# Buck and Boost

#### Boggsc71

##### Member
I need to boost 208 vac to 240vac what KVA Transformer do I need? The fla is 17,

#### electrofelon

##### Senior Member
The manufacturers have online calculators that will size the units. I'm sure you can find it with a quick google. I know square d has one.

#### gar

##### Senior Member
181911-1414 EDT

Boggsc71:

Quite obviously the quick answer is (240-208)*17/1000 = 0.544 . Other factors might modify this.

.

#### Boggsc71

##### Member
181911-1414 EDT

Boggsc71:

Quite obviously the quick answer is (240-208)*17/1000 = 0.544 . Other factors might modify this.

.
Thanks for the quick response.

#### Carultch

##### Senior Member
181911-1414 EDT

Boggsc71:

Quite obviously the quick answer is (240-208)*17/1000 = 0.544 . Other factors might modify this.

.
I'm confused. Why are the two voltages subtracted from one another? I would think it would be 240*17/1000 = 4.08 kVA, assuming the 17 Amperes is carried on the 240V side.

#### electrofelon

##### Senior Member
I'm confused. Why are the two voltages subtracted from one another? I would think it would be 240*17/1000 = 4.08 kVA, assuming the 17 Amperes is carried on the 240V side.
With an autotransformer, you essentially only need the kva for the voltage change.

#### pv_n00b

##### Senior Member
With an autotransformer, you essentially only need the kva for the voltage change.
That's not entirely accurate. The current through the common part of the winding is the difference between the input and output current but the current in the part of the winding that is used by only one side of the autotransformer is the full current for that side and needs to be sized for that current. In this case, the shared winding would carry 2.6A and the secondary only part of the winding would carry 17A. If the autotransformer was wound with a single size conductor it would have to be rated for 17A. Sometimes autotransformers are wound with two conductor sizes to take advantage of the reduced current in the shared winding.

#### Besoeker

##### Senior Member
That's not entirely accurate. The current through the common part of the winding is the difference between the input and output current but the current in the part of the winding that is used by only one side of the autotransformer is the full current for that side and needs to be sized for that current. In this case, the shared winding would carry 2.6A and the secondary only part of the winding would carry 17A. If the autotransformer was wound with a single size conductor it would have to be rated for 17A. Sometimes autotransformers are wound with two conductor sizes to take advantage of the reduced current in the shared winding.
So what do you think the kVA should be?

#### ggunn

##### PE (Electrical), NABCEP certified
I'm confused. Why are the two voltages subtracted from one another? I would think it would be 240*17/1000 = 4.08 kVA, assuming the 17 Amperes is carried on the 240V side.
Multiply the delta V by the current.

#### winnie

##### Senior Member
That's not entirely accurate. The current through the common part of the winding is the difference between the input and output current but the current in the part of the winding that is used by only one side of the autotransformer is the full current for that side and needs to be sized for that current. In this case, the shared winding would carry 2.6A and the secondary only part of the winding would carry 17A. If the autotransformer was wound with a single size conductor it would have to be rated for 17A. Sometimes autotransformers are wound with two conductor sizes to take advantage of the reduced current in the shared winding.
Most commonly for buck and boost applications under 600V, the transformer used has separate primary and secondary coils. These coils are then connected in an autotransformer configuration.

The secondary coil(s) are 12 or 16 or 24V, and often there are 2 coils which can be in series or parallel. These transformers have 600V insulation on both the primary and secondary side. In this case the transformer secondary will be rated to carry the full current needed to give the rated kVA at the secondary voltage. The secondary coil(s) will be wound with different size wire than the primary.

The other common autotransformer is the variable autotransformer. In this case there is only a single coil, generally made with a single size wire. The kVA rating will be at full output voltage, and at low voltage output the available kVA is less than the rating.

-Jon

#### electrofelon

##### Senior Member
That's not entirely accurate. The current through the common part of the winding is the difference between the input and output current but the current in the part of the winding that is used by only one side of the autotransformer is the full current for that side and needs to be sized for that current. In this case, the shared winding would carry 2.6A and the secondary only part of the winding would carry 17A. If the autotransformer was wound with a single size conductor it would have to be rated for 17A. Sometimes autotransformers are wound with two conductor sizes to take advantage of the reduced current in the shared winding.
So what do you think the kVA should be?
Most commonly for buck and boost applications under 600V, the transformer used has separate primary and secondary coils. These coils are then connected in an autotransformer configuration.

The secondary coil(s) are 12 or 16 or 24V, and often there are 2 coils which can be in series or parallel. These transformers have 600V insulation on both the primary and secondary side. In this case the transformer secondary will be rated to carry the full current needed to give the rated kVA at the secondary voltage. The secondary coil(s) will be wound with different size wire than the primary.

The other common autotransformer is the variable autotransformer. In this case there is only a single coil, generally made with a single size wire. The kVA rating will be at full output voltage, and at low voltage output the available kVA is less than the rating.

-Jon
So, like I said: "essentially" delta v times amps

#### ggunn

##### PE (Electrical), NABCEP certified
So, like I said: "essentially" delta v times amps
Um, It was I who said that, wasn't it?

#### GeorgeB

##### ElectroHydraulics engineer (retired)
Most commonly for buck and boost applications under 600V, the transformer used has separate primary and secondary coils.
I SLIGHTLY disagree with the terminology primary and secondary. In boost applications that terminology fits mostly, but in buck I don't think it does. There are, as you said, often 2 (usually identical) low voltage coils allowing a 12, 16, 24, 32, or 48 volt change (I've never run into the pair of 24V coils, but there is certainly no reason it wouldn't exist; all it offers is a 48V change.)

Of course, there is no problem in using a transformer with one primary (even if dual coils) and 1 or 2 secondaries as a buck-boost ...

George

#### Besoeker

##### Senior Member
So, like I said: "essentially" delta v times amps
That would be my take also.
I was questioning pv n00b's disagreement with that.

#### gar

##### Senior Member
181013-1446 EDT

GeorgeB:

My definition of primary is the pair of terminals to which net positive average power is input. A secondary is any pair of terminals thru which net positive average power is delivered. Thus, in an autotransformer some winding may be part of both the primary and secondary.

Where this may not be useful is in a telephone transformer where simultaneously there can be uncorrelated power transferred in both directions.

For a distribution transformer supplying a home with solar panels the role of the primary can shift back and forth between the two physical coils based on the direction of average energy flow. In this case we have synchronous correlated signals.

.

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#### Phil Corso

##### Senior Member
Biggsc...

Because load kVA is 240x17/1000, then minimum AT-Xfmr size is 4.08-kVA! I hope you didn’t purchase an AT based on the ratio provided!
Remember, an AT changes input-to-output voltages, but primary-to-secondary power transfer must remain the same!

Regards, Phil Corso

Ps: if additional detail is required contact me!

#### GoldDigger

##### Moderator
Staff member
Biggsc...

Because load kVA is 240x17/1000, then minimum AT-Xfmr size is 4.08-kVA! I hope you didn’t purchase an AT based on the ratio provided!
Remember, an AT changes input-to-output voltages, but primary-to-secondary power transfer must remain the same!

Regards, Phil Corso

Ps: if additional detail is required contact me!
The buck-boost transformer configuration delivers the load current times input voltage directly from the source without any involvement of the "primary" section of the single winding. The output power corresponding to the voltage delta times the load current comes from a corresponding input power in which current over turns ratio flows through the primary winding.
If there is only one tapped winding, then all of the primary winding must be sized for the output current, though the extra heat dissipation will not be present.
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#### Phil Corso

##### Senior Member
Biggscc, GoldDigger, Gar, et al...

I humbly apologize for my grave error.. I overlooked the basic principle of the AT... power is delivered by both transformer and conduction action. Mea Culpa! :dunce:

More tomorrow!

Phil

#### GoldDigger

##### Moderator
Staff member
Come home, Phil, all is forgiven.

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