Re: gEDA-user: Alarm clocks and switching regulators

[John Doty <jpd@xxxxxxxxxxxxx>]

On Sep 24, 2007, at 11:41 AM, Randall Nortman wrote:

> On Mon, Sep 24, 2007 at 10:34:58AM -0600, John Doty wrote:
> [...]
>> But the truth is, D401 isn't very useful where it is. L400 is going
>> to keep it in conduction, so it just acts like a resistor of value ~
>> (kT)/(qI), or ~25 milliohms at 1 amp. That's smaller than the series
>> resistances of L400 or C402, so those are where the damping's going
>> to be. And it's pretty good: the characteristic impedance of the
>> filter is sqrt(L/C), or about 250 milliohms. That's about the same as
>> the sum of the series resistances of L400 and C402, so the Q will be
>> about 1, not a serious resonance problem.
>
> Gak!  As soon as the discussion turns to characteristic impedances and
> Q's, my eyes glaze over.  It's a mental block I have.  I suppose I
> could go dig up one of my electronics textbooks and figure it out.
> You're talking about the LC filter formed by L400 and C402 -- but the
> presence of C400 turns this into a CLC filter and makes me worry about
> this being a funky multi-pole filter with strange resonance points.

C400 is much larger than C402, and it's in parallel with a bunch of  
other stuff, including (through D400) the mains adapter. It's  
therefore relatively low impedance, and for the purpose of a hash  
filter design may be ignored. The resistive component of this  
impedance should just provide more damping.

> Because, as I've said already, analog is black magic to me, and I
> don't trust myself to think too hard about it.  But you're saying
> there's nothing to worry about, even without D401?

Oh, there's plenty to worry about, see below.

>
> (Of course, it's even more complicated than a CLC filter, since other
> components are between C400 and L400: the 5V regulator, LED, and
> whatever might connect to J401.  But I'm assuming those are all small
> enough current draws that they can be ignored.)
>
> Now -- here's where I break down and cry like a baby: How should I
> size L400 and C402 for my own design?

With care. And the L and C values are the least of your worries.

>   Mine operates at ~600kHz
> switching frequency, and the goal is to attenuate that frequency (to,
> say, -3dB) upstream of the inductor, so that the bulk caps don't see
> it and my power supply lines don't broadcast it to the neighbors.

What power? This isn't a minimalist design. The LM2595 is a 500 mA  
regulator, but other components are rated well above that. DJ seems  
to have wanted to keep the parts variety down here.

Here, I'd start with ripple current rating of C402. At 500 mA out,  
the switcher will alternate between 500 mA in and zero, roughly. Duty  
cycle about 1/4, so the average is 125 mA. That's what L400 will  
carry. Thus, C402 is sourcing 375 mA a quarter of the time, and  
sinking 125 mA 3/4 of the. RMS of that is ~200 mA, comfortably below  
the 850 mA ripple current rating of C402.

This is a 150 kHz switcher, so the 375 mA lasts ~1.6 µs/cycle.  
Multiply to get 0.6 µC of charge/cycle, divide by 47 µF to get 12 mV  
of ripple for an ideal capacitor. However, the resistance of C402 is  
~0.2 ohms: that times 375 mA is 75 mV. That's the dominant ripple on  
C402: at the switching frequency it's more a resistor than a capacitor.

In general, especially at the higher switching frequencies, this is  
what you'll see with Al or Ta electrolytics: ripple current ratings  
will drive you to a design with reasonably low ripple voltage and low  
Q at the input of the switcher.

OK, so RMS ripple is roughly 40 mV on C402. At 150 kHz, the impedance  
of L402 is ~3 ohms. So, we push ~13 mA of ripple back into C400 and  
the stuff in parallel with it. That's not very much into a node with  
impedance of ~0.1 ohm.

>   All
> the equations I've been able to dig up require me to know load
> impedance -- that is, the impedance of my switcher and everything
> downstream of it.

Mostly the downstream stuff doesn't matter: the switcher turns the  
varying input voltage into a fixed output voltage. But the switcher  
itself is effectively a negative resistance because if you increase  
its input voltage, its input current decreases. And in cases like  
this, it doesn't matter: input resistance is -Vin/Iin, or >30 ohms in  
this case. This has magnitude much greater than the impedances of the  
filter components, so it's effectively infinite and may be ignored.  
The game is different with high voltage switchers using high Q  
components like polypropylene capacitors but you don't intend to go  
there today, I hope ;-)

>   I have no clue, and there's so much stuff there it
> would be pretty difficult to calculate.

You have to learn to approximate and to focus on the dominant  
effects. You'll notice my arithmetic is very rough. This isn't like  
designing a signal filter to precise requirements. Here, the  
capacitor's capacitance isn't even particularly relevant: ratings and  
parasitics are what count! That's where experience comes in. And to  
get experience, you need to charge in and try things...

>   And it will vary a lot in
> real life anyway, depending on what the electronics are doing moment
> by moment.  So where do I start?  (The bulk caps will be sized to
> handle the 120Hz ripple from my rectifier, so I just need to figure
> how the other two values.)
>
> I could just overkill it, of course, but I'm trying to save board
> space as much as possible, and L's and C's can get really big and
> expensive.
>
> -- 
> Randall
>
>
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John Doty              Noqsi Aerospace, Ltd.
http://www.noqsi.com/
jpd@xxxxxxxxx




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