Voltage Drop - MVOLT LED Drivers

[Sparks215]

Good Morning, I was looking through the forums about voltage drop for my case but couldn't find much.

How do you calculate voltage drop for multiple legs of a circuit with self adjusting LED drivers? I can calculate the voltage drop in the first leg of the circuit using amperage * resistance in the wires but when you add multiple legs, do you use the voltage 'seen' (panel voltage - volts dropped) at the 1st light as the new voltage for the second leg? The factory mentioned the driver will increase current to meet the wattage of the LED system due to voltage drop.

Also, how does this affect the voltage drop in the first leg? When you add a second leg in parallel, the current is added to those cables feeding the 1st light and the volts drop even further causing the amps to go up in the second leg and seems like an endless cycle, am I overlooking something?

 

[tortuga]

Whats the system voltage, amps of each light and the spacing?
Have you looked at a 'Mean Average Voltage Drop' calculation ?
Where:
K*2L/A = R And I*R =Vd
Therefore
K*2L*I/A = Vd
Therefore
K*2L*I/Vd = A
Then the (L) with varying distances is calculated by the amp feet.

 

[ptonsparky]

I, for one would like to hear back how it all works out.

 

[winnie]

Also, how does this affect the voltage drop in the first leg? When you add a second leg in parallel, the current is added to those cables feedingst light and the volts drop even further causing the amps to go up in the second leg and seems like an endless cycle, am I overlooking something?

You grasp the essence of the problem.

1 The voltage drop in a length of wire depends on the resistance of the wire and the total current flowing in the wire from all the loads connected to that wire.

2 The net voltage at the end of the wire is the voltage at the start minus the drop, as you describe the starting point for the second leg is the end of the first leg.

3 The current in the wire is set by the load, and with a constant power load the current goes up as the voltage goes down.

A full and accurate answer requires solving simultaneous sets of equations; but you probably don't need this sort of accuracy. If you simply assume a maximum allowed voltage drop amount, then figure how much current the loads draw at this reduced voltage, and then calculate the required wire size based on this current level, you will be close enough and conservative.

Say this is a 120v circuit, and you allow 5%. There is also utility tolerance so figure that 110v gets to the lamps. Calculate the lamp current using 110v then do your wire sizing based on that current. The real current will be a few% lower, thus the actual voltage drop slightly better, but you will know the wire is big enough, but the approximation is close enough that the wire won't be grossly oversized.

Jon

 

[Sparks215]

Appreciate the feedback.
tortuga - we're using 277V, and the lights are 4.5kVA for each leg. The first leg is spaced 300' from the panel and the second is 500' from the first leg. I haven't heard of mean average voltage drop - is the formula you posted used for that?

winnie - I was leaning towards using the reduced voltage as the new to calculate the amperage but felt like I could calculate multiple iterations and would just keep reducing the voltage. What's messing me up is that the driver increases the current based on the lower voltage which increases the voltage drop - it seems like an endless iterative problem. However assuming a max allowed voltage drop seems like a great starting point.

 

[winnie]

winnie - I was leaning towards using the reduced voltage as the new to calculate the amperage but felt like I could calculate multiple iterations and would just keep reducing the voltage. What's messing me up is that the driver increases the current based on the lower voltage which increases the voltage drop - it seems like an endless iterative problem. However assuming a max allowed voltage drop seems like a great starting point.

You are exactly right; there are ways of solving this sort 'infinite series', but true wisdom is realizing that you _don't_ need the complete accurate solution. You just need a good enough approximation, which means not too much voltage drop and wire not excessively oversized.

But with 277V you presumably have 3 phase available. By 2 legs, do you mean 2 sets of lights on the same 277V circuit, or do you mean 2 separate 277V circuits, possibly sharing a neutral, or do you mean a full 3 phases first going to 1 set of lights and then to the next?

-Jon