Mains wiring question: Sizing buck-boost transformer?

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:

- - - "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.

Reply to
Gary Walters
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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
Reply to
Phil Hobbs

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

Reply to
tm

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
Reply to
John Larkin

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.

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.

Reply to
Cydrome Leader

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

tm

Reply to
tm

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.

Reply to
Sylvia Else

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: snipped-for-privacy@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

Reply to
etpm

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.

Reply to
bud--

--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

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

Reply to
Don Kelly

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