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How can µF-range capacitors reduce AC voltage?

Discussion in 'General Electronics Discussion' started by gsrilanka, Jan 27, 2015.

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  1. gsrilanka

    gsrilanka

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    Jan 27, 2015
    i need 220V AC get into 5v DC .
    i found circuit on Google , it used 1uf capacitor to do it.
    how that work?
    [​IMG] :)
     
  2. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Hi there and welcome to Electronics Point :)

    That circuit is called a capacitor-fed power supply. The reactance of the capacitor converts the AC voltage to a current that charges the reservoir capacitor and powers the LEDs.

    These power supplies are not isolated and are an electric shock hazard. They may only be used when the entire circuit, including the load, is fully isolated and contained in such a way that it can't make contact with anything conductive, and especially, that no person can touch it.

    These power supplies also have a fairly limited output current capability. They are not suitable for general use. They are only usable in specific situations where the whole product is fully contained and isolated, and there are no external connections to it.
     
    Harald Kapp, gsrilanka and Arouse1973 like this.
  3. duke37

    duke37

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    Not a nice circuit.
    There is no need for C1 and if the leds are disconnected, you will get 310V across it - big boom.
     
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  4. gsrilanka

    gsrilanka

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    Jan 27, 2015
    Thank u all.
    duke37 sorry
    Can u explain ur point further.how is it happen?
     
  5. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    duke37 is pointing out two disadvantages of having C1 present.

    The four LEDs in series clamp the votage across C1 to about 12V (depending on the LED type). If the LED path goes open circuit (a failure of any LED or a break anywhere in that path), C1 will try to charge up to about 310~320V DC and if it's rated at 25V, as shown on the diagram, it will explode. In other words, that circuit is not just an electric shock hazard and a fire hazard, it is an explosion hazard.

    Secondly, the circuit will work almost as well without C1 anyway. C1 is there to smooth the LED voltage and reduce the flicker, and increase the average brightness somewhat, but these benefits are small and C1 may not be justified.

    The important thing about this power supply is that it is dangerous because it is not isolated from the mains, and it must only be used in a fully enclosed product with no external connections.
     
  6. Bluejets

    Bluejets

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    Woman in Aus was recently killed by a charger such as this connected to her mobile phone.
    She answered a call with the charger still plugged in, something went amiss and the result was fatal.
     
  7. gsrilanka

    gsrilanka

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    Jan 27, 2015
    ohh
    i got that point.
    we can add C1 100uf 400v capacitor
    then i think its k
     
  8. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    The presence or absence of the 100 µF smoothing capacitor doesn't change the fact that the circuit is not safe, as I described in post #2.

    Edit: Using a 400V capacitor will stop it from exploding if the LEDs get disconnected, but if they get reconnected (for example if there's a dry joint or loose wire), the LEDs will be destroyed instantly.
     
    Last edited: Feb 3, 2015
    gsrilanka likes this.
  9. HellasTechn

    HellasTechn

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    If you add a 100uf 400v capacitor then you make your circuit large (in dimensions) because theese are large capacitors.

    In that case you may as well use a standard transformer and be completely safe !!! standard 220 to 5 volt 100-200ma transformers are not big in size.

    One other solution is a really cheap switching mood power supply.
    1.they are safe
    2.they are really small in size.
    3.they can deliver large current.
     
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  10. HellasTechn

    HellasTechn

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    Apr 14, 2013
    One other thing you have to keep in mind is that theese capacitor-fed psu's also called transformerles psu's can only be used for applications that draw up to 50ma current.

    now on your schematic you could replace C1 cap with a 5 volt zener. (what do you think avout that chris?)
     
    gsrilanka likes this.
  11. duke37

    duke37

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    `C1 is not necessary, many led displays are multiplexed and have a pulsed supply. If you wish to smooth the current in the leds, then a resistor would be needed to run C1 appreciably above the led voltage.

    Putting a 5V zener in place of C1 may not work. The supply is current controlled. If the leds need more than 5V, they will not light, if they need less than 5V then the zener does nothing.
     
  12. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    LEDs do have some internal incremental resistance so the capacitor will have some effect. I ran a simulation to find out, but I had to change "CX" to about 0.47 µF (not "472", i.e. 4.7 nF, as shown in the original schematic, which I assume is a typo) to get significant LED current. LTSpice doesn't have a great range of LEDs. The LEDs I used in the simulation, type QTLP690C, appear to be pretty standard red LEDs.

    I ran the simulation with C1 = 100 µF, then again with C1 = 1 pF (LTSpice doesn't like capacitors with zero capacitance, but 1 pF is close enough!)

    epoint 272459 test schematic.png

    And here's the graph of LED currents, with and without 100 µF of smoothing. You can probably guess that the green trace is with 100 µF and the blue trace is without!

    epoint 272459 LED current graph.png

    In both cases, the mean LED current is 28.4 mA, but the ripple is obviously reduced quite a bit with the capacitor in circuit, so the blinking effect is definitely reduced. Whether or not this is visible will depend on the application.
    Agreed!
     
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  13. gsrilanka

    gsrilanka

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    Jan 27, 2015
  14. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    I can't view your Google Docs page - I don't have permission.
     
  15. Externet

    Externet

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    Hello KrisBlueNZ.
    I have seen schematics like the one on your post #12 many times and places on the web.

    I do not understand why the components D5, D6, D7, D8 are not fully erased/deleted from the circuit; and make D1, D2, D3, D4 the LEDs instead.
    Same for the original poster circuit, no C1 either. And 1 uF may be too much for 230VAC on ~20mA LEDs. I would use 0.1 to 0.33 uF

    Do you have an explanation ? Sort of that is the way I did my house ceilings 'night lights' , the LEDs as light emitting full wave rectifier, and work superb for years across the wall switch.

    And, why is there a grounding symbol ?
     
    Last edited: Feb 4, 2015
  16. Harald Kapp

    Harald Kapp Moderator Moderator

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    The main reason is that only 1/2 of all LEDs would be on per half cycle of the mains sine, thus effectively halfing light output for a given number of LEDs.


    This is required by the simulator as 0V reference point. It has no practical relevance. But voltages are differences in potential energy and since there is no absolute 0V reference potential, you need to supply one to the simulator. You can place the GND symbol (almost) anywhere in the circuit.
     
  17. gsrilanka

    gsrilanka

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    Jan 27, 2015
    [​IMG]
    this is the standard mobile phone charger circuit.
    is this technology differ from above method?
    is there a transformer or chock or inductor in yellow colored component ?
     
  18. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    What Harald said. Each LED would only illuminate for a maximum of half the time.

    Also, LEDs can't be used in a bridge rectifier generally because they will be damaged by a reverse voltage of more than about 5V. But what you're suggesting is a bridge rectifier with its output shorted out; this is really just equivalent to two LEDs in series, and another two LEDs in series, connected in reverse parallel. So that won't exceed the reverse voltage specification.
    I used 0.47 µF and got a mean LED current of 28.4 mA. That's too high for 20 mA LEDs but the application is lighting, not indication, so LEDs are normally rated for more than 20 mA.
    As Harald said, it's only needed for LTSpice, the simulator. Without it, LTSpice goes haywire and throws up all sorts of crazy errors! It definitely does not correspond to mains earth! It's not actually a ground symbol anyway.
     
  19. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Still can't see it. Save the image to your hard disk then upload it as an attachment to your post.
     
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