# Minimum size of CPC

When using the Adiabatic equation, only to calculate the minimum size CPC, I am thinking that using the fault currents given in Appendix 3 of BS7671 rather than that calculated (UO/ZS) would give a more realistic outcome.

For example: A D6 circuit breaker is protecting a circuit where the calculated fault current is 328A. The table associated with Fig 3A6 in Appendix 3 of BS7671 shows that an overcurrent of 120A will operate the 6A breaker in 0.4s. Using these values and a K value of 115 results in a minimum CPC of 0.66mm2 (1.0mm2), whereas if you use the 328A the calculation results in a minimum CPC of 1.80mm2 (2.5mm2).

I cannot see a problem with using the lower value, but would appreciate confirmation and guidance from others.

Parents
• If as 541.1 allows, I select the size of the CPC rather than use the calculation method, I would be able to select a 1mm2 conductor based on the fact that the corresponding line conductor is also 1mm2.

This then correlates with the calculation method outcome when the 120A generic value for a D6 circuit breaker (taken from Appendix 3) is used.

My assumption, maybe naively, has been that the calculation method would nearly always produce a smaller CPC than the selection method?

• I would be able to select a 1mm2 conductor based on the fact that the corresponding line conductor is also 1mm2.

How? The combination of overcurrent protective device and line and neutral conductors also need to meet Regulation 434.5.2?

So, if 434.5.2 is leading to 1.5 sq mm, this applies to L and N too. 434.5.2 applies in addition to

Usually, though, with prospective fault currents under 1.5 kA, the energy let-through I2t taken from manufacturer data (for 6 A Type B and C circuit-breakers at least) would permit selection of 1.0 sq mm. It's probably true for many D6 breakers too.

What is the manufacturer and part number/range of the D6 circuit-breaker you are using? I will see if I can get hold of the data to illustrate what  I'm saying.

My assumption, maybe naively, has been that the calculation method would nearly always produce a smaller CPC than the selection method?

Yes, it can but not always.

It's worth noting that whilst our usual selection case of IZ ≥ In ≥ Ib (Reg 433.1.1) relates to protection against overload current, wheras Reg Group 434.5 relates to protection against fault current. Selection of a conductor size S has to satisfy both conditions (overload and fault) and therefore the highest S calculated using each of the Regulations (and of course, not forgetting volt-drop) is the minimum size to be selected.

• The circuit breaker is a Hager NDN106A (6A 10kA SP Type D).

• It's worth noting that whilst our usual selection case of IZ ≥ In ≥ Ib (Reg 433.1.1) relates to protection against overload current, wheras Reg Group 434.5 relates to protection against fault current. Selection of a conductor size S has to satisfy both conditions (overload and fault) and therefore the highest S calculated using each of the Regulations (and of course, not forgetting volt-drop) is the minimum size to be selected.

Is the suggestion then, that for normal MCBs, the 2nd paragraph of 435.1 doesn't apply? (i.e. where a suitable overload protective device has suitable breaking capacity, it may be assumed that requirements for fault protection are met).

- Andy.

• Thank you,

Just as an example, from the 2007 data sheet data sheet I quickly found on the internet for this device, I believe the let-through energy for the D6 would be as follows:

• For Ipf = 3 kA, I2t = 5,900 A2s
• For Ipf = 6 kA, I2t = 10,500 A2s
• For Ipf = 10 kA, I2t = 15,000 A2s

Rearranging the formula k2S2 ≥ I2t from Regulation 434.5.2, we have S ≥ {√(I2t)}/k ... which spookily looks just like the formula in 543.1.3

k = 115 for copper conductors, so using the values above,

• For Ipf = 3 kA, S ≥ {√(5900)}/115}, or S ≥ 0.67 sq mm, so minimum S = 0.75 sq mm (flex from connection unit etc) or 1.0 sq mm (fixed wiring)
• For Ipf = 6 kA, I2t = 10,500 A2s, or S ≥ 0.89 sq mm, so minimum S = 1.0 sq mm
• For Ipf = 10 kA, I2t = 15,000 A2s, or S ≥ 1.1 sq mm, so minimum S = 1.25 sq mm (flex from connection unit etc) or 1.5 sq mm (fixed wiring)

So, if you know the prospective fault current at the DB end of the circuit is 6 kA or less, then 1.0 sq mm is OK for the D6 NDN106A, but if it's closer to 10 kA, you will need 1.5 sq mm. We already know it's lower than 3 kA at the remote end ...

This applies to cpc and live conductors equally.

For another manufacturer's product, or if Hager have provided a more up-to-date data sheet, the let-through energy values will be different, leading to potentially different results.

• Thank you,

Just as an example, from the 2007 data sheet data sheet I quickly found on the internet for this device, I believe the let-through energy for the D6 would be as follows:

• For Ipf = 3 kA, I2t = 5,900 A2s
• For Ipf = 6 kA, I2t = 10,500 A2s
• For Ipf = 10 kA, I2t = 15,000 A2s

Rearranging the formula k2S2 ≥ I2t from Regulation 434.5.2, we have S ≥ {√(I2t)}/k ... which spookily looks just like the formula in 543.1.3

k = 115 for copper conductors, so using the values above,

• For Ipf = 3 kA, S ≥ {√(5900)}/115}, or S ≥ 0.67 sq mm, so minimum S = 0.75 sq mm (flex from connection unit etc) or 1.0 sq mm (fixed wiring)
• For Ipf = 6 kA, I2t = 10,500 A2s, or S ≥ 0.89 sq mm, so minimum S = 1.0 sq mm
• For Ipf = 10 kA, I2t = 15,000 A2s, or S ≥ 1.1 sq mm, so minimum S = 1.25 sq mm (flex from connection unit etc) or 1.5 sq mm (fixed wiring)

So, if you know the prospective fault current at the DB end of the circuit is 6 kA or less, then 1.0 sq mm is OK for the D6 NDN106A, but if it's closer to 10 kA, you will need 1.5 sq mm. We already know it's lower than 3 kA at the remote end ...

This applies to cpc and live conductors equally.

For another manufacturer's product, or if Hager have provided a more up-to-date data sheet, the let-through energy values will be different, leading to potentially different results.

Children
• I have just checked and the PFC at the origin of the circuit is 1.38kA.

• So, yes , this would be the relevant calculation for you:

Ipf ≤ 3 kA, S ≥ {√(5900)}/115}, or S ≥ 0.67 sq mm, so minimum S = 0.75 sq mm (flex from connection unit etc) or 1.0 sq mm (fixed wiring)

And 1.0 sq mm would therefore seem to be OK. At the 3 kA fault level, you'll see that the 0.67 sq mm value is very close to your original calculation of 0.66 sq mm ... not so when you get to 10 kA fault current where the minimum S comes out at 1.1 sq mm.