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Can I just simply replace inductor with a choke?

Discussion in 'General Electronics Discussion' started by xelatay, Aug 30, 2012.

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  1. xelatay

    xelatay

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    Aug 30, 2012
    Hi,

    I currently doing my final year project regarding LED driver. In my design I required 300 mH (at least 0.5 A rate current) inductor to smooth current after the full diode rectifer. (hopefully u guys can understand I try to import schematic but I don't know how to do it)

    Now I facing a problem with this inductor, it is hard to get normal 2 pin inductor (I know there are a tranformer type inductor with has high inductance and high current but I try not to use them because they are huge). After some research I found out that inductor can simply replace by a choke. Therefore I bought B82733F2701B001 which is 100 mH, 0.7 A choke from EPCOS http://www.farnell.com/datasheets/1465589.pdf. I place 3 of them in series to produce 300 mH. Currently I only connected from pin 1 to pin 2 of the choke.

    After connecting three choke in series I measure the inductance of the three indutor from a LCR meter and I got around 300 mH. Then I happily turn on the power supply and think that it going to work but load current wave form does not get smoothed. P.S I try the same circuit by using an inductor bank and set the inductance to 300 mH and I got a quite reasonable waveform (small current ripple).

    Summary:
    1) Can a 4 pin choke act as normal inductor?
    2) How to connect B82733F2701B001 form EPCOS such that it will act as an inductor or can it be an inductor?

    Thanks for the help. I am just an beginner.
     
  2. john monks

    john monks

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    1
    Mar 9, 2012
    1) Yes but you must keep the DC low.
    2) Connect the coils so they are aiding, wound the same way. Either 1 & 4 or 2 & 3 should be hooked together. the other pins are what you use.

    According to the datasheet the inductance will decrease with a DC bias current.
    You must analyse the datasheet carefully to see if the decrease is within acceptable limits. (0.5amps is a lot)

    Frankly I would be worried about this. Your inductance may come in too low.
     
  3. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    With these mains filter chokes, don't you have to have equal and opposite currents flowing in the two halves? If you don't, I think the core will saturate at a pretty low current.

    That core must have a very high permeability to get 100 mH from what looks like only a few dozen turns of wire on each side. That's why I suspect that those mains chokes rely on the equal and opposite current flow to prevent saturation.

    If you have an LCR meter, try measuring the inductance while you vary the DC current through it. That should tell you pretty quickly how much current it will take before it saturates.

    Disclaimer: I don't know much about these mains chokes. YMMV.
     
  4. Electrobrains

    Electrobrains

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    Jan 2, 2012
    Kris is right. Such current compensated chokes are mainly used to filter off high frequency signals from the mains voltage.
    Both conductors (N and L) need to be brought through the choke (in the right direction) and carry the same current (could also be other supplies, eg. DC plus and minus conductors).

    The choke functions as a sensitive transformer.
    The "current compensation" means the normal current through N and L will balance out each other and you will basically have no magnetic field when the current is totally symmetrical.
    That's why you can keep them so small, with high permeability cores. The specified current of those chokes is the max allowed current flowing in both of the windings at the same time.

    If a (small) non symmetrical signal will appear on the choke's winding, it will meet the full inductance and be blocked. The test signal for the inductance in the data sheet is "10 kHz, 0.1 mA,"!. Probably a much higher current than that would allow the core not to saturate, but your 0.5A is obviously far too much.

    I don't understand why you are using inductors to smooth the LED current... Normally, if even necessary to smooth, you would use capacitors.
     
  5. xelatay

    xelatay

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    Aug 30, 2012
    Thanks for the help guys.

    Electrobrains I used inductor to smooth current instead of capacitor because we know that LED had long life time but capacitor ussually had much shorter life time. I don't want my LED driver fail before the LED.

    By the way, I want to ask is that possible to smooth current using capacitor? I always though capacitor smooth voltage and inductor smooth currennt. I might be wrong because I am new in electronic and this is my first time building stuf.
     
    Last edited: Sep 1, 2012
  6. Electrobrains

    Electrobrains

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    Jan 2, 2012
    I am sure your capacitor will outlive your LED, if it's well dimensioned!

    Without looking this up, I think the rated hours of an electrolytic capacitor, is for the maximum temperature (usually 85 or 105°C) at the maximum ripple current.
    Roughly, for each 10°C (?) lowering of the ambient temperature, the "life expectancy" of the capacitor doubles (I once looked deeply into this while developing a high temperature power supply for an oven). Using a lower ripple current than maximum will additionally lower the internal temperature and expand the "life expectancy".
    I put in those "", because the "life expectancy" doesn't mean it will stop functioning after those 2000 hours or whatever is specified.
    Usually an electrolytic capacitor with time dries out from the heat and thus looses capacitance. The easy way to overcome that problem is to over dimension a bit.

    Normally, when you "smooth current" of an LED (or whatever), you give it a constant voltage and add a known resistor in series. That will reduce your circuit to a fraction of the size (and price) of one with an inductor.

    (why smoothing at all? your eye is very slow and usually no smoothing is necessary for rectified 50/60Hz signals)
     
    Last edited: Sep 1, 2012
  7. xelatay

    xelatay

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    Aug 30, 2012
    Great Electrobrain u are able to get the whole picture of my project.

    The aim of my project is to build LED driver to drive LED that will use in VLC (Visible light comunication). I am not sure u hear about it of not basically VLC is sending "1" and "0" through light by switching on and off the light throught some sort of modulation. If my LED current is not constant I will not get a threshold for the "1" and "0". I am sure whether VLC can work even with constant light output anyways it is not my part of the project.

    Besides constant current, I have to keep the efficiency of the driver as high as posible. That why I try to use passive component. Lets say I use 100 ohm resistor place in series with my LED and the current across it is 0.5 A. There will be 50W lost throught the resistor. It is not practicle to do that since nowdays we change to LED because the consumed less power and we have the driver that consumed large power.
     
  8. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Current is measured THROUGH a component, not ACROSS it. Also, if you pass 0.5A through a 100 ohm resistor, it will dissipate 25 watts, not 50 watts.

    No one is suggesting that you use a 100 ohm resistor. I would use a few ohms, if your application is anything like what I imagine it is.

    I think you're fighting a losing battle if you want to avoid using ANY smoothing capacitor.

    I think you should tell us a lot more about the project and provide a schematic for what you have so far. We don't even know the voltage souce or the ripple frequency. And we can definitely advise you on driving an LED with a data signal.
     
  9. xelatay

    xelatay

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    Aug 30, 2012
    Ops sorry I am wrong. I forget to square the current when calculating the power dissipation by the resistor. Kris u are right it is 25 W and not 50 W.

    Now I can clearly see two path to go:

    1) Continue using inductor and lead to no where unless I get the huge and heavy inductor which will definately increse the size and the weight of my lamp by >100%.

    2) Use smoothing capacitor (of much larger size as suggest by electrobrain) and resistor. This will cause the driver to loss it efficiency for sure.

    FYI:
    The power source I use is from the main power supply (240 VAC / 50 Hz). I had a transformer to step down the input voltage to 15 VAC before the diode rectifier. Currently I able to switch on the LED with a current ripple of 100 Hz. The 100 Hz of switching cannot be seen by the human eye.
     
  10. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    A load current of 0.5A from a 15VAC mains transformer is not much.

    If you need a clean supply rail, have you considered using a regulator to remove the ripple?

    You really are wasting our time by feeding little bits of information one at a time. Please post a schematic that shows everything you plan to do, and please explain your application in detail. That way, we can avoid "playing tennis" with you.
     
  11. xelatay

    xelatay

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    Aug 30, 2012
    Sorry for wasting u guys time, Here is the circuit schematic.

    design 1.png
    (rev2)

    My project main objective is a built a intelligence ligthing for SMART home. The driver much has high reability and high efficientcy (>75%). That is the reason why i am avoiding active component and capacitor (since I know that electrolytic capacitor had short life span but accoring to ElectronicBrains electrolytic capacitor can had long life span if i design correctly by keeping the temperature low and low ripple).

    It is an extra bonus for me if I able to design a constant current source for my LED which later on VLC can be implemented on it.
     
    Last edited: Sep 1, 2012
  12. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Thanks for that. That gives us some idea what you're talking about.

    Yes, electrolytics do eventually dry out and fail, but in a design like this it's going to be difficult to avoid them. When you're working with mains frequency, you need to "fill in the gaps" in the waveform somehow (assuming you don't want a pulsating light), and this requires some kind of energy storage; capacitance is the easiest to deal with. Yes, keeping them cool, using over-rated components, and avoiding excessive ripple will all help them last longer.

    There are alternatives to aluminium electrolytics but they are expensive. I suggest you get something working first, then look at improving the circuit's lifespan.

    Your circuit seems to be driving a string of series-connected LEDs from a source that is derived from a mains transformer. There is no smoothing, which means the illumination will not be steady. As you said, the flicker isn't visible to the naked eye, but will affect any video recordings and camera pictures you take, and it would prevent you from modulating the light to transmit data, so I strongly suggest you power the LEDs from DC.

    I would give up on the inductor idea completely. Others here may be able to see a good reason to stick with it, if you can find an inductor with enough inductance that can also handle the current, but I think it's a red herring.

    The first thing you need to do is read Steve's article on driving LEDs: https://www.electronicspoint.com/got-question-driving-leds-another-work-progress-t228474.html

    Now, can you explain what the box marked "ADC1" and the circle with an "A" inside it marked "Valley fill circuit" are supposed to do, and what they will contain?

    When you say "valley fill circuit", are you referring to "filling in" the valleys between the bumps in the rectified mains waveform? If so, your "valley fill circuit" is a large-value capacitor (e.g. at least 1000 uF), an electrolytic, connected across the output of the bridge rectifier.

    You can't just connect a string of LEDs across a voltage. You need some way to limit or regulate the current, otherwise they will lose their magic smoke. Please have a careful read of that linked article and try to understand how LEDs behave and why you need to control the LED current.

    You mentioned using the LEDs to communicate information. Is that a later stage of your project? Are you planning to transmit this data using the same LEDs that are used for illumination? What kind of device would be receiving this data?

    What LEDs are you using? Do you have a part number for them? Can you link to a data sheet for them?
     
    Last edited: Sep 1, 2012
  13. Electrobrains

    Electrobrains

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    Jan 2, 2012
    Just a hint: If you want high efficiency, small size and low cost, I suggest you buy a finished SMPS (Switched Mode Power Supply) to supply that DC voltage.
    It's (sadly) not worth building them yourself. You just need an output voltage that is a little higher than the total voltage over the LEDs (at the desired current).
    With a low value resistor in series with the LEDs, you then set the current.

    If you need more accurate current regulation (which I doubt), you could use a simple 3-leg Regulator-IC as a current regulator. See for instance this LM317 example with calculator. The LM317 needs a few volt supply over that circuit to work properly. If you really look for efficiency, you could exchange it with a Low-Drop-Regulator and reduce the supply voltage.

    You can also buy SMPS Current Supplies very cheap. Here is an example (not applicable for you) of a mini supply that was included with a small IKEA LED lamp. The lamp cost me only a couple of €, so I bought several and use the power supplies as both current and voltage sources. I didn't measure, but I suppose such unit has between 80 and 90% efficiency.

    With the above suggestions you don't have to worry about capacitors or smoothing.
     

    Attached Files:

    Last edited: Sep 1, 2012
  14. xelatay

    xelatay

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    Aug 30, 2012
    The post by Steven regarding the LED is great.

    I had edit my previous picture. The A symbol is the Dc ampmeter and AC ampmeter use to measure DC current and current waveform respectively. As for the valley filled circuit I had remove it from my design because it does not filled the valley well. It suppose to look the same as http://www.element14.com/community/thread/1728.

    The led I used is from CREE (XREWHT-L!-0000-00C01) http://www.cree.com/~/media/Files/C...dules/XLamp/Data and Binning/XLamp7090XRE.pdf. It rated for 1 A but I current only allow half it rated current across it to reduce the heat and increase the lifespan.

    As for the VLC it is not part of my project. My job is to build a LED driver.

    As for Electrobrain suggestion, SMPC I think is flyback converter. I had look into them and it is too complex. Sadly I am not allow to buy the current supply I had to build them myself. I do not know much on how to use voltage regulator (LM 317) to smooth the rectified current. I will look into that.
     
  15. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    If he's so concerned about reliability, a mains-powered SMPS is a bit of a liability. They do tend to go pop in response to voltage surges. Of course if he's aiming for high efficiency, it's the obvious answer, so there's a conflict there. Perhaps there are mains-powered switching supplies that are deliberately over-engineered and designed for high reliability?

    Also for best efficiency and quality, power factor correction would be wanted, but this adds another potential point of failure in the project. As usual, tradeoffs must be made somewhere.

    Interesting. I'd never heard of that circuit before. But it seems to be a real circuit - it's described on Wikipedia: http://en.wikipedia.org/wiki/Valley-fill_circuit

    It's a very crude type of power factor correction. It reduces the ripple at the output of the bridge rectifier from 100% to 50%. I don't think it's very useful for this project. You would want a smooth DC voltage to drive the LEDs. If you use a linear voltage or current regulator with an input with 50% ripple, it will waste a lot of power. The proper way to do PFC is before the main SMPS.

    Thanks for the information.
    Right... but you want your LED driver to be able to accept a data signal input, so that in future, it can be used to transmit data on the LEDs, right?
     
  16. szhighstar

    szhighstar

    6
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    Aug 7, 2012
    Hi, you do not need buy three inductors, it is trouble fr you, a inductor is ok, select high permeability core, or EI lamination configuration.
     
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