1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Determine: The r.m.s. ripple voltage

  1. Feb 29, 2016 #1
    • moved into h/w help, so template is missing
    An aircraft electrical system operates at 115 volts at 400 hertz. It is to supply an average d.c. voltage of 28 volts and an average current of 20 A at a ripple factor of 0.005.

    Determine: The r.m.s. ripple voltage

    Here is my attempt.

    Vc= Vs-2Vd
    =28-2(0.7)
    = 26.6 Volts

    Vr= (1/2 x fs x C x Rl )(Vc)

    R=V/I 28/20= 1.4

    (1/2x400x1.4)(26.6)
    = 0.023

    Vrms= Vr/2sqrt3


    Vrms = 0.0063

    Now theres a good chance I'm way of here.....Thanks for any help!
     
  2. jcsd
  3. Feb 29, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    Hi ifan davies, Welcome to Physics Forums.

    Different authors define the ripple factor slightly differently. In most cases it's considered to be equal to the ratio of the RMS ripple voltage to the DC component of the power supply output. Others use ratio of the peak to peak value of the ripple voltage to the DC component. Still others convert the result to a percentage ("percent ripple") by multiplying the ratio by 100.

    What definition is being used in your textbook or course?
     
  4. Feb 29, 2016 #3
    Hey hey, its the first one you mentioned where it's considered to be equal to the ratio of the RMS ripple voltage to the DC component of the power supply output.
     
  5. Feb 29, 2016 #4

    gneill

    User Avatar

    Staff: Mentor

    Okay, so write it out as an equation and plug in your givens. Solve for what you want.
     
  6. Mar 1, 2016 #5
    right lets start again...

    Vrms= Vripple/2sqrt3

    Vripple= I(load)/2fc

    Vripple= 2/(2x400)

    = 0.0025

    Vrms= 0.0025/(2sqrt3)

    vrms= 0.00072

    this seems low..
     
  7. Mar 1, 2016 #6

    gneill

    User Avatar

    Staff: Mentor

    You're making it much more complicated than it needs to be. Work with the definition of the ripple factor. What terms in that defining equation are you given in the problem statement?
     
  8. Mar 1, 2016 #7
    Vpk= Vaverage x (pi/2)

    Vpk = 28 x (pi/2)

    Vpk= 43.98v

    Vrms= Vpk/sqrt(3)

    Vrms = 43.98/ sqrt(3)

    Vrms= 25.39
     
  9. Mar 1, 2016 #8

    gneill

    User Avatar

    Staff: Mentor

    Again, you're all over the map. Write out the defining equation for the ripple factor. It's one equation.
     
  10. Mar 1, 2016 #9
    Kv= rms value of the a.c. voltage component/ average value of the load
     
  11. Mar 1, 2016 #10

    gneill

    User Avatar

    Staff: Mentor

    What specifically do you mean by "average value of the load"?

    Hint: See your own post #3.
     
  12. Mar 1, 2016 #11
    its the average load voltage according to my notes? Thanks for sticking with me!
     
  13. Mar 1, 2016 #12

    gneill

    User Avatar

    Staff: Mentor

    Okay, it's the average DC component of the load voltage. So the equation is:

    ##K_r = \frac{V_{rms}}{V_{DC}}##

    where ##K_r## is the ripple factor, ##V_{rms}## is the RMS ripple voltage, and ##V_{DC}## is the average DC voltage.

    Which values do you know from the given information?
     
  14. Mar 1, 2016 #13
    Vrms = 115/sqrt2

    Vdc= 28

    Kr= 81.3/28

    Kr = 2.9

    Please be right
     
  15. Mar 1, 2016 #14

    gneill

    User Avatar

    Staff: Mentor

    You're given the ripple factor (##K_r = 0.005##) and the DC voltage (##V_{DC} = 28~V##). You want to find the ##V_{rms}##.
     
  16. Mar 1, 2016 #15
    this has to be it ....

    Kr x Vdc = Vrms

    0.005 x 28 = 0.14
     
  17. Mar 1, 2016 #16

    gneill

    User Avatar

    Staff: Mentor

    Yup.

    Be sure to include units on your answer.
     
  18. Mar 1, 2016 #17
    When you put it like that its easy! Your awesome!

    Now to calculate a suitable value of smoothing capacitor

    C= i load / frequency x V

    20/ (400x0.14)

    0.357 F
     
  19. Mar 1, 2016 #18

    gneill

    User Avatar

    Staff: Mentor

    I'm not sure about your capacitor calculation. For a full-wave rectifier you should have something like

    ##V_{pp} = \frac{I}{2 f C}##

    where ##V_{pp} = 2 \sqrt{3} V_{rms}## is the peak-to-peak ripple voltage assuming a sawtooth waveform.

    That would yield a somewhat smaller value for the capacitor.
     
  20. Mar 1, 2016 #19
    0.48= 200/2x400x C

    0.00024= 2x 400 x C

    0.0012= 400 x C

    C= 3 uF
     
  21. Mar 1, 2016 #20

    gneill

    User Avatar

    Staff: Mentor

    What does the "200" value represent? The load current was given as 20 Amps.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Determine: The r.m.s. ripple voltage
Loading...