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Chip Amps Amplifiers based on integrated circuits 

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16th June 2013, 04:14 PM  #11 
diyAudio Member
Join Date: Jun 2013

Ok I follow and understand all of your math except how you came up wih the capacitor value.
Could you explain this again. Thanks 
16th June 2013, 09:32 PM  #12 
diyAudio Member

Yeah, I didn't explain that, the first time, actually.
I derived it in the thread entitled "Power Supply Reservoir Size", which was started by Nico Ras. Knowing the maximum load current and the desired maximum peaktopeak ripple amplitude, the minimum required reservoir capacitance can be calculated. Since for an ideal capacitor, i = C dv/dt, we can make some assumptions and linearize in a neighborhood and say (ignoring ESR): C = i Δt / Δv where Δv is the desired maximum dip in the supply rail voltage when i amps are drawn for Δt seconds, and C is in Farads. We can assume that Δt can, at worst, be the time between charging pulses, i.e. 1 / (2 fmains). And we get to pick Δv, the pp ripple amplitude. And we can calculate the maximum load current. Most people use the RMS current. But that isn't really good enough, because it only accounts for the possibility of a single sine wave and we should try to account for the worst case load. And that would be a DC voltage and current at the Vpk and Ipk level. The Δv should be chosen carefully, taking into account the specified Vpk and the Vclip of the amplifier itself, to leave room for the Vripple, relative to the peak rail voltage. So Δv = Vrail  Vclip  Vpk I've done the algebra, and also took into account the ESR, using the approximation ESR = 0.02 / ( C x Voltage_Rating) and the overall equation for the minimum value of the required capacitance is Cmin (in uF) = 1000000(Vpk/(Rload(VrailVclipVpk)))( (1/(2 fmains)) + (0.02/Voltage_Rating) ) I just guesstimated the 22000 uF, from a spreadsheet I looked at that uses that equation. But we should check to see what the equation gives, for the 80 Watt into 8 Ohms case (assuming 60 Hz fmains): Cmin = 1000(35.8/(8(46.2535.8))) ((1/120)+(0.02/63)) Cmin = 7169 uF That agrees with my spreadsheet, too, but it also results in over 5 Volts pp ripple. If we want ripple around 1.9 Volts, we get 22000 uF (and the max power without clipping also rises to about 96 Watts, with Vpk a little over 39 Volts). BUT NOTE that if you use a 10% lower rail voltage, then 21800 uF would give you 1.73 V ripple and only 76 Watts RMS max, at the onset of clipping. So that's why I picked 22000 uF. You can find my Excel spreadsheet and download it from the thread I mentioned. Sorry I don't have the exact link at hand right now. Last edited by gootee; 16th June 2013 at 09:35 PM. 
20th June 2013, 04:42 AM  #13 
diyAudio Member

Links to my Power Supply solver and minimum reservoir capacitance Excel spreadsheets are in the post at:
Amp design attempt number 2 (simpler) 
21st June 2013, 08:17 PM  #14 
diyAudio Member
Join Date: Jun 2013

THANK YOU
Alright, Today I thought I would start to design my enclosures, but I am having a lot of trouble. I've tried various software and online calculators, but all of them have many differant paramateres for the speaker that I do not know.
I have: 1 10" MTX sub, 8 Ohm, 125RMS 2 4.75" (12cm), Panasonic, 6 Ohm, 45RMS 2 2.375' (6cm), Panasonic, 6 Ohm, 45RMS (a "supertwetter" piezoelement is attached to this tweeter which I hope to use" How would I go about beginning to design this? I am thinking of a ported enclosure for the sub. I would like indidvidual boxes for each speaker. 
22nd June 2013, 08:18 AM  #15 
diyAudio Member
Join Date: Jul 2004
Location: Gone on holiday again. back soon.

start a new Thread in the Speakers section
__________________
regards Andrew T. Sent from my desktop computer using a keyboard 
22nd June 2013, 12:28 PM  #16 
diyAudio Member
Join Date: Jun 2013

Alright, thank you.

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