A few corrections have been made to this file on 9/20/12

1. Formulas Review for
NABCEP PVI Exam Part
2
Sarah Raymer
Director of Education and Training Services
SolPowerPeople, Inc.

2. Notice to viewers:
Photovoltaic (PV) Installer Resource Guide
(Revised August 2012, version 5.3) Please click here
to download a copy.
The PV Resource Guide is an excellent resource for
those planning to take the NABCEP Solar PV Installer
Certification Exam.
While it remains a useful study resource it, as always,
is NOT the definitive guide to the exam. Also, we
would like to remind readers that unlike the exam
questions in the NABCEP Certified PV Installer
Exam, sample questions in this guide were not
written under the supervision of a professional
psychometrician.

3. Notice to viewers:
The newly revised NABCEP Installer Resource
Guide Version 5.3 includes the Solar Installer Job
Task Analysis WITH links and suggested
readings on where to locate information to help
you learn and understand those concepts!
Verify System Design, Confirm
String Sizing Calculations [Sect. 5]

4. http://www.solarpaneltalk.com/showthread.php?
727-Any-NABCEP-certified-members-around
Post:
I wonder if this certification is as tough to
acquire as some folks gossip it to be...
Comment:
Why, yes, yes it is.
I have been in the industry for a few years as a
full time professional & also teach the NABCEP
Cert of Knowledge and PV Advanced courses at
a local community college & I had trouble
passing it.

5. Some Need-to-Know
Formulas
http://www.sengpielaudio.com/calculator-ohm.htm
Basic Electrical
Formulas
Voltage Drop
Calculations
Maximum Circuit
Current
Calculations
Conductor and
OCPD Sizing
It’s actually neither 

6. Basics
The Formula Wheel

7. The Power Equation
The Magic Triangle
The Electric Power Law

8. Ohm’s Law
Ohm’s Magic Triangle
Note: V is often an E
in equations. They are
interchangeable.

9. It’s neither!

10. Voltage Drop

11. Voltage Drop
NEC 2011 Code references - additional notes:
210.19[A][1]

12. Voltage Drop
NEC 2011 Code references - additional notes:
215.2[A][4]

13. Voltage Drop
Ch , Table 8 2011 NEC Codebook

14. Voltage Drop
The formula:
CM = [25 x I x L] ÷ Vdrop
CM – Circular mills
I – single phase current [amps]
L – one-way length
Vdrop – % voltage drop

15. Voltage Drop
CM = [25 x I x L] ÷ Vdrop
CM is the circular mills diameter of the conductor [see
NEC Chapter 9, Table 8]
I is the Imp/amps [Are you looking at 1 string, 2
strings, max power conditions or STC?]
Length is ONE-WAY distance
Vdrop is the % of the Voltage of the system that is
lost, so:
If 240V, < 1%: 240V x .01 = < 2.4V
If 48, < 2%: 48V x .02 = < .96V

16. Voltage Drop
Formula approach:
What do we know?
What do we need to know?
CM = [25 x I x L] ÷ Vdrop
Look at the problem
What is given
What is asked
Clarify
Refer to diagrams and codebook

17. Voltage Drop
Example 1- # 47 of the NABCEP™ 2009 Study
Guide:
• D = 60 Feet
• System is a 24 V system
• Vdrop = < 2% [2% of 24V = 24V x .02 = .48]
• I =7A Imp – one string “when Im is flowing”
• What wire size [CM] could we use to stay under 2%?

18. 24 volt system, 2 strings of 2 modules in
series

19. Voltage Drop
WHAT DO WE KNOW?
• D = 60 Feet
• System is a 24 V system
• Vdrop = < 2% [2% of 24V = 24V x .02 = .48]
• A = Question involves junction box to combiner box circuit: 2
modules in series, 7 amps each, “when Im is flowing”
• We are looking at one string!
• SVA, PAA- series volts add, parallel amps add
CM = [25 x I x L] ÷ Vdrop
WHAT DO WE NEED TO KNOW?
• What wire size [CM] could we use to stay under 2% [<.48V]
CM = [25 x 7x 60] ÷[24V x .02]

20. Voltage Drop
CM = [25 x 7x 60] ÷[24V x .02]
Solve:
25 x 7 x 60 = 10,500
10,500 ÷ .48 = 21,875
Now- what wire size is at least 21,875 CM?
#8 is only 15,510, not big enough
#6 is 26,240- bigger than 21,875!
Answer is C, #6!

21. Voltage Drop
CM = [25 x I x L] ÷ Vdrop
Example 2- # 49 of the NABCEP 2009 Study
Guide:
We know:
I = 7A
L = 5ft
V = 24V
Vdrop = < 1%

22. Voltage Drop
CM = [25 x I x L] ÷ Vdrop
We know:
#16 is only
I = 7A
2580, not big
L = 5ft
enough
V = 24V
Vdrop = < 1%
#14 is 4110-
bigger than
3,646- OK!
Solve:
CM = [25 x 7 x 5] ÷ [.01 x 24]
CM = 875 ÷ .24 A, #14!
CM = 3645.8 or 3,646

23. Voltage Drop
CM = [25 x I x L] ÷ Vdrop
Example 3- # 50 of the NABCEP 2009 Study
Guide:
Note- this question is stating “under maximum power conditions at
STC”. The Max power voltage for this circuit which is 2 modules in
series would be 34.2V [Vmp x 2].
We have to be careful about what is being asked- this is tricky!

24. • SVA- series volts
add!
• MAX power
conditions!

25. Voltage Drop
CM = [25 x I x L] ÷
Vdrop
We Know:
I = 7A
L = 60ft
V = 34.2V
Vdrop = < 2%
So- Vdrop is 34.2V x .02 =
.684
Solve:
CM = [25 x 7 x 60] ÷ [.684]
CM = 10500÷ .684 How does it work? Forums are great!
CM =15351 http://www.physicsforums.com/showthread.ph
So B, # 8 would be it! p?t=577708

26. Voltage Drop
CM = [25 x I x L] ÷
Vdrop
What size wire would carry a load of
40A at 240V a distance of 500 feet with
2% or less voltage drop?
CM = [25 x 40 x 500] ÷ [240 x .02]
CM = 500,000 ÷ 4.8
CM = 104,167
What Size AWG?
Where do we find that?

27. Voltage Drop
CM = [25 x I x L] ÷
Vdrop
What size wire would carry a load of
40A at 240V a distance of 500 feet with
2% or less voltage drop?
CM = [25 x 40 x 500] ÷ [240 x .02]
CM = 500,000 ÷ 4.8
CM = 104,167
What Size AWG? #1/0
Where do we find that?

28. Voltage Drop
CM = [25 x I x L] ÷
Vdrop
What size wire would carry a load of
40A at 240V a distance of 500 feet with
2% or less voltage drop?
CM = [25 x 40 x 500] ÷ [240 x .02]
CM = 500,000 ÷ 4.8
CM = 104,167
What Size AWG? #1/0
Where do we find that?

29. Voltage Drop
CM = [25 x I x L] ÷
Vdrop
What size wire would carry a load of
40A at 240V a distance of 500 feet with
2% or less voltage drop?
CM = [25 x 40 x 500] ÷ [240 x .02]
CM = 500,000 ÷ 4.8
CM = 104,167
What Size AWG? #1/0
Where do we find that?
Chapter 9, Table 8 2011 NEC Codebook- the only
reference you’re allowed during the NABCP PVI Exam

30. Ch. 9 Table 8 2011 NEC Codebook

31. Voltage Drop
What voltage drop would arise from carrying 50A
a distance of 250 feet with #8AWG?
CM = [25 x I x L] ÷ Vdrop
16510CM = [25 x 50A x 250] ÷ Vdrop
16510CM= 312500 ÷ Vdrop
16510CM x Vdrop = 312500
Vdrop÷ 16510CM = 312500 ÷ 16510CM
Vdrop = 18. 92V
• Remember: Vdrop is the % of the
Voltage of the system that is lost, so:
If 240V, < 1%: 240V x .01 = < 2.4V
If 48, < 2%: 48V x .02 = < .96V

32. Last one…
What wire size would you use to design for less
than 1% voltage drop in a circuit carried over a
160 ft. distance from a 3000W/240V inverter to
the main distribution panel?
D = 160
V = 240, V drop is < 1% of 240 = 2.4V
I = 3000 ÷ 240 = 12.5A
CM = [25 x 12.5 x 160] ÷ [2.4]
CM = 50,000 ÷ 2.4
CM = 20,833
So - # 6 AWG at 26,240CM – is big
enough

33. Max Circuit Current, Conductor and
OCPD Sizing
2011 NEC Code
For this review, we are primarily dealing with:
Article 240 - Overcurrent Protection
240.4 - OCPD/AWG imitations
240.6 - OCPD sizes
Article 310 - Conductors for General Wiring
Table 310.15[B[2][a] - ambient temp. correction
Table 310.15[B][3][c] - # of current carrying
conductors in conduit adjustment
Table 310.15[B][3][c] – Rooftop temp. adjustment
Article 690 – Solar PV Systems
690.8 – Circuit Sizing and Current

34. Chapter 3 of the NEC 2011 Codebook
SOME PV References
Download complete file on our webpage:
http://solpowerpeople.com/2011-nec-2011-study-guide-
database/

35. Max Circuit Current,
Conductor and OCPD Sizing
MAX Circuit Current:
Use the Isc
Multiply for 1.25 for Irradiance spikes
If more than one, multiply by # strings: PAA
[parallel - amps add]
Minimum Conductor Ampacity and OCPD:
Use the Max. Circuit Current and multiply by
another 1.25 to calculate continuous load
ampacity.
Use the Max Circuit Current and apply
corrections and adjustments for conditions
of use.
USE THE HIGHER of these 2 calculations
for ampacity.

36. Circuits
Inverter Output
Circuit
[Inverter Input
Circuit]
Combiner box
PV Output Circuit
DC disconnect
PV Source Circuit
Inverter
AC disconnect

37. Step 1: Max. Circuit
Current
690.8[A] Max. Circuit Current:
690.8[A][1] - Isc x 1.25 – to account for high
irradiance fluctuations
690.8[A][2] - Multiply by # strings if more than one
in parallel
PAA [parallel- amps add]
690.8[A][3] – Inverter output circuit current =
continuous inverter output rating [on spec sheet-
or calculated]
2000W ÷ 240V = 8.3A

38. Step 1: Max. Circuit
Current
Inverter Max. Cir. Current is the Max Output Current from
the spec. sheet. Make sure you look at the right one for
your nominal system voltage and inverter size.
2000 ÷ 240 = 8.3A

39. Step 1: Max. Circuit
Current
Module Isc is 12.5A, what is the max. circuit
current?
12.5 x 1.25 = 15.63A
Isc of 12.5A, 3 series strings in parallel. What is
the PV output max circuit current [same as
Inverter input circuit]?
12.5 x 3 x 1.25 = 46.88A
Inverter’s max circuit current is the Max Cont.
Output Rating [spec sheet]
8.3A [2000 ÷ 240 = 8.3A]

40. Step 2: OCPD and
MINIMUM Ampacity
690.8[B][1]- Overcurrent Devices
690.8[B][1][a] – OCPD must carry at least 125%
of max circuit current
Isc x 1.25 x 1.25 or Isc x 1.56
690.8[B][1][b] – must be installed in accordance
to listing/labeling {110.3[B]}
690.8[B][1][c] – consider temp. rating of terminal.
You must use the 75ºC column for sizing
ampacity of conductor to match that allowed by
the terminal. You can use the 90ºC column for
ampacity adj. and corr. with a 90ºC rated
conductor {110.14[C]}

41. Step 2: OCPD and
MINIMUM Ampacity
690.8[B][2]- Conductor Ampacity
690.8[B][2] – conductors must be the larger of
the two
690.8[B][2][a] – Must be 125% of 690.8[A] without
corrections for conditions of use
Isc x 1.25 x 1.25
690.8[B][2][b] – Must be max. circuit current
calculated in 690.8[A] after conditions of use
corrections have been applied
690.8[B[[2][c] – The conductor must still be
protected by the OCPD

42. Step 2: OCPD and
MINIMUM Ampacity
690.8[B[[2][c]
OCPD must protect the wire that comes into it.
The wire coming into the terminal must have an
ampacity HIGHER than the OCPD rating.
Think-
BAD: If the conductor could only carry 18A before
failing, and your OCPD was a 20A breaker- it wouldn’t
trip and prevent a failure in the conductor.
GOOD: If you have a conductor that could carry up
to18A under conditions of use before failing and your
OCPD is a 15A fuse, the conductor will never get up to
failure point of 18A because the fuse will trip and open
the circuit at 15A- before the conductor fails.

43. Step 2: OCPD and
MINIMUM Ampacity
Minimum Conductor Ampacity and OCPD:
Use the Max. Circuit Current and multiply by another 1.25
to calculate continuous load ampacity.
690.8[B][1][a] – OCPD must carry at least 125% of max
circuit current
690.8[B][1][b] – consider temp. rating of terminal. For
example, if you are using a 75ºC rated terminal, you must
use the 75ºC column for sizing ampacity of conductor to
match that allowed by the terminal. You can use the 90ºC
column for ampacity adj. and corr. with a 90ºC rated
conductor {110.14[C]}
Isc x 1.25 x 1.25 or Isc x 1.56
Use the 75ºC column for this conductor calculation
because it must be designed to not exceed the terminal
temp rating

44. Step 2: OCPD and MINIMUM
Ampacity
What size conductor would be required for the
PV output circuit from the combiner box to the
disconnect in a system with 3 strings of 225 Watt
modules with an Isc of 6.7A?
(3) strings of Scheuten 225W Modules
Isc = 6.7A
Step 1- Determine Maximum Current-
Isc x 1.25 x 3
6.7 A x 1.25 x 3 = 25.125 A
690.8[B][2][a] Step 2- Determine MINIMUM
Conductor Ampacity and OCPD-
25.125 A x 1.25 = 31.4 A minimum

45. Step 2: OCPD and
MINIMUM Ampacity
6.7 A x 1.25 x 3 = 25.125 A
690.8[B][2][a]Step 2- Determine
MINIMUM Conductor Ampacity and
OCPD-
25.125 A x 1.25 = 31.4 A minimum
The next size OCPD is A 35A OCPD
[NEC 240.6]

46. OCPD sizes- NEC 240.6
Minimum Ampacity = 31.4 A minimum
OCPD = 35A (next standard size – NEC 240.6)

47. Step 2: Ampacity and
Minimum AmpacityOCPD
= 31.4 A
OCPD = 35A (next standard size – NEC 240.6)
Step 3: Choose Conductor-
Choose from Table 310.15(B)(16) from 75°C column (because
of terminal rating) = #10 AWG (max 35A) {690.8 [B}[1][b]}
690.8[B][1][b] – consider temp. rating of terminal. You must
use the 75ºC column for sizing ampacity of conductor to match
that allowed by the terminal. Later, you can use the 90ºC
column for ampacity adj. and corr. with a 90ºC rated conductor
{110.14[C]}
However, due to NEC 240.4(D)** which states the max. OCPD
for #10 AWG is 30A, you must go up to #8 AWG which can
handle up to 50A, and is not limited by the OCPD as per
240.4[D]**.

48. Table 310.15(B)(16) and **

49. **240.4(D)
At the bottom of Table 310.15(B)(16), there is a notation that must be
considered for wire sizes #10 through #18 (marked with 2 asterisks **).

50. Step 3: Conditions of Use
690.8[B][2][b]
Minimum Ampacity = 31.4 A minimum
OCPD = 35A (next standard size – NEC 240.6)
#8 AWG which is not limited by 240.4D
as#10AWG is- to be protected by no less than a
30A OCPD.
Step 4: Based on 90°C column for ampacity,
apply your derates based on conditions of use.
8 AWG allows for 55 AMPS.

51. Conductor Sizing - Conductor Ampacity
Under Conditions of Use
8 AWG allows for 55 AMPS.
Ambient Temperature is 40.5°C
Conduit is 4” off roof
3 strings have come to 1 in the PV
output circuit, so we have 2 current
carrying conductors

52. Conductor Sizing Under
Conditions of Use
In conduit 4” above the rooftop [Table
310.15(B)(3)(c)] add 17°C to the 40.5° to
get to 57.5°C , which derates to .71 (Table
310.15(B)(2)(a)).
There are only 2 conductors in conduit ,
so there is no deration multiplier for how
many conductors in conduit (Table
310.15(B)(3)(a)).

53. “Conditions of Use”
NEC Table 310.15(B)(2)(c) (2008)
NEC Table 310.15(B)(3)(c) (2011)
Temperature Adder is added to record high
temperature then derated using Table
310.15(B)(3)(c) Roof to
Distance From
Bottom of Conduit Temperature Adder
0 to ½ inch 33°C
½ to 3 ½ inches 22°C
3 ½ to 12 inches 17°C
12 to 36 inches 14°C

54. “Conditions of Use”
Correct for Number of Conductors in Raceway
NEC Table 310.15(B)(2)(a) (2008)
NEC Table 310.15(B)(3)(a) (2011)
Number of Current-
Carrying Conductors Correction Factor
4 to 6 0.80
7 to 9 0.70
10 to 20 0.50
21 to 30 0.45
31 to 40 0.40
Over 40 0.35

55. Ambient
Temperature
Correction Table
310.15(B)(2)(a)

56. Conductor Sizing Under
Condition of Use
8 AWG allows for 55 AMPS (90° column)
We are allowed to use 90ºC column as per 110.14C.
55 A x .71 = 39.05 A
Because 39.05 A is greater than max current of
31.4A, then #8 AWG is GOOD!
If you did the calculation with 90ºC column and a
#10AWG wire
40A x .71 = 28.4 Ampacity after COU. This is not OK
for the expected Max current of 31.4A.
Conductor can handle 39.05A, and so is protected by
the 35A OCPD.

57. More materials are available on our website:
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58. More materials are available on our
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59. Thank you!