Voltage drop calcuations for PV systems.

1. Voltage Drop Problems from
2009 NABCEP Study Guide
SolPowerPeople, Inc.
Austin, TX
March 19, 2012

2. Problem 47

3. Problem 47

4. Problem 47
 Use Chapter 9, Table 8 of the NEC to determine wire resistance.
Choose from the stranded, uncoated copper wire.

5. CHOOSE
FROM
UNCOATED
COPPER IN
ohm/kFT
INDICATES
STRANDE
D

6. Problem 47
Distance* x Imp x wire resistance/1000 ft.
VD% =
System voltage
*Distance is round-trip One way distance = 60 feet
Im = 7 amps
System voltage = 24 V
2 x 60ft x 7 amps x 0.491 ohms/1000 ft.
VD% = = 0.017
24 V
You can guess at the size conductor that will
work and then plug in values for wire
resistance from ch9 table 8. Here we have
guessed that #6 will work, which it does.

7. Problem 47
Distance* x Imp x wire resistance/1000 ft.
VD =
System voltage
One way distance = 60 feet
Im = 7 amps
*Distance is round-trip System voltage = 24 V
2 x 60ft x 7 amps x 0.778 ohms/1000 ft.
VD = = 0.027
24 V
While we have determined that #6 is less than 2% (1.7%
actually), we still don’t know if that is the smallest conductor
that will stay below 2%. We must plug in the resistance of the
next smallest conductor, #8 at 0.778 ohms. Result is that #8
has too much resistance and puts us over 2% (at 2.7%) so #6
is the correct answer.

8. Problem 47
 Another thing to consider with this problem is that the
answer implies that we are to be using the nominal
system voltage to calculate the %VD. Later, in problem
50, the voltage drop calculation is done with the actual
Vmp of the modules instead of system voltage. This,
presumably, is done because the question states “under
maximum POWER conditions” instead of just when “Im is
flowing” or “when current is the maximum power current”.
 I believe this distinction is made to illustrate the difference
in voltage drop calculations between systems that have
MPPT tracking versus those that do not (hence being
held down closer to the nominal system voltage).

9. Problem 49

10. Problem 49

11. Problem 49

12. Problem 49
Calculating the necessary Ω/kFT, using V = 24 V, I = 7 A, %VD = 1, and d = 5
gives 3.42, which corresponds to 14 AWG copper, which has the next lowest
value.
Referring the Figure provided, if the length from the junction box to the circuit combiner is five
feet, the smallest wire size needed to keep the voltage drop in this circuit less than 1% when
the current in the circuit is the maximum power current, is
a. 14 AWG copper
b. 12 AWG copper
c. 10 AWG copper
d. 8 AWG copper

13. Problem 49
System Voltage 24 VDC
1% VD in our 24 VDC system is 0.01 x 24 = 0.24 VDC
Imp = 7 amps
One Way Distance = 5 feet
VD = 2 x one way distance x Imp x wire resistance/1000 ft.
Wire Resistance/1Kft (Rc) = VD
x 1000
(2 x one way distance x Imp)
0.24
Rc = = 0.24 x 1000 = 3.42 Ω/kFT
2 x 5ft x 7A 70
Therefore, conductor resistance needs to be < 3.42 ohm/1000 ft.

14. Problem 49
Referring to the Figure, if the length from the junction box to the circuit
combiner is five feet, the smallest wire size needed to keep the voltage drop in
this circuit less than 1% when the current in the circuit is the maximum power
current, is
a. 14 AWG copper
b. 12 AWG copper
c. 10 AWG copper
d. 8 AWG copper

15. Problem 50

16. Problem 50

17. Problem 50
If the distance from the junction box to the combiner box is 60 feet, to keep the
voltage drop between the module junction box and the source circuit combiner
box less than 2% under maximum power conditions at STC, the smallest wire
size that can be used for each source circuit in the system in the Figure
provided is
a. 10 AWG copper
b. 8 AWG copper
c. 6 AWG copper
d. 4 AWG copper
 In this case, the problem is asking you to use the maximum power
voltage as the system voltage, likely because of the use of a MPPT
charge controller.

18. Problem 50
VD = 2 x one way distance x Imp x resistance/1000 ft.
VD
Conductor Resistance = X 1000
(2 x one way distance x Imp)
Voltage Drop of 2% of 34.2 V = 34.2 x 0.02 = 0.684 V
0.684 0.684
Rc = = x 1000 = 0.814
2 x 60 x 7 840
Wire resistance < 0.814 Ω/kFT

19. Problem A
 What conductor should you use in the PV output circuit in
a two string system with 12 modules per string, each
module with a Vmp of 34.2V and Imp of 8.2A? The
distance from the combiner box to the DC disco is 200
feet. Keep the voltage drop under 2%.

20. Problem A
 What conductor should you use in the PV output circuit in
a two string system with 12 modules per string, each
module with a Vmp of 34.2V and Imp of 8.2A? The
distance from the combiner box to the DC disco is 200
feet. Keep the voltage drop under 2%.
Distance x Imp x wire resistance/1000 ft.
VD =
System voltage
2(200)x 2(8.2)x 1.24/1000=8.1
34.2 x 12 = 410.4
8.1/410.4 = 1.9%
2(200) x 2(8.2) x 1.98 / 1000 = 12.98
12.98/410.4 = 3.1%
Therefore #10 is the smallest conductor that will
work.

21. Problem B
 What is the smallest size conductor you can have from
an AC disconnect of a 2000W/120V Inverter to the main
panel if it is 150 feet away? Assume maximum voltage
drop of 1%.

22. Problem B
 What is the smallest size conductor you can have from
an AC disconnect of a 2000W/120V Inverter to the main
panel if it is 150 feet away? Assume maximum voltage
drop of 1%.
VD
Conductor Resistance = X 1000
(2 x one way distance x Imp)
2000/120=16.666 amps
2AWG with a resistance
.01 x 120 = 1.2 V (voltage loss at 1%) of .194
(1.2 / (16.66 x 300)) x 1000 = .240 ohms/kFT

23. Problem C
 What AWG would you use to design for less that 2%
voltage drop between a PV output circuit and DC
disconnect, assuming the distance between is 14 feet
and the system consists of 4 modules in series with an
Imp. 8.2A and a Vmp of 27.6V?

24. Problem C
 What AWG would you use to design for less that 2%
voltage drop between a PV output circuit and DC
disconnect, assuming the distance between is 14 feet
and the system consists of 4 modules in series with an
Imp. 8.2A and a Vmp of 27.6V?
VD
Conductor Resistance =
(2 x one way distance x Imp)
2% voltage drop = 0.02 x (4 x 27.6) = 2.2 V
Imp = 8.2 amps
Distance = 28 feet
System Voltage = 110.4
(2.2/(28 x 8.2)) x 1000 = 9.58 ohms/kFT
18AWG at 7.95 is the smallest conductor in
the table.

25. Problem D
 In a system consisting of 3 series strings of 14 modules,
each with a Vmp of 29.5V and an Imp of 7.8A, what is the
smallest conductor available to allow for less than 1.5%
voltage drop between a PV output circuit and the DC
disconnect that is 32 feet away?

26. Problem D
 In a system consisting of 3 series strings of 14 modules,
each with a Vmp of 29.5V and an Imp of 7.8A, what is the
smallest conductor available to allow for less than 1.5%
voltage drop between a PV output circuit and the DC
disconnect that is 32 feet away?
VD
Conductor Resistance = X 1000
(2 x one way distance x Imp)
(6.195V/(64 x 23.4A)) x 1000 = 4.13 ohms/kft
29.5 x 14 = 413 V or
Imp = 3 x 7.8A = 23.4 A 64 x 23.4 x 3.14 / 1000 / 413 = 1.13%
Distance = 32 x 2 = 64
1.5% x 413 = 6.195V Answer: 14AWG with resistance of 3.14

27. Problem E
 What wire size would you use to design for less than 1%
voltage drop over a 160 foot distance from a 3000W/240
V inverter to the main distribution panel?

28. Problem F
 Which size conductor would be required to design for
less than 1% voltage drop from the 8A PV output circuit
to the DC disconnect stand alone system with a 24V
battery bank, assuming the distance is 22 ft?