Buck Boost for Various Load Currents

[kwired]

What do you mean by "...resulting in more VA than the load itself is rated for..." ??

There is resistance in the conductors, even on short conductors where you disregard voltage drop. Passing current through a resistance gives up heat, that is a real load to the source, it is wasted energy in relation to your application. More current you have in same resistance the more VA is lost to this "side effect".

Increasing conductor size reduces conductor resistance, means less voltage drop and less lost VA.

Increasing voltage but keeping intended load VA the same like for a constant watt output type of load would lower current, lower current through same resistance will lessen lost VA.

Transforming from 120 to 240 will halve the current, lessen how many volts are dropped in same resistance and also lessen VA lost in that same resistance.

You would have to add the loss in the conductors to the total VA if you still want rated output from the load, which is what will increase current in the run if the BB is at the load end of the run

If you put the BB at beginning of run, you will be boosting voltage to whatever level is needed to have desired volts at the load, after losses in the conductor. Sort of works out similar losses either way but is going to be some differences.

 

[Jerramundi]

Ha ha – Jerramundi you are not crazy

Well it's good to hear that from an objective source other than the Leprechaun perched on my shoulder.

 

[Jerramundi]

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

Yea, I just ran it a few times and realized the same thing. It was just the repetitive, circular logic that made me feel like I was doing something wrong.

 

[Jerramundi]

Transforming from 120 to 240 will halve the current, lessen how many volts are dropped in same resistance and also lessen VA lost in that same resistance.

Yea. I understand. I was just holding off on looking at the 240v option until I had my head wrapped on the buck boost scenario. The only thing with the 240V option is that to get the most efficient result, I should upgrade to #10's and on top of that, the 240v/120v step down is more expensive than the buckboost.

Whereas if I stick with the existing #12's and install a buckboost, I've saved on costs.

Then again given the variable loads, the 240v may be a better option. On the flip side, the variance is so minuscule in my application that I wonder if spending that extra $50-$75 to both (1) upgrade the conductors and (2) pay for the step down is worth it.

But that's on me to decide. I was just trying to get my head around the buckboost, increased line current, and repetitive VD calculations... and appreciate the mathematical assistance.

 

[wwhitney]

This is to entertain the hypothetical problem posed by the OP, not to suggest what the best practical solution is.

If you have a fixed supply voltage V, and a long conductor run (a resistor R), and a load that draws a constant power P0, then calculating the conductor current I is just algebra. There are a couple way to get there, but I think the simplest is just power accounting:

Power leaving the panel is I * V (assuming power factor 1)
Power dissipating as heat in the conductor run is I^2 * R
Power used at the load is P0

So I * V = I^2 *R + P0, a quadratic equation in I. In standard form, it's just R * I^2 - V * I + P0 = 0. Use the quadratic formula to find I as (V + sqrt(V^2 - 4*R*P0))/2R

Also, an idealized buck-boost transformer (no power loss) at the load doesn't change the conductor current at all, the above analysis still applies. If the voltage at the load (V0 - I * R) is within the LED driver's operating window, then the idealized buck-boost is useless; and a real buck-boost just makes the situation worse by adding some power losses. If (V0 - I*R) is outside the LED driver's operating window, then a buck-boost at the load could be used, but for a proper analysis, you'd want to know its power loss as a function of line side current. Then you could add that to the power accounting and get a new equation for I.

For a buck-boost transformer at the supply end of the conductors, the voltage V is now a variable depending on what buck-boost we use. Say you want the load voltage V - I * R = 120V. Then you can plug the formula for I above into that equation for load voltage and get a more complicated equation for V to solve for the boosted voltage. Choose a buck-boost appropriately. This is only useful if (a) without it the load voltage is outside the LED driver's window or (b) the power loss in the buck-boost is less than the reduction in I^2*R power losses in the conductors.

Cheers, Wayne

 

[oldsparky52]

Yea. I understand. I was just holding off on looking at the 240v option until I had my head wrapped on the buck boost scenario. The only thing with the 240V option is that to get the most efficient result, I should upgrade to #10's and on top of that, the 240v/120v step down is more expensive than the buckboost.

Whereas if I stick with the existing #12's and install a buckboost, I've saved on costs.

Then again given the variable loads, the 240v may be a better option. On the flip side, the variance is so minuscule in my application that I wonder if spending that extra $50-$75 to both (1) upgrade the conductors and (2) pay for the step down is worth it.

But that's on me to decide. I was just trying to get my head around the buckboost, increased line current, and repetitive VD calculations... and appreciate the mathematical assistance.

That's why you make the big bucks.

 

[kwired]

Yea. I understand. I was just holding off on looking at the 240v option until I had my head wrapped on the buck boost scenario. The only thing with the 240V option is that to get the most efficient result, I should upgrade to #10's and on top of that, the 240v/120v step down is more expensive than the buckboost.

Whereas if I stick with the existing #12's and install a buckboost, I've saved on costs.

Then again given the variable loads, the 240v may be a better option. On the flip side, the variance is so minuscule in my application that I wonder if spending that extra $50-$75 to both (1) upgrade the conductors and (2) pay for the step down is worth it.

But that's on me to decide. I was just trying to get my head around the buckboost, increased line current, and repetitive VD calculations... and appreciate the mathematical assistance.

I will say it again, buck boost is not a good way to fix voltage drop other than maybe to supply a load that doesn't vary in current, and even worse for multiple loads that can turn on/off pretty much at random. It will leave utilization voltage all over the place every time total load changes and possibly no better than just living with the voltage drop in many cases.

 

[hillbilly1]

This would be about as close of a solution for varying loads you can get short of upsizing the wire.

Voltage Boosters - Hughes Autoformers

hughesautoformers.com

 

[winnie]

Yea. I understand. I was just holding off on looking at the 240v option until I had my head wrapped on the buck boost scenario. The only thing with the 240V option is that to get the most efficient result, I should upgrade to #10's and on top of that, the 240v/120v step down is more expensive than the buckboost.

I don't understand the connection to upgrading to #10.

If you bring out 240V and step down to 120V, you will make better use of the existing #12 conductors. Double the voltage, half the current, 1/4 the % voltage drop for the same power.

Yes #10 would be better, but you could do #10 or 240V or both.

I know you are focusing on understanding the voltage drop issue with bb transformers in this thread; perhaps start a new thread about the 240V operation. There is lots to explore there as well, eg different LED drivers for the sign or autotransformers to get the 120V

 

[Jerramundi]

I don't understand the connection to upgrading to #10.

If you bring out 240V and step down to 120V, you will make better use of the existing #12 conductors. Double the voltage, half the current, 1/4 the % voltage drop for the same power.

Yes #10 would be better, but you could do #10 or 240V or both.

I know you are focusing on understanding the voltage drop issue with bb transformers in this thread; perhaps start a new thread about the 240V operation. There is lots to explore there as well, eg different LED drivers for the sign or autotransformers to get the 120V

It's just as you said. The #10's would be better.

I'm just practicing being cognoscente of the voltage drop "recommendations" of 3% on branch circuits, especially given that this circuit will be feeding a $16,000 LED sign. In other words, #12 on 240V post step-down would get me ~115V, which could be considered acceptable, but is still greater than 3%.

Even if I were to accept the 115V, the step-down is still more expensive than the buck boost.

 

[synchro]

It would be helpful to know what the specified voltage requirements are for the sign and the other loads. Some LED fixtures can operate over a 120V to 277V range or more, but even so you may be limited to a much narrower range by other loads that share the same circuit.

 

[winnie]

It's just as you said. The #10's would be better.

I'm just practicing being cognoscente of the voltage drop "recommendations" of 3% on branch circuits, especially given that this circuit will be feeding a $16,000 LED sign. In other words, #12 on 240V post step-down would get me ~115V, which could be considered acceptable, but is still greater than 3%.

Even if I were to accept the 115V, the step-down is still more expensive than the buck boost.

Hmm. 11.5A at 120V is the same power as 5.75A at 240V. #12 can carry 7.5A 271 feet at 240V with 3% voltage drop.

synchro has a good point on the LED drivers; you may be able to operate at 240V without a step down transformer. On the other hand if this expensive sign requires a narrow regulated voltage range, then buck-boost will bite you.

I'd expect an expensive LED sign to do things like have changing images...and that means changing power consumption, which means changing voltage drop. If you boost your voltage up and the sign shows a dark image, the supply voltage will rise and cause problems. This goes back to the original question you were asking: how do you understand the current drawn by a load given conditions of voltage drop, and how does boosting the voltage change this. But now you have the additional factor to consider: how the load itself changes current consumption based on use.

-Jon

 

[kwired]

Hmm. 11.5A at 120V is the same power as 5.75A at 240V. #12 can carry 7.5A 271 feet at 240V with 3% voltage drop.

synchro has a good point on the LED drivers; you may be able to operate at 240V without a step down transformer. On the other hand if this expensive sign requires a narrow regulated voltage range, then buck-boost will bite you.

I'd expect an expensive LED sign to do things like have changing images...and that means changing power consumption, which means changing voltage drop. If you boost your voltage up and the sign shows a dark image, the supply voltage will rise and cause problems. This goes back to the original question you were asking: how do you understand the current drawn by a load given conditions of voltage drop, and how does boosting the voltage change this. But now you have the additional factor to consider: how the load itself changes current consumption based on use.

-Jon

Yes why try to save $2-500 here when if you don't get it right you possibly compromise something worth much more.