Supplementary Protective Bonding Conductors ( Where Required )
Supplementary Bonding is Required where Circuit Disconnection Times Cannot be Met ( Regulation 411.3.2.6 ) or
Where there is an Increased Risk of Electric Shock ,
Those Areas would be in Bathrooms , Swimming Pools & Other Special Locations ,
Where Disconnection Times Cannot be Met and the Effectiveness of Supplementary Bonding is Required to be Checked
( Regulation 415.2.2 ) a Simple Test is Required ,
Automatic Disconnection is by a Protective Device , Regulations give a Formulae -
( R ≤ 50V / Ia ) if the Circuit is Protected by an RCD , the Formulae Becomes ( R ≤ 50V / I∆n )
When this Formulae is Used , it will Ensure that Any Touch Voltage in the Bonded Area will Not Rise above ( 50 Volts a.c. )
Before the Protective Device Operates ,
The Formulae where a Protective Device is in Place for Automatic Disconnection of Supply ( ADS )
Step 1 , find ( Ia ) for a 5 seconds Disconnection Time ,
Device 32Amp BS- 88 fuse
Appendix 3 / BS-7671 fig 3.3A
32A Device a Current of 125 Amp is Required to Operate the Fuse ,
This Value Can also be Found by Using the Maximum ( Zs ) Value Found Table 41.4 ( 1.84Ω )
Calculation : Ia = Uo ÷ Zs ( Ia = 230 ÷ 1.84 = Ia = 125A
This Method Can be Used for all Protective Devices :
Now the Values Can be Used to Verify that the Area does or does Not Require Supplementary bonding to be Installed ,
( R ≤ 50V ÷ Ia ( 50 ÷ 125A = 0.4Ω ) is the Maximum Value Permitted between Exposed or Extraneous Conductive Parts ,
If when Measured it is found that the Résistance is Higher than 0.4Ω then Supplementary bonding will be Required
The Values of Résistance will Not be the Same for Different Ratings or Type of Protective Devices ,
Where an RCD is Installed to Protect the Circuit the Calculation to find out if Supplementary bonding is Required ,
R = 50v ÷ I∆n ( R = 50v ÷ 0.03 R = 1666Ω , ( 50v ÷ 30mA = 1.666Ω
Bonding is Not well-Understood by Joe Public ,
You have Travelled 20/30 Miles to Fit Kitchen / Completed Everything to Comply with Required Regulations ,
A few days later, however, and before you have been paid for the Work ,
You Receive Phone Call from your Customer Informing you that his Next door Neighbour has Spotted that you
Have Not Bonded the Sink , of Course your Customer will Believe that his Neighbour is Right and you have Forgotten Something ,
The Choice is your / Do you Try and Convince your Customer that his Neighbour is Wrong ,
Do You Travel back to the Job to Carry Out the Bonding to Ensure Big Bucks ,
Just Put a Couple of Earth Clamp and a Short Length of 4mm2 Earthing Round the Pipes ,
( Cheaper just to Use the Supplementary Protective Bonding in the First Place )
For the Apprentice :
2.5mm2 / 1.5mm2 / Twin & Earth Cable 22meters long ,
OSG 9A ( Résistance of Copper ) 2.5mm2 / Copper : 7.41
The Résistance / Phase Conductor , ( R1 )
R1 - 7.41 x 22m ÷ 1000 = 0.163Ω *
Divide the Largest Conductor by the Smallest to find the Ratio of Conductors ,
( how much Bigger is the Largest Conductor , ( 2.5 ÷ 1.5 = 1.67
2.5mm2 Conductor is 1.67 x Larger than 1.5mm2 Conductor , therefore , it must have 1.67 x less Résistance than 1.5mm2 Conductor ,
0.163Ω * x 1.67 = 0.27Ω ( this is the Resistance of 1.5mm2 Conductor ,
OSG 9A ( Résistance of 1.5mm2 Copper : 12.10mΩ ( 22meters of 1.5mm2 Copper ,
22 x 12.10 ÷ 1000 = 0.266Ω
Final Check : OSG 9A
Résistance 2.5mm2 / 1.5mm2 Cable , we will See that it has a Résistance ( 19 51mΩ ) per meter
22 meters of it will have a Résistance of ( 22 x 19.51 ÷ 1000 = 0.429Ω
The Résistance Value of 2.5mm2 ( 0.163Ω ) and the Résistance Value of the 1.5mm2 is ( 0.266Ω )
Add them Together : ( 0.163Ω + 0.266Ω = 0.429Ω ( Finally , 0.429Ω is the Résistance of our 2.5mm2 / 1.5mm2 -
Measured as One Cable ,
This beats tons of paper work ,
“ Insulation Résistance “
This is a Test that can be Carried Out on a Complete Installation , or Single Circuit
For This Case ,
Domestic , ( Single Circuit or Ring ) New Installation ,
The Test is Necessary to find Out if there is Likely to be any Leakage of Current Through the Insulated Parts of the Installation ,
A Leakage could Occur for Various Reasons ,
Good Way to Think of this Test is to Relate it to a Pressure -Test
We Know that Voltage is the Pressure where the Current is Located in a Cable ,
On a Low Voltage Circuit , the Expected Voltage would be around 230v a.c the Voltage Used in an Installation Test on a 230V -
Circuit is 500v d.c , Which is More than Double the Normal Circuit Voltage , therefore , it can be Seen as a Pressure Test Similar
To a Plumber Pressure Testing the Central Heating Pipes ,
Circuits between 0V – 50v a.c
Required Test Voltage , ( 250v d.c. )
17th Edition : ( 0.5MΩ )
Circuits between 50V a.c – 500v a.c
Required Test Voltage , ( 500V d.c. )
17th Edition : ( 1MΩ )
Circuits between 500V & 1000V a.c.
Required Test Voltage , ( 1000V d.c. )
17th Edition : ( 1MΩ )
Domestic Installations ,
Remember that Testing should be Carried Out from the Day the Installation Commences ( Regulation 610.1 )
“ Insulation Résistance “
If for Some Reason , there is a Piece of Equipment Connected to the System that Cannot be Isolated from the Circuit Under Test ,
Do Not Carry Out the Test Between Line Conductors – Only Test Between live Conductor & Earth
This is to Avoid Poor Readings and Possible Damage to Equipment , this Test should Only be Carried Out on Individual Circuits ,
Not the Whole Installations , as it is Important to Test as Much of the Installation as Possible ,
Three Phase Sub-Mains is Tested the Results are :
L1 to Earth is 130MΩ
L2 to Earth is 80MΩ
L3 to Earth is 50MΩ
N to Earth is 100MΩ
If these Conductors were now Joined and Tested to Earth the Value would be as Given ,
Calculation ,
1 1 1 1 1
--- + --- + --- + --- + ---
R1 R2 R3 R4 Rt / N
( 1 1 1 1 1 )
--- + --- + --- + --- + --- = 19.92MΩ
130 80 50 100 0.05
Enter it this way into a Calculator : ( Remember the Button 1/X )
130X-1 + 80X-1 + 50X-1 + 100X-1 = X-1 = 19.92MΩ
This Value is Still Acceptable but Lower because the Conductors are in Parallel ,
Iceman , this is what you can face ,
Swindon Massive , send email to Kev , 20 pounds C/D It’s a Must
A Ring Circuit is Protected by a 30A BS-3036 Semi-Enclosed Rewirable Fuse
Measured ( Zs ) 0.96Ω
( Final Circuits Not Exceeding 32A )
This is a Ring Final Circuit , Disconnection time has to be 0.4sec ( table 41.2 BS-7671
Maximum ( Zs ) / 30A Rewirable Fuse – 1.04Ω
Eighty per Cent of this Value must Now be Calculated , this can be Achieved by Multiplying it by ( 0.80. )
1.04 x 0.80 = 0.83Ω
The Measured Value for the Circuit must now be Lower than the Corrected Value if it is to Comply : BS-7671
Measured Value 0.96Ω
Corrected Value 0.83Ω
The Measured Value of ( Zs ) is Higher , therefore , the Circuit will Not Comply ,
Option one is the Preferred Method because it will give an Accurate Value whereas the Test in -
Option two will Include Parallel Paths and because of this will Often give Lower Readings ,
Option one should Always be Used for an Initial Verification as the First Reading will be Used as a Benchmark -
To be Compared with Results taken in Future Periodic Tests ,
On a Periodic Inspection Using Option two , a Higher Test Results is Obtained than on the Initial Verification ,
This would Indicate that the Circuit is Deteriorating and that Further Investigation would be Required ,
The Methods Described here must be Fully Understood by Anyone who is Intending to sit the ---- & ------
---- Exam on Inspection & Testing of Electrical Installations ,
For the Apprentice
BS 7671 lists Five Types of Earthing System :
TN-S, TN-C-S, TT, TN-C, and IT.
T = Earth (from the French word Terre)
N = Neutral
S = Separate
C = Combined
I = Isolated ( The source of an IT system is either )
Connected to Earth through a Deliberately Introduced Earthing Impedance or is Isolated from Earth. All Exposed-Conductive-Parts of an Installation are Connected to an Earth Electrode. )
When Designing an Electrical Installation, One of the First things to Determine is the Type of Earthing system.
The Distributor will be Able to Provide this Information.
The System will Either be TN-S, TN-C-S (PME) or TT for a Low Voltage Supply given in Accordance with the
Electricity Safety, Quality and Continuity Regulations2002. Appendix 2 This is because TN-C Requires an Exemption
from the Electricity Safety, Quality and Continuity Regulations, and an IT system is Not Permitted for a
Low Voltage Public Supply in the UK because the Source is Not Directly Earthed. Therefore TN-C and IT
Systems are both Very Uncommon in the UK.