Mains wiring question: Sizing buck-boost transformer?

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

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

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.

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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.

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

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.

sed

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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.

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.

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

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Yep. Power is power, and if your secondary side load is pulling a certain amount of power, you can bet the primary side is consuming exactly that much, plus a bit more, in order to provide it.

Goddamned unformatted web based access retards should all get a clue about line length in Usenet too.

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

Of course there are ways of looking at it. It's a device that operates under the laws of physics, in the analysis and understanding of which one can take any perspective that is physically valid, including such things as superposition.

You appear to have absorbed the knowledge about how the voltages and currents on the various windings are related, but not why they are related that way, which depends on how currents produce fields, and how changing fields produce emfs.

There are various takes on exactly what the OP was concerned about. My take was that he was concerned about how connecting a conventional transformed in an auto-transormer configuration can manage to increase its power rating.

Sylvia.

The answer is it doesn't increase the power rating of the transformer. Anal ysis is what tells you the required transformer power handling capability i n the buck/boost configurations a lot more than any fundamental physics. Ma ybe take it up a notch and declare a N:1 stepdown xfrmr can be used in buck /boost circuits with loading up to Nx KVA rating of the transformer.

How finely do you want to split that particular hair?

Sylvia.

Using obvious notation: Actually, KVAXfmr= VLine/N x ILoad and KVALoad= Vline( 1+ 1/N)x ILoad s o that Iload=KVALoad/[Vline(1+1/N)], making KVAXfmr=VLine/N x KVALoad/[ VLine(1+1/N)]= KVALoad/(N+1) so you can use it with loads up to (N+1)x KV AXfmr , for boost. For buck the factor is VLine(1-1/N), so it can be used u p to loads of (N-1)xKVAXfmr. Which makes sense because for the same KVA loa d, the current is greater for buck, but the transformer voltages always run at line.

A 1kVA 1:1 transformer can be used as a 2kVA stepdown auto-transformer.

I think the formul is actually a XkVA N:1 transformer has a (N+1)x kVA rating as an autotransformer.

The Nx rating you give above seems to ignore the current in the common leg.

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** Only true if used in a system with DOUBLE the supply voltage the tranny is rated for. ** Simply a result of how the parameters are defined - not physics.

For one thing - the "power rating" of a tranny is proportional to the applied primary voltage.

For example: raising the supply frequency allows more voltage to be applied ( before core saturation ) so up goes the VA rating too. A 50Hz tranny run at 400Hz has 8 times the previous VA capacity - long as the insulation can stand it.

Auto-transformers can be given very large VA ratings for their size - but it is a nonsense as the tranny is suppling only a small part of the output.

The AC supply delivers the rest - directly to the load.

... Phil

Schematic?

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John Larkin                  Highland Technology Inc 
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A 1:1 buck delivers 0 KVA to the load :-) He must be talking about using the center tap of secondary for 2:1.