Question : LC filter in switching power supply

Discussion in 'General Electronics Chat' started by binvic, Mar 24, 2006.

  1. binvic

    Thread Starter New Member

    Mar 24, 2006
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    Currently I'm working on switching power supply circuit (buck) step down, and I noticed in most of their application notes from various IC producer that they put an optional LC filter at the end of the circuit before the load. Yes, I'm aware the purpose of that LC filter is to reduce the voltage ripple at the output which is a good thing.

    Now the question is how to calculate for the "suitable" value for the L and C ? What other consideration or parameter in determining it ? Switching freq ? Max. current ? Voltage ? etc.. ?

    Fyi, I'm no EE, so please be gentle ! :)

    Thanks in advance.
     
  2. binvic

    Thread Starter New Member

    Mar 24, 2006
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    Anyone please ?
     
  3. hgmjr

    Moderator

    Jan 28, 2005
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    Switching power supply design is a weighty subject to be approached very carefully by EEs and non -EEs alike. Since you stated that you are a non-EE then I have to assume that you aspire to be an EE or perhaps an electronic technician.

    Linear Technology, Inc, who makes a number of switching power supply controller devices has a very useful software package titled "SwitcherCAD". It is free for download at their website www.linear.com. You may want to download a copy of the software and play around with it as a means of exploring the subject of switching power supply design.

    hgmjr
     
  4. binvic

    Thread Starter New Member

    Mar 24, 2006
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    Yes, switching is no simple stuff like linear one and I'm just a hobbyist, thanks for the software info, will take a look at it.

    So are you suggesting that there is no such "simple" solution or just an equation for LC filter design for the purpose of reducing output ripple ?

    I mean like we have to consider so "many other things & factors" that simply using switching freq, voltage and current consumption is not adequate for it ?

    Although I haven't got any clue yet, really appreciate and thanks hgmjr for your reply.
     
  5. Papabravo

    Expert

    Feb 24, 2006
    10,170
    1,797
    I don't know if there will be a sufficient payoff for adding an LC filter. If you use LTSpice to simulate the switching regulator then you should be able to add the LC filter to see what effect it has.

    Generally speaking you want the L to have low DC resistance so it can pass the DC output of the regulator without dropping the output voltage. You can compute the inductive reactance, X_sub_L, at the regulators switching frequency with the formula
    Code ( (Unknown Language)):
    1.  
    2. X_sub_L = 2*pi*f*L
    3.  
    4. where
    5. pi = 3.1415926...
    6. f = regulator switching frequency
    7. L = inductance in henries
    8.  
    9.  
    If the capacitor has a reactance, X_sub_C, of say 0.1 times X_sub_L at the regulators switching frequency.
    Code ( (Unknown Language)):
    1.  
    2. X_sub_C = 1/(2*pi*f*C)
    3.  
    4. where
    5. pi = 3.1415926...
    6. f = regulator switching frequency
    7. C = capacitence in Farads
    8.  
    9.  
    Then you have an AC voltage divider at the switching frequency of the regulator with a ratio of
    Code ( (Unknown Language)):
    1.  
    2. 0.1/(0.1 + 1.0) = 0.0909..
    3.  
    This will attenuate the AC ripple by that factor, but let the DC voltage pass through with minimal attenuation from the DC resistance of the wire in the inductor.

    Did this help?
     
  6. binvic

    Thread Starter New Member

    Mar 24, 2006
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    Hi Papabravo,

    About your final calculation -> 0.1/(0.1 + 1.0) = 0.0909..

    Got lost here, could you elaborate further please ? Looking at the 0.1/(0.1 + 1.0) numbers, which is which ? Is it X_sub_L / (X_sub_C + ??) ?

    Regarding your concern about the wire resistance on the inductor, don't worry, got lots of ferrite cores (any shapes n sizes) all were hand wounded at sub 100 mOhms resistance with LCR meter handy. But no scope though ... sigh....

    Yes, it helps ! Thanks for the start & hoping that you will not get bore with this noob question ! ;)
     
  7. hgmjr

    Moderator

    Jan 28, 2005
    9,030
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    I did not mean to infer that the use of an LC filter could not be tackled with straightforward calculation as papabravo has illustrated. Rather, that there are numerous design factors that need to be considered in order to insure that the performance of the resulting design will meet your expectations.

    I do wish you success in your design effort and look forward to your feedback regarding the improvements you realized by adding the LC filter to your design.

    Good Luck,
    hgmjr
     
  8. Papabravo

    Expert

    Feb 24, 2006
    10,170
    1,797
    If the capacitative reactance is one tenth the inductive reactance then
    Code ( (Unknown Language)):
    1.  
    2.  
    3. Vo = Vi * (X_sub_C)/(X_sub_L + X_sub_C)
    4. = 0.1*X_sub_L/(1.0*X_sub_L + 0.1*X_sub_L)
    5. = 0.1/(1.0 + 0.1)
    6. = 0.090909...
    7. for all X_sub_L
    8.  
    9.  
    exactly like a resistor voltage divider. The critical difference is that this reactance divider is frequency dependent, and you have chosen the frequency where the computed ratio applies.
     
  9. binvic

    Thread Starter New Member

    Mar 24, 2006
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    Well, it turns into more complicated than I though before, here is my story and the plan and hoping you masters can help this noob apprentice ! :lol:

    First, I sucessfully made a step down buck converter with this chip from National Simple Switcher family -> LM2576 PDF datasheet (click it to download) and I did supply the LC filter as in page 18 and it delivers the result as expected, which is low enough ripple (below 100mV), but I knew it from a friend who has a scope to measure it.

    Now, since my requirement is now double (in term of current), I'm looking into this beast, again from the same family LM2678 PDF datasheet (click to download)

    Here some brief summary so you don't have to go through the datasheets :
    LM2576 -> Out put current : 3 Amp, Switching Freq : 52 KHz
    LM2678 -> Out put current : 5 Amp, Switching Freq : 260 KHz

    Both are designed as buck converter with minimal external components to work with (and cheap too). Now the problem is before I started to build the LM2678, I want to ensure it delivers the same ripple as it's brother LM2576, but in the LM2678 datasheet, there is no explanation or comments about the LC filter like the one in LM2576.

    Now my friend is gone moved to other city (of course with his scope) and I'm worry that the ripple from the LM2678 is not what I'm expecting if I don't put the LC filter.

    Am I making any sense ? or worry too much ? :)

    I hope I'm not asking too much, TIA.
     
  10. n9352527

    AAC Fanatic!

    Oct 14, 2005
    1,198
    4
    In designing a buck controller, you actually aim not to use LC output filter if possible. The inductor and the capacitor used in a buck controller are already an LC filter. The way to get specified output voltage ripple (and maintain output regulation under maximum and minimum load) is to start from the beginning and choose appropriate values for your main inductor, capacitor, frequency and inductor current. Only after optimising these parameters and if you still need to lower the output ripple you should consider putting an LC output filter.

    Duty cycle (D) = Vout/Vin

    On time (dT) = D/frequency (f)

    Higher frequency is desirable to reduce inductance and capacitance sizes and costs.

    Inductor ripple current (dI) = (dV/L)*dT

    where dV is voltage across inductor (L) (Vin - Vout)

    Choose an appropriate inductor value so that you have enough swing for max and min load conditions and as low dI as possible (for better output ripple voltage). These would also have an effect on inductor value (costs more and larger dimension).

    Output voltage ripple is due to two components, inherent on/off switching and ESR of output capacitor.

    Vc = dI*dT/Cout

    Vesr = dI*ESR

    Output ripple voltage = Vc + Vesr

    You could optimise these back and forth to get the output ripple and load regulation that you need. Failing that, then you could add an LC output filter, which is undesirable.

    The main considerations against an output LC filter are it costs more, add to the size of the circuit and degrade the transient response. Basically, it is a voltage divider/attenuator as Papabravo already mentioned. You could calculate the attenuation at your frequency of interest bearing in mind that there is no one unique solution. For example, you could put a large inductor and a small capacitor which would give you the same ripple attenuation as a smaller inductor and a larger capacitor. The distinction is on the transient response (how fast the output adjust to a sudden current surge), size and cost. Large inductor and small capacitor would give you pretty poor transient response, large physical inductor size which costs more. Smaller inductor is desirable, as larger capacitor is cheaper and easier to obtain. So in calculating your LC filter, aim for smallest inductor as possible and then calculate the capacitor that you need.

    Three things that you need to note:
    1. Not all ferrite cores are the same. These are rated with frequencies, Al and saturation current. Choose one that is appropriate for your frequency, the highest Al you could find (minimise the number of turns, minimise coil resistance/loss) and rated at higher saturation current than your design. It is important to distinguish between maximum saturation current (due to saturated core magnetic field) and maximum current (due to power dissipation on coil resistance), some unscrupulous manufactures quote the maximum current instead of maximum saturated current to make their products look better.

    2. Use _low_ ESR output capacitor. This is more expensive than normal capacitor, but it has pronounce effect on output voltage ripple and absolutely worth the extra cost. If you can afford it, you could use a few in parallel.

    3. Mount these (transistor switch, inductor and capacitor) as _close_ to each other as possible. Avoid long tracks. Use thick tracks. These are very important, especially if you aim for high frequency switching.

    As hgmjr pointed out, good switching power supply design is quite an involved subject. Some people even said that it is a black art! If you are interested then nothing's stopping you from delving further. I'm sure with practices and a few good articles/books you could gain the necessary experiences. If you just want to build one switcher, then just put an hefty LC filter at the output, the effort needed to learn these is not worth the few bucks saving gained from perfecting the switcher and the LC filter.
     
    joshzstuff likes this.
  11. sharath_412

    Active Member

    May 7, 2007
    32
    0
    Hello,
    I have designed a switching regulator which generates 5 V DC from 15 V DC supply. For this i used LM2678 switching regulator. It has 7 pins (1 -7) and a tab which is ground. In my circuit the tab is left open as pin 4 is grounded ( pin 4 and tab are short internally). The inductor which i used is 15 uH( DELEVAN ). The output capacitance is 300 uF(three 100 uF in parallel). In datasheet of LM2678 he recommended to use two 180uF capacitors in parallel. Input capacitance is three 15uF in parallel. The recommended is three 33 uF in parallel. (At the input section there are two 1000 uF capacitors also .1 uF capacitor)
    My circuit is PCB which also generates other voltages 3.3, 12, -5, -12. For 3.3, 5, 12 LM2678 fixed output IC's are used. For -5 and -12 MAX774, MAX775 are used.

    These are the problems in my circuit

    1. At the output of the inductor (also across output capacitors) there a ripple around 1 V peak to peak which has a frequency of around 55.55M Hz.

    2. At the input pin of LM2678 (pin no 2) there is a ripple around 1.2 V peak to peak which has a frequency of around 55.55 MHz.


    I would be very happy if someone could help me with some solution.
     
  12. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    Do you have a nice wide groundplane, as reccommended in the datasheet? And is your feedback conductor well clear of you inductor, as reccommended in the datasheet? Pin 7 is not floating, is it?
     
  13. gootee

    Senior Member

    Apr 24, 2007
    447
    50
    "I'm not an expert, but..."

    You probably need a snubber across your diode (see farther below, after reading the rest, first).

    The PCB layout IS absolutely critical, for switchmode power supplies, as has already been quite-rightly mentioned. There is an appnote at just about every switcher IC manufacturer's site, about PCB layout considerations for switchmode power supplies. If you haven't read one, you "might" want to do that. With a "less than optimal" layout, the parasitic R, L, and C of the PCB traces (and/or current-induced voltages from R and L of shared traces, if your layout is really boneheaded) can completely ruin a switchmode design.

    But sometimes/often it can help (tremendously) to put a snubber network in parallel with your diode. Usually, just a resistor and capacitor in series will do it. There are ways to calculate the R and C. But, often, you can just pick an R that's value is low-ish but high-enough to not allow too much current to flow, and then find a C, experimentally, that gives the best result.

    If you have a simulation of the circuit, you could add the parasitic inductances and resistances and capacitances of the PCB traces that have large and/or quickly-changing currents, and add some ESR to your caps and DCR to your inductances (from specs), and maybe a small C across each resistor, and you should SEE the 55.55 MHz, there. You MIGHT not even HAVE to add the parasitics, to see it.

    ANYWAY, there's usually (always?) a tendency for a switching circuit to "ring", at around the time that switching occurs. Sometimes (if you're lucky, or good), the ringing is a decaying-amplitude high-frequency oscillation, with a very short duration. But sometimes it doesn't decay much, or at all, and just oscillates, basically all the time (which just might be what you're seeing, with your circuit). Even in that case, a proper snubber network might be able to almost-completely eliminate it (if your PCB layout isn't TOO hopeless. :)

    You might also want to "chuck" that Delevan inductor (Bobbin type, isn't it? Nasty.) and get something like a J.W.Miller toroid. They're cheap. And they work well for switchers. See mouser.com, digikey.com, et al.

    If your power supply needs to be extra-clean, you might also want to consider using an L-C post-filter and/or post-regulator(s), such as LM338.

    Good luck.

    Tom Gootee

    http://www.fullnet.com/~tomg/index.html
     
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