Boggsc71
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- Kingsport Tennessee USA
I need to boost 208 vac to 240vac what KVA Transformer do I need? The fla is 17,
Thanks for the quick response.181911-1414 EDT
Boggsc71:
Quite obviously the quick answer is (240-208)*17/1000 = 0.544 . Other factors might modify this.
.
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.
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.
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.
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.
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
Um, It was I who said that, wasn't it?So, like I said: "essentially" delta v times amps
Most commonly for buck and boost applications under 600V, the transformer used has separate primary and secondary coils.
That would be my take also.So, like I said: "essentially" delta v times amps
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.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!