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Ripple factor

  1. Aug 22, 2014 #1
    I am stuck up at finding the ripple factor for a half wave rectifier.
    According to the formula- V(rms)/Vdc.
    But how to find Vrms?
    To find it I would require the equation of the half wave rectified sine wave, right? How do I form that?
  2. jcsd
  3. Aug 22, 2014 #2


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  4. Aug 22, 2014 #3
    But for half wave rectifier Vrms=Vpeak x 0.385
    How did that come?
  5. Aug 22, 2014 #4
    This pic shows a different version of the formula
    What is the exact difference between Iac and Irms?

    Attached Files:

  6. Aug 22, 2014 #5
    RMS stands for root mean square and it is a special average of any periodic function that can be integrated. Wiki for RMS.

    For your other qustion, let's say that we have I(t) = 5A + 5*sin(2*pi*t)A. The first term5A is obviously a DC current and 5*sin(2*pi*t)A is an AC current. It's common to have a DC offset mixed in with an AC signal.
  7. Aug 22, 2014 #6
    So if we have to find out the ripple factor of the equation you mentioned, how should I proceed?
    What is Iac ?
  8. Aug 22, 2014 #7
    Idc is going to be a simple average of the function that defines your current. Did they give a formula for that or do you need to integrate to solve it? Irms is going to be Vrms/R. The formula for Vrms is given for a half rectified sine wave. You asked about it in one of your posts above. So, Iac is going to be the square root of Irms^2-Idc^2.

    This is your homework right? Take a deep breath and read your material again.
  9. Aug 22, 2014 #8
    This is not my homework,I just came by the term ripple factor when I was studying rectifier circuits and then they just mentioned that ripple factor for a half wave rectifier is 1.21, just the value no derivation,nothing.So I started exploring the topic and I had these questions.The above picture which I posted was some pdf I found online.
    One more doubt by the way-
    The equation Irms=√(iac^2 + idc^2) results from something called fourier transform?
  10. Aug 22, 2014 #9
    You don't have to derive it with Fourier analysis but you can. Have you studied Calculus and the integral? I'm trying to think of a way to explain the root mean square and how it relates to DC and AC components of a signal without math but I'm at a loss.
  11. Aug 22, 2014 #10


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    The pulsating output of a rectifier consists of DC component and an AC component ( also known as ripple). The AC component is unwanted and accounts for the pulsations in the rectifier output. The efficiency of a rectifier depends upon the magnitude of AC component in the output. The smaller this component, the more effective is the rectifier.

    The ratio of rms value of AC component to the DC component in the rectifier output is known as ripple factor

    so r ( ripple factor)= Iac/Idc

    the image you posted above tells you how the ripple factor of 1.21 was achieved
    what part didn't you understand ?

    Last edited: Aug 22, 2014
  12. Aug 22, 2014 #11

    jim hardy

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    RMS stands for a simple arithmetic procedure.
    In EE we usually apply it to a repetitive waveform like a sine wave.
    Here's the concept:
    Slow down your thinking like a slow-motion movie. Slow it all the way down to one single frame at a time, that is instant by instant. At any instant AC has some definite value.

    Now imagine yourself doing these steps:
    Write down the value of instantaneous voltage at each instant of time for one cycle(choose some increment - microsecond by microsecond? )
    That'd be 16,666 values for a single cycle of 60 hz AC, 20,000 for 50 hz.

    Now you're ready to do RMS on that series of values.
    R stands for square Root
    M stands for Mean , which is just the average
    S stands for Square .

    We work it backwards. S then M then R.

    First you'd Square every one of your several thousand voltage readings giving you a series of Squares.
    Next you'd average those squares, giving you the Mean of the Squares
    Then you'd take the square Root of that mean giving you the square Root of the Mean of the Squares.

    Calculus lets you do the calculation in one line not thousands of them.

    That process has been worked out already for a lot of common waves .
    see also wikipedia for Root Mean Square

    It so happens that RMS gives the effective heating value of a waveshape, that is a DC current of RMS amps carries same power as the complex wave. That's why they thought it up.

    Another neat feature is this - average value of a sine wave is zero because it is symmetrical above and below zero. That's why a DC meter connected to AC reads zero.
    When you square each individual reading you make them all positive before taking the average.

    Now it's your turn - work out some non sinewave RMS's.
  13. Aug 22, 2014 #12

    Yes I have studied calculus and integrals but not fourier analysis.
    Can I interpret the equation like -
    The rms value is a resultant of ac component and the dc component,similar to resultant of vectors,like we do it in a phasor diagram?
  14. Aug 22, 2014 #13

    Thank you.
    Now if we go by the exact mathematical definition of rms-
    Rms value of a function f(t) with period T
    Some formula in which we integrate the function with limits 0 to T and take the avg over T
    In case of the sine wave of the output half wave rectifier -
    We integrate with limits 0 to T/2 and take the avg over T or T/2? because if I use T/2 I get the wrong result.
  15. Aug 22, 2014 #14


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    the output of a half wave rectifier isn't a sine wave its just a series of pulses

  16. Aug 22, 2014 #15
    You should read jim's post carefully. His explanation is brilliant as he didn't rely on mathy explanations. Don't worry about Fourier analysis yet. You need a firm grasp of what the RMS is first.

    To answer your question, no, rms values are scalars not vectors.

    Jim touched on this but let me reiterate. The RMS value tells you about the power capability of a signal.

    Think of a light bulb. You can power it with AC or DC. You could power the light bulb with a sine wave such as sin(2*pi*t) Volts AC or you could power it with 0.71 Volts DC and the light bulb will be equally bright either way. The RMS value of sin(2*pi*t) Volts AC is going to be amplitude/sqrt(2) = 0.71 Volts RMS and the RMS of 0.71 Volts DC is also going to be 0.71 Volts RMS. They deliver equivalent power. Here's a question for you, what happens if we apply sin(2*pi*t) Volts + 0.71 Volts to the light bulb? Will it be brighter, dimmer, or no change? What is the total RMS of the combined signal?

    Many devices are not like the light bulb. They can only be operated by DC power. A DC motor is a good example. If we apply a voltage defined as sin(2*pi*t) Volts + 1.5 Volts then that sine term is useless to to the motor. The motor can only make use out of the 1.5V DC term. We can find the total RMS to be:

    RMS{sin(2*pi*t) Volts + 1.5 Volts} = sqrt{RMS{sin(2*pi*t)}^2 + RMS{1.5}^2} = sqrt{0.71Vrms^2 + 1.5Vrms^2} = 1.65Vrms

    ... but the motor can't use that total power capability of the voltage. It can only use the DC term of 1.5V.

    It would be nice to have a figure to describe the quality of the voltage being supplied to the DC motor so we define the ripple factor as Vrms_ac/Vrms_dc. In our case it's going to be 0.71/1.5 = 0.47. Lower is better. A ripple factor of 0 would mean that there is no AC component at all and our DC motor can use all the signal.

    You asked earlier how to find the DC component. You could simply average the signal using calculus. The RMS of the DC component is then the absolute value of the average. Remember that the RMS can never be negative even if the average of the signal is negative. Note that the average of sin(2*pi*t) Volts + 1.5 Volts is going to be 1.5V.

    The Fourier transform is a much more powerful mathematical tool that can separate nearly any signal into frequency components. That includes any component at the 0 frequency, DC in other words.
  17. Aug 22, 2014 #16
    The answer to your posted question-
    It will be √(0.71)^2 + (0.71)^2 = 1 volt
    So the bulb will glow brighter compared to individual ac and dc voltages.
  18. Aug 22, 2014 #17

    Or will it be just 0.71+0.71? because in the above example of motor you added up the two rms values.
  19. Aug 22, 2014 #18
    I blundered it. I fixed it quickly but not before you noticed it.

    edit: your post #16 is correct
  20. Aug 22, 2014 #19

    Oh,okay. Thanks by the way now I am atleast clear with what rms is.

    Now if I want my bulb to glow brighter I should not use pure ac nor pure dc but a combination of both(equally).
  21. Aug 23, 2014 #20

    jim hardy

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    good - your calculus is probably fresher than mine (1964).

    Take average over T not T/2. One whole line cycle. If you use T/2 you haven't accounted for the second half cycle where the function has value zero and your mean will be too high.

    And i think you implied integrating the function^2 ?

    Well done ! Thanks ! We all love to see "the light go on" .


    I woke up worried i might've been ambiguous, or wrong...

    We only integrate to T/2 because after that the function is no longer a sinewave.
    Maybe you're sharp enough to write the equation of a halfwave rectified sine but i'm not.

    So we integrate to T/2 accumulating the mean of sin^2 curve for that first half cycle
    and realize that mean is spread over twice the period of integration making it half as big
    so there's a 1/2 underneath the radical
    which puts a √2 into our result
    that's why:
    RMS output of a full wave rectified sinewave is Vpeak/√2
    but a half wave is Vpeak/2
    less by only √2 not half. Root comes after Mean.

    So T1 to T/2 are limits of integration
    but T1-T2 is the interval over which we average.

    You said it exactly right.

    Bravo ! And Thanks Again for closing the loop .
    Last edited: Aug 23, 2014
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