How to figure the voltage drop at different points along a circuit?

[The Electrician]

This wire table says 12 ga wire is .001588Ω/ft.

https://www.powerstream.com/Wire_Size.htm

 

[Babadag]

It seems that actually it is a little more complicate.

The supply voltage is AC -as usual- and a driver AC/DC is provided in order to supply DC. Then 350 mA it could be the DC output [18-42 V].Let’s say Vdc=20 V Pdc=0.35*20=7 W.

In the interval of 115 to 125 Vac the LED current is 350 mA, but the Iac current is variable from 7/115=0.06087 to 7/125=0.056 A neglecting the losses and power factor.

If we consider 0.8 as power factor and 95% as efficiency we get

S=7/0.8/0.95=9.21VA and the currents Iacmin=9.21/125=0.07368 A and 36.87 degrees

Iacmax=9.21/115=0.08 A and 36.87 degrees.

 

[The Electrician]

@alan2023, are you still around?

 

[berkeman]

@alan2023, are you still around?

From their profile page:

Last seen:
Yesterday, 10:03 AM

 

[alan2023]

Yes, I'm still here.

Without getting into all the nitty-gritty with the inner workings of an LED driver/fixture, I believe I'm starting to understand a bit more how the loads after a point on a circuit affect the voltage drop.

Looking at this original question from an electrical install perspective, my challenge to someone would be how to build a spreadsheet to plug in the distances and any other necessary values (voltage, wire size/resistance, amps, etc.). As an electrician, I use custom calculators usually in a spreadsheet to determine voltage drop, conduit fill, cable tray fill. I first understand the code/math requirements, then build the sheet/formulas.

Wanting to add the feature of multiple loads on the same circuit, (which is something very common, and voltage drop becomes an issue in long distance applications like exterior lighting), I believe I could just build the circuits including the wire as a resistor. So basically a combo series/parallel circuit. I know there are probably some great software tools you guys use, but I believe this could built in a spreadsheet and in doing so, I have a calculator to speed up future calculations, but also understand how it works.

Thank you all for the input, and I'm sorry if sent us down a rabbit hole.

 

[Baluncore]

Looking at this original question from an electrical install perspective, my challange to someone would be how to build a spreadsheet to plug in the distances and any other necessary values (voltage, wire size/resistance, amps, etc.).

An estimate of the voltage drop can be made by assuming equal loads, distributed equally along the cable.

The number of loads is, n = 13.
The current drawn by each load is, i = 0.35 A.
The length of the cable is, d = 580 ft.
The resistance per unit length is, r = 1.6 milliohm/ft.

The average current is, i * n / 2.
The return resistance per unit length is, 2 * r.
The twos will cancel.

V drop = n * i * d * r = 4.2224 volt.
V far end = 120.0000 - 4.2224 = 115.7776 V.

 

[The Electrician]

Looking at this original question from an electrical install perspective, my challenge to someone would be how to build a spreadsheet to plug in the distances and any other necessary values (voltage, wire size/resistance, amps, etc.). As an electrician, I use custom calculators usually in a spreadsheet to determine voltage drop, conduit fill, cable tray fill. I first understand the code/math requirements, then build the sheet/formulas.

Here's how you could setup a spreadsheet using 7 columns, A,B,C,D,E,F,G. This setup assumes
a constant load current of .35 A per load.

Think of the length of wiring between loads as constituting a "zone". There are 13 zones also counting the distance from the 120V feed to the first load as a zone.

Column A is the distance of each load from the 120 volt feed.
Column B is the distance between loads, including the distance from the feed to the first load.
Column C is just a reverse integer sequence starting with the number of loads.
Column D is .35 times column C (element by element). These are the currents in each zone.
Column E is .0032 (2 * the resistance per foot of the wire used; the approximate value used here is
the one given by Berkeman. For other than 12 ga, use 2 * the value for that wire) times column D.
This is the voltage drop per foot of wire in a zone.
Column F is column B times column E (element by element). At this point the elements of column F are the voltage drops within each zone; they are the decrease in voltage from one load to the next.

Column G is the voltage at each load. Creating column G requires the use of a procedure. Start with 120V and subtract the first element of column F; this becomes the first element of column G. Now subtract the second element of column F from the just calculated first element of column G and this will be the second element of column G. Then subtract the third element of column F from the second element of column G and this will be the third element of column G. Continue this to the end.

 

[alan2023]

Here's how you could setup a spreadsheet using 7 columns, A,B,C,D,E,F,G. This setup assumes
a constant load current of .35 A per load.

View attachment 331825

Think of the length of wiring between loads as constituting a "zone". There are 13 zones also counting the distance from the 120V feed to the first load as a zone.

Column A is the distance of each load from the 120 volt feed.
Column B is the distance between loads, including the distance from the feed to the first load.
Column C is just a reverse integer sequence starting with the number of loads.
Column D is .35 times column C (element by element). These are the currents in each zone.
Column E is .0032 (2 * the resistance per foot of the wire used; the approximate value used here is
the one given by Berkeman. For other than 12 ga, use 2 * the value for that wire) times column D.
This is the voltage drop per foot of wire in a zone.
Column F is column B times column E (element by element). At this point the elements of column F are the voltage drops within each zone; they are the decrease in voltage from one load to the next.

Column G is the voltage at each load. Creating column G requires the use of a procedure. Start with 120V and subtract the first element of column F; this becomes the first element of column G. Now subtract the second element of column F from the just calculated first element of column G and this will be the second element of column G. Then subtract the third element of column F from the second element of column G and this will be the third element of column G. Continue this to the end.

What application is that in the snip?

 

[The Electrician]

It's just a math app that I used to format the data in columns, but I didn't do the calculations with it; I did the actual calculations on my HP50g calculator. I don't have Excel on my computer so I couldn't actually give an example of the calculations done in Excel, but I wanted to show what it probably would look like in Excel.