How come plugged in transformers don't consume power?

Discussion in 'Power Electronics' started by Amit Ika, Jul 4, 2016.

  1. Amit Ika

    Thread Starter New Member

    Jul 4, 2016
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    Hey guys,

    This has boggled my mind for a bit: How come a plugged in transformer does not consume power? Here's a link to a circuit diagram of a phone charger:
    http://circuitdigest.com/sites/default/files/circuitdiagram/Cell-Phone-Charger-Circuit.gif
    Obviously as its plugged in, the primary coil circuit of the transformer is completed with the wall which means ac power is flowing through it regardless of whats going on with the secondary coil. What am I missing here? Is it consuming power if it is plugged in and not used? Any clarification would be great!
     
  2. MaxHeadRoom

    Expert

    Jul 18, 2013
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    Of course it consumes power.
    The primary or other SMPS device is presenting a load, albeit small to the service power supply.
    Max.
     
  3. Amit Ika

    Thread Starter New Member

    Jul 4, 2016
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    So then how does a plugged in phone charger that is not connected to a phone not consume power? I mean plugging it in to the wall is basically shorting the circuit. Is it not?
     
  4. crutschow

    Expert

    Mar 14, 2008
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    It most definitely is not.
    A transformer with no load is not a short, instead it presents a high impedance (high inductance) to the line.
    Why would you think it's a short?
    For an ideal transformer the input power equals the output power so, no output power equals no input power.
    A real transformer has some small loses in maintaining the magnetizing inductance so it will draw a small amount of power (typically a few percent of its rated power).
    Suggest you read up on how transformers work.
     
  5. BR-549

    Well-Known Member

    Sep 22, 2013
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    They do consume power. I recall a study, that concluded saving a ridiculous amount of power by unplugging wall-warts.
     
  6. dannyf

    Well-Known Member

    Sep 13, 2015
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    an ideal transformer is essentially an impedance transformer: it "maps" the secondary windings impedance onto's the primary side. An unloaded transformer has an infinite impedance on its seconday winding. Thus infinite impedance onto the primary winding -> no power consumption.

    In reality, a transformer isn't perfect so it does consume a little bit of power, even when unloaded.
     
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  7. Amit Ika

    Thread Starter New Member

    Jul 4, 2016
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    The reason I think it's a short is because one end of the primary coil is in contact with one terminal of the outlet and the other end of the primary coil is in contact with the other terminal of the outlet. I understand that there is an impedence but is that really enough to prevent the power consumption of a short? Also this then means that a primary coil with low impedence will act as more of a short. Is this not true?
     
  8. nsaspook

    AAC Fanatic!

    Aug 27, 2009
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    How about a tiny bit of deductive reasoning. Transformers are plugged into the power grid everywhere, the power grid has not been shorted out because of these transformers. So what would make you think an unloaded transformer is a short?
     
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  9. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    Actually, for a very short period of time (microseconds) the transformer is just a piece of copper wire.
    Once current flow starts a magnetic field is established in the iron core.
    This creates a "resistance" to current flow. This kind of resistance is called impedance.
    It is the magnetic field that prevents the short circuit you are envisioning
     
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  10. hp1729

    Well-Known Member

    Nov 23, 2015
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    It does consume some power, just no where near as much as when in use. The key is the characteristic impedance of an inductor. Impedance changes with the density of the magnetic field which depends on the current. As it starts drawing current impedance decreases so primary current increases. (in 100 words or less)
     
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  11. DGElder

    Member

    Apr 3, 2016
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    It is not a short for the same reason an inductor across and AC voltage is not a short. The current into the coil is inversely proportional to the frequency of the AC signal and the inductance, 1/(2*pi*f*L). Since a transformer has a very high magnetization inductance the current through the primary coil, with the secondary coil open, is quite small. And most of the energy that goes into the primary is returned to the mains since an inductor can not dissipate energy, just momentarily store the energy and return it to whence it came. There will be small IR losses due to resistance in the wire, magnetization of the core and eddy currents, so a few watts will be lost as heat.
     
    Last edited: Jul 4, 2016
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  12. Amit Ika

    Thread Starter New Member

    Jul 4, 2016
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    Thank you all for your replies! This makes a lot more sense now.
     
  13. nsaspook

    AAC Fanatic!

    Aug 27, 2009
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    To the OP please read this:
    http://www.allaboutcircuits.com/tex...chpt-9/mutual-inductance-and-basic-operation/
    [​IMG]
     
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