On Fri, 29 Mar 2013 12:59:34 -0700, Gary Walters
wrote:

In USA.

Source: 208v, 60 hz, 2-wire (2 phases from 3 phase "Y" supply). Load: 240v,
20A.

I presumed that sizing a buck-boost transformer is simple KVA math (source
volts * load amps). But...

This PDF document:

http://www.acmepowerdist.com/pdf/Page_104-109.pdf

on the last page says:
- - -
"An example of an everyday application is always a good way to explain the
intent of the ³Code.² Example: A 1 kVA transformer Catalog No. T111683 has a
primary of 120 x 240V and a secondary of 12 x 24V. It is to be connected as
an autotransformer at the time of installation to raise 208V to 230V single
phase.

When this 1 kVA unit is connected as an autotransformer for this voltage
combination, its kVA rating is increased to 9.58 kVA (may also be expressed
as 9,580 VA). This is the rating to be used for determining the full load
input amps and the sizing of the overcurrent protect device (fuse or breaker)
on the input.

Full Load Input Amps =
9,580 Volt Amps / 208 Volts = 46 Amps"
- - -
I'm puzzled by the 10x increase of KVA rating. When and how is this true?

What size B-B transformer do I need?

Thanks.
I recently wired in a couple Buck/boost xmfrs in a buck configuration.
To see if the xmfrs I had were the correct size and to find out how to
wire the xmfrs to get the amount of voltage reduction I wanted I
looked online for info from Square D. If you google "Jefferson
Electric Buck-Boost Application Manual" you should be able to find the
pdf version of it. This is what I did. Or, since I still have the pdf
version I will email it to you if you want. Just reply to my email
address: . The manual tells you how figure out what
size xmfr you need and all the various ways to connect the xmfr the
get the voltage out that you want.
ERic