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17th September 2012, 08:18 PM  #1241 
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Join Date: Nov 2007
Location: near the House of the Mouse

It is easy, don't think I am that smart
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17th September 2012, 09:48 PM  #1242 
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Join Date: Nov 2007

See the interesting file that Tom attached to post 1219.
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17th September 2012, 09:57 PM  #1243 
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Join Date: Nov 2007
Location: near the House of the Mouse

I looked at the spread sheet, but I could not directly see the formulas.
My display made some of the fields tiny. When Tom "Plainly" states what they are I will put it on an app. I follow most of what has been detailed in this thread, but I wish someone would state the max max of 110 at 60 htz with 15 or 20 amps or 220 at 50. There should be some magic power output and rail voltage combo to be derived a great sounding amp. Maybe I just want to build the biggest one I can based on 110, then it will loaf along never being stressed, never hitting the rails, over kill my way into the sweet spot.
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17th September 2012, 10:21 PM  #1244  
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Join Date: Nov 2007

Quote:
P.S. Well, the 60hz is wishful thinkingthere's a lot of bonus noise content, but the higher pitched signals don't have as much strength as the 60hz; yet the bonus noise content is enough to drive an ordinary transformer, rectifier, caps, setup to a few volts higher unloaded and that makes more rail bounce.
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17th September 2012, 11:06 PM  #1245 
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Join Date: Nov 2007
Location: near the House of the Mouse

I am saying 110 for the sake of talking, I have not measured any voltage on my home.
Maybe I need to do a SMPS, no chance of droop
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18th September 2012, 06:47 AM  #1246  
diyAudio Member

Quote:
I understand your point but I think that most people here know how to calculate the reservoir capacitance using the standard approximate formulas, based on an RMS or DC load current. (For example, see Unregulated Power Supply Design, or, for another method, my post #1061 at Power Supply Resevoir Size , but with the + sign changed to a minus sign in step 5. (However, note that in the example at the end, the resulting capacitance was too LOW, by a factor of just over 2. The actual answer should have been 6650 uF instead of 3000 uF. But also note that the calculated value was not "doubled" arbitrarily. The method calculates the C of both rail capacitances as one capacitance, but since they are in series as far as the calculation is concerned, the value has to be doubled to get the two individual rail capacitances.) I think that I was originally wondering about the possibility of those types of standard methods not accounting for nonconstant load currents and the worstcase timing of loadcurrent peaks and charging pulses, while I was investigating just the very first step of finding a general rule for "What C value could and should we use?", with that first step being to see if I could find a way to know the "true" lower bound of the C value, for sinusoidal load currents at the rated maximum output power. I have had limited time and that first step has seemed quite belabored. Anyway, I think that the part where you say "Then double it." would need to be justified. I'm sorry if you already did that. I know there were a couple of posts of yours, with equations, that I didn't get to study and that I meant to go back to, but haven't yet. At any rate, my original plan was to try to use simulations to find the ACTUAL lower bound for the C value, for different cases, and then compare that to what the standard equations give, and then try to move forward from there. I think that I am just now finally able to compare the simulation results to the standard equations' results, which I attempt to do, below or in another post. I do realize that you pointed out, correctly, that it will also depend on the amplifier topology, since the actual absoluteminimum C depends on the minimum voltage that can exist across the amplifier, i.e. between the rail and the high side of the load, with the minimum C being the one that allows the rail voltage to have a worstcase drop that doesn't quite reach Vload_max_peak + Vamp_minimum, for the worstcase timing of a load current peak and the first quarter of a charging pulse. But I can either leave the minimum Vamp as a variable, or, just add a few volts (for a minimum Vce plus a small series resistor), for now, and be close enough for this purpose. (Yes, it appears that a railvoltage incursion into the output voltage only occurs during the upwardcurving portion of the beginning of a current pulse from the transformer, between zero and the first inflection point. Made me wonder if the second derivative of the transformer current was important.) On to the standard equations: (I HOPE that someone will check my algebra and my arithmetic!) From Unregulated Power Supply Design : Ripple Voltage (peaktopeak) = Vr = i_load_rms / (2 ∙ f ∙ C) Peak Rail Voltage = Vc_pk = (Vrms_secondary x 1.414)  1.4 Average rail voltage = Vc_avg = Vc_pk  (Vr / 2) = Vc_pk  ( i_load_rms / (4 ∙ f ∙ C) ) For the absolute minimum C, we would need: Vc_avg  (Vr / 2) ≥ Vout_pk + Vamp_min in order to not have any rail voltage dips gouging into the output voltage. But Vc_avg  (Vr / 2) = Vc_pk  Vr = Vc_pk  ( i_load_rms / (2 ∙ f ∙ C)) . So we can reexpress what we need as: Vc_pk  ( i_load / (2 ∙ f ∙ C)) ≥ Vout_pk + Vamp_min Solving for C, we need to have: C ≥ i_load / (2 ∙ f ∙ (Vc_pk  Vout_pk  Vamp_min)) That's actually a little better than the "standard" formula, since it also accounts for the minimum voltage across the amplifier. This should give slightly higher capacitance values than the usual standard formulas. We already know: i_load_rms_max = √(Power_rms_max / R_load) Vout_rms_max = √(Power_rms_max ∙ R_load) Vout_pk = (√2)(√(Power_rms_max ∙ R_load)) Vc_pk = (Vrms_secondary x 1.414)  1.4 Vamp_min = 3 (Min Vce plus worstcase voltage across the 0.22Ohm R)  To compare to the four cases for which I recently posted simulation results, Power_rms_max would be one of 25, 50, 75, and 100 Watts RMS, giving i_load max rms values of 1.77, 2.5, 3.06, and 3.535 Amps RMS, and Vout_pk values of 20.00, 28.28, 34.04, and 40.00 Volts peak. With the 44044 transformer output, we get Vc_pk = 60.83 Volts peak. Plugging in those values and turning the crank yields C_minimums of C = 390 uF, 705 uF, 1100 uF, and 1650 uF, for Power = 25 W, 50 W, 75 W, and 100 W. But, with a 240 VA secondary for each rail (a 480 VA transformer), simulations found the "actual" C_minimums to be: C = 330 uF, 770 uF, 1540 uF, and 6650 uF  It appears that the standard equations are assuming an ideal transformer, and that they also don't account for the occasional badlytimed load current peak relative to a charging pulse. So the standard equations do well when the transformer is ridiculously oversized compared to the output power spec. HOWEVER, as I've been saying repeatedly, everything in my simulations depends on how good the transformer models are. And Terry Given just mentioned that it looks a little wacky. Does anyone have some "typical" values for such a model? (The transformer ratings don't have to be the same as my original ones.) Hey Terry, I have an LCR meter that I did not have, back when I was doing the original transformer measurements. Could that help to check them? I think that I still have the original 120 VA 25 Vrms transformer that was measured. Cheers, Tom Last edited by gootee; 18th September 2012 at 06:59 AM. 

18th September 2012, 07:13 AM  #1247  
diyAudio Member

Quote:
I temporarily used a 44044 transformer model in order to try to remove the effects of having a rail voltage that was already relatively close to the maximum peak output voltage (with a possiblytooweak transformer). I just wanted to see how the numbers came out, for the highest C that would still make Vrail droop until it corrupted the output. And I also wanted to see if any trends came into focus better, when there was not so much of a problem with the headroom. But I'm starting to suspect that the particular transformer model parameters I am using might not be giving very good load regulation. However, regarding the load regulation, we should also keep in mind that I am simulating with square waves that have the same peak amplitude as would a sine wave that was producing the maximum rated output power. i.e. The square waves produce 2X the maximum rated output power. (But still, the transformer seems like it should handle it way better than it is.) Regards, Tom 

18th September 2012, 07:19 AM  #1248  
diyAudio Member

Quote:
If you have a VARIAC, and a lowvalue power resistor, that's all you need besides a multimeter. The complete (and very simple) instructions are printed right in the diagram, at Power Supply Resevoir Size . You might want to read it to see if my safety warning was adequate, for people playing around with exposed mains connections and no safety devices whatsoever, except their breaker box. Regards, Tom 

18th September 2012, 10:25 AM  #1249 
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Join Date: Jul 2004
Location: Scottish Borders

I'll try to fit in the whole test today.
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18th September 2012, 10:52 AM  #1250  
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Join Date: May 2007

Quote:
Quote:
The calculations you have done show good agreement with simulations for lower powers. The discrepancy for higher power mainly comes from Vmin being a bit too small. This has little effect at low power (i.e. high ripple voltage) but is critical at high power (low ripple voltage). As you have included the 0.22ohm drop in Vmin (not a good idea, in my opinion) your Vmin should be bigger for greater power. Try repeating your calcs for Vmin=4V instead of 3V. Including transformer resistance in the calculations gets messy, unless the resistance is sufficiently low that the cap can recharge more or less fully on each mains halfcycle. As I said in an earlier post, this gets worse for really big caps which may be a reason to avoid really big caps. So we are in a position where we can calculate, with some confidence, the minimum cap value. We can also estimate, roughly, the maximum. Isn't that 'job done'? 

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