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Old 1st January 2007, 08:01 AM   #1
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Default Parallel Inductors in PSU

Hi all,

I'm trying to source inductors to implement a PI Filter in my Aleph 2 amps. Suppliers in Australia are as rare as hen's teeth and international shipping is prohibitively expensive. I happened to find a local supplier carrying Solen inductors. They only stock 16AWG versions, which may get a little hot and steamy. I was looking for 14AWG or larger guage.

Now the good news this local supplier is having a sale on 4.8mH inductors (about 1/3 retail price, and cheaper than buying direct from Solen). Would anyone advise against putting two of these inductors in parallel in place of a single 14AWG inductor? The resulting inductance and DCR would be fine are there any other issues to take into account?

Hope that's not too silly a question...

Cheers,

Dan
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Old 1st January 2007, 08:26 AM   #2
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Since they're so affordable, perhaps you CAN run two pair with each pair in a parallel configuration. CLCLC

I like the CLCLC configuration because I run horns and don't want any hiss what-so-ever.
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Old 1st January 2007, 08:57 AM   #3
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They're not *that* affordable

PSU Designer is showing a ripple voltage under 8mV at full load, which I'm pretty comfortable with.

Given your suggestion, I assume that there is no reason not to put inductors in parallel in the PSU. I was kinda hoping that someone would suggest that it's advantageous to use a number of inductors in parallel (as many would suggest for filter caps).


Thanks,

Dan
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Old 1st January 2007, 10:23 AM   #4
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Two inductances in parallel
share the input current half and half

Advantage: half heat
Disadvantage: half inductance

What else?
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Old 1st January 2007, 12:27 PM   #5
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I do not believe the inductance is /2 You have the same number of turns in paralell. The DCR will be /2 but I think the L will stay the same
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Old 1st January 2007, 12:49 PM   #6
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Putting 2 equal inductors in parallel is fine. The resistance and inductance of the pair will be 1/2 that of a single inductor. The total power loss will be 1/2 of the original loss, so each inductor has a loss of 1/4 of the original.

Note that the parallel inductors should be the same type. If they are not, then they won't share the current as well.
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Old 1st January 2007, 01:26 PM   #7
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Quote:
Originally posted by flg


I do not believe the inductance is /2 You have the same number of turns in paralell. The DCR will be /2 but I think the L will stay the same

I believe you are confused

emf = - L x di/dt
Actually each L remains the same
But, di reduces half
so emf (voltage drop) reduces half, too

It's the same as the parallel-resistors

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Old 1st January 2007, 02:01 PM   #8
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inductances in parallel
same algebra as for resistors
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Old 1st January 2007, 04:28 PM   #9
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I was thinking the inductors are physiacally coupled, like wire tied together, vs. two seperate inductors? Would the 1/2 part still be the same math in that case???
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Old 1st January 2007, 07:21 PM   #10
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I just tried an experiment with two inductors. When paralleled, the inductance meter read half the value of a single inductor; when coupled in series, the inductance doubled.


Good to know, thanks guys.
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