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Capacitor for voltage regulator

Discussion in 'General Electronics Chat' started by MrNobody, Feb 7, 2008.

  1. MrNobody

    MrNobody New Member

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    Just wondering.. what is the general rule tor selecting capacitor for voltage regulators..? is there a formula for calculating it..? is it like, the higher the capacitor value is the, the better it is because it can filter out more noise..?
     
  2. dknguyen

    dknguyen Well-Known Member

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    Real capacitors have a parasitic inductance in series with the capacitance which makes the capacitor appear as an open-circuit at high frequencies, and large capacitors have larger inductances so larger capacitors don't filter out high frequency noise better than small capacitors.

    They do, however, filter out low frequency noise better so they reduce ripple (but not necessarily noise). There are formulas and such but it's quite a bit more than most people want to know. What are you trying to do? Usually for "electronics" 100uF is really big, unless you are dealing with motor filter capacitors. 4.7uF is also quite big.

    The type of capacitor can also make a difference since some kinds are available in larger sizes, but other kinds have better high frequency characteristics and thus can filter out higher frequency noise.

    Linear regulars sometimes need a minimum capacitor (and sometimes with a parasitic resistance called ESR) that needs to be high enough so the regulator can be stable. It should be in the datasheet if it does with recommended values.

    Going bigger is usually okay though, just stick small high frequency decoupling capacitors along other things due to the inductance of the large capacitor, among other parasitic inductances.
     
    Last edited: Feb 7, 2008
  3. mneary

    mneary New Member

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    General rule? Sorry there is none.

    On some regulators lower ESR (equivalent series resistance) is better, but on others it upsets the feedback loop.

    Follow the guidance in the manufacturer's data sheet or application notes.

    If there is no guidance,
    (1) choose another device
    (2) use the guidance from a similar part
    (3) take the risk that you guessed wrong.
     
  4. dave

    Dave New Member

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  5. MrNobody

    MrNobody New Member

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    dknguyen:
    Actually, its for the ENC28J60 ethernet controller..
    The requirement is 3.3V..
    I am using LM1117T voltage regulator..
    According to the datasheet, the V(in) capacitor is 10uF and V(out) capacitor is 10uF.. however, according to the ENC28J60 board schematic, the V(in) capacitor is 0.47uF and V(out) capacitor is 47uF. The V(in) voltage is 5V..
    If I don't have 0.47uF and 47uF capacitor, can I use other capacitor..? I have 0.1uF (many), 10uF (only 1) and 100uF (only 2)..

    Another thing is, currently, i am feeding the LM1117 (3.3V regulator) from the output of LM7805 regulator.. At the same time, the 7805 regulator is also supplying voltage to a microcontroller circuit.. Is that OK or should I feed the 3.3V regulator directly from power 9V supply/battery..?
     
  6. dknguyen

    dknguyen Well-Known Member

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    I don't think it'd be a problem. The value of decoupling capacitors usually isn't critical as long as you don't stray too far. Try and stay in the same order of magnitude. You could try feeding it directly, but it's almost always better to have a capacitor some sort. You can probably get away with a smaller capacitor on the regulator input than on the regulator output.

    With what you have I'd use the 47uF on the input and a 100uF on the output. Or I might use the 47uF on the output and whatever on the input (I don't think I remember seeing a regulator where the input capacitor needed to meet some requirements for the regulator to be stable but I could be wrong). There may be a reason the output capacitor is that size due to power draw of the load or for some other reason.
     
    Last edited: Feb 7, 2008
  7. MrNobody

    MrNobody New Member

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    dknguyen:
    Actually i don't have 47uF. can I use 10uF on the input and 100uF on the output instead..?
     
  8. dknguyen

    dknguyen Well-Known Member

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    It's not critical. THat sounds okay to me. 100uF probably isn't too different from 47uF to cause a problem. If you stuck a 1000uF capacitor on the output, then it might cause a problem (or it might not). It's kind of a "it feels close enough to work" kind of thing.
     
  9. MrNobody

    MrNobody New Member

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    oh.. ok.. thanks..
     
  10. dknguyen

    dknguyen Well-Known Member

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    If this is on a breadboard, you can always swap it out if it doesn't work. If it's a PCB, choose a footprint that you can get other capacitor sizes for.

    That's one of the reasons I try to use 1206 package capacitors and resistors everywhere.
     
  11. MrNobody

    MrNobody New Member

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    1206 package..? u mean SMT footprint..?
     
  12. dknguyen

    dknguyen Well-Known Member

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    Yeah, 1206 is an SMD foot print. Easier to handle than a 0805, handles more power, almost the same price, and common.
     
  13. MrNobody

    MrNobody New Member

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    hmm.. jst wondering.. can i monitor the output voltage on a oscilloscope..?
    i mean.. if the capacitor is no good, while operating the microcontroller or the board, the waveform should appears noisy like rite..? i mean, with ripples etc instead of a ideal straight line..
     
  14. dknguyen

    dknguyen Well-Known Member

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    Sure you can. You can even try different caps to see how it affects the output noise.
     
  15. kchriste

    kchriste New Member Forum Supporter

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    It should be OK because the LM1117T is a low dropout regulator and will work down to 4.5V input with 3.3V out at 800ma in the worst case scenario. Now if it was a typo and you meant LM117, then no, you'd need to run it directly from the 9V supply.
    I would go by the LM1117Ts datasheet as far as the input and output decoupling caps are concerned. You can always go higher in uF than the datasheet, but don't skimp on the low side.
     
  16. Hero999

    Hero999 Banned

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    Are you talking about the input or output?

    If it's the input and you're running it from rectified AC power, you'll need a huge capacitor. As a general rule, it should be > 1/(2*FC).

    On the output, 100nF will usually do for an LM7805 or an LM317 but some more exotic regulators need more or less and a specific ESR.
     
  17. Honduras

    Honduras Member

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    Oddly enough, I got curious about this the other day, so I collected some data.

    My test setup consisted of a 6 volt wall wart rated at 0.6mA, an assortment of capacitors, and a 1.5 kOhm resistor to act as the load. I didn't use an output filter. The capacitors were hooked across the load resistor one at a time. The assumption was that an increase in the voltage available at the load was good.

    Results were:

    6500 uF 0.120 VAC 5.19 VDC
    4300 uF 0.120 VAC 5.17 VDC
    3800 uF 0.119 VAC 5.23 VDC
    2200 uF 0.120 VAC 5.20 VDC
    570 uF 0.115 VAC 5.22 VDC
    470 uF 0.120 VAC 5.27 VDC
    100 uF 0.115 VAC 5.27 VDC
    10 uF 0.110 VAC 5.20 VDC
    1 uF 0.110 VAC 5.21 VDC
    .047 uF 0.110 VAC 5.20 VDC
    .01 uF 0.110 VAC 5.21 VDC

    It appears that the size of the capacitor, within my ability to test them, has very little effect on the voltage regulator, especially when you consider that the numbers are 0.120 VAC and 5.20 VDC with no capacitor.

    I ended up going with the 100uF cap, but mainly because I have a lot and it did seem to be on the high end if my observations were correct.

    Now I have to do this for the output. ;(
     
  18. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    hi H,
    This thread is over 5 years old.:rolleyes:

    Is your 6V wall wart a DC output.? and do you mean 0.6Amp not .6mA

    Your load resistor of 1k5 is too high to give you a meaningful result.

    E
     
  19. MikeMl

    MikeMl Well-Known Member Most Helpful Member

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    To come close to loading the wall-wart at near its rated output, your load resistor should be R=E/I = 6/0.6 = 10Ω. Its power rating would have to be P > E*I = 6 * 0.6 = 3.6W

    If you use that, you will find that to get smooth DC (with an amount of ripple ΔV), your filter cap will need to be C = I*t/ΔV. If you want ΔV to be less than 1V, and t=8.33ms at 60Hz, then C = 0.6*0.0083/1 = 0.005F = 5,000uF, which is a more realistic way of looking at a practical filter cap...
     
  20. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Why don't you read the datasheet for your LM1117 regulator?
    It recommends a 10uF tantalum input capacitor that MUST be used if the main filter capacitor is at a distance.
    Since it is a low dropout regulator then it MUST have an output capacitor and they say it is minimum 10uF tantalum (not electrolytic and not ceramic). It can be 22uF, 47uF or 100uF.
     
  21. levent84

    levent84 New Member

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    Ok, first please establish what the purpose of a certain capacitor is, if it is being used in a filter or in series in a circuit to smooth out edges of any signal, the value of the capacitor is critical, do not used any other value than the one mentioned. however, if the capacitor is being used just for the purpose of noise filtering between a DC signal and the ground you can play around with the value of the capacitor a bit without expecting huge consequences on the functionality of the circuit. I'd suggest sticking to the given values but well, if you can identify what's critical and what isn't, that's an added advantage. And talking about which capacitors are better, it's mostly the ones with higher voltage rating and lesser tolerance bands.

    prototype assembly
     
    Last edited: May 20, 2015

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