Page 1 of 2 12 LastLast
Results 1 to 10 of 20

Thread: Buck and Boost

  1. #1
    Join Date
    Oct 2018
    Location
    Kingsport Tennessee USA
    Posts
    2

    Buck and Boost

    I need to boost 208 vac to 240vac what KVA Transformer do I need? The fla is 17,

  2. #2
    Join Date
    Jun 2004
    Location
    Cherry Valley NY, Seattle, WA
    Posts
    4,719
    The manufacturers have online calculators that will size the units. I'm sure you can find it with a quick google. I know square d has one.
    Ethan Brush - East West Electric. NY, WA. MA

    "You can't generalize"

  3. #3
    Join Date
    Apr 2008
    Location
    Ann Arbor, Michigan
    Posts
    7,029
    181911-1414 EDT

    Boggsc71:

    Quite obviously the quick answer is (240-208)*17/1000 = 0.544 . Other factors might modify this.

    .

  4. #4
    Join Date
    Oct 2018
    Location
    Kingsport Tennessee USA
    Posts
    2
    Quote Originally Posted by gar View Post
    181911-1414 EDT

    Boggsc71:

    Quite obviously the quick answer is (240-208)*17/1000 = 0.544 . Other factors might modify this.

    .
    Thanks for the quick response.

  5. #5
    Join Date
    May 2013
    Location
    Massachusetts
    Posts
    2,395
    Quote Originally Posted by gar View Post
    181911-1414 EDT

    Boggsc71:

    Quite obviously the quick answer is (240-208)*17/1000 = 0.544 . Other factors might modify this.

    .
    I'm confused. Why are the two voltages subtracted from one another? I would think it would be 240*17/1000 = 4.08 kVA, assuming the 17 Amperes is carried on the 240V side.

  6. #6
    Join Date
    Jun 2004
    Location
    Cherry Valley NY, Seattle, WA
    Posts
    4,719
    Quote Originally Posted by Carultch View Post
    I'm confused. Why are the two voltages subtracted from one another? I would think it would be 240*17/1000 = 4.08 kVA, assuming the 17 Amperes is carried on the 240V side.
    With an autotransformer, you essentially only need the kva for the voltage change.
    Ethan Brush - East West Electric. NY, WA. MA

    "You can't generalize"

  7. #7
    Join Date
    Nov 2015
    Location
    CA, USA
    Posts
    648
    Quote Originally Posted by electrofelon View Post
    With an autotransformer, you essentially only need the kva for the voltage change.
    That's not entirely accurate. The current through the common part of the winding is the difference between the input and output current but the current in the part of the winding that is used by only one side of the autotransformer is the full current for that side and needs to be sized for that current. In this case, the shared winding would carry 2.6A and the secondary only part of the winding would carry 17A. If the autotransformer was wound with a single size conductor it would have to be rated for 17A. Sometimes autotransformers are wound with two conductor sizes to take advantage of the reduced current in the shared winding.

  8. #8
    Join Date
    Sep 2008
    Location
    UK
    Posts
    12,009
    Quote Originally Posted by pv_n00b View Post
    That's not entirely accurate. The current through the common part of the winding is the difference between the input and output current but the current in the part of the winding that is used by only one side of the autotransformer is the full current for that side and needs to be sized for that current. In this case, the shared winding would carry 2.6A and the secondary only part of the winding would carry 17A. If the autotransformer was wound with a single size conductor it would have to be rated for 17A. Sometimes autotransformers are wound with two conductor sizes to take advantage of the reduced current in the shared winding.
    So what do you think the kVA should be?
    Si hoc legere scis nimium eruditionis habes.

  9. #9
    Join Date
    Oct 2009
    Location
    Austin, TX, USA
    Posts
    10,013
    Quote Originally Posted by Carultch View Post
    I'm confused. Why are the two voltages subtracted from one another? I would think it would be 240*17/1000 = 4.08 kVA, assuming the 17 Amperes is carried on the 240V side.
    Multiply the delta V by the current.

  10. #10
    Join Date
    Apr 2006
    Location
    Springfield, MA, USA
    Posts
    3,258
    Quote Originally Posted by pv_n00b View Post
    That's not entirely accurate. The current through the common part of the winding is the difference between the input and output current but the current in the part of the winding that is used by only one side of the autotransformer is the full current for that side and needs to be sized for that current. In this case, the shared winding would carry 2.6A and the secondary only part of the winding would carry 17A. If the autotransformer was wound with a single size conductor it would have to be rated for 17A. Sometimes autotransformers are wound with two conductor sizes to take advantage of the reduced current in the shared winding.
    Most commonly for buck and boost applications under 600V, the transformer used has separate primary and secondary coils. These coils are then connected in an autotransformer configuration.

    The secondary coil(s) are 12 or 16 or 24V, and often there are 2 coils which can be in series or parallel. These transformers have 600V insulation on both the primary and secondary side. In this case the transformer secondary will be rated to carry the full current needed to give the rated kVA at the secondary voltage. The secondary coil(s) will be wound with different size wire than the primary.

    The other common autotransformer is the variable autotransformer. In this case there is only a single coil, generally made with a single size wire. The kVA rating will be at full output voltage, and at low voltage output the available kVA is less than the rating.

    -Jon

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •