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rhythmsandy
diyAudio Member

Join Date: Aug 2012
Quote:
 Originally Posted by palstanturhin Yes. But: The one on my drawing board right now the bank holds 2500W of capasitor power!!!! Amplifier is only 2*100W.
can you explain how did u store 2500W of cap power?

 8th December 2012, 06:42 PM #62 palstanturhin   diyAudio Member     Join Date: Jun 2011 :-) Yes, easily: I stored it into 100 volts... :-)
 8th December 2012, 07:09 PM #63 rhythmsandy diyAudio Member   Join Date: Aug 2012 how about the calculations that I have done... i was looking at the nelsons article where he states about for 200watts he recommends using 700w and for stereo to have a 2kva and sounds pretty sufficient.. but i have a practical example for 300 I=sqrt(P/load) sqrt(300/8) = 6.12amps now here the voltage ratings of the trafo is 56-0-56 so considering this when compared to the psu ratings it has to be 2 x 300 = 600W as common mode of selection which will result in Arms in trafo = 600/(56*2) = 5.35 amps... now the question is when there is 6.12 amps in speaker then where does the extra current come from? or 50% has to convert into heat so we need 6.12 amps more to burn into heat... so the ratings of the trafo has to be atleast min of 12.24amps which results as (56*2)x12.24 = 1370VA and with 5-10% losses its roughly 1500VA as standard size... if that is right? when we take a 4 ohm load into 500W the same amplifier which works in 300watt in 8 ohm.. the the current in speaker is 11.18 amps and when we double this its 22.36 so (56*2)x 22.36 = 2504VA now it looks like its 5 times the ratings are the right ones to deliver the continuous rated output.. are these calculations are right? my big question is that the current in load may be 5Arms so even to generate heat 5Arms is spent right? i even observe this in Nelson Class AB X350 amps where if output power in any AB nelson amp is 200W then the psu is 1000VA considering all losses and decent headroom and also whatever the transient requirement.... ( if caps are added more than 1,00,000uf ) then probably we can expect better transients... but im purely talking on continuous power ( i agree music is very random in pow requirements ) but my app is more towards PA amps. what is the Effect of R-core trafo? I heard that Rcore delivers instantaneous current peaks...?
dmills
diyAudio Member

Join Date: Aug 2008
Location: High Wycombe
Quote:
 Originally Posted by rhythmsandy what is the Effect of R-core trafo? I heard that Rcore delivers instantaneous current peaks...?
How can it?
For most of the cycle, the transformer in a capacitor input power supply is delivering NO CURRENT!

The diode bridge only conducts during that part of the cycle where the transformers secondary voltage exceeds the voltage on the energy storage caps by the diode forward voltage drop, so as the caps get bigger the conduction angle for a given load current decreases as the ripple voltage decreases.

Making transformers and smoothing caps very oversized just reduces the proportion of the cycle for which the diodes are conducting and power is being deliverd from the transformer, while increasing the RMS/Average ratio of currents in the transformer and diodes, and making more harmonics on the transformer primary.

It is instructive to consider that except during diode conduction (Roughly half that part of the cycle during which the transformer output exceeds the lower point of the ripple on the caps) ALL the power is supplied from the capacitors and the transformer plays no part at all.

Even if we accept very short high current transients as being likely, the impedance that matters is the ESR and ESL of the caps provided the total enery in such a transient is small compared to the energy required to hold up the cap voltage for one half cycle of the supply).

Oh, BTW energy in a capacitor bank is measured in Joules, not Watts (One Joule per second equals one watt) , and yes the difference matters. It is vital to keep energy and power as separate distinct concepts, especially when dealing with pulse and transient load applications.

Sit down with scilab or ltspice sometimes and plot some graphs, it is interesting and quite informative.

Regards, Dan.

 9th December 2012, 02:20 AM #65 benb   diyAudio Member   Join Date: Apr 2010 You can store enough power (actually, energy) in a capacitor to shrink a quarter.
Art M
diyAudio Member

Join Date: Oct 2010
Location: SF Bay Area
Quote:
 Originally Posted by MarianB @AndrewT with all do respect, that is ridicoulus, you really should research more. best of wishes. Marian.
Andrew is trying to show us the Basic mathematics of using 2 amplifiers in a Bridged Tied Load mode. This is the Practical application and attendant calculations. You need to pay attention to learn some simple mathematics as applied to this configuration.

MarianB
diyAudio Member

Join Date: Jul 2011
Location: Romania
Quote:
 Originally Posted by Art M Andrew is trying to show us the Basic mathematics of using 2 amplifiers in a Bridged Tied Load mode.
Is he now? Well than you agree that 50A minimum is needed in he's example of 1,2kW bridge mode, wright? well we can almost say we have an arc welder machine at that current, and i would hate to be the guy that builts that power transformer for just 1,2kW... that's a laugh
__________________
Have a go at my latest set => Dj Marik -
Trance session live ep. 94

 9th December 2012, 10:45 AM #68 AndrewT   R.I.P.   Join Date: Jul 2004 Location: Scottish Borders Marian, you still seem to be refusing to accept that transient currents exist. And worse you also refuse to accept that currents can be measured and/or expressed in ac, or rms, or pk, or pk to pk units. Take for example the 50A Marian has stated in the above post. Is it 50Apk, or 50Adc, or 50Arms, or some other? I'll assume for this that Marian meant 50Apk even though Marian claims that transient currents can't exist. Power (in the load) is defined as the current squared times the resistance (of the load) it passes through. Idc^2*R = Power for a DC current. For an AC current the definition changes to Irms^2*R = Power for an alternating current. Instantaneous power (power at one instant of time) = Ipk^2*R For a sinewave, the Power = Ipk^2*R/2 Let's look at that 50Apk Power = 50*50*1r/2 = 1250W and would need a waveform that develops 50Vpk across the 1r0 load. If one wants 1250W to be delivered to that 1r load then the load needs to see 50Vpk and simultaneously pass 50Apk. Nothing impossible in that, or is there? __________________ regards Andrew T. Last edited by AndrewT; 9th December 2012 at 10:58 AM.
 9th December 2012, 11:27 AM #69 MarianB   diyAudio Member     Join Date: Jul 2011 Location: Romania @AndrewT i am not asumming anything, those 50 amps are your statement, not mine ( actually 49, witch it would be just as well ), read your previous posts and you will see it. On the other hand i do not refuze to accept transients, if that's what you understud thus far then i am even more dissapointed by you, transients exist ofcourse they do, but they cannot be predicted with exactity and by no means you shoud take them into account when calxculating the current needed from the PSU, the verry nature of transients tells you that they are a transitory phenomena, extreemly short, too short to affect the PSU! I guess you could say that transients could affect the power devices of the amplifier but i doubt it. In the end to me it makes no difference who is wright and who is wrong, i've seen enough in you to know what i need to know, what is important is that everyone here get's the wright ideea of ahat is to be done and i strongly recommend everyone should make a print screen of @Mooly's post #51 on page no.6, it is the only thing one needs to correct calculate the PSU for a given power and load. Anything else is as Douglas S. says just subjectivism witch is sort of a plague that infested audio electronics, with no end in site for the near future. All the best. Marian. __________________ Have a go at my latest set => Dj Marik - Trance session live ep. 94 Last edited by MarianB; 9th December 2012 at 11:30 AM.
 9th December 2012, 04:35 PM #70 jlind54   diyAudio Member   Join Date: Nov 2012 Location: Kansas This whole pk current thing got me thinking. So I set out on a little adventure with one of my high powered audio subwoofer amps. In the car audio world do they take these peak currents into account? I would be inclined to say they do or come very near it. I have an amp that is capable of 1000 watts into 2 ohms rms continuous power. Generally speaking it will pop a 100 amp fuse when presented with a sine wave input in the 30 hz region. The amp itself packs 120 amps of fuse protection on the battery input. So if one assumes that the smps inside the amp must produce around 51 vac rms to give a Vpk around 71 so the rails are at +-63 Vpk after some losses. In this example the pk current would equate to near 31 amps. Taking the supply voltage into account and the battery supply voltage its possible to see that 56volts divided by nominal 13.8 battery is approximately 4 times larger so multiply the 31 by this 4 yields 124 amps. So my question now would be does car audio operate in a different realm? or should smps setups really have a higher rating than its simple 60hz step down transformer counter part? I've not had the opportunity to tear into any high end amps using a smps to see how they do it.

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