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Interpreting a transformer diagram

Discussion in 'General Electronics Discussion' started by Moschops, Apr 4, 2010.

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  1. Moschops

    Moschops

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    Apr 4, 2010
    Mispost to get to posting images!
     
  2. Moschops

    Moschops

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    Apr 4, 2010
    Not one of these - [​IMG]

    But this: [​IMG]

    Not a great image, I know. That's photocopies of photocopies for you. I'm unfamiliar with anything but the simplest of transformers and I'd like to check that I'm interpreting this correctly. To me, it appears to be a transformer plugging into a 220VAC/50Hz wall socket (i.e. UK mains) and presenting eight output terminals.

    The bottom two terminals are simple enough - they must have a potential difference of 190V AC (presumably rms) when an rms current of 5.5 mA is drawn from them. I'll drop saying rms from here, but I mean rms in every case.

    The others I'm not so sure of - 6 and 7 must be able to provide 0.650A of current at 10.25V AC, but so must 7 and 8. Why are there two options (i.e. 6 & 7, or 7 & 8) for the same voltage, able to source the same current? Is the idea that I could also connect to 6 and 8, and in doing so draw current at 20.5 V AC? If so, how much current does the diagram dictate those terminals must be able to source?

    Terminals 3, 4, 5 are the same again with different values. Any explanation anyone can provide would be gratefully received.
     
  3. jackorocko

    jackorocko

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    Apr 4, 2010
    afaik, one of 3 & 4 and one of 6 & 8 must be negative voltage. 4 & 7 will be common ground? zero volts.
     
  4. Moschops

    Moschops

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    Apr 4, 2010
    Firstly, do you mean "one of 3 & 5"?

    Negative compared to what? Do you mean that terminal 8 must be negative whatever terminal 6 is, such that when terminal 6 is at +10V terminal 8 will be at -10V, both measured relative to terminal 7? I ask for clarification as the simple concept of negative and positive gets a bit more interesting with AC.
     
  5. Resqueline

    Resqueline

    2,848
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    Jul 31, 2009
    jacko is presumably thinking about instantaneous voltages, and this is actually a correct way of presenting the issue. When 6 is positive then 8 is negative.

    Between 6 & 8 you get 20.5V @ 0.65A - with a centertap 7 (which is usually grounded).
    The reason behind using centertapped windings is that you get away with only two diodes to get a full wave rectifier, which also translates to better efficiency.
    A full wave bridge rectifier and two capacitors will furthermore give you a +/- supply (dual voltage) which is often what is used in amplifiers.
     
  6. Moschops

    Moschops

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    Apr 4, 2010
    Okey dokey, that's all very helpful.

    To recap, then, 4 and 7 are probably some kind of common ground, and in either case are a reference voltage we can call ground or zero for measuring the voltages at the pins adjacent to them.

    At any given time t, using pin 4/7 as a zero reference voltage, V3 = - V5 and V6 = -V8

    Furthermore, measuring the voltage (for a time period, obviously, rather than at an instant) across pins 3&5 (not the ground) will reveal an AC voltage of whatever that badly printed value is, and will remain that voltage with a current draw of 0.3A.

    Measuring the voltage across pins 6&8 (not the ground) will reveal an AC voltage of 20.5 volts, and will remain that voltage with a current draw of 0.65A
     
  7. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    Yes, that's all correct. (Pin 3 to 5 will give you 2x18V=36V since they are series-connected/wound and therefore 180 degrees out of phase with each other.)
    Likewise, for example, if you have a transformer with one 6V winding and one 18V winding, you can get either 12V or 24V depending on the way you series-connect them.
     
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