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DC power jack

Discussion in 'General Electronics Discussion' started by vick5821, Feb 4, 2012.

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  1. vick5821

    vick5821

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    Jan 22, 2012
    [​IMG]

    For this DC power jack, how do I connect them ?
    It has 3 legs and what does each legs means ?

    First time dealing with this :)
     
  2. Resqueline

    Resqueline

    2,848
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    Jul 31, 2009
    The rear connection is the center pin. It's usually positive for chargers and negative for power supplies.
    The front connection is the circumference or outer sheath (shield).
    The side connection is in contact with the front connection - until the jack is inserted, then it "breaks" and becomes isolated.
    The side connection is usually connected to the (dry cell) batteries in the equipment, disconnecting them when using external power..
     
  3. vick5821

    vick5821

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    Jan 22, 2012
    Any source for me to have a look ? I tried search didnt manage to get :(
     
  4. vick5821

    vick5821

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    Jan 22, 2012
    If I didnt use battery, I can just connect two legs?
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Yep.

    It pays to double check the polarity with your choice of power supply plugged in.
     
  6. vick5821

    vick5821

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    Jan 22, 2012
    Oh..so if I wan to connect 2 legs only, so one legs is positive and one leg is negative ?

    The two side legs either one can be used ?
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    No

    However a wise man would verify this with his multimeter and not need to ask the question.
     
  8. vick5821

    vick5821

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    Jan 22, 2012
    Ok..I am not wise enough then.thanks :)
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    OK.

    Then do this...

    1) plug in your power supply

    2) use your multimeter to determine which connection is +ve and which is -ve

    3) Use these 2 connections.
     
  10. vick5821

    vick5821

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    Jan 22, 2012
    Cool..thanks :)
     
  11. vick5821

    vick5821

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    Jan 22, 2012
    Hey, I plan to do a circuit where I get the power supply from the wall and I connected it to a dc power jack, then I output it to two wire where the two wire can supply the reduced volatge from the adapter.How would be the circuit be ? Any capacitor needed ?
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    OK, there are a number of things we need to know.

    The first is, what sort of plugpack is it?

    If it's one that contains a transformer, then the output can be both variable under load, and poorly filtered. In that case you may require additional filtering and regulation.

    If its a switchmode device (as almost all modern ones are) then the output voltage is likely to be well regulated,

    If you can't tell, the differences are:

    1) a plugpack with a transformer will feel like it has something really heavy in it, a switchmode power supply will not.

    2) a switchmode power supply may operate over a wide range of voltages (e.g. 90-260V) whereas one with a transformer will not (with a switch it *may* operate from one of two distinct voltages, e.g. 110V or 220V)

    3) When measuring the voltage output by them, a switchmode power supply wil give almost exactly the same output from zero (or very nearly zero) load through to the maximum load. A plugpack with a transformer in it will exhibit a significantly higher voltage under no load conditions that will reduce gradually as load is increased. The rated voltage may not be observed until nearly a full load is present.

    So, what do you have?
     
  13. vick5821

    vick5821

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    Jan 22, 2012
    I think I have an unregulated adapter.Means plugpack.
    Link : http://cytron.com.my/viewProduct.php?pcode=AD-TMC-500PM
     
  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    If that weight is correct, at 24 grams it couldn't be anything other than a switch-mode regulator.

    24 grams? really?
     
  15. vick5821

    vick5821

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    Jan 22, 2012
    However, it stated there unregulator power supply and I tested it, I swtich to output 7.5V but from what I measure the voltage higher than 7.5V.

    Yes 24g
     
  16. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Well, what is it? Saying it is higher than 7.5V tells me nothing more than it is between 7.50000000000000001 and infinity volts.

    Also you will note that I didn't say that a SMPS would always give a regulated voltage off load, however with a small load it will get to its regulated voltage very quickly. Also note that component tolerances may mean that the actual regulated voltage may vary by 5% or so (although it's normally less).
     
  17. vick5821

    vick5821

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    Jan 22, 2012
    So now how ? Should I use capacitor in my circuit ? I think no harm :)
     
  18. vick5821

    vick5821

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    Jan 22, 2012
    Hey there,
    I plan to build a circuit where I can power it either using battery or sources from the wall through adapter which inserted into DC power jack.

    However, can anyone tell me how will be the circuit be ? Is it just extend the circuit to have two supply terminal ?

    Thank you.

    [​IMG]
     
  19. jackorocko

    jackorocko

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    Apr 4, 2010
    You don't need a capacitor if you are using a battery because there is no ripple.

    With an SMPS, you will only need filter caps if the output ripple is more then your circuit can safely work under.

    As long as your two sources are very similar and you want to use a dc adapter, then you could wire a male dc adapter to the battery to plug into the female dc adapter on the board. Then you only need one connector on the board. There is a few ways to skin a cat...
     
    Last edited: Feb 6, 2012
  20. vick5821

    vick5821

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    Jan 22, 2012
    I do not get what you means..Is it means that from the power jack, one is positive terminal which shared by both the jack and the battery and the pin 2 is the negative for battery and pin 3 is negative for jack and from supply ?
     
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