Smoothing circuits after rectification

Discussion in 'General Electronics Chat' started by big_bird24, Jul 27, 2014.

  1. big_bird24

    Thread Starter New Member

    Nov 7, 2009
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    Looking at some power supply circuits, I've noticed different types of smoothing circuits after the diode bridge.
    Some include just include a series inductor with the load resistance
    Others a shunt/parallel capacitor with the load resistance.
    And other include a LC filter with the load resistance.

    I understand the operation of the 2nd order LC filter to block out higher frequency harmonics. I understand the derivation of the transfer function for the filter aswell.

    My confusion comes from the qualitative explanations I've read online, that state the inductor smoothens current ripples and the capacitor smoothens voltage ripples. Can someone explain why those explanations are used? Isnt the capacitor providing both current and voltage smoothing?

    Below I've typed out my though process and where my confusion lies.

    What doesnt make sense is that current is a function of voltage, so by smoothing the voltage across the load we should also be smoothing the current. For example, if we used a shunt capactior for smoothing the voltage across the load then it should also smooth the current through the load since I=V/R.

    I tried thinking about this in the frequency domain to no avail.
    Mathematically, we know the voltage across the capacitor in frequency domain is V=(1/wC)*I . From that we see that higher voltages dropped across the capacitor (and therefor the parallel load) for lower frequencys/harmonics which is our voltage smoothing action. But we can re order the equation such that I=V*wC in which case we see that I (the current shunted/bypassed through the capacitor) is larger for higher frequencies. So larger magnitudes of current are bypassed though the cap away from the parallel load for higher harmonics, thus also providing current smoothing for the load.
     
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Here is a comparison between two versions of a full-wave rectifier driving a 1A load, one without an inductor and one with.

    Compare the various branch currents, especially the ripple current through the filter capacitor, and the ripple voltage across the load resistor.

    Note the current that flows in the transformer (V1) and diodes. Without the inductor, the ratio of peak to average current in the transformer is very high (6:1), and require a huge derating factor. With the inductor, the transformer current is almost a square wave with a 1:1 peak to average...
     
  3. alfacliff

    Well-Known Member

    Dec 13, 2013
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    look up "smoothing choke" and "swinging choke" for choke type filters. the shunt caps and series inductors work together to smooth the output of the rectifier, these days designers just use a "brute force" filter with a lot of capacitance for the filtering.
     
  4. MaxHeadRoom

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    Jul 18, 2013
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    One down side of increasing the capacitance to achieve a less than required percentage ripple when the supply is operated at maximum load, the VA rating of the transformer has to be increased, due to the higher energy of the charge pulse.
    Max.
     
  5. alfacliff

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    also, the initial surge when power is first applied is increased. soft start is needed.
     
  6. MrChips

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    Also the peak diode current increases dramatically when you increase the capacitance of the reservoir capacitor.
     
  7. big_bird24

    Thread Starter New Member

    Nov 7, 2009
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    I did not realise that although the capacitor can help reduce ripple voltage (and by affect reduce the ripple current in the resistive load), its downside is that it generates source/diode current spikes on the line side because the diodes now can no longer conduct until the a.c source voltage is greater than the capacitor voltage. I neglected to consider the diodes only conduct when there is a forward voltage drop.

    When the diodes do finally conduct in Mike's simulation without the inductor, why are there such high current spikes (7A) through the diodes?
    Does this have to do with the effects of the transformer? If so i will read up more on that effect.
     
  8. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    You seem to be looking at the whole design problem in a very limited way. You are mostly looking at the LOAD and asking questions about the effects of different topologies on the LOAD, when really you should be concerned with both what happens on the load side AND what happens at the output of the transformer or even just the plain old line itself (designs that dont have transformers).

    In particular, the current though the inductor, not just the current in the load.

    If the design has sufficient capacitance then the output will be a fairly constant voltage, and with resistive load that means fairly constant current too.
    But meanwhile the inductor is smoothing out the current through the diodes so that the peak current is not so high, and also the peak current in the capacitor is no so high too.

    So the inductor also "smooths" but when we say that it is more about the smoothing of the current getting to the capacitor than about current getting to the load. The load is fine with the capacitor, but the capacitor is not as fine without the inductor. The load doesnt care, but the cap does care about the current getting to it from the diodes/inductor. Capacitors heat up and burn out or explode when they get too much ripple current, and a higher ripple current for the cap usually means a higher price for the cap or the need to use multiple caps instead of just one.
     
  9. MrChips

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    The diodes conduct only when the transformer voltage exceeds the load voltage, i.e. when the diode anode voltage exceeds the cathode voltage.

    [​IMG]


    The charge consumed by the load must be replenished during the time the diode conducts. As you increase the capacitance of the reservoir capacitor, the phase angle during which time the diode conducts diminishes and hence the diode current increases.

    For more on unregulated power supply design, see this:

    http://www.zen22142.zen.co.uk/Design/dcpsu.htm
     
  10. MaxHeadRoom

    Expert

    Jul 18, 2013
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    In the dia shown, the 're-charge' pulse becomes narrower and higher amplitude as the load current approaches maximum, this has the effect of placing extra load on the transformer and in some cases where higher values of capacitance are used to reduce the ripple, can double the VA requirements of the transformer that would normally be calculated for the maximum load current itself.
    Max.
     
  11. MrChips

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    Good point. The peak diode current will also be limited by the internal resistance of the windings of the transformer, i.e. its VA ratings.
     
  12. wayneh

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    Sep 9, 2010
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    I was looking at the price of inductors, for instance a 0.5H inductor as shown in Mike's #2 post. For a current of ~1A, Mouser has just one and it costs $76.50. (albeit 1H at 1A)

    I don't see how that saves money?
     
  13. MrChips

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    Yes, an inductor with those values will be as big as your power transformer.
     
  14. MikeML

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    Oct 2, 2009
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    I still have a KiloWatt HF RF Amplifer that uses a "swinging choke" between the rectifier and the HV filter capacitor in its power supply. The choke weighs about 50lbs. It has an inductance of ~12H at low DC current. Its winding will stand off 6kV to its own frame.

    Wonder what I would have to pay for it today?
     
  15. #12

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    Nov 30, 2010
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    Yes, wayneh, inductors are expensive and capacitors are cheap. A hundred years ago, that wasn't true. That's why there were so many transformers in early 1900's designs. I find the history of how thin film technology made inductors almost obsolete very interesting. Now that we have high speed switching power supply designs, inductors have a new reason to exist as a practical solution.
     
  16. wayneh

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    I'm wondering how you specify the inductor in a circuit such as Mike posted. Is "Ipeak" of an inductor really for a single peak transient, a level of current you can never, ever exceed? Or does that spec mean an average current over some short period of time, say 100ms. There is a huge difference in the cost of the inductor between those two interpretations!

    My little 1N4007 diodes can tolerate a 30A surge for 8ms. That's a lot more than the 1A continuous rating. I'm wondering how inductors behave.
     
  17. MikeML

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    Note that rectifiers have three ratings:

    1. Non-recurrent Peak Inrush rating.
    2. A cycle-by-cycle ripple rating
    3. A DC rating.

    Most folks overlook 1 and 2 when choosing rectifiers for a no-inductor, capacitor-input power supply.
     
  18. wayneh

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    What I was after was how to add an inductor to the typical old wall wart configuration of transformer + rectifier + filter cap. I'm in the middle of designing a PS for my EL tape project and it employs a large cap to stabilize the HV rail. I modeled the added inductor in LTspice and I can see the point you made earlier; it dramatically reduces peak currents. So I went looking for an inductor. But everything inductor I find that might make a difference is more costly than the entire wall wart. I guess the lack of an inductor in every wall wart I've ever seen tells me the answer - it's not economical. Just trying to reconcile all this.
     
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