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18th January 2013, 01:18 PM  #1701  
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Kindest regards Nico 

18th January 2013, 02:01 PM  #1702 
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I don't trust the doctors that do charge either.
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It doesn't count how one deals with winning, but how to handle a loss (© DjT) https://www.youtube.com/watch?v=FJEzEDMqXQQ 
18th January 2013, 03:22 PM  #1703 
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Since I am doing all of this for free, anyway, I would prefer to see the results applied to the development of a commercial product. I am happy for all of the diyers too. But helping someone make a better product, and a profit, is even more gratifying. And if we can also help Nico to look better professionally, and make more money, I can't see any downside to that. Nico did mention, a long time ago, what his intentions were.

18th January 2013, 03:32 PM  #1704  
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18th January 2013, 03:40 PM  #1705  
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18th January 2013, 05:38 PM  #1706 
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"The total harmonic distortion is not a measure of the degree of distastefulness to the listener and it is recommended that its use should be discontinued." D. Masa, 1938 
18th January 2013, 09:55 PM  #1707 
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Nico,
Did you try it with speakers and music, yet??? :) What is the gain, or what was the amplitude of the 20 Hz sine wave input? What peak and RMS voltages do you see across the MOSFET and resistor, total, with the 20 Hz sine input? It would be useful if you could measure the transformer parameters, per the procedure at the beginning of the transformer tab in the spreadsheet, for your 42V case. You would probably need a variac (variable transformer). Do you have a way to measure the ESR of the 4700 uF caps at different frequencies (e.g. 20 Hz), or have some spec? I can't think of anything else I might need to be able to simulate certain aspects of it, at the moment. It would be interesting to see the rest of the uF vs Vrms data points that you found, and maybe the AB measures for them. Regards, Tom 
31st January 2013, 01:55 AM  #1708  
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Unfortunately, the largest 50V value that's 10mm is 470 uF. And their ESR is 2X the ESR of the 1000 uF. Anyway, I thought that I might as well see what I come up with when calculating the capacitance needed to fullysupport perfect bass down to 20 Hz. My capacitor arrays will function as both the PSU and decoupling caps, because they will fill the space between the rectifiers and the amplifier's output stage. I will mostlikely be mounting the output devices (probably a chipamp for initial testing) on a parallel PCB, as close as possible to the rear side of the cap array PCBs (like, 3 mm, but I might try to make it 0 mm, or use some extremely lowinductance connection method; I would hate to get 0.5 nH inductance from the array and then throw any of it away with a stupid connection method). I will place the amp pcb over the junction of the adjacent edges of the two array pcbs (one for each rail), so that the power and ground pins can go directly into the cap array boards' power and ground planes. If I calculate what would be needed for a full 20 Hz halfcycle without any charging pulses available for 25 ms, then it should be more than enough to be sufficient for most real music, and even for continuous DC at the full peak output voltage, when there ARE charging pulses. OK. For a 20 Hz sine, if we use: (17a): C ≥ Δi / ( πf∙(Δv  (ESR∙Δi))) which is 2x equation (17), to consider the whole halfcycle, and we use an estimate for the ESR: ESR = 0.02 / (C x VR), where VR = Voltage Rating of capacitor and substitute that into (17a), we get: (18a): C ≥ Δi ((0.04πf / VR) + 1) / (πf∙Δv) [approximate, for electrolytic cap, only] If we assume that we want Dpct percent voltage droop, then (19a): Δv = Vsupply∙Dpct/100 and if we leave Vamp Volts for the minimum voltage across the power amplifier, the peak usable output voltage would be: (20a): Vpk = (Vsupply  Vamp)(1  (Dpct/100)) and then, with Rload, (21a): Δi = Vpk/Rload = ((Vsupply  Vamp)(1  (Dpct/100)))/Rload We could get an equation for the minimum required capacitance in terms of Vsupply, Dpct, Rload, and the cap's voltage rating VR, by substituting (19a) and (21a) into (18a). EXAMPLE: Say we want a 100 Watt RMS sine output power spec, into 8 Ohms. We would need the usable peak voltage to be (22a): V_usable_pk = sqrt(2∙Rload∙Power_rms) The minimum power supply peak voltage would then need to be (23a): Vmin_psu >= (V_usable_pk + Vamp)/(1  Vripple_percent) Vmin_psu >= (sqrt(2∙8∙100) + 4)/(0.975) [assuming 2.5% max ripple] Vmin_psu >= 45.13 V (peak) = 31.91 V RMS minimum transformer secondary rating for 100 Watts into 8 Ohms (or more, if more than 2.5% ripple were allowed) [NOTE that transformer regulation and AC Mains deviations were NOT taken into account.] So, we will, for now, assume that we can use capacitors rated at VR = 50 VDC, which we will do because lower VR gives a worse case, i.e. larger required C, than higher VR does. V_usable = 40V peak = 28.28 V RMS The load current will swing 5 Amps when the voltage across the load swings 40 V. So use Δi = 5 Amps. Using equation (18a): C >= (83577 uF) / Δv Estimated ESR was .02/(.083577 x 50) = 0.0048 Ohm. That appears to be reasonable since an 8200 uF 50V cap was quoted (at mouser.com) as having ESR = 0.04 Ohm and ten of those in parallel would be about 0.004 Ohm. For 2.5% voltage droop (ripple) with 40 V, Δv = 1 V, giving C >= 83577 uF. Allowing 10% ripple would give C >= 20900 uF. Just to check the worstcase upper bound, a square wave, or DC, for 25 ms at 40 V output would require C >= Δi Δt / Δv (without taking ESR into account) C >= 5(0.025)/1 = 125000 uF [ = 1.5 * 83577 uF ] Or, with the same ESR approximation as was used above: C ≥ (Δi / Δv)∙(Δt + (0.02 / Vrating)) (approximation, for electrolytic capacitors ONLY; NOTE: Used 2x, for DC instead of linear ramp.) C >= (5/1)(.025+(.02/50)) = 127000 uF That means that with charging pulses every 10 ms, only 127000(10/25) = 50000 uF should be sufficient, to handle continuous DC output at the rated peak voltage of 40V, without allowing more than 2.5% railvoltage droop. So 82000 uF should be enough for a single 20 Hz sine of 100 Watts into 8 Ohms, with <= 2.5% ripple, without being recharged for a full 25 ms halfcycle of 20 Hz. If being recharged every 10 ms, 33431 uF should be sufficient (i.e. (10ms/25ms)*83577 = 33431), for a single sine, at the maximum rated output power. So my planned 10x10 array of 1000 uF caps (100000 uF) will be fairlymassive overkill, for 100 Watts into 8 Ohms. But what about 4 Ohms? It turns out that 100000 uF is exactly the minimum required C to supply a continuous 10 Amps of DC current at 40 Volts DC, into 4 Ohms across the output, with less than 2.5% rail voltage droop, if charging pulses occur every 10 ms, and 200 Watts into 4 Ohms only requires 7.07 Amps RMS. So the 10Amp peaks required for that should be effortless, for these arrays. The cost of the caps is still worrying me, a little. If I used two arrays per channel, that would be 400 caps. Might as well go for the Qty 500 price break and use 11x11 arrays, in that case. Still, it's a few hundred dollars. Maybe I will think a little longer about either using smaller arrays or smallervalue caps. Cheers, Tom Last edited by gootee; 31st January 2013 at 02:23 AM. 

31st January 2013, 02:03 AM  #1709 
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31st January 2013, 02:44 AM  #1710  
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