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Old 17th May 2007, 02:44 AM   #11
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Quote:
Originally posted by GRollins
Pretend the filter is in a speaker instead of an amp. Suddenly that CLC snaps into focus as an 18dB/oct lowpass filter as opposed to a C filter being--obviously--a 6dB/oct lowpass filter.



Quote:
Originally posted by GRollins
Wanna bet? Why do do think I "invented" water cooling for audio?
No heatsinks here in the hinterlands.
I've only ever mixed water and power once, and I didn't enjoy it. So I'm going to try to cool it another way...
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Old 17th May 2007, 03:39 AM   #12
ssmith is offline ssmith  France
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Quote:
Originally posted by wuffwaff

Up to 3A of bias you can use a siple 1.8mH, 1.4mm air choke (less than 15)
1.4mm is about 15 gauge.

I'm having trouble working out how the DC current handling of an air core inductor can be calculated. Is there any formula to figure this out?

For example, after thinking of CRC from my Mini-A stereo amp project, i've been looking at CLC -- having found 17 gauge inductors available locally. How can I work out how much current these will take?
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Old 17th May 2007, 03:58 AM   #13
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Quote:
Originally posted by ssmith


1.4mm is about 15 gauge.

I'm having trouble working out how the DC current handling of an air core inductor can be calculated. Is there any formula to figure this out?

First calculate the total length (L) of the wire in cm, and calculate
the cross sectional area (A) of the wire in cm2. As the copper has
the resistivity (RO) of 1.724 (microhm.cm), the DC resistance of
coil is

R = RO*L/A/1000,000 ohms

Or, simply measure resistance using DMM ^^.

With the calculated or measured DC resistance, you could know
the power dissipation in Watt. Your thickness of wire depends on
how much hot (warm) coils in service you want to have. Good luck!


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Old 17th May 2007, 05:37 AM   #14
ssmith is offline ssmith  France
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Quote:
Originally posted by Babowana

With the calculated or measured DC resistance, you could know
the power dissipation in Watt. Your thickness of wire depends on
how much hot (warm) coils in service you want to have.
this is the bit I'm having trouble figuring out -- more googling is at hand...
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Old 17th May 2007, 06:25 AM   #15
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Quote:
Originally posted by Nelson Pass
Either a CLC or CRC is still a lot better than C. As an example,
I have here a supply with 30,000 uF after the rectifier, then
.12 ohms, then another 30,000 uF. The difference on either side
of the resistor is about 10 dB worth of ripple, and that's a lot.

thanks a lot for the suggestion Mr Pass.
I was using the following topology CLCLCRCC with R=3.3ohm.
So i'll try to lower this value

I would like to ask you guys, this things:


Since i was suggested that a regulated power supply is, in general if well done, better for a preamplifier

in case, would it be convenient to put the stab stage after clc... section or before?
To me, it lools more convenient t oput it after.

Do you guys have any suggestion on a good topology to implement to stabilyze the rail voltage (without using LM or any chip)?

The rail voltage that i should have is about 60V (the supply for the P1.7).

What ripple value can be considered to be VERY good?


Thanks a lot in advantage.


Best,
Stefano.
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Old 17th May 2007, 11:32 AM   #16
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Quote:
Originally posted by ssmith
this is the bit I'm having trouble figuring out -- more googling is at hand...
Not sure, but isn't this simply an application of P=I^2*R (current squared times DCR resistance)? So, for me, 8.8amps (squared) times resistance (between .2 and .3) gives somewhere between 15 and 23 Watts.

With the size of the coils I'd be looking at, that would be "not a problem".

Corrections, please...
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Old 17th May 2007, 03:36 PM   #17
RoboMan is offline RoboMan  Hong Kong
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There is a good explanation of Capacitance Multiplier for those interested:

http://sound.westhost.com/project15.htm
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Old 17th May 2007, 07:16 PM   #18
Magura is offline Magura  Denmark
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Quote:
Originally posted by Cloth Ears




The penny has dropped!

So how do you find these wonderful inductors to do this for you? I guess there is always "roll your own" (now where is that inductor calculator?). As big air-cored inductors seem to cost about $50 each (for a 2mH) in Australia. Maybe Mundorf has a iron-cored one available here for less. And I can't afford to get them sent from the US any more as it's too expensive now .

Thanks, Babowana!

The "roll your own" solution is not that much trouble, and hands you a very flexible solution for future projects as well.

Take a look here for a little inspiration.....most of the inductor-stuff you see there can be made on the kitchen table

www.briangt.com/gallery/magura


Magura
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Old 17th May 2007, 07:45 PM   #19
steenoe is offline steenoe  Denmark
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Considering the sound quality alone, and nothing else; there is nothing like a big CLC supply The second best is the regulated supply! A very good example is to be found here: http://www.passdiy.com/pdf/zen-ver3.pdf Do mind, though that if you make a dual-rail from that PDF, the lower fet must be turned a 180 degrees!! I never thought that I would catch NP in making a mistake..he-he

Steen
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Old 17th May 2007, 09:51 PM   #20
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how can i match a big CLC with a regulated pwer supply without using zeners? ... mmm... i will read very carefully that pdf attached...seems very interesting.
Now it's just too late.....
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