Transformer loads

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wch

Member
If a 240/208 buck/boost T/F has a "load rating" of 13.3% of the connected
appliance kva, why would a 208delta/208 wye isolation T/F need a "load rating"
matching the kva of the downstream connected load?
 

Volta

Senior Member
Location
Columbus, Ohio
The buck/boost will only carry the va associated with the approximately 32 volts added or dropped. The isolation will carry the va of the entire load.

240 volts - 208 volts = 32 volts
32 volts / 240 volts = 13.333 percent of the load

208 to 208 will carry the whole load, thus, must be rated 100%.
 

Chamuit

Senior Member
Location
Texas
The buck/boost will only carry the va associated with the approximately 32 volts added or dropped. The isolation will carry the va of the entire load.

240 volts - 208 volts = 32 volts
32 volts / 240 volts = 13.333 percent of the load

208 to 208 will carry the whole load, thus, must be rated 100%.
Look at the red selection again.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Let me see if I can help.

A buck-boost transformer works in boost by adding its secondary voltage to the primary by being wird in series with it in an additive way.

But, the added voltage doesn't supply free energy. It comes from the primary being energized because it's paralleled with the incoming voltage.

So, whatever load demand is made by the load is manifest by adding 13.3% more current at the lower voltage, which is the line side of the BB.

The current at the lower (incoming) voltage is 13.3% higher than it is at the higher (outgoing) voltage. That power has to come from somewhere.

So, the BB only has to carry the power its secondary adds to the load. This is unlike a standard transformer, which must be sized to carry the entire load.
 

Chamuit

Senior Member
Location
Texas
Let me see if I can help.

A buck-boost transformer works in boost by adding its secondary voltage to the primary by being wird in series with it in an additive way.

But, the added voltage doesn't supply free energy. It comes from the primary being energized because it's paralleled with the incoming voltage.

So, whatever load demand is made by the load is manifest by adding 13.3% more current at the lower voltage, which is the line side of the BB.

The current at the lower (incoming) voltage is 13.3% higher than it is at the higher (outgoing) voltage. That power has to come from somewhere.

So, the BB only has to carry the power its secondary adds to the load. This is unlike a standard transformer, which must be sized to carry the entire load.
Yeah, what he said! :grin:
 

rcwilson

Senior Member
Location
Redmond, WA
Draw the buck-boost's two transformer windings with the 208 V winding H1-H2 and the 32 volt winding X1-X2. Connect X2 to H1. Connect the 240V load to X1 and H2. Run wires from H1 and H2 to your 208V service, H2 on the neutral. Your volatges are H1-H2 = 208V, X1-X2 = 32V, X1-H2 = 208+32 = 240V.

Now chase the current through the transformer to your load. Assume it is 2400VA so the current at 240V is 10 amps and 11.53 amps at 208 V.

10 amps flows from the X1 to the load and back to the H2 junction and on to the neutral of the panel. That means 10 A has to flow X2-X1 since it is in series with the load.

11.53 amps flows from the panel to the H1/X2 junction. 10 amps of that goes to X1 and the load, the remaining 1.53 amps goes through the winding to H2, where it adds to the 10 amps from the load to make the 11.53 amps going back to the panel neutral.

Now check the KVA needed. X1-X2 = 32V x 10 A = 320 VA. H1-H2 = 1.53 A x 208 V = 320 VA. Load KVA is still 240V x 10 A = 2400 VA . 320/2400= 13.3%.

The transformer is only having to make the 32 V boost, not the full 240V. 32V/240V = 13.3%


Draw the same diagram with your 208-240 V isolation transformer. Ouput = 2400 VA so amps on the 240V winding = 10 A. Amps on the 208 V winding = 11.53 A. The transformer is having to make the full 240V so it has to be a 2400 VA minimum size.
 

Smart $

Esteemed Member
Location
Ohio
I still believe, but I need to wrap the concept.
Let's start with an isolation scenario...

The main thing to consider here is the full load currrent passes through both secondaries.



However, we do not need to isolate the 208V source from the 208V portion going to the load, so we simply bypass/remove it. The full load current still passes through the B/B secondary, so it has to be rated for its voltage times the full load current.

 

LarryFine

Master Electrician Electric Contractor Richmond VA
Your volatges are H1-H2 = 208V, X1-X2 = 32V, X1-H2 = 208+32 = 240V.
Technical side note: If I'm not mistaken, the secondary is actually lower, because the 32v secondary is based on the primary receiving 240v, and not the actual 208v.

That means the secondary is really ((208/240) * 32) 27.73v, if I'm correct. It doesn't change the theory, but the output would actually be (208 + 27.73) 235.73v, not 240v.
 

wch

Member
When I get tired of working 9-5 I'll try to understand what y'all just told me for free. It's amazing work you do--nice to know somebody understands. Now
if I can just connect the dots....
Regards, and I appreciate the help.
wch
 

wch

Member
smart$

smart$

You are tapping the 208 leads to a T/F that will make a 32 volts secondary.
You run the 208 leads across the 32v. secondary to the load. The 32v adds and the load uses it up like it sees 240v. If I can't take it from there,
I'll study it some more.
That's pretty slick. How come I couldn't get that from the Square D diagram?
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Well, if you want details, make sure you grasp a couple of things:

Most buck/boost transformers have dual-voltage primaries and secondaries, so you must make sure each set is connected in parallel (low v) or in series (hi v).

The primary is easy, because it's always wired for the supply voltage range. You wire them in parallel for 120v and in series for either 208v to 240v or 240v to 208v.

The secondary can be either parallel series, depending on the supply voltage (actually measured is best), the desired voltage, and whether you're boosting or bucking.

Every BB I've seen comes with a chart of voltages and loads, which point to a diagram or chart that shows how to connect everything for the desired voltage output.


Just to add, you're merely supplying the primary with the circuit supplying the load. The BB is being connected as an auto-transformer, which is a single winding with taps.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
These are excellent pics with which to explain! :)

The first one is made up of a 208v to 208v unit on bottom and a 208v to 32v unit on top. You can see how the two primaries are in parallel, and the secondaries are in series:


The second pic shows how the lower (imaginary) unit is eliminated, and the 32v secondary is now in series with the supply itself, so its voltage adds to the source's voltage:


Just to mention, bucking is done by again connecting the primary and secondary in series, applying the supply across the entire string, and taking the lower voltage across the 'primary' alone.
 
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