Correcting V.D. with Buck Boost Transformers

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RPEC_Muns

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We are working on a project where the electrical engineer is using a Buck Boost Transformer to correct the voltage due to the distance from the electrical panel. We feel that it will not work but unable to prove or disprove that we will have the correct voltage at the receptacle.

The incoming voltage is 120-volt, 1-phase circuit, distance from the panel to the buck boost transformer is 1,300-ft and they are specifing #10AWG wire. They indicate to provide 120V/120V 2KVA buck boost transformer and showing a electrical load of 720VA. From the transformer to the receptacle is another 300-ft with #10AWG.

I've read some of the buck boost transformers can boost/correct the voltage by up to 25% but I've was unable to determine/confirm if this will work.
 
A problem with using a buck-boost transformer for correcting voltage drop is the higher voltage that will occur when less load is present. If you make a correction for a 10%VD you will have a 10% overvoltage when no .old is present.

Buck-boost is usually a small transformer with a 12, 16, 24, or 32V secondary which is connected in an autotransformer configuration. Because this arrangement only needs to "transform the voltage change" a much smaller transformer is used. A 2kVA buck-boost arranged to correct a 10% VD could handle a 20kVA load.

Is your load really only 720VA at 120V?
 
A buck/boost transformer can absolutely be used to boost voltage. While commonly used for less than a 25% boost, they can certainly do more.

The _serious_ problem with this is voltage stability. Imagine you have a 25% voltage drop under load, and then you compensate with a 33% voltage boost (75% * 133% = 100%). What voltage do you get when you don't have a load present? Your 120V receptacle is now delivering 160V.

There are 'voltage regulating transformers' which will compensate for voltage drop in a dynamic fashion to maintain a steady 120V output. But these are rather more expensive beasties then an ordinary buck/boost setup.

If you can split the load up you might be able to run a multi-wire branch circuit and get the benefit of higher supply voltage without transformers.
 
ptonsparky said:
I think it’s 12/24 or 16/32 volts typically. I didn’t do the math to see how that works out for you. Do a VD calc and see what you get.
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When you take the lenght 1300-ft, figure 6.25amps with #10AWG you end up droping 20-volts or 16.83% and that is just to the primary side of the transformer.
I took the 103.17-volts and multiplied it by 25% which is the maxium output from the transformer, I ended up at 128.96-volts on the secondary side of the transformer.
So from the secondary side of the transformer, the lenght to the outlet is 300-ft, 6.25amps with #10AWG you end up with 3.61% V.D.

If the load increases and/or decreases we could end up with a higher or lower voltage reading.

I'm unsure if my calcuations are right, but also don't think this is the correct application for the busck boost transformer
 
RPEC_Muns said:
When you take the lenght 1300-ft, figure 6.25amps with #10AWG you end up droping 20-volts or 16.83% and that is just to the primary side of the transformer.
I took the 103.17-volts and multiplied it by 25% which is the maxium output from the transformer, I ended up at 128.96-volts on the secondary side of the transformer.
Click to expand...
There is a bit of subtly here that will catch you off guard. Say your load is 6.25A. But you put in the buck/boost transformer that is trading voltage for current. The current for the 1300 foot run is actually going to be 6.25 * 1.25 = 7.8A. If you were to go down the buck/boost path, your best result is to put the transformer at the beginning of the run.

As a separate issue, the '25%' maximum boost number is just an approximate limit. The real boost depends upon the specific transformer selected; if you use a typical 120/240V : 16/32V transformer (like this one: https://www.automationdirect.com/ad..._buck-boost_transformers_(nema_rated)/t181058 ) you would have a boost ratio of 32/120 = 26.6%

RPEC_Muns said:
So from the secondary side of the transformer, the lenght to the outlet is 300-ft, 6.25amps with #10AWG you end up with 3.61% V.D.

If the load increases and/or decreases we could end up with a higher or lower voltage reading.

I'm unsure if my calcuations are right, but also don't think this is the correct application for the busck boost transformer
Click to expand...

Your calculations are close enough to see the problem. Redo the calculation with no connected loads and see how high the voltage gets.

RPEC_Muns said:
The outlets are for a softball field, so it would be a pitching machine, coach's cell phone or computer, radio. No specific loads
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The pitching machine will likely be very unhappy. That is a motor load where starting draws lots of current (huge voltage drop) and then the current drops considerably. The voltage will be all over the place. The cell phone and computer would probably be just fine, because the power supplies are probably rated for 100-240V. The radio? Might be fine or might let the smoke out.

If the pitching machine will work at 240V, I'd consider running a 10/3 circuit to provide a 240V outlet for the pitching machine and 120V outlets for the others.
 
For a load this small, I wouldn't bother boosting up to 480V. Remember that transformers have their own impedance/voltage drop, and that an energized transformer consumes power even if the load isn't connected.

But before using any transformers at all, I'd see if the pitching machine could run at 240V. This same circuit run at 240V has a voltage drop less than 5%.
 
This is a lot of engineering and calculating for no good reason. The pitching machine ain't gonna go on strike if it only gets 108V instead of 120V.

A Multi Wire Branch circuit will help with the voltage drop, put two plugs out there, upsize the wire to #4, or #3 if you're worried about it, and leave the transformers on the drawing board.
 
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