# Mains wiring question: Sizing buck-boost transformer?

Started by March 29, 2013
```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

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 &sup3;Code.&sup2; 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.

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.

```
```On 03/29/2013 03:59 PM, 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 &sup3;Code.&sup2; 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.
>

IANAE.  However, it sounds like they're talking about the VA rating of
the circuit rather than the transformer itself.  9580 VA = 1000 VA *
230/24 .

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net
```
```"Gary Walters" <gwprez@yahoo2.cz> wrote in message
news:0001HW.CD7B40B60050CF9CB01029BF@news.eternal-september.org...
> 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 &sup3;Code.&sup2; 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.
>

Those are common devices, to go from 208 to 230/240 volts. The secondary
current determines the rating in KVA.

```
```On Fri, 29 Mar 2013 12:59:34 -0700, Gary Walters <gwprez@yahoo2.cz>
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
>
>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 &sup3;Code.&sup2; 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.
>
>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.

The secondary of an autotransformer carries the load current but only
supplies the voltage *difference*. 240-208 is only 32 volts, so at 20
amps the boost secondary is delivering 640 VA.

True *if* the boost transformer is being used as an autotransformer.

You could use a 208-to-32 volt transformer rated 640 VA or so.

--

John Larkin         Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom laser drivers and controllers
Photonics and fiberoptic TTL data links
VME thermocouple, LVDT, synchro   acquisition and simulation
```
```In sci.electronics.repair Gary Walters <gwprez@yahoo2.cz> 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 ?Code.? 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?

It's true because you're only using the transformer to "create" 24 volts at the current you wish
to draw at 230v. This extra 24 volts is added back into the line voltage.

You can switch flip the leads and subtract voltage too, then the transformer is in buck mode.

> What size B-B transformer do I need?
>
> Thanks.

If you need 20 amps at 230v and start with 208, you need to boost 22volts (208+22=230) x 20 amps
= 480VA transformer. A 24 volt transformer rated over 480VA should be fine.

Autotransformers can be confusing, so pretend it's just DC and some batteries.

Let's say you need 24 volts at 10 amps and have a 12 volt battery that can already output 10
amps.

what size power supply do you need to run in series with this battery to get the 24 volts?

just another 12 volts, at at least 10 amps, or a 120 watt power supply. Those wired in
series (your battery and the new power supply)  will provide 240 watts.

If you already had an 18 volt battery, you'd just need a 6 volt, 10 amp or 60 watt power supply.

The less the voltage adjustment, the smaller then buck/boost transformer rating becomes as it's
really not doing all that much work.

```
```"John Larkin" <jlarkin@highlandtechnology.com> wrote in message
news:4vvbl8dup3vr40m6f39alljljkd4qneon8@4ax.com...
> On Fri, 29 Mar 2013 12:59:34 -0700, Gary Walters <gwprez@yahoo2.cz>
> 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
>>
>>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 &sup3;Code.&sup2; 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.
>>
>>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.
>
> The secondary of an autotransformer carries the load current but only
> supplies the voltage *difference*. 240-208 is only 32 volts, so at 20
> amps the boost secondary is delivering 640 VA.
>
> True *if* the boost transformer is being used as an autotransformer.
>
> You could use a 208-to-32 volt transformer rated 640 VA or so.
>
>
> --

Any electrical supply house will have them in stock. They are not too
expensive either.

tm

```
```On 30/03/2013 6:59 AM, 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 &sup3;Code.&sup2; 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.
>

You can look at it this way: All the input current flows through the
primary, and all the output current flows through both the primary and
the secondary. But the output current is in antiphase with the input
current, so most of the current in the primary is cancelled. The primary
has to handle only the difference between the input and output current.

Sylvia.
```
```On Friday, March 29, 2013 5:01:02 PM UTC-4, John Larkin wrote:

> >When this 1 kVA unit is connected as an autotransformer for this voltage=
=20
>=20
> >combination, its kVA rating is increased to 9.58 kVA (may also be expres=
sed=20
>=20
> >as 9,580 VA). This is the rating to be used for determining the full loa=
d=20
>=20
> >input amps and the sizing of the overcurrent protect device (fuse or bre=
aker)=20
>=20
> >on the input.
>=20
> >

> The secondary of an autotransformer carries the load current but only
>=20
> supplies the voltage *difference*. 240-208 is only 32 volts, so at 20
>=20
> amps the boost secondary is delivering 640 VA.

Wiring the primary of 10:1 step-down across the line "pins" its secondary v=
oltage at 10% line voltage. So obviously if that secondary is in series wit=
h the load it only delivers 10% of the load power with the line delivering =
the other 90%. If the max transformer KVA is X then the maximum load that c=
an be safely driven is.... wait for it....wait for it... 10 x X. Makes no d=
ifference if it's buck or boost. But in boost mode, the line supplies 1.1 x=
Load current, and in buck mode it supplies 0.9 x Load current.
```
```On Friday, March 29, 2013 11:17:28 PM UTC-4, Sylvia Else wrote:

>=20
> You can look at it this way: All the input current flows through the=20
>=20
> primary, and all the output current flows through both the primary and=20
>=20
> the secondary. But the output current is in antiphase with the input=20
>=20
> current, so most of the current in the primary is cancelled. The primary=
=20
>=20
> has to handle only the difference between the input and output current.
> Sylvia.

Huh? In boost mode , the transformer high side is the primary, and the low =
voltage side is secondary . In buck mode, the transformer low voltage side =
is the primary, the high voltage side is the secondary. In each case, trans=
former high voltage side current is 1/10 transformer load voltage side curr=
ent. There's no looking at it this or that way, there's only comprehension =
of what a transformer is.
```
```On Sat, 30 Mar 2013 14:15:00 -0700 (PDT),
bloggs.fredbloggs.fred@gmail.com wrote:

>On Friday, March 29, 2013 11:17:28 PM UTC-4, Sylvia Else wrote:
>
>>
>> You can look at it this way: All the input current flows through the
>>
>> primary, and all the output current flows through both the primary and
>>
>> the secondary. But the output current is in antiphase with the input
>>
>> current, so most of the current in the primary is cancelled. The primary
>>
>> has to handle only the difference between the input and output current.
>> Sylvia.
>
>Huh? In boost mode , the transformer high side is the primary, and the low voltage side is secondary . In buck mode, the transformer low voltage side is the primary, the high voltage side is the secondary. In each case, transformer high voltage side current is 1/10 transformer load voltage side current. There's no looking at it this or that way, there's only comprehension of what a transformer is.

Now, now, Freddy!  To fit in here you need to learn to treat the
ignorant in a politically correct manner >:-}

Otherwise it'll be your turn to be attacked and told to change your
diaper ;-)

...Jim Thompson
--
| James E.Thompson                                 |    mens     |