For those who understand rectifiers

Discussion in 'General Electronics Chat' started by studiot, Dec 14, 2009.

  1. studiot

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    Discuss the following statement

    It is however important to realise that the reservoir capacitor could not be charged without ripple current, nor the load current supplied.
     
  2. Paulo540

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    well, before the point of the capacitor, all you have is ripple, no?
     
  3. studiot

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    Most textbooks start with a statement like

    'The capacitor charges up to the peak value of the input voltage.'

    If this were simply true then the capacitor would thenceforth always be equal or greater than the input voltage. Therefore no more curent could flow into the capacitor. Therefore no current could flow out of the capacitor into the load.

    And no, the waveform before the capacitor is not a ripple waveform.
     
  4. mik3

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    Maybe the books say that there is no load on the rectifier.
     
  5. studiot

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    A pretty useless circuit that has no load?
     
  6. GetDeviceInfo

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    of course you could have DC flowing into a rectifier, but the common use is to rectify AC into DC. Your cap will charge to Vpeak, and it will discharge to some value, as the input falls away from peak.

    In a typical AC rectifier circuit, your rectifier output is a 'ripple' that rises from zero to some peak value, then returns to zero. Remove this 'ripple', and your left with zero potential that can neither charge a cap or supply a load.
     
  7. studiot

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    Not sure I understand your comments get device info?

    My original is not a trick question, by the way.
    Most texts describe in some detail the voltage waveforms in a transformer/rectifier/reservoir capacitor lineup. Many offer some version of the formula

    ΔV = 1/2fRC or ΔV = 1/4fRC

    But few address the much more difficult issue of the current waveforms, or the plain fact that the thing would not work without the ripple. Or to put it another way - it is impossible to smooth out all the ripple. The greater the smoothing the greater the peak current.
     
    Last edited: Dec 14, 2009
  8. Ron H

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    You need to have V in the numerator of the right side of the equation.
    And what type of circuit has a 4 in the denominator?
     
  9. studiot

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    Yes Ron the delta is relative so to get an absolute value you need to multiply the factor by the average output voltage, which for most purposes can be taken equal to the capacitor peak voltage.

    The different formulae refer to half wave (2) and full wave (4) rectification. The ripple on full wave being half that of half wave.

    But I am trying to move away from the voltage equations to get people to acknowledge that the capacitor is charged by a series of very short duration very high current pulses, not by a continuous waveform as with the voltage.
     
  10. Ron H

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    OK, I totally agree that the cap is charged by a series of very short duration, very high current pulses. In fact, it is possible to make the cap so large that the peak repetitive current spec of the rectifier is exceeded. All the coulombs that are removed from the cap by the load during the time the rectifier is reverse biased must be replaced during the time it is forward biased.
    Now, back to the ΔV equation. Half wave should have a (1) in the denominator, not a (2). Full wave has a (2), not a (4).
     
  11. studiot

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    I suppose that depends upon your Δ doesn't it?

    The factor of 2 comes in because to a first approximation the average voltage is halfway between the peak and minimum of the ripple so

    Vav = Vpk - ΔV or Vav = Vmin + ΔV

    Ripple = 2Δ
     
  12. Ron H

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    I see what you mean. I was thinking peak-to-peak ripple, you were thinking peak deviation from average. This makes sense for delta, although ripple, in my experience, is almost always specified as p-p.
     
  13. studiot

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    That's why I said 'some variation on the formula'.

    The most useful one I've seen actually includes current

    Vav = Vpk - I/4Cf (for full wave)
     
  14. Ron H

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    Which is particularly useful if the rectifier is powering a voltage regulator, because they look like a constant current load (if the regulator load resistance is constant).
     
  15. loosewire

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    It would help If you used a series of drawing to this
    demonstate this process. So you could see the ac
    change to dc and how the ripple works in steps. also
    the steps that follow for near pure dc. Show the whole
    regulated process for steady dv voltage. Sometime a
    new student is taken to the end of the cliff with a curcuit
    such as your discussion,you talk about the ripple with
    out drawing and stop,without simple and complex curcuit.
    A simple curcuit,the plug in adaptor for all kind of things.
    A sine wave to dc from the wall outlet to devise.
    To a new student it would be like evolution.
     
  16. studiot

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    I agree with you Loosewire a picture is often worth a thousand words.

    However creating and placing said picture on AAC involves me in considerably more effort than spewing out a few words so I only do it if I am convinced of the interest.
     
  17. Ron H

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    A simulation with waveforms wouldn't be too difficult.
     
  18. loosewire

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    Agreed to a e-book developer for what purpose do you serve to
    the students and uninformed. (for thoughs you don't understand
    rectifiers) why would you leave out the uninformed if you are a
    E-guy. We are at the edge of the cliff,where some post end with
    no drawings,no ripple,no bucking the ac hump.
     
  19. loosewire

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    Again to the edge of the cliff on a post that could be helpful.
     
  20. The Electrician

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    You raise an interesting point by saying this. For the theoretical case usually described in the textbooks, this is true. But the impedance of the source driving the rectifier is crucial. Looking through several texts of mine, I only found one that includes the source impedance in the analysis.

    There was a paper published in the IEEE Transactions on Industrial Electronics in November 1974 that pointed out that the increasing use of rectifiers powered directly from the grid was leading to problems caused by the extremely large amplitude current pulses drawn by such rectifiers.

    In this paper, the author also mentions that these large pulses are very sensitive to the exact waveshape of the grid voltage. Nowadays, the grid waveform is rather flat-topped due to all the capacitor input rectifiers in the world, and that flat top actually helps reduce the magnitude of the pulse.

    But, if there is a significant source impedance driving a capacitor input rectifier, the magnitude of the pulse does not continue to increase as the capacitance is increased; it reaches a maximum.

    I performed an experiment to show this. I looked through my transformer collection and found a couple of transformers rated for 24 VAC output, the difference being that one of them is over-designed and provides substantially less series impedance than the other (about a factor of 10 less). I hooked up a bridge of fairly high current Schottky diodes with 40,000 μF of filter capacitance and added a resistive load drawing 3 amps DC. I then captured on a scope the secondary voltage and current under these conditions.

    The first attached image shows the voltage and current for both transformers. The traces in color show the performance with the low source impedance transformer; the orange trace is the voltage at the transformer secondary, and the magenta trace is the secondary current.

    The high source impedance transformer performance was also captured and its waveforms saved by the scope, then recalled and overlaid over the color traces; the recalled traces are the gray traces.

    You can see that the lower source impedance leads to higher magnitude, narrower, current pulses.

    I then captured the color traces from the first image and then added another 100,000 μF in parallel with the already present 40,000 μF, giving a total of 140,000 μF.

    The second attached image shows in color the performance of the low source impedance transformer with 140,000 μF, and in gray the previous performance with only 40,000 μF. It's not obvious, but if you look carefully, you can see a few gray pixels behind the color traces. The performance with 140,000 μF is essentially identical to the performance with 40,000 μF; the source impedance (the transformer impedance) has become the limiting factor. The magnitude of the current pulses won't get any bigger no matter how much additional capacitance you add.

    The ripple voltage does continue to decrease as you add more capacitance, but the ripple current in the capacitors and diodes doesn't get any larger once you have enough capacitance to reach the point where it's the source impedance that limits the current pulses.

    It is commonly believed that adding more capacitance to a capacitor input filter (to decrease the ripple voltage) will increase the stress on the rectifiers. But there is a value of capacitance such the source impedance determines the current pulse magnitude, and adding more capacitance then reduces the ripple voltage without further stressing the diodes. This shows that having a transformer with poor regulation can be a benefit in reducing diode stress. If the rectifier if followed by a voltage regulator IC, then the poor regulation of the raw DC from the rectifier may not be a problem.

    The low impedance transformer I used is better in that respect than most transformers. This means that a rectifier circuit that isn't operating directly from the grid, but uses a transformer, just won't have extremely large current pulses into the diodes.

    There is a very interesting paper available on the web dealing with this very issue: powerelectronics.com/images/archive/311pet23.pdf
     
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