# autotransformer question



## a4u2fear (Jul 13, 2018)

Ideal 25kVA 480/120V 1PH transformer is to be used as an autotransformer supplying load from a 480V source

If it is in Buck mode, and the coils are carrying rated current, the max KVA load the transformer can accomodate is?

75kVA answer

I came up with 100kVA

Is somehow "coils carrying rated current" the key?  Or is the answer a mistake?


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## Szar (Jul 14, 2018)

I've seen this and will respond.  But working the weekend and life issues at home...

I'll try and respond before Monday.


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## Szar (Jul 14, 2018)

I believe the answer key is correct.   

Single Phase transformers can be wired as Auto transformers in one of three (technically two) ways.  In one way, the coils are additive (boost) and increase the supply voltage higher then what would otherwise be capable if configured normally while the other two (really one...) configurations have the coils wired "subtractively" (buck) and produce a lower and different  voltage then would otherwise have been possible. 

_Not really important, but it is:  The two bucking configurations are really the same wired transformer, just having the supply connected to one side or other alters the circuit effect.  If connected to the primary side, the rated current flows across the load.  if connected to the secondary side, a reduced current flows across the load.  The rated current of the coil cannot be exceeded when wiring a single phase transformer in auto configuration, so that is what limits the KVA output.  _

The question is asking for maximum apparent power, so we can safely ignore the other secondary side "buck" configuration since it also tells us the supply is 480V and you would need 120V to connect it to the 120V coil.         

Due to how the coils are configured and wired,  in this configuration with the supply on the primary side...


360V is developed across the load (the 120V induced voltage in the secondary subtracts from the 480V supply). 

As such, the transformer develops rated current in the secondary, 208A (= 25 kVA / 120) . 

As X1 and H1 are wired together, the secondary current produced is split across the primary coil and through the "source". 

(Primary coil current cannot be exceeded just like the secondary) 


[*]The apparent power delivered to the load is *75 kVA* (= 360V * 208A)



If the source was 120V and connected the other way (rather then 480V supply we have here)"


360V is developed across the load (the 120V induced voltage in the secondary subtracts from the 480V supply). 

Rated current that follows on the primary side is only 52A. 

Since the primary coil current rated current cannot be exceeded, instead of 208A flowing across the load and only 52A @ 360V... the apparent power is limited to 18,720 VA.

Notice... smaller then rated output of the transformer in single phase configuration.  

Only reason to do this is if you need the lower voltage and only have the secondary voltage available.  It can be done, but you take a hit to delivered power.


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## a4u2fear (Jul 16, 2018)

Thanks Szar for your response.  However, I'm still not getting it.

I've gone through all the examples I can find, and all the literature on it I can find.

I have no issues calculating the right answers for Boost configurations.  For Buck, sometimes I get it and sometimes I don't.  The books and literature seem to show

Sauto = Sw(Nc+Nse)/Nse

This formula works when I'm doing a buck circuit, but not when the input voltage is the same as the original transformer.

i.e. That formula doesn't work when 480/120 used as a 480/360, or when a 600/120 is used as a 600/480

It does work when a 480/120 is used as a 600/480, or a 480/120 is used as a 600/120

And it has worked for every boost problem (Sauto = Sw(Nc+Nse)/Nc

Maybe I'm not understanding the fundamentals?  There's not enough  examples on the internet or if there are the solutions are weak.  Maybe I need photos to see what I'm missing.


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## Szar (Jul 16, 2018)

a4u2fear said:


> Thanks Szar for your response.  However, I'm still not getting it.
> 
> I've gone through all the examples I can find, and all the literature on it I can find.
> 
> ...


I would agree it sounds like a fundamentals thing.  From what I can tell your looking to plug and chug the formula without the full understanding where the equation applies (or more important what the terms area). 

The "N" terms are not the voltage.  Its the number of turns for that coil.  A true Auto transformer, for which this equation was derived, has a series coil "se" and a common coil "c" and they are "wired" in series.  Really its just an adjustable center tap and the the ratio between the the two develops the equation referenced.  So while I get what your doing, its just a terminology thing I'm point out.    

While a single phase transformer can be connected like an Autotransformer to produce an electrically "equivalent" , its still physically a single phase transformer and is NOT a strait forward ratio between the primary and secondary.  Draw out the single phase transformer circuits to visualize.  The Single phase transformer wired as an auto-transformer has source voltage (480V) delivered to BOTH coils.  The secondary coil (120V) actually acts against this supplied voltage to drop the voltage across the load to 360V.

So the easy answer after all is said and done (and I just came to this myself since I never drew out the circuits in advance and only established what needed to be done with each to avoid math for this on the test...) one of the terms must be considered negative.  This is not an additive configuration.  The auto-transformer equation assumes the coils are in series and additive, where-as that is not the case here and they counteract each other.

That is also why the "boost" version works.  The equation assumes additive, boost is additive... so the result works.


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## Szar (Jul 16, 2018)

Szar said:


> I would agree it sounds like a fundamentals thing.  From what I can tell your looking to plug and chug the formula without the full understanding where the equation applies (or more important what the terms area).
> 
> The "N" terms are not the voltage.  Its the number of turns for that coil.  A true Auto transformer, for which this equation was derived, has a series coil "se" and a common coil "c" and they are "wired" in series.  Really its just an adjustable center tap and the the ratio between the the two develops the equation referenced.  So while I get what your doing, its just a terminology thing I'm point out.
> 
> ...


I'd like to add...

in 11 years I have worked with Commercial, Industrial, and utility clients in the largest metropolis in the world... I have had ZERO jobs involving auto-transformers.  These things are a purple unicorn used in specialty circumstances.


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## Surf and Snow (Jul 17, 2018)

We use lots of auto transformers at my work, but all at transmission voltage (500/230kV, 230/115kV etc..)

FWIW Engineering Pro Guides technical study guide has some good study examples.


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## justin-hawaii (Jul 17, 2018)

Sorry but my content currently doesn't include buck and boost auto transformers. But I will add this concept into my next update.


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## a4u2fear (Aug 6, 2018)

Double checking on this - For any conventional transformer, used as an autotransformer, there are a total of FOUR configurations that could be used.

i.e. a 480/120 could be used as a 480/600 or 480/360 or 120/600 or 120/360

Anyone disagree?


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## rg1 (Aug 6, 2018)

a4u2fear said:


> Double checking on this - For any conventional transformer, used as an autotransformer, there are a total of FOUR configurations that could be used.
> 
> i.e. a 480/120 could be used as a 480/600 or 480/360 or 120/600 or 120/360
> 
> Anyone disagree?


Perfect! There can be 4 combinations. The boundary limits are, do not exceed the rated current of the coils, rated voltage of the coil, see for addition or subtraction (dot convention) of voltage depending on buck/boost. The transformation ratios of currents and voltages (make sure directions are correct) still holds good between the two coils as was there when conventionally connected. The input VA is always equal to output VA.

The last two sentences are very important. 

With above items in mind one should be able to do all AT questions. If it does not help, let me understand your specific gap in the understanding; I will help you fill that gap.


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