Smoothing Filter Capacitance Equation

Discussion in 'General Electronics Chat' started by Lightningbolt, Mar 1, 2013.

  1. Lightningbolt

    Thread Starter New Member

    Feb 4, 2013
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    I have been studying about the capacitance required for the smoothing filter of a rectified circuit and I am having difficulty understanding the following.

    In a post by thingmaker3 [ 3rd post down @ http://forum.allaboutcircuits.com/showthread.php?t=12878 ] he states "Rule of thumb for calculating power supply filter capacitors: C = 0.7(I)/DE(f)". Other posts I have read also indicate similar formula.

    However, I cannot figure out where the "0.7" factor in his equation comes from.

    If I understand the derivation provided in the textbook I am studying on the subject, there should be no factor at all in the equation.

    Can anyone help me understand the discrepancy?

    Thank you in advance for your help!
     
  2. SPQR

    Member

    Nov 4, 2011
    379
    48
    I "grew up" on the old type of power supply (xformer/bridge/cap/inductor/cap), and thought that your question was interesting.
    Been reading a bit today, and it seems that the classic formula (dV = I/fC)
    does not take all circuit issues into consideration -
    for example, the type of rectification (full/half wave).

    THIS is a nice reference.

    So let's write down the formula but add an "adjustment" constant.
    If the constant = 1, then we are under ideal conditions.

    dV (ripple) = K * (I / fC)

    Ripple is large for a half-wave rectifier, and smaller for a full-wave rectifier.
    So I believe that they empirically adjust for ripple with this constant.

    K=0.5 for a full-wave, and >.5 for half-wave.
    That 0.7 in thingmaker3's post looks suspiciously like 0.707 which would be close to the calculation of rms of ripple.

    Let's see what the experts say.
     
  3. #12

    Expert

    Nov 30, 2010
    16,298
    6,811
    Disagree with the last part.
    The form I use is: root2 C Er F = I
    and I have tested it to 1% accuracy on my workbench.
    Mnemonic for California surfers: "I am radical to surf peak to peak" (I=Radical 2 * C * Eripple * Frequency)
    This formula gives the peak to peak value, which is critical in regulator circuits.
    I don't know how the square root of 2 got in there, but it's dead accurate, and somebody else will be along and explain that part. (We have some math guys here that smoke me.)
     
  4. #12

    Expert

    Nov 30, 2010
    16,298
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    Actually, that equation is about the smoothing capacitor on a DC power supply where I is the load current of the design, it is almost never sinusoidal, and the answer comes out in peak to peak terms.

    This might explain why you wrote several hundred words and did not address the question, which is: Why is the square root of 2 a valid constant for this equation?
     
  5. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    It is simpler to solve for a supply filter cap value in the time domain. For a cap we start as always from:

    I = C dv/dt

    Solving for C:

    C = I dt/dv

    In a conventional bridge rectifier the filter cap is charged in short bursts of current at the line frequency (half wave rect) or twice the line freq (full wave rect). In between these charge bursts the voltage droops, so the line period is the dt.

    The current due to the load is assumed to be constant, and is in fact constant if a linear regulator is involved.

    Lastly we have the voltage the cap is allowed to droop, that is a value for the designer to choose.

    Let's start with a full wave bridge on a 60Hz system:

    C = I dt/dv = I (1/120)/dv = .0083 I/dv

    if we pick a 1A load, and a 2V voltage droop we get:

    C = .0083 1A/2V = 4,150 uF

    So if you want a rule of thumb calculation you can use:

    C = .0083 I/dv (full wave bridge)

    C = .0166 I/dv (half wave bridge)

    Going back to the form used in the original post:

    C = I/(f * dv)

    (Thus I have no idea where the 0.7 factor comes from)
     
  6. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    An assumption can either be applicable or non applicable to some situation. It cannot be incorrect.

    A statement such as "a sinusoidal load current" being produced "when the load is an audio amplifier that is reproducing a single tone" may be true or false. This statement happens to be false, as the current from the supply capacitor is the summation of a sinusoid and a dc current. For a non-saturating output the average sinusoid over a period is zero when the frequency is much larger then the line frequency, and a dc value when the frequency is much larger then the line frequency. The dc component is of course a dc value, so the summation is a dc value.

    I would be hard pressed to state any examples of a sinusoidal load.

    It is curious that you are disputing the conclusion (which agrees with mine) in the commentary you posted. It leads me to believe you either did not read or understand the work you copied and pasted.
     
  7. Lightningbolt

    Thread Starter New Member

    Feb 4, 2013
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    Member #12, I assume your formula is for a Half-Wave Rectifier. And, for a Full-Wave Rectifier there would be an additional factor of (2) to account for there being twice as many pulses.

    Therefore, for a Full Wave Rectifier operating at 60 hz, your formula would be similar to that promulgated by Member thingmaker3:

    C=(1/√2) / [(2 for full wave) * (60 hz)] * I / Vripple where Vripple=peak-to-peak voltage difference
    or
    C=0.005893 * I / Vripple

    From the responders so far there appears to be two schools of thought on how to calculate the smoothing filter capacitance for a full-wave rectified circuit.

    School #1:
    Member SPQR (along with the referred to reference), Member ErnieM, and the textbook I have been studying [Electronic Engineering by Charles L. Alley & Kenneth W. Atwood, Wiley 1966, pages 100-105 ( http://archive.org/details/ElectronicEngineering )] all promulgate that

    C1={1 / [(2 for full wave) * (60 hz)]} * I / Vripple
    or
    C1=0.008333 * I / Vripple

    School #2
    Promulgated by Member thingmaker3, and Member #12, but without any explanation nor mathematical justification (so far):

    C2=0.005893 * I / Vripple

    Comparing C1 to C2:

    C1=1.4 * C2 or (√2) * C2
    or
    C2=0.7 * C1 or (1/√2) * C1

    This is a significant difference. Especially since I believe under sizing the capacitor would lead to greater problems than over sizing it [am I correct on this? your thoughts please]. The only downside for providing a larger than needed capacitor would be perhaps a slightly greater cost.

    And, if the performance tolerance of the capacitor is to also be considered (±20% min. variation as per reference: http://www.radio-electronics.com/in...er-filtering-smoothing-capacitor-circuits.php ) then using School #2 formula for sizing the capacitor may under size it by as much as 43% or more.

    So, is there someone out there that can provide a reasonable explanation for the derivation of the School #2 formula or an authoritative reference supporting it?

    Thanks again to all for your responses and help.
     
    Last edited: Mar 6, 2013
  8. Jaguarjoe

    Active Member

    Apr 7, 2010
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    90
    If you don't already have one, get a copy of the "Art of Electronics" by Horowitz and Hill. I got a paperback version for ~$20. It's old, but all of the basics don't change.

    Ernie's derivation is in there along with much, much else. Their writing style makes it fun to read.
     
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