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#1 |
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diyAudio Member
Join Date: Apr 2011
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Hi guys, I'm trying to design a little PSU for my mini-A.
The whole point of the excersize was the mini part. The amp is in class A, Dissipates 30W at idle ( +-15V @1A), and can push 10W output. So its by no means one of those big nasty class-a's. The problem is I am not sure how to interpret the ripple rateing for capacitors. I have done a duncan's PSU plot (attached), now if my eyes are right, it shows 14.27A on C1, but it also shows RMS (5.8A) and mean (-244uA). Which of these rateings, i.e. max, RMS or mean, do we use to check if a capacitor has an approriate ripple rateing? It is quite important for me to know, not just for the sake of reliability, but also I am trying to keep the case small (3U), and that rules out alot of the realy high ripple rateing caps, so I wont be able to try something small and then just swap in something big if it fails. I don't want to use snap in caps because they have such short lifespans, add to that the ambient temp will be about 50C on a 25C day... What causes the big current, the transformer or the capacitor bank/load? I have found one cap I like and can sorta afford, it is 22000uf ESR at 100Hz is 17m Ohm, but ripple current is rated at just 9.9A. Will my PSU destroy these caps? I ordered a thermistor similar to CL60, which I believe can be connected in series between the two 115V primaries (we are on 230V). Will this help? The transformer specs is as follows incase someone wants to doublecheck my duncan psu calcs. 2x115V to 2x12V. Primary; DCR 2x4 ohm rated, measured was 3 point something. Secondaries 112.5VA each, 9.375A per secondary, 13.07V No load, and DCR 2 x 0.0515 ohm. Last edited by digits; 18th July 2011 at 06:23 AM. |
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#2 |
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diyAudio Member
Join Date: May 2007
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I believe ripple current is RMS. Ripple current rating is not a cliff edge, but best to keep below if you want long life from your caps.
Ripple current is increased by: low transformer impedance high cap value low cap ESR high load current |
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#3 |
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diyAudio Member
Join Date: Jun 2011
Location: Johannesburg
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It is better to use two or more smaller capacitors rather than one big one. As an example: 2 x 4700uF of 3300mA ( = 6600mA) ripple is better than 1 x 10000uF with 4800mA ripple. Also caps in parallel have a lower combined ESR than a single cap.
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#4 |
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diyAudio Member
Join Date: Apr 2011
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Hmmmm I might have to do that, I'll see what I can find, It looks like a single cap could handle the 2nd part of the LC or RC (Whitch ever I use).
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#5 |
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diyAudio Member
Join Date: Jun 2011
Location: Johannesburg
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Digits, have you considered a capacitance multiplier rather as a supply?
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#6 |
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diyAudio Member
Join Date: Apr 2011
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Yeah, but I don't have that much voltage headroom.
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#7 |
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diyAudio Member
Join Date: Nov 2003
Location: Brighton UK
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Hi,
For class A you are best off with at the most basic a CRC supply. Assuming a 1A standing current is both rails its easy to work out the R voltage drop, 1V for 1R, but 0.22R will still have an effect. As long as you have decent capacitors don't worry about ripple. In your application its not an issue, with CRC it becomes less. 0.47R for CRC for both channels seems about right. rgds, sreten. Split the total C value for CRC, or make the initial capacitor slightly higher, e.g. 4,700uF, 1R, 3,300uF for one channel. The above will be far better than a single 10,000uF.
__________________
There is nothing so practical as a really good theory - Ludwig Boltzmann When your only tool is a hammer, every problem looks like a nail - Abraham Maslow Last edited by sreten; 18th July 2011 at 09:18 PM. |
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#8 |
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diyAudio Member
Join Date: Nov 2005
Location: PA
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If you're going to mess with an inductive section at all, why not make it the first secton? Then you don't have to think much about capacitor ripple rating.
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#9 |
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diyAudio Member
Join Date: Sep 2010
Location: Melbourne
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The ripple current of a transformer is not just dependant on the DCR, the leakage inductance of the transformer and the mains impedance will reduce it somewhat.
To get a real value, power up your transformer measure the output voltage then place a resistive load on it say one or two amps then measure the output voltage. The reduction in voltage divided by the current gives the transformer impedance. This with the DCR can be used to calculate the leakage inductance if you want to be thorough. Or just put the impedance into the simulator as a series resistance in place of the DCR and see what sort of ripple you get. If the transformer is a toroidal type, consider putting a few more turns of plain hookup wire around the core to give 16V and use a low dropout series pass regulator on each supply rail . Class A amplifiers have poor ripple rejection and need a very quiet power supply, the money saved on two caps should pay for the regulators and heatsink. |
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#10 | |
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diyAudio Member
Join Date: Apr 2010
Location: seattle, wa
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Would you please define what that is.
Quote:
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