondary  problem solved! Frank 

18th January 2013, 12:52 AM  #1699 
diyAudio Member
Join Date: Nov 2005
Location: East Coast of South Africa

Hi Frank see my comment above in Edit which I repeat here:
Seems that ˝CV˛ should be considered against wire cost (in a few extra transformer windings) compared to additional Capacitor cost. Keep in mind I am looking at it from a commercial point of view. My conclusion is that 48 J would could be an optimal reservoir energy required in the light of the test performed on a practical 100 watt into 8 ohm amplifier together with a 15% increase in expected rail voltage. Hope this makes sense. This at the moment is applicable to 100 watt into 8 ohm since there have not been tests carried out on higher or lower rated LMOSFET amps, it may be also somewhat different for BJTs.
__________________
Kindest regards Nico Last edited by Nico Ras; 18th January 2013 at 12:57 AM. 
18th January 2013, 11:20 AM  #1700 
diyAudio Member
Join Date: May 2007

Increasing the cap size means you droop less. Increasing the transformer voltage means you start higher up. Both mean your minimum voltage is raised. Once the capacitor droop becomes a small proportion of the total voltage drop, as is the case here, then a small increase in transformer voltage does more than a big increase in cap value. This was made clear in the thread, on several occasions.
Am I providing free consultancy for a commercial product development? 8) 
Thread Tools  Search this Thread 


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Valve power supply  How to size transformer?  SanderW  Power Supplies  25  4th January 2013 04:12 PM 
How do you calculate choke size in a power supply?  Burnedfingers  Tubes / Valves  25  5th January 2012 12:23 AM 
power supply bypass cap size  BigE  Power Supplies  11  5th July 2011 02:59 PM 
Power Supply Case Size  diymixer  Power Supplies  1  10th October 2010 05:47 AM 
What size power supply should I get for repair work?  spooney  Car Audio  3  6th December 2007 11:50 PM 
New To Site?  Need Help? 