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Chip Amps Amplifiers based on integrated circuits 

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17th August 2012, 05:38 PM  #21  
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Join Date: May 2010
Location: Skokie Il

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For many applications a properly designed circuit (careful attention to the poles) will circumvent the need for fancy pants capacitors. I know a lot of you guys get a woodie for expen$ive caps, but the truth is that in many applications they don't make a lick of difference. Walter Jung did an excellent essay on choosing capacitors some years back. You can look it up if you're interested. He explains in detail what I only briefly mentioned, and a whole lot more. 

17th August 2012, 06:03 PM  #22 
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Join Date: Aug 2012

instead of using 10 x Keltron( which is designed by sparague ) 1,00,000uf caps can I use two 10,000 Nichicon KG?

17th August 2012, 06:14 PM  #23  
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Join Date: May 2010
Location: Skokie Il

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http://www.diyaudio.com/forums/solid...voirsize.html What is your power supply voltage going to be? 4 or 8 ohm load? You can experiment too. Try 4700, 10,000, and 20,000 uF and see if you can find the point of diminishing returns. 

17th August 2012, 06:51 PM  #24 
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it says more about rejection ratio where the capacitors value doesnt make any sense. I do understand that but the equation between the ripple current and the capacitance is not understood. Example I need 10 amps continuous at the output considering Im having 3 LM3886 in parallel mode. So If I have a trafo at 24024 ( 2 amps ) so its roughly 100VA then with bridge rectification it will give 34 volts so each chip can delver 3.79 amps of current at the output so how much capacitor to be used considering for each chip?
Taking the capacitor voltage is about 50volts Ex: Nichicon KG 10000uf 50v ( rated ripple Arms is 4 amps ) so how can I choose the capacitor limit in psu? 
17th August 2012, 07:13 PM  #25 
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Join Date: Aug 2012

I found one thing for the Decoupling caps with the equations considering slew rate and change in the voltage when I is peak... for the LM3886 to be 220uf paralleled with 100 to 200nf film cap right at the input pins and according to the layout principles its better to connect the input wires right beside the input caps with 1uf series DC blocking cap. But regarding psu i need certain equations...

17th August 2012, 07:22 PM  #26  
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17th August 2012, 07:26 PM  #27 
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Join Date: Jul 2004
Location: Scottish Borders

you don't need equations to mount the local decoupling "right at the input pins".
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17th August 2012, 08:17 PM  #28 
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I was going through that psu reservoir and I found these equations to calculate the psu capacitor value...
OK. Here is a nice webpage about unregulated power supply design, with just the bare basics: Unregulated Power Supply Design Their equation for the peaktopeak amplitude of the ripple voltage is equivalent to the equation I gave for calculating capacitance based on Δi, Δt, and a desired Δv: (1) Δv = i / (2fC) where f is the supply frequency (60 Hz, in our case). Solving for C, we get: (2) C = i / (2fΔv) which is equivalent to the equation I gave in post # 61: (3) C ≥ (Δi Δt) / Δv if we realize they are using 1 / 2f in place of Δt, because the rectified AC mains (and ripple) frequency is twice the mains supply frequency, f, and the time for one cycle at frequency f is 1/f, AND we realize that in (3), the Δi is usually PEAK current, not RMS, since it's usually applied for transients or short time periods. First we can note that the "3300 uF per Amp" rule would give, using equation (1), with C = 3300uF x 3.54A rms = 11682uF, a pp ripple amplitude of Δv = 2.52 Volts. (Or was that rule for Amps PEAK? That would be 16500 uF, and 1.79 Volts pp ripple.) But the ripple voltage is not the only rail disturbance. And the ripple calculation in (1) only took into account a constant load current. Our load will pull current from the caps that will not be constant. It will be sinusoidal, probably a halfcycle sine at a time, in the simplest case of one pure tone. It might be worse than that but that means that the capacitance requirement for that can be a lower bound, i.e. a minimum. So I started out thinking that the lowest frequency would need the longest current draw and would be the worst case. I looked at 15 Hz as an example. The period for one cycle of 15 Hz can be found using (4) T (sec) = 1 / f (Hz) as 1 / 15 = .0667 sec. So we know that just to supply the current for the rising half of the positive part of a 15 Hz sine wave, we would have to go from 0 Amps to 5 Amps Peak (worst case) in onefourth of a cycle, or 0.0167 sec. Using equation (4), we can get C in terms of the Δv we decide we can tolerate: C = (5A)(0.0167 sec) / Δv, or C = .083500 / Δv , for 15 Hz, which can also be rearranged to give Δv = .083500 / C for 15 Hz. BUT THEN I realized that since the charging pulses are faster than 15 Hz, and would actually recharge the caps up to four times during a halfcycle of 15 Hz, that the equations in the previous paragraph were not valid because of that, and that 60 Hz would probably be a better "worst case". 60 Hz is 0.00417 sec per 1/4th cycle. So, C ≥ (5A)(0.004170 sec) / Δv, or (5) C ≥ 0.020850 / Δv , for 60 Hz (note that it's 20850 uF for 1V Δv), and (6) Δv = 0.020850 / C , for 60 Hz. Note that the Δv in (5) and (6) is in addition to the mains ripple. If we used the 11682 uF given by the "3300 uF pr Amp" rule, that would give Δv = 1.78 Volt of ADDITIONAL railvoltage disturbance, which is 71% more (1/√2 more) than what we thought the 3300 uF rule would give us. (Or if that rule was supposed to be for PEAK amps, it would have given 16500 uF and 1.79 V pp 120 Hz ripple, which would give an additional 1.26 Volts of rail disturbance due to the positive half of our 60 Hz tone, or 42% more.) That might be especially inconvenient if we were also using a linear regulator, since the current humps pulled by the load would cause voltage sags that would essentially pull the troughs of the ripple voltage downward farther than planned, by up to 71% of its originallycalculated amplitude, which might be likely to cause the regulator's input minus output voltage difference to become less than it's dropout voltage spec, during parts of the troughs, which would get really ugly. It would only possibly be 71% [or 42%, if the rule was for peak amps] worse when we were driving the amp all the way to the rail. But the bottom of the ripple waveform would come down lower than we thought, by some percentage, at every volume setting. If we were happy with the mains ripple that the "3300 uF per Amp" rule was going to give us, then it looks like we should ADD 71% (or 42% if the rule used peak current instead of RMS) to the capacitance value it gave us, AND to the rule, in order to keep the worstcase rail disturbance the same. In any case, we might as well come up with a general equation that will take into account BOTH the worstcase mains ripple AND the transient voltage sags that our signals might induce. There might be some "overlap" between those causes and effects but this should give a safe minimum capacitance: For the transient component, from (3) we can substitute 1/4f for Δt, assuming that 1/4 cycle at the AC Mains frequency is the worst case, and get (7) C = ipeak / (4fΔv) and in order to take into account both the DC and transient components of Δv we can combine (7) with (2) to get the total C needed: (8) C = [ iRMS / (2fΔv) ] + [ ipeak / (4fΔv) ] Then we can solve for Δv: (9) Δv = [ iRMS / (2fC) ] + [ ipeak / (4fC) ] If we convert iRMS to ipeak, pull out the common denominator factor, then clean up the remaining fractions, we get: (10) Δv = (0.6036)(ipeak)/fC or (11) C = (0.6036)(ipeak) / (f Δv) where f is the AC Mains frequency, ipeak is the power supply's peak (rail) voltage (not RMS), i.e. ipeak = Vrail/Rload, and Δv is your choice of an acceptable worstcase voltagerail variation. (So it looks like the new rule should be "≥ 5600 uF per Amp (peak)".) So, assuming that the "3300 uF per amp" rule meant PEAK amps, and without considering whether or not the 1.79 Volts worstcase peaktopeak ripple amplitude that it gave is "good enough", the minimum total reservoir capacitance needed in order to REALLY get a worstcase of less than or equal to 1.79 Volts Δv should be, using (11): C = [(0.6036)(5)] / [(60)(1.79)] = 28100 uF We might also want to add 15% or more to that, to account for variations in mains voltage and transformer regulation. So maybe 33000 uF would be a better number. (I hope that someone will check my math and logic.)  but what actually makes as 3300uf as 1 amp? what is Ipeak? is it current in the speaker load? 
17th August 2012, 08:32 PM  #29 
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Join Date: Aug 2012

for a chipamp where if Ipeak is considered as output peak current in the speaker load then the value with 15% of voltage regulation is considered as .. 24561.7uf which can be taken as 30000uf as nearest value per chip. Are the calculations right?

17th August 2012, 08:35 PM  #30 
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Join Date: Aug 2012

Lets consider that if im getting 10 amps as I peak then its 60000uf. Once I remembered a mail response from Jeff Rowland stating that if an amplifier is capable of delivering 15 amps of continuous current in the speaker load then its a serious pa speaker or a big subwoofer you can drive any driver in 12 inch range. Hence using 3 or 6 chips in Bridgeparallel can drive any speaker i think...

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