# Buck Boost for Various Load Currents

#### synchro

##### Senior Member
In your previous thread you mentioned that the 11.33A load is for a sign and that other 120V loads could add up to a total of as much as 20A:

If you can fit four wires in the conduit for L1, L2, N, and EGC on a MWBC then you could put the 11.33A load on L1 and the other loads on L2.
If they were all #12s then the voltage drop you'd get on L1-N and L2-N vs. their currents would be:

IL1IL2Itot = IL1+IL2IN = |IL1-IL2|L1-N dropL2-N drop
11.33A0A11.33A11.33A9.75V 8.13%0V
11.33A8.67A20A2.66A6.02V 5.02%4.88V 4.07%

#### Jerramundi

##### Senior Member
I'm aware. It's because of your suggestion in the previous thread that I ran the numbers for a MWBC for several different conductor sizes and it was a good suggestion (aside from the present day arguments pushing us away from MWBC's all together).. but right now I'm not interested in other solutions. I want to understand the buck boost scenario and I mostly have it. It's just the additional current on the line side of the transformer that's tripping me up.

I understand it in theory, but I don't understand how you can apply that additional current, which is ultimately the result of a VD calculation because the VD calculation determines what percentage of a boost voltage is required, to the VD calculation...

#### LarryFine

##### Master Electrician Electric Contractor Richmond VA
Just to make it simple (including for myself), I'll use easy numbers.

If you need to boost a voltage by 12v, and the load is 20a, the BB needs to produce 12v at 20a, or 240w. Ignoring losses, etc., that means the primary needs 240w at its primary voltage, let's say 120v. So, the BB will add 2a to the conductors between the source and the BB.

So, you have to supply 22a to the BB at 120v to cover the load and the BB. If you locate the BB at the load, the entire long run will suffer the voltage drop at 22a instead of at 20a. If you need additional boost because of the additional drop, the BB will need more than 200w.

#### winnie

##### Senior Member
If you want to solve that question exactly, you end up with a differential equation and a headache.

That is why you probably want to approximate. Make a guess as to the expected voltage drop. From the load characteristics figure out what the load will draw. Now use that current for your voltage drop calculation. If your initial guess was reasonable then the answer you get will be close.

How the applied voltage changes the load current depends on the specific load. Reduced voltage will reduce the current to a resistor. But a constant power load such as an LED driver will draw more current as the voltage drops. There is no general rule, it depends on the load.

If you invert the voltage drop question the boost transformer at the front of the circuit might make more sense. Assume that the circuit delivers 120V at the load. That gives you the load current. Do the voltage drop calculation to figure out what the voltage needs to be at the head of the circuit to get 120V at the load.

Say you need 120V and 11A at the load, and to get this you need 132V at the head of the circuit. You need a 10% voltage boost. For a transformer to boost output voltage by 10% it has to draw 10% more input current.

If the transformer were near the load then the increase in current would mean additional voltage drop. Another differential equation: )

Jon

#### Jerramundi

##### Senior Member
Just to make it simple (including for myself), I'll use easy numbers.

If you need to boost a voltage by 12v, and the load is 20a, the BB needs to produce 12v at 20a, or 240w. Ignoring losses, etc., that means the primary needs 240w at its primary voltage, let's say 120v. So, the BB will add 2a to the conductors between the source and the BB.

So, you have to supply 22a to the BB at 120v to cover the load and the BB. If you locate the BB at the load, the entire long run will suffer the voltage drop at 22a instead of at 20a. If you need additional boost because of the additional drop, the BB will need more than 200w.
Okay. I'm beginning to understand the benefits of placing the BB nearer the source than the load.

#### Jerramundi

##### Senior Member
...How the applied voltage changes the load current depends on the specific load. Reduced voltage will reduce the current to a resistor. But a constant power load such as an LED driver will draw more current as the voltage drops. There is no general rule, it depends on the load...

...Say you need 120V and 11A at the load, and to get this you need 132V at the head of the circuit. You need a 10% voltage boost. For a transformer to boost output voltage by 10% it has to draw 10% more input current...
Understood, but this 10% Boost Voltage = 10% Increase In Line Current is only true if the load is a CPL, correct?

#### Jerramundi

##### Senior Member
Just for the sake of mathematical competency, entertain the following scenario in which I place the BB near the load and the subsequent calculations if you will...

I have (2) loads: (1) An LED ECM that the manufacturer states is 11.33A Continuous @ 120VAC. (2) LED Outline Lighting that totals 77.5W Continuous @ 12VDC (~6A @ 12VDC).

Given both loads are LEDs on Constant Power Drivers (i.e. 77.5W / 120VAC = 0.65A), the total Load Current should be 11.33A + 0.65A = 11.98A Continuous @ 120VAC, which for simplicity sake I will round up to 12A. So I have my Load Current.

Now I enter this Load Current into the VD Calculations (which may not be accurate due to the temporary calculator I'm using) for a 12A Load Current on #12AWG @ 271 Feet from Source to Load. I get a 10.33V drop and an available voltage at the load of 109.67V.

I choose a 10% BB for 120V on the load side of the BB. Therefore 12A * 120V = 1440W.
Again, given this is a CPL, 1440W = I * 109.67V... and I (w/ increase in Line Current due to BB) = 13.13A actual.

This is where I get tripped up...
Now I take the 13.13A in total and run ANOTHER VD calculation???

#### kwired

##### Electron manager
Just for the sake of mathematical competency, entertain the following scenario in which I place the BB near the load and the subsequent calculations if you will...

I have (2) loads: (1) An LED ECM that the manufacturer states is 11.33A Continuous @ 120VAC. (2) LED Outline Lighting that totals 77.5W Continuous @ 12VDC (~6A @ 12VDC).

Given both loads are LEDs on Constant Power Drivers (i.e. 77.5W / 120VAC = 0.65A), the total Load Current should be 11.33A + 0.65A = 11.98A Continuous @ 120VAC, which for simplicity sake I will round up to 12A. So I have my Load Current.

Now I enter this Load Current into the VD Calculations (which may not be accurate due to the temporary calculator I'm using) for a 12A Load Current on #12AWG @ 271 Feet from Source to Load. I get a 10.33V drop and an available voltage at the load of 109.67V.

I choose a 10% BB for 120V on the load side of the BB. Therefore 12A * 120V = 1440W.
Again, given this is a CPL, 1440W = I * 109.67V... and I (w/ increase in Line Current due to BB) = 13.13A actual.

This is where I get tripped up...
Now I take the 13.13A in total and run ANOTHER VD calculation???
What is VA rating of your BB?

Your main load is 1440 VA, the BB will draw 12 amps x the 12 volts boosted as the bulk of additional VA, may be some added inefficiencies that are much less significant.

#### Jerramundi

##### Senior Member
The reason this trips me up is because it seems like circular logic...

If I run that second VD calculation, same scenario, new load current due to BB, I get an increased 11.45VD or an available voltage of 108.55VAC.

Since the 108.55VAC is less than the above 109.67VAC, would I not see ANOTHER increase in line current due to BB?
(i.e. 1440W = I * 108.55... and I = 13.27A)

Which according to the above logic of running a 2nd VD at the new line current, should dictate I run a THIRD VD...

And round and round we go? See where I'm getting tripped up?

#### Jerramundi

##### Senior Member
What is VA rating of your BB?

Your main load is 1440 VA, the BB will draw 12 amps x the 12 volts boosted as the bulk of additional VA, may be some added inefficiencies that are much less significant.
500 or 0.5kVA. But I'm less concerned about having the appropriate rated BB than I am understanding this seemingly circular logic of running a 2nd VD calculation after getting my increased line current due to BB. That's what I'm hung up on and I am not good at letting a lack of mathematical understanding go, even if it's in the best interest of my bottom line, lol.

My brain thirsty, yo lol... more than my wallet is.

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#### kwired

##### Electron manager
The reason this trips me up is because it seems like circular logic...

If I run that second VD calculation, same scenario, new load current due to BB, I get an increased 11.45VD or an available voltage of 108.55VAC.

Since the 108.55VAC is less than the above 109.67VAC, would I not see ANOTHER increase in line current due to BB?
(i.e. 1440W = I * 108.55... and I = 13.27A)

Which according to the above logic of running a 2nd VD at the new line current, should dictate I run a THIRD VD...

And round and round we go? See where I'm getting tripped up?
Presuming resistance of the circuit remains constant, VD is totally dependent on current, so any time you change volts current is going to change or any time you change current volts (dropped, the source is presumed to remain constant) is going to change.

If your load is pure resistance the calculations are easier, if your load tends to at least try to maintain constant power it can be trickier trying to get an exact calculation.

#### winnie

##### Senior Member
Understood, but this 10% Boost Voltage = 10% Increase In Line Current is only true if the load is a CPL, correct?
A transformer always trades voltage for current. If the transformer doubles the voltage from primary to secondary, then the current going into the primary must be half that going out of the secondary. (Approximately...losses and magnetizing current come into play)

This trade of voltage for current is separate from the load characteristics.

The transformer is not itself a load consuming power but rather a device that trades voltage and current as it coupled power to a load.

Jon

• Jerramundi

#### winnie

##### Senior Member
And round and round we go? See where I'm getting tripped up?
The loop that is tripping you up is real; you are running smack dab into the concept of infinite series and limits.

If you actually did the calculation that you described (and had a variable BB transformer so that you could pick the exact voltage ratio), you would find at each step you would get closer and closer to a 'limiting' value. That limiting value is the true voltage drop as corrected by the variable BB.

There are ways to calculate this sort of infinite series, without going around in circles forever.

Jon

• Jerramundi

#### kwired

##### Electron manager
500 or 0.5kVA. But I'm less concerned about having the appropriate rated BB than I am understanding this seemingly circular logic of running a 2nd VD calculation after getting my increased line current due to BB. That's what I'm hung up on and I am not good at letting a lack of mathematical understanding go, even if it's in the best interest of my bottom line, lol.

My brain thirsty, yo lol... more than my wallet is.
If you are trying to maintain same load level you would have if VD were not a concern, then you will need to pull more VA to overcome the lost VA in the resistance of the circuit, which unfortunately does result in more current and more lost VA due to resistance in the circuit.

Keeping load to simple resistance load for now - if you have 100 VA rated load but are losing 10 VA in the conductors then the load is only seeing enough voltage that it only puts out 90 VA.

You then find appropriate BB transformer to get correct voltage at the load, but total VA in the circuit is going to be the 100 VA load plus whatever is lost in the conductors, which will be more than it was initially as the conductor resistance remained the same but either voltage or current increased depending on whether the BB was near source or near load.

Ohm's law is not violated.

#### Jerramundi

##### Senior Member
The loop that is tripping you up is real; you are running smack dab into the concept of infinite series and limits.

If you actually did the calculation that you described (and had a variable BB transformer so that you could pick the exact voltage ratio), you would find at each step you would get closer and closer to a 'limiting' value. That limiting value is the true voltage drop as corrected by the variable BB.

There are ways to calculate this sort of infinite series, without going around in circles forever.

Jon
Okay, so the circular logic is real and you would just keep running the VD calculation with a smaller and smaller increase in line current as you approach infinity?

I'm not crazy. Whew. So, this is what you meant by "differential equation and a headache?"

I did really well with math in school. Took AP courses in high school and everything... but Calculus made me it's b*tch, lol.

#### kwired

##### Electron manager
A transformer always trades voltage for current. If the transformer doubles the voltage from primary to secondary, then the current going into the primary must be half that going out of the secondary. (Approximately...losses and magnetizing current come into play)

This trade of voltage for current is separate from the load characteristics.

The transformer is not itself a load consuming power but rather a device that trades voltage and current as it coupled power to a load.

Jon
But in his application he still has to overcome the losses in the long conductors, resulting in more VA than the load itself is rated for, plus any lesser significant inefficiencies in the transformer.

#### winnie

##### Senior Member
Okay, so the circular logic is real and you would just keep running the VD calculation with a smaller and smaller increase in line current as you approach infinity?

I'm not crazy. Whew. So, this is what you meant by "differential equation and a headache?"

I did really well with math in school. Took AP courses in high school and everything... but Calculus made me it's b*tch, lol.
One approach to solving these sort of equations is to iterate until the change from one step to the next gets small enough.

There are other techniques.

But at some point the real world says hello and you realize that it doesn't pay to make the calculation any more accurate because error caused by things like manufacturing tolerances is greater than the calculation errors.

Look up 'convergent series ' . Fun math to play with.

Jon

#### Jerramundi

##### Senior Member
But in his application he still has to overcome the losses in the long conductors, resulting in more VA than the load itself is rated for, plus any lesser significant inefficiencies in the transformer.
What do you mean by "...resulting in more VA than the load itself is rated for..." ??

I plan to listen to the advice of installing the BB closer to the source. I'm just giving the scenario of it at the load to help better understand this circular logic of multiple and repetitive VD calculations.

#### winnie

##### Senior Member
The load consumes VA. The conductors also consume VA.

The VA lost in the conductors depends on the current flowing in the conductors. By putting the BB transformer at the head of the circuit you reduce the current flowing in the circuit and thus the VA lost in the conductors

Jon

#### MD Automation

##### Member
Okay, so the circular logic is real and you would just keep running the VD calculation with a smaller and smaller increase in line current as you approach infinity?

I'm not crazy. Whew. So, this is what you meant by "differential equation and a headache?"

Ha ha – Jerramundi you are not crazy And Winnie is spot on – it’s all math yo! And he is also correct when he indicates it’s not worth trying to be perfect here (Perfect is the enemy of Good) so knowing when to say “Good Enough” is important!

I think you see correctly that your constant power load will draw more current to compensate for the voltage drop, then that slightly higher new current means more voltage drop, which means more current drawn which means more voltage drop… and on and on.

You will find, that for your application, the “limit” that Winnie is describing is approached quite rapidly. We are not doing rocket science and I’d argue anything much beyond that first digit to the right of the decimal point (tenths) is not to lose any sleep over.

If you iterate, that is keep calculating, your new voltage drop calculations you will see that after the 2nd or 3rd iteration you settle quickly on ~ 13.26A

For fun…

1440 W load draws 12A at 120V

At 271 feet and 12A, 120V drops to 109.7V............................Voltage drop calc #1

Load now draws 13.13A at 109.7V

At 271 feet and 13.13A, 120V drops to 108.7V.......................Voltage drop calc #2

Load now draws 13.25A at 108.7V

At 271 feet and 13.25A, 120V drops to 108.6V........................Voltage drop calc #3

Load now draws 13.26A at 108.6V

You could keep recalculating this, over and over, and you’d get a new answer every time. But after the 3rd calculation, it’s silly to keep doing because the answer is changing only slightly. So iterate a few times to watch how fast you approach a limit and then know when to say, That’s good enough!

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• Jerramundi