Diode Capacitor

Discussion in 'General Electronics Chat' started by qitara, Nov 25, 2013.

  1. qitara

    Thread Starter Member

    Jan 18, 2013
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    Hi guys

    Would be great if some one could explain what is going on here

    I made this diagram so it would be much easier to understand what i got in front of me


    [​IMG]


    I am feeding it with 220VAC and getting 309VDC out, How is the voltage stepped up to 309V ?
     
  2. MrChips

    Moderator

    Oct 2, 2009
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    220VAC specification is RMS value.
    Multiply this by 1.41 to get the peak value.
    Subtract the diode drop.
     
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  3. tubeguy

    Well-Known Member

    Nov 3, 2012
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    (US voltages shown)

    [​IMG]
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    MrChips has answered your question quite well. I would just like to add that you need to be careful playing around with mains voltages. It only takes one slight mistake and you will find yourself just part of a number in this year's vital records statistics.
     
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  5. qitara

    Thread Starter Member

    Jan 18, 2013
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    The peak voltage makes sense, but why is my voltmeter showing 309 V and not 220 V as the feed voltage ?


    Would be very thankful if this could be explained in more details :)


    I thought that this setup could be used as a rectifier, i mean instead of using 4 diodes, the three diodes could be replaced with a capacitor


    Enlighten me
     
  6. qitara

    Thread Starter Member

    Jan 18, 2013
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    Did a run trough the mains line two times, now i am running 60V trough a power supply

    much safer this way :D
     
  7. tindel

    Active Member

    Sep 16, 2012
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    Only 'True RMS' volt meters will display your RMS voltage. Look at the Fluke 87 series if you need this functionality... they can be found for about $100 on craigslist. Your cheep harbor freight or radio shack meter's won't work.

    You have a half-wave rectifier that will charge your cap on the positive half-cycle of you source voltage.
     
  8. tubeguy

    Well-Known Member

    Nov 3, 2012
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    The diode charges the cap to approximately the peak voltage, the cap stores that voltage (it discharges very little) between cycles. If measuring on DC you will get this peak reading with the cap connected.
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    How, exactly, are you measuring "the feed voltage"?
     
  10. qitara

    Thread Starter Member

    Jan 18, 2013
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    I am using a Fluke 17B and a ocilliscope both shows the same result

    The feed voltage is measured at the input leads
     
  11. qitara

    Thread Starter Member

    Jan 18, 2013
    87
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    But talking about RMS voltage, is it still the same as the input voltage ?,
    i am not able to measure the voltage with my multimeter set to AC
     
  12. #12

    Expert

    Nov 30, 2010
    16,257
    6,757
    220√2 is 311 volts. The diode is going to use up about .6 volts so the answer should be 310.527 Volts. 309 volts is less than 1/2 of 1% error.
    I don't see any problem
     
  13. MrChips

    Moderator

    Oct 2, 2009
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    The original intent of the AC VOLTS range of your voltmeter is to measure 50Hz or 60Hz AC sine wave. You cannot measure this with the voltmeter set to DC VOLTS.

    To measure the voltage across the capacitor after the rectifier diode you must set the meter to DC VOLTS.
     
  14. tubeguy

    Well-Known Member

    Nov 3, 2012
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    Yes. The 220VAC input voltage is RMS voltage. The RMS spec. is left off as it has become 'understood' that a given AC voltage is RMS unless specified otherwise.
    (I'm not sure if spec. is the proper term..:rolleyes:)

    The RMS spec. is used because (for example) it results in the same brightness with a lamp load as does the equivalent DC voltage.
     
    Last edited: Nov 25, 2013
  15. qitara

    Thread Starter Member

    Jan 18, 2013
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    0
    Got it clear now thanks to you guys.

    Is it possible to use this setup as a rectifier ?
     
  16. vk6zgo

    Active Member

    Jul 21, 2012
    677
    85
    Let's recap for a moment:-


    RMS voltage:-
    Back at the dawn of the Age of Electricity,the early experimenters amassed a lot of knowledge about DC voltages & currents.


    There were accurate & repeatable methods of measuring power,such as finding how long it took for a known volume of water to be raised in temperature a given number of degrees C (or F,or K).


    The heat source could be a gas flame,coal fire,etc,or in this case,a resistive element.
    This allowed the verification of the formulas :- P= VI & P = Vsquared/R.


    With the advent of AC,it was logical to use the same method to measure the power available from a sine wave.

    With P now known,& R known,it was convenient to quote an “effective voltage”,or what DC voltage would be needed to produce the same power.

    In Mathematics,“finding the area under the curve” was a common process,using the “root mean square” method.

    Knowing the characteristics of a sine curve,the “effective voltage”(Vrms) of an AC sinewave was determined to be:-
    Vpeak/√ 2,or approx 0.7071 Vpeak



    Inversely,knowing Vrms,the value of Vpeak can be calculated as Vrms/0.7071.or 1.414 Vrms.


    Meters:-
    Early analog AC meters mainly read average values of voltage & current,with a correction factor added to the scale to show RMS.

    Many DMMs also read average values,with a hardware or software correction factor added. after
    the analog/digital conversion & calculation is performed.

    “True RMS” meters calculate the RMS values “on the fly”,& hence show the real RMS value.


    OP's circuit:-

    This shows a 220v RMS AC supply,halfwave rectified ,followed by a capacitor C across the output line.
    When the rectifier conducts on the positive half cycle,C will begin to charge,until if it were an ideal component,it would reach V peak which is 1.414 Vrms.

    In real life,it reaches 1.414 Vrms- Vd ,where Vd is the drop across the rectifier diode.
    It is still “very close to”Vpeak.


    As the half cycle reduces to zero & reverses in polarity C cannot discharge via the diode as it is reverse biased,so its charge remains at very close to Vpeak.


    Your DMM is very high impedance so ,it does not appreciably discharge C,so reads again ,very close to Vpeak


    The DC reading (309v) you obtained is very close to V peak-Vd,for a 220v AC source.

    As I said,this is really just a recap of what all the other posters have said,many of which could probably have phrased the above better.



     
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  17. qitara

    Thread Starter Member

    Jan 18, 2013
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    Fantastic explanation TX


    But does this setup do the job as a Rectifier ?
     
  18. vk6zgo

    Active Member

    Jul 21, 2012
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    Yes,but!

    This circuit is a half-wave rectifier,so you only get to use one half cycle of the applied AC supply.

    As soon as you apply a load to the circuit,C will discharge between each rectified half cycle,so the DC at the output will have a 50Hz "ripple" superimposed upon it.

    To minimise this ripple,you need to increase the capacitance of C.

    If you use a full wave rectifier,you are using both half cycles of the applied AC supply.
    The "ripple" in this case will be at 100Hz,so the value of C can be lower than with the half wave rectifier circuit.
     
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  19. WBahn

    Moderator

    Mar 31, 2012
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    I would expect your scope to show ~309 volts as the amplitude (somewhere around 622V peak to peak). I don't know why or how the DMM is reading peak voltage. What does it read if you remove the diode an measure the voltage just across the source?
     
  20. WBahn

    Moderator

    Mar 31, 2012
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    But this only explains it (and I think it IS the explanation), if the OP is measuring the voltage at the output (as labeled in his diagram) and not the input, as he states.
     
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