1. Insulation Resistance Calculations of Airfield Lighting Circuits
35th annual Hershey conference
Insulation Resistance Calculations of Airfield Lighting Circuits
Joseph Vigilante, PE

2. Presentation Objective
To develop a formula to calculate insulation resistance for an airfield lighting circuit and provide theoretical and real-life examples.

3. Presentation Agenda Cable insulation resistance (IR) background
FAA IR guideline recommendations
Formula development
Airfield lighting circuit calculations
Summary
Open Q&A

4. Anatomy of insulation current flow
Capacitance charging currents, C
Absorption current, RA
Conduction current, RL
Circuit model
DC VOLTAGE
SOURCE RA C RL S

5. Anatomy of insulation current flow
CURRENT - MICROAMPERES
100 90 80 70 60 50 40 30 25 20 15 10 9 8 7 6 5 4 3
2.5 2
1.5 1
0.1
0.15
0.2
0.25
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
SECONDS
CAPACITANCE CHARGING CURRENT

6. Anatomy of insulation current flow
CURRENT - MICROAMPERES
100 90 80 70 60 50 40 30 25 20 15 10 9 8 7 6 5 4 3
2.5 2
1.5 1
0.1
0.15
0.2
0.25
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
SECONDS
CAPACITANCE CHARGING CURRENT
ABSORPTION CURRENT

7. Anatomy of insulation current flow
CURRENT - MICROAMPERES
100 90 80 70 60 50 40 30 25 20 15 10 9 8 7 6 5 4 3
2.5 2
1.5 1
0.1
0.15
0.2
0.25
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
SECONDS
CAPACITANCE CHARGING CURRENT
ABSORPTION CURRENT
CONDUCTION CURRENT

8. Anatomy of insulation current flow
CURRENT - MICROAMPERES
100 90 80 70 60 50 40 30 25 20 15 10 9 8 7 6 5 4 3
2.5 2
1.5 1
0.1
0.15
0.2
0.25
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
SECONDS
CAPACITANCE CHARGING CURRENT
TOTAL CURRENT
ABSORPTION CURRENT
CONDUCTION CURRENT

9. Types of insulation resistance testing
Short-time/spot reading
TIME
60 sec
MEGOHMS
VALUE READ AND RECORDED
TIME (MONTHS)
MEGOHMS
VALUES CHARTED

10. Types of insulation resistance testing
Time-resistance method
TIME
10 Min
MEGOHMS
INSULATION SUSPECT
INSULATION PROBABLY OK

11. Testing airfield lighting circuits
AC 150/ F, Design & Installation Details for Airport Visual Aids - Chapter 12, Equipment & Material, Section 13, Testing
50MΩ Non-grounded series circuits
FAA-C-1391, Installation & Splicing of Underground Cable

12. Resistance values for maintenance
AC 150/ B, Maintenance of Airport Visual Aid Facilities
Suggested minimum values
10,000 ft. or less - 50MΩ
10,000 ft. – 20,000 ft. - 40MΩ
20,000+ ft. - 30MΩ

13. FAA-C-1391 Installation & Splicing of Underground Cables
Spot Test
Take readings no less than 1 minute after readings stabilized
Cable IR values = 50MΩ, 40MΩ & 30MΩ
Loop IR reduced due to parallel summation
Cable IR never less than above values

14. Insulation resistance formula
Constant current series lighting circuit
Provides for parallel summation of circuit components
3 components
L-824 cable
L-823 cable connectors
L-830 isolation transformers RT RN RC I V

15. FAA minimum IR values L-824 cable
AC 150/5345-7E, Specification for L-824 Underground Electrical Cable for Airport Lighting Circuits
Table 1, Test #9 > ICEA S , Section
Corresponding to IR Constant 50,000 MΩ - k-ft. at 15.6°C

16. FAA minimum IR values L-823 cable connectors
AC 150/ D, FAA Specification for L Plug and Receptacle, Cable Connectors
Section 5.1 – Type I Connectors 75,000 MΩ

17. FAA minimum IR values L-830 isolation transformers
AC 150/ C, Specification for Series to Series Isolation Transformers for Airport Lighting Systems
Table 3 Insulation Resistance 7,500 MΩ

18. Base formula
1/IR = 1/RC + 1/RN + 1/RT RT RN RC I V IR

19. Cable equation
The insulation resistance of cable for a certain length can be calculated by the following formula:
Where:
RC = Insulation resistance in Megohms of cable
K = Specific IR in Megohms - k ft at 60˚ F of insulation
D = Outer diameter of insulation
d = Outer diameter of bare copper wire
L = Length of airfield cable in feet
Value of K for EPR insulation = 50,000 Megohms
𝑅𝐶=𝐾 ∗𝐿𝑜𝑔 𝐷 𝑑 ∗ 𝐿

20. Cable equation D= 9.18 mm d= 4.58 mm D d CONDUCTOR JACKET INSULATIONOD

21. Cable equation 𝑅𝐶=𝐾 ∗𝐿𝑜𝑔 𝐷 𝑑 ∗ 1000 𝐿
𝑅𝐶=𝐾 ∗𝐿𝑜𝑔 𝐷 𝑑 ∗ 𝐿
RC = 50,000 MΩ-k ft * Log ( 9.18/4.58) * (1,000/L)
RC = 15,098,860 MΩ / L

22. Connector equation
The insulation resistance of L-823 connector splices can be calculated by the following formula:
RN = Rc/Nc
Where:
RN = Insulation resistance in Megohms of all connectors
Rc = Insulation resistance of L-823 connector splice
Nc = Quantity of L-823 connector splices
RN = 75,000MΩ/Nc

23. Isolation transformer equation
The insulation resistance of L-830 isolation transformers can be calculated by the following formula:
RT = Rt/Nt
Where:
RT = Insulation resistance in Megohms of all transformers
Rt = Insulation resistance of L-830 isolation transformers
Nt = Quantity of L-830 isolation transformers
RT = 7,500MΩ/Nt

24. IR Base formula 1 𝐼𝑅 = 𝐿 15,098,860 + 𝑁𝑐 75,000 + 𝑁𝑡 7,500 MΩ⁻¹
1/IR = 1/RC + 1/RN + 1/RT
1 𝐼𝑅 = 𝐿 15,098, 𝑁𝑐 75,000 + 𝑁𝑡 7,500
MΩ⁻¹ RT RN RC I V IR

25. Developing IR calculation formula
L-823 Connector
L-830 Isolation Transformer
L-824 Series Lighting Cable
Supply
Return
Section 1
Section 2
Section 3

26. Developing IR calculation formula
L-823 Connector
L-830 Isolation Transformer
L-824 Series Lighting Cable
Section 1
Section 2
Insulation Resistance to Earth
Earth Ground
Section 3

27. Developing IR calculation formula
L-823 Connector
Supply
L-824 Series Lighting Cable
Insulation Resistance to Earth
Earth Ground
1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 1 = 𝐿 15,098, 𝑁𝑐 75,000
MΩ⁻¹

28. Developing IR calculation formula
1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 2 = 𝐿 15,098, 𝑁𝑐 75,000 + 𝑁𝑡 7,500
MΩ⁻¹
L-823 Connector
L-830 Isolation Transformer
L-824 Series Lighting Cable
Insulation Resistance to Earth
Earth Ground

29. Developing IR calculation formula
1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 3 = 𝐿 15,098, 𝑁𝑐 75,000
MΩ⁻¹
Earth Ground
Insulation Resistance to Earth
L-823 Connector
Return
L-824 Series Lighting Cable

30. Developing IR calculation formula
1 𝐼𝑅𝑡𝑜𝑡𝑎𝑙 = 1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛3
𝐼𝑅𝑡𝑜𝑡𝑎𝑙= 1 1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 3 MΩ
IR total
IRsection 1
IRsection 2
IRsection 3

31. Calculation example Vault SECTION 1 SECTION 2 SECTION 3 CCR Handhole

32. Calculation example Cable = 5,000 feet Section 1: Section 2:
Connectors = 2
Isolation XFMRs = 0
Section 2:
Cable = 20,500 feet
Connectors = 224
Isolation XFMRs = 112
Section 3:

33. Calculation Example
Section 1:
Cable = 5,000 feet
Connectors = 2
Isolation XFMRs = 0
Vault
CCR
Handhole
Handhole
1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 1 = ,098, ,000 = 𝑀Ω⁻¹

34. Calculation Example
Section 2:
Cable = 20,500 feet
Connectors = 224
Isolation XFMRs = 112
1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 2 = 20,500 15,098, , ,500 = 𝑀Ω⁻¹

35. Calculation Example
Section 3:
Cable = 5,000 feet
Connectors = 2
Isolation XFMRs = 0
Vault
CCR
Handhole
Handhole
1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 3 = ,098, ,000 = 𝑀Ω⁻¹

36. Calculation Example
1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 1 = ,098, ,000 = 𝑀Ω⁻¹
1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 2 = 20,500 15,098, , ,500 = 𝑀Ω⁻¹
1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 3 = ,098, ,000 = 𝑀Ω⁻¹
IR total = 50 MΩ Circuit length is 30,500 ft, so the circuit IR is well over the recommended minimum value.

37. Modifying a circuit Vault SECTION 1 SECTION 2 SECTION 3 CCR Handhole

38. Modifying a circuit
Each circuit modification warrants a reevaluation of the IR for that circuit
Assume:
Segment 1:
Cable = 5,000 feet
Connectors = 2
Isolation XFMRs = 0
Segment 2:
Cable = 20,500 feet
Connectors = 430
Isolation XFMRs = 215
Segment 3:
Total circuit:
IRsection1 = 2,795 MΩ
IRsection3 = 2,795 MΩ
IRsection2:
RC = 50,000 * Log ( 9.18/4.58) * (1,000/L) = 15,098,860/20,500 = 755 MΩ
RN = 75,000/Nc = 75,000/430 = 174 MΩ
RT = 7,500/Nt = 7,500/215 = 35 MΩ
IRsection2 = 1/(1/ /35 + 1/174)
IRsection2 = 28 MΩ
IR = 1/ ( 1/2, /28 + 1/2,795 )
IR = 27.4 MΩ

39. Modifying a circuit Total circuit IR = 27.4 MΩ
Circuit Length = 30,500 feet
Recommended value for +20k feet circuit = 30 MΩ
If you remove the transformer component,
the circuit IR = 128 MΩ

40. Case study example: Measuring IR of a TDZ Circuit
Section 1:
Cable = 1,615 feet
Connectors = 5
Isolation XFMRs = 0
Section 2:
Cable = 13,200 feet
Connectors = 365
Isolation XFMRs = 180
Section 3:
Total circuit IR = 33.2 MΩ
Circuit Length = 16,430 feet
10k-ft – 20k-ft = 40 MΩ
W/O Transformers; IR = 164.3
Measured value = MΩ

41. Case study example: Measuring IR of a REL Circuit
Section 1:
Cable = 1,540 feet
Connectors = 5
Isolation XFMRs = 0
Section 2:
Cable = 27,777 feet
Connectors = 180
Isolation XFMRs = 90
Section 3:
Total circuit IR = 61.6 MΩ
Circuit Length = 30,857 feet
+ 20k-ft = 30 MΩ
Measured value = 998 MΩ

42. Summary
Good engineering practice to perform circuit load and insulation resistance calculations
Best practice – establish field test baseline and track results
Standards provide recommended values - reductions are allowed
Check & verify transformers, connectors and cable types and sizes
Minimum allowable component values – higher factory test values
High initial field value does not necessarily indicate a good circuit

43. Questions