Buck Boost for Various Load Currents

Jerramundi

Senior Member
Location
Chicago
Occupation
Licensed Residential Electrician
Just hoping to get some feedback on the following scenario - a bucks boost transformer with various load currents.

I'm planning to install a buck boost transformer on a longer run of #12 w/ 20A OCPD, but the circuit is potentially going to have various load currents.
At the low end, it will be pulling 11.33A consistently. At the high end, for calculation purposes, I'm inputting the full 20A.

I quickly ran the numbers through an online voltage drop calculator (which may not 100% accurate, but just go with me on this) and got the following results...

VoltageDrop1_MinA.JPG VoltageDrop2_MaxA.JPG

Then I took the results for both scenarios and plugged them into Square D's / Schneider Electric's Buck Boost Calculator and got the following...

BuckBoost1_MinA.JPG BuckBoost2_MaxA.JPG

My plan is to go with the highlighted #500SV46B Transformer and the highlighted Wiring Diagram #1 on the top right for both scenarios.

Obviously the Schneider Electric Buck Boost Calculator makes different recommendations for different scenarios, but am I correct in assuming the highlighted transformer model numbers (i.e. #250SV46B and #500SV46B) are the same (if wired the same), just rated for different VA's?

So theoretically if I go with the #500SV46B for both scenarios, I should the get the same boost voltage (125v - 117v) as displayed in the calculators above in both scenarios despite the calculator recommending a 250 on the left? Correct?
 

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synchro

Senior Member
Location
Chicago, IL
Occupation
EE
Your calculations are with the buck boost placed on the load side of the #12 run. You'd be better off putting the buck boost on the supply side of the run as discussed here:
 

synchro

Senior Member
Location
Chicago, IL
Occupation
EE
Your calculations are with the buck boost placed on the load side of the #12 run.
Also with the buck boost on the load side of the run the current drawn on the wires is 13.3% higher than your load current. And so that higher current needs to be used in the voltage drop calculation.
 

Jerramundi

Senior Member
Location
Chicago
Occupation
Licensed Residential Electrician
Your calculations are with the buck boost placed on the load side of the #12 run. You'd be better off putting the buck boost on the supply side of the run as discussed here:
This question may make me come across as ignorant, but gosh golly darnit I don't have a clue because for my purposes, I use this forum to become a better electrician.

When you say "your calculations are with the buck boost placed on the load side of the #12 run," do you mean the far end of the branch circuit closer to the actual load, as opposed to being near the source / panel (i.e. supply side)?

I've just never looked at a single stretch of wire as having both a supply side and a load side unless there is an intermittent device, splice, etc... but I suppose these terms would make sense.

If that's the case and you're recommending placing the buck boost nearer the source / panel (i.e. supply side), wouldn't that affect the input voltage into the buck boost and thus change recommended equipment?
 
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synchro

Senior Member
Location
Chicago, IL
Occupation
EE
When you say "your calculations are with the buck boost placed on the load side of the #12 run," do you mean the far end of the branch circuit closer to the actual load, as opposed to being near the source / panel (i.e. supply side)?
Yes, that's what I meant.
I've just never looked at a single stretch of wire as having both a supply side and a load side unless there is an intermittent device, splice, etc... but I suppose these terms would make sense.
Yeah, I don't think those are official terms in the NEC. But it's important to distinguish the two if we are concerned about voltage drop, because the load side of the wire will have a lower voltage than the source / supply side.

If that's the case and you're recommending placing the buck boost nearer the source / panel (i.e. supply side), wouldn't that affect the input voltage into the buck boost and thus change recommended equipment?
Yes that's possible, you just have to go through the numbers. As an example if you put an available voltage of 120V from the POCO into the Square-D calculator, select a 13.3% boost, and a 20A load current it gives us the following result:
boost voltage: 136
transformers 500SV46B or 750SV46F (first one same as you had)

Now your on-line calculator showed a voltage drop across the #12 at 20A of 17.22V. And so the voltage on the load at the end of the run would be 136V - 17.22V = 118.7V.

By the way, with your on-line voltage drop calculator the percentage of voltage drop depends on the value of the input voltage to the wire, but the actual drop in volts across the wire does not (it only depends on the current). That's why I could just take the same 17.22V drop from your calculation even though the input voltage is the boosted 136V in this case vs. the 120V that you input to the calculator.
 

Jerramundi

Senior Member
Location
Chicago
Occupation
Licensed Residential Electrician
Yeah, I don't think those are official terms in the NEC. But it's important to distinguish the two if we are concerned about voltage drop, because the load side of the wire will have a lower voltage than the source / supply side.
I'm not trying to be a stickler about NEC terminology, at least not this time, haha.

I've just never thought about it that way using those terms... but you're absolutely right in regards to it being relevant to voltage drop (i.e. the voltage available at the source end of a branch circuit will be greater than the voltage available at the load end of the same branch circuit).
 

Jerramundi

Senior Member
Location
Chicago
Occupation
Licensed Residential Electrician
By the way, with your on-line voltage drop calculator the percentage of voltage drop depends on the value of the input voltage to the wire, but the actual drop in volts across the wire does not (it only depends on the current).
Humor me for a second... because I'm an inquisitive fella.

I follow you that the "actual drop in volts" is dependent upon the current only (i.e. in regards to the Ohm's Law Method of VD)
(i.e. VD = I * R)... but could you not argue that the "actual volts dropped" IS dependent upon the input voltage because the amount of volts dropped depends on the load current which is relative to the load voltage?

In other words, my 11.33A load current used in the voltage drop calculation is only 11.33A @ 120V.... If that voltage changes from 120V, my load current used in the voltage drop calculation will change, correct?

If my understanding is correct... and I'm by no means a master yet, the only "constants" we have to rely on in regards to VD is (1) Wattage and (2) Resistance... that is as long as you stick to the Ohm's Law Method and not the VD = 2 * K * Q * I * D/CM formula, which introduces other constants.
 
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curt swartz

Electrical Contractor - San Jose, CA
Location
San Jose, CA
Occupation
Electrical Contractor
BB transformers are a horrible choice to deal with voltage drop issues. If your load is 100% fixed they can be used but should not be used when the loads may change.

For a circuit like this you should just increase wire size. If that is not possible then run the circuit at 240 volts then step it down to 120 at the load.

What are the loads?
 

Jerramundi

Senior Member
Location
Chicago
Occupation
Licensed Residential Electrician
BB transformers are a horrible choice to deal with voltage drop issues. If your load is 100% fixed they can be used but should not be used when the loads may change.

For a circuit like this you should just increase wire size. If that is not possible then run the circuit at 240 volts then step it down to 120 at the load.

What are the loads?
Understood and I appreciate the input. That's why I've written this post - to discuss the use of a buck boost transformer with varying loads.

But with all due respect, for reasons I'd rather not get into because it will just take us off topic, I'd rather not digress from discussing this particular set of circumstances.

Basically, given financial constraints, the 240V circuit is not an option because I'm trying to utilize the existing conduit and the largest wire I can fit, even at 240V, will still result in a over a 9% VD once stepped back down to 120V.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Humor me for a second... because I'm an inquisitive fella.

I follow you that the "actual drop in volts" is dependent upon the current only (i.e. in regards to the Ohm's Law Method of VD)
(i.e. VD = I * R)... but could you not argue that the "actual volts dropped" IS dependent upon the input voltage because the amount of volts dropped depends on the load current which is relative to the load voltage?
You are correct, load current will change with available voltage. How it changes depends on the load characteristics and current could go up or down depending on the specifics.

Usually voltage drop calculations assume a constant load current as a simplifying assumption, but this probably is not accurate for the large voltage drop you are considering.

Jon
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Basically, given financial constraints, the 240V circuit is not an option because I'm trying to utilize the existing conduit and the largest wire I can fit, even at 240V, will still result in a over a 9% VD once stepped back down to 120V.
Would you mind a double check on that?

At 20A you calculated a 14% voltage drop.

The same power delivered at 240V requires only 10A, and the voltage drop is relative to a higher voltage. I would expect only 3.5% + transformer drop if you supplied 240V and stepped down.

Jon
 

Jerramundi

Senior Member
Location
Chicago
Occupation
Licensed Residential Electrician
Also with the buck boost on the load side of the run the current drawn on the wires is 13.3% higher than your load current. And so that higher current needs to be used in the voltage drop calculation.
This part I'm still struggling with a little bit. I understand that the boost voltage on the load side of the transformer will be seen as increase in current on the line side of the transformer because as voltage goes down from one side of the transformer to the other (load to line), the amount current will increase to maintain the constant wattage...

...but how can you apply this increase in current flow to the voltage drop... when prior to figuring out what your boost voltage will even be and thus the increase in current on the line side of the transformer, you need a pre-voltage drop load current amount to begin with??

Are you saying figure out what the increase in load current will be because of the buck boost and then redo the voltage drop calculation with that new current amount? That seems like circular logic.
 
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kwired

Electron manager
Location
NE Nebraska
Understood and I appreciate the input. That's why I've written this post - to discuss the use of a buck boost transformer with varying loads.

But with all due respect, for reasons I'd rather not get into because it will just take us off topic, I'd rather not digress from discussing this particular set of circumstances.

Basically, given financial constraints, the 240V circuit is not an option because I'm trying to utilize the existing conduit and the largest wire I can fit, even at 240V, will still result in a over a 9% VD once stepped back down to 120V.
What is the load and how does this voltage drop impact it?

Buck boost not the greatest way to deal with voltage drop unless maybe it is a fixed continuous load. When the load is off or at a low level of draw, voltage will be high.

Also if voltage drop is still an issue if you went with 240 then transformed to 120 at the load -- you really need to increase conductor somehow or maybe step voltage up to 480 or even 600 and then step down at load. Either way going to cost you, unless the load can take the drop you are having without any serious effects.
 

Jerramundi

Senior Member
Location
Chicago
Occupation
Licensed Residential Electrician
When the load is off or at a low level of draw, voltage will be high.
I understand. That exact same issue is depicted in the OP. That's why I wrote this post...

... but I'm stuck on what Synchro said about the increased current on the line side of the transformer and applying it to the VD which is why I'm trying to (respectfully) shoo off other ideas until I've worked through this one.
 

Jerramundi

Senior Member
Location
Chicago
Occupation
Licensed Residential Electrician
Initially I was just looking for confirmation that the boost voltage would be the same for both a #250SV46B and #500SV46B if wired the same way, with the same available voltage, and the same load current, and that they are just rated for different VA's...

... but then Synchro hit me with the suggestion about putting the transformer nearer the source and the issue about increased line current. So that's what I'm trying to work through before entertaining different solutions.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
In other words, my 11.33A load current used in the voltage drop calculation is only 11.33A @ 120V.... If that voltage changes from 120V, my load current used in the voltage drop calculation will change, correct?
No. A fixed resistance will drop a fixed voltage with a fixed current.

What changes with voltage change is the percentage the drop is relative to the overall voltage.

Added: Don't confuse comparing 120v to 240v circuits, where the equipment is built to consume the same power at different voltages. We're discussing a dynamic analysis of a system with given equipment.
 
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LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
Are you saying figure out what the increase in load current will be because of the buck boost and then redo the voltage drop calculation with that new current amount? That seems like circular logic.
It is, but that's how it works. If you energize such a set-up in your mind, the voltage at the load will settle at a steady state, and re-settle every time the load changes.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
Initially I was just looking for confirmation that the boost voltage would be the same for both a #250SV46B and #500SV46B if wired the same way, with the same available voltage, and the same load current, and that they are just rated for different VA's...
Yes, as long as the smaller unit is large enough, they should behave (almost) identically.

... but then Synchro hit me with the suggestion about putting the transformer nearer the source and the issue about increased line current. So that's what I'm trying to work through before entertaining different solutions.
The idea is to avoid having the energy necessary to energize the transformer (primary) be an additional load on the long run.

Added: In other words, let's say the BB would use 5a to provide the load with the boosted voltage at its current at that voltage. Why add the 5a to the long run, when you can energize it at the source and simultaneously reduce the voltage drop?
 

Jerramundi

Senior Member
Location
Chicago
Occupation
Licensed Residential Electrician
No. A fixed resistance will drop a fixed voltage with a fixed current.

What changes with voltage is the percentage.
But the current is not fixed....

If I'm installing a set-up that has a total wattage of 1359.6W and manufacturer's specs call for an ideal 120V... and I plug in 120V for the "ideal operating voltage" in order to calculate my load current I get 11.33A.

But after I plug that load current of 11.33A into my VD and realize that the operating voltage will actually be somewhere around 110V, the load current is no longer 11.33A. Per P=I*E, my load current will actually be 12.36A when running on the true, post VD calculation operating voltage of 110V... no?

It seems to me that the 120V is just an ideal number that we use for calculations, and obviously strive to hit for the purposes of proper operation and efficiency... but as we discover the true operating voltage via VD calculations, it stands to reason via P = I * E that the load current then changes.

Or I am just allowing VD to run circles around me? lol
 
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LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
But after I plug that load current of 11.33A into my VD and realize that the operating voltage will actually be somewhere around 110V, the load current is no longer 11.33A. Per P=I*E, my load current will actually be 12.36A... no?
For motor loads, yes, because they tend to operate as constant-power loads. It takes a given amount of power to produce a given torque. That's why a 208/230v motor uses more current on 208v.

For resistive loads, no, because the current will rise and fall proportionately with applied voltage. You must keep in mind which values are the constants and which are the variables when calculating.
 
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