208V to 230V buck boost

[david luchini]

Size the wire and OCPD as if the BB were not there.

That might get you into trouble. The ocpd can't be more than 125% of the full-load input current of the transformer, or the next size up.

 

[kwired]

That might get you into trouble. The ocpd can't be more than 125% of the full-load input current of the transformer, or the next size up.

What do you mean by input current to transformer? The so called primary coil will be well below the current of the final load. 125% of the rated current of the lower voltage coil used for buck/boosting? that makes sense at least for non motor loads.

 

[david luchini]

What do you mean by input current to transformer?

I'm not sure what you are asking. 450.4(A) says the autotransformer ocpd shall be rated or set at not more than 125% of the rated full-load input current of the autotransformer...

 

[kwired]

I'm not sure what you are asking. 450.4(A) says the autotransformer ocpd shall be rated or set at not more than 125% of the rated full-load current of the autotransformer...

That should be for the load carried by the supply side conductors then. I was (probably mistakenly) thinking of the rated current of the higher voltage coil which would be well below the rated current of the lower voltage coil. But the whole thing and how it is connected is what makes it an autotransformer. Same thing can be connected as a separately derived system in most cases, like 120 x 12/24 volts.

 

[david luchini]

That should be for the load carried by the supply side conductors then. I was (probably mistakenly) thinking of the rated current of the higher voltage coil which would be well below the rated current of the lower voltage coil. But the whole thing and how it is connected is what makes it an autotransformer. Same thing can be connected as a separately derived system in most cases, like 120 x 12/24 volts.

Yes, it would depend if you are bucking or boosting. But it's not the load current that the OCPD is base on, but the rated input current.

See post 12. In that case, the load current is 10A, but the rated input current is 11.5.

You might have the load on a 20A circuit, but the transformer noted in Post 12 would require a max 15A ocpd. You would either need a smaller ocdp, or a larger transformer.

 

[wwhitney]

I'm not sure what you are asking. 450.4(A) says the autotransformer ocpd shall be rated or set at not more than 125% of the rated full-load input current of the autotransformer...

If I understand correctly, you are saying that for an isolation transformer configured as buck/boost by putting the two coils in series, the "rated full-load current of the autotransformer" should be understood as the sum of the isolation configuration rated primary and secondary currents?

Cheers, Wayne

 

[david luchini]

If I understand correctly, you are saying that for an isolation transformer configured as buck/boost by putting the two coils in series, the "rated full-load current of the autotransformer" should be understood as the sum of the isolation configuration rated primary and secondary currents?

Depends if you are boosting or bucking. For bucking, yes, the rated full-load current is the sum of the rated "primary" and "secondary" currents.

For boosting, the rated full-load current is the rated "secondary" current.

For example: A 250va, 240-24V transformer, boosting 208V to 229V, the rated load current is 250va/24v=10.41A. The rated full-load input current 1would then be 10.42A*229V/208V=11.46A

A 250va, 240-24V transformer, bucking 230v to 209V, the rated load current is 250va/24v + 250va/240v = 11.46A. The rated full-load input current would then be 11.46A*208v/229v=10.41A

 

[wwhitney]

Depends if you are boosting or bucking.

Thanks for the correction that it differs.

For bucking, yes, the rated full-load current is the sum of the rated "primary" and "secondary" currents.

For boosting, the rated full-load current is the rated "secondary" current.

These statements are true on the output side. I left the word "input" out of my quote, so perhaps you thought I was asking about the output side. But 450.4(A) refers to the "rated full-load input current", so that is what I meant.

In your subsequent examples you calculated the input currents, and I agree with that.

Cheers, Wayne

 

[cppoly]

@david luchini , your calculations were very help.

What are your thoughts on skipping the calcs when the manufacturer gives you the kVA rating in a chart format (see attached)?

Simply take 4.79 kVA listed and divide by the 208V input to get the input current = 23 amps.

Or the long way (using David's calculations)
500 VA / 24V = 20.8A
20.8A * 229V = 4.77 kVA
4.77 kVA / 208V = 23 amps.

The spec sheet from the catalog number gives you the specs of 500 VA, primary voltage 120V / 240V, secondary 12/24V.

 

[david luchini]

@david luchini , your calculations were very help.

What are your thoughts on skipping the calcs when the manufacturer gives you the kVA rating in a chart format (see attached)?

Yes, I would just use the manufacturer's kva, if the give it to you in a chart.

That info might not be available in all cases, however. I seem to remember on those Eaton charts, I couldn't find a 208V required output.

Maybe it just wasn't in the catalog I looked at.

 

[kwired]

Yes, I would just use the manufacturer's kva, if the give it to you in a chart.

That info might not be available in all cases, however. I seem to remember on those Eaton charts, I couldn't find a 208V required output.

Maybe it just wasn't in the catalog I looked at.

Any such chart or on line calculator works for selecting any brand as long as they offer same transformer sizes and configurations. Some may only return a catalog number, you just need to cross reference to find the equivalent in the other brand.