|Voltage Drop Calculations
|By Mike Holt for EC&M Magazine
you know how and why to calculate voltage drop?
Contrary to common
belief, the NEC generally does not require you to size conductors to accommodate
voltage drop. It merely recommends that you adjust for voltage drop when sizing
conductors. The recommendations are in the Fine Print Notes to Sections
210.19(A), 215.2(A)(4), 230.31(C) and 310.15(A)(1). Fine Print Notes are
recommendations, not requirements [90.5(C)].
The NEC recommends that the
maximum combined voltage drop for both the feeder and branch circuit should not
exceed five percent, and the maximum on the feeder or branch circuit should not
exceed three percent (see Figure 8-10). This recommendation is a performance
issue, not a safety issue.
Note: Graphic are not included with this
If the NEC doesn't require you to size for voltage drop, why
even think about it? Here are some reasons:
System efficiency. If a
circuit supports much of a load, a larger conductor will pay for itself many
times over in energy savings alone.
System performance. Lighting loads
perform best when voltage drop is minimal. You get the light of a higher-watt
system simply by running larger wires.
Troubleshooting. If you follow the
NEC voltage drop recommendations, you don't have to guess whether your field
measurements indicate a problem or if the voltage is low due to not
accommodating voltage drop in the design.
Load protection. Undervoltage for
inductive loads can cause overheating, inefficiency, and shorter life span. When
a conductor resistance causes the voltage to drop below an acceptable point,
increase the conductor size (see Figure 8-9).
What is it?
drop of a circuit is in direct proportion to the resistance of the conductor and
the magnitude of the current. If you increase the length of a conductor, you
increase its resistance-and you thus increase its voltage drop. If you increase
the current, you increase the conductor voltage drop. Thus long runs often
produce voltage drops that exceed NEC recommendations.
Here's a pop quiz.
What is the minimum NEC-recommended operating voltage for a 115V load connected
to a 120V source (see Figure 8-11)?
(a) 120V (b) 115V (c) 114V (d)
Answer: (c) 114V
The maximum conductor voltage-drop
recommended for both the feeder and branch circuit is five percent of the
voltage source (120V). The total conductor voltage drop (feeder and branch
circuit) should not exceed 120V x 0.05 = 6V. Calculate the operating voltage at
the load by subtracting the conductor voltage drop from the voltage source: 120V
- 6V = 114V.
Doing the calculations
You can determine conductor
voltage drop by the Ohm's Law method or by the formula method. You can use the
Ohm's Law method (I x R) for single-phase only. Regardless of which method you
use, observe the following:
For conductors 1/0 AWG and smaller, the
difference in resistance between dc and ac circuits is so little that it can be
ignored. In addition, you can ignore the small difference in resistance between
stranded and solid wires.
VD = Voltage Drop
I = The load in amperes at
100%, not at 125%, for motors or continuous loads.
R = Conductor Resistance,
Chapter 9, Table 8 for dc or Chapter 9, Table 9 for ac.
Let's do an Ohm's
Law Method sample calculation. What is the voltage drop of two 12 AWG THHN
conductors that supply a 16A, 120V load located 100 ft from the power supply
(see Figure 8-12)?
(a) 3.2V (b) 6.4V (c) 9.6V (d) 12.8V
The math is straightforward:
VD = I x R
I = 16A; R =
2V per 1,000 ft, Chapter 9, Table 9: (2V/1,000 ft) x 200 ft = 0.4V
VD = 16A
x 0.4V = 6.4V
This method is a bit more involved that
the Ohms Law method, but the big advantage is you can use it for three-phase or
single-phase. Here are some additional items to observe:
= 2 x K x I x D/CM.
Three-phase VD = 1.732 x K x I x D/CM. The difference
between this and the single phase formula is you replace the 2 with 1.732.
= Direct-Current Constant. K represents the dc resistance for a 1,000-circular
mils conductor that is 1,000 ft long, at an operating temperature of 75ºC. K is
12.9 ohms for copper and 21.2 ohms for aluminum.
Q = Alternating-Current
Adjustment Factor: For ac circuits with conductors 2/0 AWG and larger, you must
adjust the dc resistance constant K for the effects of self-induction (eddy
currents). Calculate the "Q" Adjustment Factor by dividing the ac
ohms-to-neutral impedance listed in Chapter 9, Table 9 by the dc resistance
listed in Chapter 9, Table 8.
I = Amperes: The load in amperes at 100% (not
at 125% for motors or continuous loads).
D = Distance: The distance the load
is from the power supply. When calculating conductor distance, use the length of
the conductor-not the distance between the equipment connected by the conductor.
To arrive at this length, add distance along the raceway route to the amount of
wire sticking out at each end. An approximation is good enough. Where we specify
distances here, we are referring to the conductor length.
Circular-Mils: The circular mils of the circuit conductor as listed in NEC
Chapter 9, Table 8.
Let's do a three-phase example. A 3Ø, 36 kVA load rated
208V is wired to the panelboard with 80 ft lengths of 1 AWG THHN aluminum. What
is the approximate voltage drop of the feeder circuit conductors (see Figure
(a) 3.5V (b) 7V (c) 3% (d) 5%
Answer: (a) 3.5V
did we do that? Applying the three-phase formula, where:
K = 21.2 ohms,
I = 100A (36,000/(208 x 1.732))
D = 80 ft
83,690 (Chapter 9, Table 8)
VD = 1.732 x 21.2 x 100A x 80/83,690 = 3.51
Using basic algebra, you can apply the
same basic formula to find one of the other variables if you already know the
voltage drop. For example, suppose you want to know what size conductor you need
to reduce the voltage drop to the desired level. Simply rearrange the formula.
For three-phase, it would look like this: CM (3Ø) = 1.732 x K x I x D/VD. Of
course, for single-phase calculations, you would use 2 instead of
Suppose you have a 3Ø, 15 kVA load rated 480V and 390 ft of
conductor. What size conductor will prevent the voltage drop from exceeding
three percent (see Figure 8-17)?
K = 12.9 ohms, copper
I = 18A
(15,000/(480 x 1.732))
D = 390 ft
VD = 480V x 0.03 = 14.4V
1.732 x 12.9 x 18 x 390/14.4V = 10,892, 8 AWG, Chapter 9, Table 8
also rearrange the formula to solve a problem like this one: What is the maximum
length of 6 AWG THHN you can use to wire a 480V, 3Ø, 37.5 kVA transformer to a
panelboard so voltage drop does not exceed three percent (see Figure
D (3Ø) = CM x VD/1.732 x K x I
CM = 26,240, 6 AWG Chapter
9, Table 8
VD = 480V x 0.03 = 14.4V
K = 12.9 ohms, copper
I = 45A
(37,500/480 x 1.732))
D = 26,240 x 14.4/1.732 x 12.9 x 45 = 376
Sometimes, the only method of limiting voltage drop is to limit the
load. Again, we can rearrange the basic formula algebraically: I = CM x VD/1.732
x K x D. Suppose an installation contains 1 AWG THHN conductors, 300 ft long in
an aluminum raceway, to a 3Ø, 460/230V power source. What is the maximum load
the conductors can carry, without exceeding the NEC recommendation for voltage
drop (see Figure 8-21)? Let's walk it through:
I = CM x VD/1.732 x K x
CM = 83,690 (1 AWG), Chapter 9, Table 8
VD = 460V x 0.03 = 13.8V
K = 12.9 ohms, copper
D = 300 ft
I = 83,690 x 13.8/1.732 x 12.9 x
300 = 172A
Note: The maximum load permitted on 1 AWG THHN at 75ºC is 130A
[110.14(C) and Table 310.16].
As you can see, voltage drop is something
you calculate. And calculating it is easy. The NEC doesn't require you to do
these calculations, so voltage drop isn't a safety issue. However, a system that
meets NEC requirements may not be efficient in terms of power consumption or
optimum in terms of equipment longevity and performance. As the old saying goes,
"Wire is cheap, but performance loss is costly." First ensure your system meets
NEC requirements. Then, look at it for voltage drop so you get the right cost
trade-offs between performance and cost.