Mains wiring question: Sizing buck-boost transformer?

[Gary Walters]

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 łCode.˛ 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.

 

[Phil Hobbs]

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 łCode.˛ 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

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

 

[tm]

"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 łCode.˛ 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.

 

[John Larkin]

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
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 łCode.˛ 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.

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

 

[tm]

"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
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 łCode.˛ 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.

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

 

[Cydrome Leader]

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.

 

[Sylvia Else]

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 łCode.˛ 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 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
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 łCode.˛ 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: etpm@whidbey.com . 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

 

[bud--]

On 3/29/2013 3:01 PM, John Larkin wrote:

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
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 łCode.˛ 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.

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.

All the buck-boost transformers I have seen are 120 or 240V primary.
They are available with a secondary voltage of 24 or 32V (or half that -
12 or 16V).

If you want to go from 208 to 240 that is 32 volts. If the boosted
circuit is 20A the rating of the secondary is 20A. The transformer power
rating is 32V x 20A = 640VA. That is the same as what John L said. This
is simple math. IMHO the pdf is much less simple.

The circuit rating is 240V x 20A = 4,800VA. The 640VA transformer
supplies a 4,800VA circuit because it only supplies 32V of the load. A
calculation like that is how the article comes up with 9580VA (for a 24V
transformer).

If you connect the 240V primary to the 208V supply you will get 27.8V
boost instead of 32V, for a boosted voltage of 236V. You can get the
full 32V boost by connecting the primary to the 240V boosted side of the
circuit, but then the primary current of 2.7A will be supplied by the
32V winding and the available boosted circuit supply will be 17.3A.

When you turn the transformer on there can be an inrush current that can
trip the protection for the primary winding. That is why the pdf article
talks about protection higher than the current rating of the primary.

 

[Don Kelly]

"Cydrome Leader" wrote in message news:kj4vp4$6m7$1@reader1.panix.com...

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?
--QUOTE

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.
QUOTE

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.
=======================================
lookup "autotransformer".
In this case you have a 2 winding transformer with a 10:1 ratio
With 240V applied the secondary will be 24V with a rated current of
1000/24=41.7A
If this is connected as a boost autotransformer- the total output voltage
would be
240+24 =264V so the output, without exceeding rated output current would be
11KW. only 1 KW (24V*41.7A) is supplied through transformer action and the
rest through a direct connection
Adjusting to 230V output leads to 11*(230/264)=9.58KW
The input voltage would be 207V excluding any voltage drops in the
transformer-so 208/230 is close enough.
Autotransformers are great for turns ratios near one as there are size and
cost advantages.
Disadvantage--no isolation between primary and secondary.

Excuse the lack of "quoting" as I am using windows live mail in an
emergency- Thunderbird downloads news but then deletes
the downloads immediately! New problem- correction not yet found.

Don Kelly
cross out to reply