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Old 24th January 2010, 12:33 PM   #1
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Default How do you calculate choke size in a power supply?

I can remember reading something here a few years back I think but I cannot find anything now. Can someone spoon feed me on this?
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Old 24th January 2010, 12:41 PM   #2
tvrgeek is offline tvrgeek  United States
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By your wallet thickness.
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Old 24th January 2010, 01:11 PM   #3
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How about an answer that would be help
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Old 24th January 2010, 02:04 PM   #4
tvrgeek is offline tvrgeek  United States
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Engineering has many constraints. When I looked up what iron costs at Allide, it selected for me reall quick.

OK, when I get a chance I will look it up in Jones book if someone else does not beat me to it. I think he had some guidelines. I was also told the best reference was the old ARRL handbook.

It matters how many poles in the filter. C-L-C, or L-C-L-C etc, if droppers or not and so on. Of course you need sufficient current ratings.
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Old 24th January 2010, 02:05 PM   #5
Loren42 is offline Loren42  United States
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Like math? This is not a simple answer or question.

There is a minimum current load that must be met to make the choke work and then there is the output capacitor size, as well as maximum load currents.

The minimum current is aproximately equal to:

Imin (mA) = [Vin(RMS)] / [L(H)]

You can calculate the ripple of the choke-cap system as follows:

[Vrip(RMS)] / [VDC] = 1 / [ 6 * sqrt(2) * 2 * PI * f * L * C]

L = inductance
f = AC line frequency
C = capacitance
VDC = DC voltage at cap
PI = 3.14...

There is more to the process. Maximum peak current through the choke is:

Iac (pos peak) = [0.544 * Vin (RMS)] / [2 * PI * f * L]

You need to consider other things with the design, such as snubber networks.

When I built my amp I bailed on the idea I had of originally adding a choke. This is just my opinion, but chokes were very popular decades ago and for good reason. Banks of large capacitance for filters simply were not practical. Today you can buy huge capacitors which simply did not exist in the 1940s, '50s, and '60s. The solution was a filtering choke (and a slide rule).

Again, my opinion... Today a brute force capacitance approach is easy and it offers another advantage; reserve power.

AC Ripple, the ultimate target we are trying to slay, can be calculated by:

Delta V = I / [2 * f * C]

You can design a multi-stage filter network and work out the AC ripple at each stage. My power supply is well under 1% and the cost to build it is cheaper than using a choke (and uses less real estate).

I am sure others will chime in and perhaps correct me if I have strayed down the wrong path somewhere.
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Old 24th January 2010, 02:41 PM   #6
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Further to Loren42 above:

Total current through the choke = Idc +Iac (positive peak)

Since L is in the denominator of the Iac term above, smaller L value chokes have larger Iac and need to be sized for larger current than larger L value chokes.
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Old 24th January 2010, 02:51 PM   #7
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Old 24th January 2010, 02:55 PM   #8
Loren42 is offline Loren42  United States
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Quote:
Originally Posted by hidnplayr View Post
By using PSUD2 offcourse
Slick!

I did it the hard way with a pencil, calculator, and confirmed it with some simple Spice analysis.
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Old 24th January 2010, 03:12 PM   #9
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"How do you calculate choke size in a power supply?"

It depends on where the choke is located. If you're designing a choke-input filter, the choke must be the first filter element and must have a certain minimum value called the "critical" value. This value is calculated by:

L (henries) = E (volts) / I (ma)

If the choke is located after the first filter cap, or if more output voltage will be required than is available from a choke-input filter, the choke can be substantially smaller.

Incidentally, there are more criteria necessary to the design of a good supply than simply output voltage and ripple. PSUD can predict those things. A supply can also ring and exhibit high Z resonances at its output when excited by the amplifier's dynamic current demands. Examining these characteristics requires looking into the supply from the output side, an exercise for which I use LT Spice (free from National).

Last edited by Triode Kingdom; 24th January 2010 at 03:14 PM.
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Old 24th January 2010, 03:18 PM   #10
llwhtt is offline llwhtt  United States
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For a power supply with a fairly constant load, preamp or class A amp, this formula works great and is easy to remember. Figure out the load resistance, E/I, then divide that by 1200. the result is in Henries. Example; 400VDC with 20ma of current needs a 16.6666666H choke.

Craig
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