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10th February 2013, 06:23 PM  #11  
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10th February 2013, 07:17 PM  #12 
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i saw the info reg slew rate over 100v/us using sikorel
Solid State Power Amplifier Supply Part 1 
10th February 2013, 07:37 PM  #13 
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Location: San Antonio

I thought that a capacitor resists a change in its voltage, and would create current to maintain it. An inductor resists a change in current, and would create voltage to maintain it. Is this just an oversimplification?
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10th February 2013, 07:40 PM  #14 
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check the post no 31: stating about how Epcos is stating on the slew rate of the caps..
http://www.diyaudio.com/forums/chip...inclone4.html 
10th February 2013, 08:11 PM  #15  
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10th February 2013, 08:22 PM  #16  
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Sorry but the guy who wrote http://www.tntaudio.com/clinica/ssps1_e.html has no clue about capacitors ... or tries to sell you some Mojo (or profit) filled stuff.
Reservoir Capacitors (that's what PSU capacitors are) have no "Slew Rate" or rather, they are the exact opposite of that concept: a perfect, huge PSU capacitor would have no voltage variation between its terminals under load, behaving as a perfect voltage and current source. While the Slew Rate concept is based on voltage variation . Ok, ok, but these are *real* capacitors, and *their* voltage across terminals does vary with load. Can we calculate it? Yes, of course. >>>>>>>>>>>>>>>>>>> Capacitors and calculus Capacitors do not have a stable "resistance" as conductors do. However, there is a definite mathematical relationship between voltage and current for a capacitor, as follows: <<<<<<<<<<<<<<<<<<< Crunching some numbers and using values supplied by your link Quote:
I (Amperes)=C (Farads) 40 (Volts)/1 x 10^6 (Second)= I= (10000/10^6) x 40 x 10^6 (Amperes). Simplifying 10^6 we end up with: I = 10000 x 40 (Amperes)=400000 Amperes. So yes, you *can* Slew 40V/uSec across a 10000 uF capacitor; you "only" need 400000 Amperes to do it. Source or sink, your choice. As of the Sikorels, you'd need 1000000 Amperes to achieve that. Good luck with that. Last edited by JMFahey; 10th February 2013 at 08:25 PM. Reason: Cleared whick link I was referring to 

10th February 2013, 09:07 PM  #17 
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Gotcha. Thanks. I knew my perspective here is/was fuzzy.
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It is error only, and not truth, that shrinks from enquiry.  Thomas Paine 
11th February 2013, 02:03 AM  #18 
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agreed its all about the current which goes through the capacitor and for such a super crazy Amperes I really dont think so. So what makes Sikorel so different as people say that its supposed to be the best... nelson doesnt use it.. neither krell I see that only Halcro using it... on the reviews its said that the amp has the speed....

11th February 2013, 03:08 AM  #19 
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Actually, you can get all kinds of crazysounding numbers for capacitors' voltage slew rates. But if the equations you use take into account ESR and maybe ESL, the crazysounding numbers will be pretty close to correct. However, it's often only for a VERY short time.
Even just the reservoir caps in a PSU behave in ways that were, to me, somewhat unexpected. You can see for yourself by either running LTSpice simulations or running the PSU spreadsheet that I uploaded as an attachment to the post at Power Supply Resevoir Size . Attached is a small sample of the numbers that the spreadsheet plots after performing the numerical solution of the differential equations for a transformerrectifiercapacitorload power supply circuit (including transformer winding leakage inductance and resistance). The attachment shows the numbers during the initial part of the first startup pulse of a particular PSU. It would be very easy to get almost any VOLTAGE slew rate you wanted, out of almost any cap (although they might explode if you overdo it). With smallvalue caps, it's quite easy. Since i = C dv/dt, and dv/dt is the voltage slew rate, just solve for that: dv/dt = i / C Obviously, for SMALLER values of C, the same current, i, would cause a larger voltage slew rate. If you put 10 amps into a 0.1 uF cap, it would give 10 A / 0.0000001 F = 100 million volts per second, which would be 100 Volts per microsecond. But if you're talkiing about a larger capacitance, the ONLY way to get a high voltage slew rate to occur would be to put in or take out a VERY LARGE current. BUT, I don't know why anyone would care about a cap's VOLTAGE slew rate. We use caps to provide CURRENT TRANSIENTS, by lowering the voltage across the cap by a very small amount (for example, when an output transistor is commanded to lower its channel resistance in order to let more current flow, which will lower the rail voltage slightly, making the decoupling and/or reservoir caps provide exactly the right surge of current). INITIALLY, for a short time, the cap will obey the impedance equation: ∆v / ∆i = ESR or ∆i = ∆v / ESR (but other circuit series resistance would need to be added to ESR, if the cap was in a circuit) That equation means that for a lower ESR, the cap will have an initial transient current with a greater amplitude, but will reach that higher amplitude in the same amount of time. Or, you could say that it would reach the same current as before in less time. So it would be faster, in terms of current tranients. Just playing around: Actually, since initially, for a linear (ramp) change in voltage, versus time, the impedance relationship holds: ∆v / ∆i = ESR But we also know that ∆i = C ∆v / ∆t which can be rearranged to ∆v / ∆i = ∆t / C (impedance in seconds per Farad?) we could also say ∆v / ∆i = ∆t / C = ESR (i.e. largervalue caps have lower ESR) or ∆t = ESR ∙ C (an RC time constant?) Also, an approximate ESR equation is ESR = 0.02 / (C ∙ VR) where VR is an electrolytic cap's voltage rating. That gives ∆t = ESR ∙ C and ∆t = 0.02 / VR which seems to imply that electrolytic caps with higher voltage ratings should have lower ESR, for the same capacitance value. Anyway, for a linear current ramp versus time of ∆i Amps in ∆t seconds, into a capacitor with ESR = Resr, the change in the cap's voltage would be: ∆v = ∆i ( ( ∆t / ( 2 C ) ) + Resr ) If it was a constant current instead of a ramp, for ∆t seconds, that would change to: ∆v = i ( ( ∆t / C ) + Resr ) or ∆v = i ∙ ( ∆t / C ) + Resr ∙ i which also implies that caps with higher ESR would have a larger ripple voltage for a given PSU load current. For the same load current, the ripple voltage amplitude with a cap with a higher ESR would be larger by the average load current times the difference in ESR valus. 
11th February 2013, 03:24 AM  #20 
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Location: Buenos Aires  Argentina

Yes.
Truth is, and going to the bottom of the problem, that those Sikorel are very good. No doubt about that. In fact, I use EPCOS exclusively, they are the best available in my market, period. It's just that some parameter, measured in V/uSec and referred to overheating , was mixed with Amplifier slew rate, which does use same units, but for a different purpose. Nothing beyond that. By the way, I use "regular" EPCOS, no need to approach "Computer Grade " capacitors or similar spec. I'd have to order the full box through ARROW or AVNET. But if you want to go for the best of the best, be my guest . Certainly can't hurt 
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