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Describing the voltage out of a rectifier

  1. May 27, 2012 #1

    davenn

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    hi guys...

    OK... I have been caught out with a question I cant answer whilst discussing rectifiers and smoothing

    what is the correct term for describing the value of the voltage out of a fullwave rectifier with no smoothing capacitor present ?

    example... so 10VACrms out of the secondary minus 1.4V dropped in the 2 conducting diodes
    what is that resultant 8.6VDC voltage called ? I dont know if that is an avg (mean) DC volts or some other name. RMS value only applies to the AC voltage I think (pretty sure) :smile:

    I understand that after a smoothing cap is added, then the avg(mean) DC Voltage is the value midway between the peaks of the ripple voltage

    cheers
    Dave
     
  2. jcsd
  3. May 27, 2012 #2
    Good one!!! I draw out the circuit. My guess the wave shape should be half the sine wave ( see 0 to 180 deg) repeating over and over with a slight flat at 0V between the adjacent peak. The amplitude should be the peak to peak amplitude minus 1.4V.

    But my question is what is the peak to peak voltage of a 10Vrms? I though it is (2X10)/0.707 but that sounds too large.

    Don't quote me on this, I just join in.

    When you add a smoothing cap, it really depends on the load. If you have no load, the DC would be the peak of the waveform. Which is the Vpp-1.4.
     
  4. May 27, 2012 #3
    In your example, I think that the 8.6VDC is the average value of the pulsating DC.
     
    Last edited: May 27, 2012
  5. May 27, 2012 #4

    davenn

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    Hi Youngman
    good evening/morning (tis Sunday evening here in eastern Australia)

    The peak voltage of a 10VACrms value is Vrms x 1.414 = 10 x 1.414 = 14.14 VAC

    now, just thinking out loud here....

    with no capacitor to reduce the ripple, the p-p ripple voltage would be 14.14VAC
    I think that would be correct.
    So going by my comments in my first post...

    it would sort of imply that the avg DC voltage would still be midway between the peaks of the ripple voltage ie. 14.14V / 2 = 7.07VDC

    Hopefully VK6KRO will jump in here somewhere and set us straight :)

    well that thought has crossed my mind :smile: but I'm not convinced that it would be specifically called the avg DC voltage

    I have gone through so many www sites talking about AC - DC PSU's and dealing with the calcs for Vrms Vpeak ripple etc but not one of them have specifically commented on a name for that voltage

    Dave
     
    Last edited: May 27, 2012
  6. May 27, 2012 #5

    davenn

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    even reading some comments from vk6kro in another active thread at the moment still leaves me unsure.
    His image is showing a fullwave rectifier without a smoothing cap and that the VDCavg = Vp x 0.636

    Now I'm assuming he's referring to the Vp as the AC peak voltage ?
    or is he referring to the Vp as the peak DC voltage?

    referring to my discussion with my friend that started all this.....

    in his simulation he has a VACrms of 6V and with a fullwave rectifier he's got a DC voltage (pulsed) of 4.6VDC ie. 6V - 1.4V drop in the 2 diodes

    VACp = 6VAC + 1.414 = 8.484 VAC
    8.484 x 0.636 = 5.395 VDCavg which is above that measured 4.6VDC

    this is giving me a headache haha

    cheers
    Dave
     
  7. May 27, 2012 #6

    NascentOxygen

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    I wouldn't label it as VAC, since it isn't a sine wave. It's just a waveform with a peak value of ....

    If you overlook the diode drops, then the RMS value of the rectified sine wave is equal to the RMS value of the AC before the rectifiers. But rarely would we be interested in knowing the RMS value of the rectified sine wave there.
    That would be true only if a sketch of one cycle of the waveform showed as much area under the halfway mark as above it. That is not true for a rectified sinusoid, so your hunch is not going to be valid. :wink:

    You will have to use calculus to determine the area, hence to find the average level.
    You could call it average, or you could label it as DC, but not both together because they are one and the same thing here. Maybe to make it clear to the reader, denote it in the manner of 47VDC (avg).
     
  8. May 27, 2012 #7

    NascentOxygen

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    No. He has a train of pulses each with a peak value of 6√2 - 1.4 volts.
     
  9. May 27, 2012 #8

    davenn

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    Thanks for your thoughts, appreciated

    its more of an academic thing, we both became interested in if there was a specific name definition for the unsmoothed rectified voltage at that point.


    Thats fair comment

    yes true

    again ... fair comment


    calculus and I are not very good friends haha

    cheers
    Dave
     
  10. May 27, 2012 #9

    vk6kro

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    The average value of a sinewave is zero, but you can take the average of half a sinewave using Excel if you find that more friendly than calculus.

    You can make a lot of little rectangles, each 1 degree wide and whatever the sine of that angle is as the height. Add them all up and divide by 180 and you should get 0.636 times the peak voltage.

    This comes out to 0.636 times the peak value of the sinewave.
    However it is not a very useful figure as we are usually interested in RMS which is the DC voltage which has a heating effect which is the same as that of the sinewave.
    This is the more familiar 0.707 times the peak voltage of the sinewave.

    You can check this with Excel, too by taking the squares of all the heights above, adding them up divide by 180 and taking the square root.
    Like this:
    13ff7a17de460cd4c28938d9fd08e730.png

    If the voltages are high enough, you can neglect the diode drops or include them if you want an exact value.

    I would just refer to that waveform as partly filtered DC if it has a capacitor which is unable to deliver pure DC under load. Almost any capacitor will deliver pure DC off load.

    I guess you have seen the waveforms you get with a capacitor:

    http://dl.dropbox.com/u/4222062/rectifiers%202.PNG [Broken]
     
    Last edited by a moderator: May 6, 2017
  11. May 27, 2012 #10

    sophiecentaur

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    The classic diagram (like the one above) is what you will get with an 'average sort of value' for the capacitor. However, there are two time constants at work. If you try to smooth that waveform even more, you can achieve much nearer DC, but this will also involve the rising edges of the red triangular looking wave no longer 'hugging' the rising edge of the black semi-sinusoid. If the added capacitor is large enough then the time constant of a high series source resistance and the capacitor can become significant and the phase shift becomes apparent and the output 'sags' below the maximum of the sinusoid. In circuits using seriously cheap transformers, made with thin wire, this effect can be striking, where the load resistor is low value and the two slopes can be similar. Smoothing can be compromised in power supplies for power amplifiers when you demand high power at low frequencies. It then comes down to the transformer rather than just the capacitor.
     
  12. May 27, 2012 #11
    OK, If the peak to peak is 14.14V, with the filter cap on and no load. the DC is definitely 14.14-1.4V( two diode). When you start drawing current, the you will see a discharge waveform that follow to 12.74V and then exponential down from the current drawn until the next peak like Vk6kro's diagram.
     
  13. May 27, 2012 #12

    sophiecentaur

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    "Average" is not a defined quantity. 'Mean' of a half sinusoid would not be "half way up" because the slope is not uniform. Mean of a rectified and partly smoothed rectified AC would be anyone's guess and would depend upon source and load impedance and capacitor value. You would need the dreaded Calculus or some signal processing to determine it accurately.
     
  14. May 27, 2012 #13
    I contend that their should be a flat spot at 0V between the two adjacent hump. The reason is when the differential sine wave is below 1.4V peak to peak, both diodes of the rectifier should be off and the output remains at 0V. It last until the differential voltage reach -1.4V then it'll start swinging again. But I never seen graph drawn this way. Am I missing something?
     
    Last edited by a moderator: May 6, 2017
  15. May 27, 2012 #14

    vk6kro

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    No, that does happen. I have seen it on an oscilloscope.

    LTSpice shows it too. I used an input of 3 volts peak to get the following:
    http://dl.dropbox.com/u/4222062/Bridge%20rectifier%20flat%20spot.PNG [Broken]

    The first graph is the current from the transformer and the second is the output voltage.
     
    Last edited by a moderator: May 6, 2017
  16. May 27, 2012 #15
    Thanks.
     
  17. May 28, 2012 #16

    sophiecentaur

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    I notice that your maximum voltage is only 2V, which accentuates the effect of the diode. Btw, the effect of reverse leakage would be a measurable but tiny negative excursion on the second graph.

    2V is good for showing the effect but if you are using higher voltages, the 'details' tend to become less significant. By 9 or 12V, the diode is more of an on/off device and the graphs would look more like the the earlier pictures. The effect of source resistance would also modify the picture, as I pointed out, earlier.
     
  18. May 29, 2012 #17

    davenn

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    Thanks guys :smile:

    I appreciate the additional input

    Its amazing how many years we can work with various electronics and not really know what is happening till much later in life.
    tis always a learning process

    cheers
    Dave
     
  19. May 29, 2012 #18
    I would say that ideally pure DC V avg = V rms - and the Ripple is an undesirable component. The DC Bus is defined by Vdc + Ripple, calculating the true average or the true RMS of the bus with significant ripple is not trivial. You then filter to reduce the ripple as much as $ and the spec allows. As for the Current waveform, it can get ugly, particularly for the line.
    BTW - to calculate the losses in the rectifier - you need to know both RMS and Average of the current because there are resistive losses depending on RMS and Vf x I losses of the diode depending on the average. Then to calculate the current in the DC Bus caps - practically takes an frequency domain ( fft ) calculation.
     
  20. May 29, 2012 #19

    sophiecentaur

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    But the ripple may not matter at all, in some cases.
     
  21. May 29, 2012 #20

    vk6kro

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    That's right.

    Power supplies are seldom just a transformer, a rectifier and a capacitor these days. Usually they are followed by a voltage regulator and the only concern then is that the input to the regulator is always above the minimum input requirement for the regulator.

    Linear regulators will do a very nice job of giving a pure DC output at the regulated voltage as long as the bottom of the ripple waveform does not dip below the minimum input voltage of the regulator.

    Switching regulators tend to be lighter and cheaper and more efficient than linear regulators but they often have output which is DC with a lot of noisy AC superimposed on it. Maybe there are good ones out there at a price, but the only ones I have seen were just noise boxes.
     
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