Buck Boost (Auto) transformers and section 450.4(A)

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fandi

Senior Member
Location
Los Angeles
Hello All,
One engineer designs a 7.5kVA 240-208 1P, 3W Buck Boost transformer with 225A breaker, #4/0 CU on the 240V primary side and 225A breaker, #4/0 CU on the 208V secondary side. The 225A main breaker of the panel on the secondary side is within 10ft of the transformer.
Per section 450.4, the OCPD shall be rated not more than 125% of the rated full load input current of the autotransformer.
To me, the rated full load input current is the primary current which is 7.5/.24 = 31.25A. The max OCPD is 39.06A or 40A.
The engineer said the Buck Boost transformer can produce: 7.5/(.24-.208) = 7.5/0.032 = 234.375A.

Can you tell me who is right?
Thank you.
 

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
Autotransformer is not the same thing as a buck boost transformer.
Sorry, isn't a transformer being used for buck/boost necessarily field-connected as an autotransformer?

But if a 7.5 kVA 240V:32V single phase isolation transformer (not sure if that correctly describes the transformer in the OP) is connected as an autotransformer to buck 240V to 208V, it's rating in that configuration is greater than 7.5 kVA, as far as 450.4(A) goes. So the engineer's computation is correct.

Cheers, Wayne
 

fandi

Senior Member
Location
Los Angeles
Sorry, isn't a transformer being used for buck/boost necessarily field-connected as an autotransformer?

But if a 7.5 kVA 240V:32V single phase isolation transformer (not sure if that correctly describes the transformer in the OP) is connected as an autotransformer to buck 240V to 208V, it's rating in that configuration is greater than 7.5 kVA, as far as 450.4(A) goes. So the engineer's computation is correct.

Cheers, Wayne
Correct me if I'm wrong but this is the way I see it:
- rated I pri = xfmr rating/V pri = 7.5/.24 = 31.25A. Per section 450.4, the max OCPD is 39.06A or 40A. So the upstream 225A feeder breaker on the primary side is oversized. The transformer is not protected.
- rated I sec = xfmr rating / Vsec = 7.5/.28 = 26.79A, not 234.375A as the engineer stated.

Quote from the above website:
'Buck-boost transformers are typically wired as autotransformers.'
'According to NEC® Section 450.4, Autotransformers 600V, Nominal, or Less, each buck-boost autotransformer 600 volts, nominal, or less shall be protected by an individual overcurrent device installed in series with each ungrounded input conductor
. '
 

petersonra

Senior Member
Location
Northern illinois
Occupation
engineer
Correct me if I'm wrong but this is the way I see it:
- rated I pri = xfmr rating/V pri = 7.5/.24 = 31.25A. Per section 450.4, the max OCPD is 39.06A or 40A. So the upstream 225A feeder breaker on the primary side is oversized. The transformer is not protected.
- rated I sec = xfmr rating / Vsec = 7.5/.28 = 26.79A, not 234.375A as the engineer stated.

Quote from the above website:
'Buck-boost transformers are typically wired as autotransformers.'
'According to NEC® Section 450.4, Autotransformers 600V, Nominal, or Less, each buck-boost autotransformer 600 volts, nominal, or less shall be protected by an individual overcurrent device installed in series with each ungrounded input conductor
. '
read this

 

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
Correct me if I'm wrong but this is the way I see it:
- rated I pri = xfmr rating/V pri = 7.5/.24 = 31.25A. Per section 450.4, the max OCPD is 39.06A or 40A. So the upstream 225A feeder breaker on the primary side is oversized. The transformer is not protected.
- rated I sec = xfmr rating / Vsec = 7.5/.28 = 26.79A, not 234.375A as the engineer stated.
[In my previous post I referenced a 240V:32V isolation transformer, I think I might have that wrong for the buck configuration and you'd actually use a 208V:32V transformer.]

The only way the 7.5 kVA rating makes sense, given the breaker sizes, is if that is the rating when used as an isolation transformer 208V : 32V. That means the primary coil has a current rating of 7500/208 = 36A, and the secondary coil has a current rating of 7500/32 = 234A.

But when configured as an autotransformer, there is no primary/secondary, there's just one long coil from putting the primary/secondary coils in series. You end up with a 240V coil, with a 208V tap. For bucking the two 240V points are the input, and the 208V points are the output, and for boosting the two 208V points are the input, and the 240V points the output. [If I've got that right.]

Note that 450.4 doesn't refer to primary/secondary, it refers to the "full load input current." Only the former secondary coil portion of the single coil sees the full load input current, and that has a 234A rating. The former primary coil sees a lesser current, in line with its rating (from Kirchoff's law, it's the difference in currents on the 240V side and the 208V side).

So wired in this way as an autotransformer, the "full load input current" rating matches the secondary coil current rating.

Cheers, Wayne
 

fandi

Senior Member
Location
Los Angeles
The former primary coil sees a lesser current, in line with its rating (from Kirchoff's law, it's the difference in currents on the 240V side and the 208V side).
So an upstream 225A feeder breaker feeding this 7.5kVA buck boost transformer complies with 450.4 ?
 

jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
450.4 Autotransformers 1000 Volts, Nominal, or Less.

Autotransformer is not the same thing as a buck boost transformer.
Probably 99% of bucks boost transformers are 2-winding transformers connected in an autotransformer configuration. The NEC rules for autotransformers apply to these installations.

The engineer is correct for this application.

I know of no manufacturer, providing buck boost wiring diagrams, that suggests putting the OCPD 'inside' of the autotransformer.
 

fandi

Senior Member
Location
Los Angeles
Probably 99% of bucks boost transformers are 2-winding transformers connected in an autotransformer configuration. The NEC rules for autotransformers apply to these installations.

The engineer is correct for this application.

I know of no manufacturer, providing buck boost wiring diagrams, that suggests putting the OCPD 'inside' of the autotransformer.
I never think the OCPD is put inside the transformer. Can you advise me how the transformer is protected by the upstream 225A breaker on the primary side per 450.4? Thanks.
 

jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
I never think the OCPD is put inside the transformer. Can you advise me how the transformer is protected by the upstream 225A breaker on the primary side per 450.4? Thanks.
A buck-boost does not have a primary and a secondary, instead it has a high voltage and a low voltage. Think of it as a black box with input and output terminals.

The series or parallel connection of the two windings determines the autotransformer resultant kVA. Once you know this value you use standard transformer formulas to determine the input and output amps.
 

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
Thanks but I still don't understand how to come up with this formula: 7.5/(.24-.208) = 7.5/0.032 = 234.375A
In the diagram Joe Stillman posted, call the Input side voltage A (240V in your application) and the Output side voltage B (208V in your application). So B is the voltage across the H portion of the coil, A is the voltage across H+X, and A-B is the voltage across X only. As an isolation transformer, the transformer would be a B : B-A transformer.

Then if the transformer power rating in the usual isolation configuration is P, the allowable current in the X coil is P/(B-A), since B-A is the voltage across that coil. But note that all of the input side current is flowing through X, so P/(B-A) is also the allowable input side current. Which is the formula the engineer used.

Cheers, Wayne
 

JoeStillman

Senior Member
Location
West Chester, PA
I find it helps to visualize the currents with these. You can see that, even though the X1-X4 coil carries the full primary current, it is at a much lower voltage than the primary voltage. So the actual transformer load is only a fraction of the connected load.
1643120621317.png
 
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