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How to calculate reactance of this capasitor

  1. Jul 27, 2016 #1
    Hi,
    I am a little lost here.
    psu-fig-1-3-3  test.jpg
    What is the reactance of C2 at the ripple frequency? Can someone show me the steps to make this calculation?

    Thanks,

    Billy
     
  2. jcsd
  3. Jul 27, 2016 #2

    berkeman

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    1/(2*Pi*100Hz*C), but it's more complicated than that. What are you wanting to calculate? :smile:
     
    Last edited: Jul 28, 2016
  4. Jul 27, 2016 #3
    Hi Berkeman,
    A little background is in order. I am studying power supply basics and typical filter circuits used in those power supplies. There was a test at the end of the learning module and the above question was the only one I could not answer. The correct answer is 3.2 ohms according to the answer sheet.

    Given only the information in the schematic one is supposed to be able to be able to compute the reactance of capacitor C2.

    I think the writer of the module expected one to use .6 volts as the voltage drop across a diode as that was mentioned somewhere in the text.That would give a value of 1.2 volts.

    That would give a voltage out of the bridge rectifier of 10.8 volts.

    I don't know how to calculate the ripple frequency. My uneducated guess would be that the ripple would be about 10% of the rectifier output value of 10.8 volts.

    So...the bottom line, given the above, by what process does one arrive at the 3.2 ohm answer?

    Thanks,

    Billy
     
  5. Jul 28, 2016 #4

    jim hardy

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    Here's "Dr Hardy's Painless Method" ( not highly academic bur practical)

    Observe that rectification is basically just flipping the bottom half of the wave up above zero.
    d101_06.gif
    That changes its average from zero to some substantial positive value. Area below zero was flipped above so it no longer cancels, adds instead.
    Notice also that it now reverses direction twice as often.... suggesting that frequency is doubled. That's your memory aid. More later.

    So what was a pure line frequency sinewave is now a badly distorted sine wave .
    If rectification was perfect there's nothing anymore at line frequency because everything now repeats twice per line cycle
    So
    The lowest frequency present is now 2X line frequency and you can use that to calculate reactance, Xc= 1/(2πfc) as Berkeman said.

    ^^^^And that's the Painless Answer to your question.^^^
    ......................................................................................................................


    Now, because you have that nifty new 'scope with FFT....

    Distortion as you know is rich in harmonics. So if you were measure voltage across and current through the capacitor and divide V by I , you might get a somewhat different number of ohms than by 1/(2πfc). That's because there's an Xc for each harmonic.... but most folks just figure Xc at twice line frequency and go with that.
    The way to remember is this
    it used to reverse direction only at its peaks , it blew right through zero,
    but now its direction reverses at peaks and at zero crossings!
    So it doubled frequency .

    It also picked up harmonics that change it away from a nice pure sinewave, as you already know distortion does..

    The math behind those harmonics is called Fourier Analysis.
    Basically Fourier says ANY repetitive waveform can be written as the sum of many sinewaves,
    the first sine being at the repetition rate , called fundamental
    the second being at twice the repetition rate , called second harmonic
    the third at thrice , third harmonic
    and so on....
    Each sinewave in that sum gets multiplied by its own scaling factor let's just call it An where n is the harmonic's # 1, 2, 3,,etc... (1 being the basic repetition rate, line frequency before rectification for your case)

    So you're looking at a math expression A1sinwt + A2sin2wt + A3sin3wt and so on, it's just addiition...
    (There's also an A0 at the beginning that's the just average value, which is what a DC voltmeter would show, as i mentioned above )
    you can draw them on graph paper and add them together to check me. Google Fourier.
    Now
    Here's a nifty simulated frequency analyzer that lets you select a frequency, full wave rectify it and look at the harmonics.
    It's a great demonstration of Fourier Analysis:
    http://www.falstad.com/fourier/e-fullrect.html

    play with it a while
    it doesn't have 50 hz but it has 52
    PlanoFourier.jpg

    you'll notice right away that all the odd harmonics are zero, including the 1st(52hz) remember i said perfect rectification doubles the frequency..
    Imperfect rectification would retain a small 1X line frequency component (and other odd harmonics).

    The math looks scary, this is for perfect rectification

    PlnoFourier2.jpg

    but all it's doing is making that sum of sinewaves
    play around with that simulator
    then with the FFT function in your fancy scope (FFT - Fast Fourier Transform)
    when it begins to feel comfortable you have acquired a powerful tool for your "bag of tricks"
    you should be able to see imperfect rectification using its FFT , odd harmonics will be nonzero
    it'll be fun to analyze clipping in guitar amps too.



    and there are plenty of folks here on PF who have way more finesse than i with that math.

    old jim
     
    Last edited: Jul 29, 2016
  6. Jul 28, 2016 #5
    Hi Jim,

    I am reading through what you said. I need to install Java on this computer to get the simulation to run.

    In the mean time I figured out the process to get to the test question answer.
    Xc=1/(2piFC)
    2xpi=6.28
    6.28 X 100 = 628.30
    628.30 X 500 = 314,159.26
    1/314,159.26 = 3.18 ohms...rounded up = the 3.2 test question answer

    So...here is the process and what is needed to answer the question without anything other than a pin and paper
    1. Know the formula for Xc
    2. Know what is meant by the words "ripple frequency" (Actually I knew the approximate frequency out of the rectifier was 100Hz, I just did not know that meant "ripple frequency")
    3. Know the value of pi
    4. Know that the frequency of 50Hz will be approximately doubled to 100Hz by a full wave bridge rectifier
    5. Get the capacitance of C2 off the schematic

    So...I was missing two things. One, I did not know the formula for Xc and two I did not understand the meaning of "ripple frequency" as it was being used in the test question.

    Now...I will dig into the more precise answer to the question by the use of FFT.

    Thanks for the heads up.

    Billy
     
  7. Jul 28, 2016 #6

    jim hardy

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    capacitance is measured in Farads
    and you have 500micro-Farads

    so that step should read
    628.30 X 500 X 10-6= 314,159.26 X 10-6 = 0.314
    poweroften.jpg

    and ##\frac{1}{0.314}## = 3.18 as you said

    You've looked at rectified filtered AC before with your scope
    th?id=OIP.Mc65f33b724a7d9c50584ce8b85cf1687o0&w=302&h=139&c=7&rs=1&qlt=90&o=4&pid=1.1.jpg
    ripples is what it looks like hence the name...


    the farther you go the faster the doors open !

    keep having fun

    ........how'd you fix that Grouper?

    old jim
     
    Last edited: Jul 28, 2016
  8. Jul 28, 2016 #7
    Hi Jim,

    Fried the grouper and baked some. It is really good!! Still have some in the freezer. You should come over for dinner!!

    Yes, I have looked at the ripple on the scope. I understand what it looks like. I did not and still do not know what the effect of C1 and R1 would have on the frequency at C2 if any. Is it still 100Hz? If the 9V 100mA load changed would that change the frequency at C2 also?

    Cheers,

    Billy
     
  9. Jul 28, 2016 #8

    jim hardy

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    can't change the frequency, it's twice line just gets smaller amplitude as you add filter stages.

    The filter reduces the higher order harmonics more because Xc is less for them, so you get closer to DC which is intuitive - more filtering reduces the ripple.
    you'll see that with your scope'sFFT, i dont think that simulator will let you add filtering

    If rectification was imperfect, say one diode had slightly higher voltage drop
    then the negative peaks that got inverted to become positive
    wouldn't be same height as the ones that were positive to start with
    so there'd be some distortion at 1X line frequency
    and odd harmonics 1,3,5 would appear but their A1,3,5,... terms will be small because the difference in diodes isn't drastic...

    You'll love that FFT function
    in my day a FFT was a separate instrument the size of a big 'scope
    lots of fun to put a mike on one and wander around the house, that's how i found i can no longer hear aTV horizontal oscillator. As a kid i could hear it from the front yard.
     
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