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

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fandi

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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.
 
petersonra said:
Autotransformer is not the same thing as a buck boost transformer.
Click to expand...
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
 
wwhitney said:
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
Click to expand...
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.

eepower.com

Buck-Boost Transformer Operations and Limitations - Technical Articles

A buck-boost transformer is a small transformer designed to buck (lower) or boost (raise) line voltage. The major advantages of buck-boost transformers are their low cost and small size in comparison to general-purpose transformers.
eepower.com eepower.com
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. '
 
fandi said:
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.

eepower.com

Buck-Boost Transformer Operations and Limitations - Technical Articles

A buck-boost transformer is a small transformer designed to buck (lower) or boost (raise) line voltage. The major advantages of buck-boost transformers are their low cost and small size in comparison to general-purpose transformers.
eepower.com eepower.com
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. '
Click to expand...
read this

americas.hammondpowersolutions.com

When a Buck-Boost transformer is connected as an autotransformer, what is the procedure for determining the current rating of the over-current protective device, such as the fuse or circuit breaker?

The NEC Article 450-4 outlines over-current protection for autotransformers. It is reproduced as follows: “NEC 450-4 – Autotransformers 600 Volts, Nominal, or Less (a) Over-current Protection. Each autotransformer 600 volts nominal, or less shall be protected by an individual over-current...
americas.hammondpowersolutions.com americas.hammondpowersolutions.com
 
fandi said:
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.
Click to expand...
[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
 
wwhitney said:
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).
Click to expand...
So an upstream 225A feeder breaker feeding this 7.5kVA buck boost transformer complies with 450.4 ?
 
petersonra said:
450.4 Autotransformers 1000 Volts, Nominal, or Less.

Autotransformer is not the same thing as a buck boost transformer.
Click to expand...
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.
 
jim dungar said:
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.
Click to expand...
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.
 
fandi said:
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.
Click to expand...
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.
 
fandi said:
Thanks but I still don't understand how to come up with this formula: 7.5/(.24-.208) = 7.5/0.032 = 234.375A
Click to expand...
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
 
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.
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