I have a Megger 1741 and I could pay Megger to upgrade it to being a Megger 1741+ presumably with the latest version of the software to give a reasonably accurate test RDC-DD ramp test result, but I still would not know how long it took, only the tripping current at the end of the test period, not the tripping current that actually activated the RDC-DD and started the tripping process.

If I have that right, I’m not sure exactly what I would prove, other than it does trip eventually, not the actual tripping current and not the tripping time.

Megger may state test the RDC-DD, but they are trying to sell you the new all singing, all dancing black tester with red buttons and a colour screen for £1600, if the EVSE manufacturer is telling you to test the RDC-DD then it seems appropriate to do so, even though the Code of Practice is telling you don’t need to, but doing so because you are being told to do so by a tester manufacturer trying to sell you a new expensive tester may not be the best of ideas.

Not that I have anything against Megger, I have five of their testers and one of their EVSE testing adapters, I used my MFT1741 with the Megger testing adapter when I took the C&G exam, but I will never tell anyone they have to buy a new tester when they don’t need to.

Thank you

I slipped another post in that I was typing whilst you posted that last comment.

Enjoy your evening. GKenyon seems to  seems to know his stuff. Personally I would test the life out of a publicly accessible product that conducts. You can do the minimum if you like. Legally you will be sound. I need to sleep without worry.

I should hope so, I have the Fifth Edition of the IET EVSE COP of Practice in front of me, have a look at the top of page six.

Also, we were told that on the EVSE installers course I did that legally the Code of Practice has higher legal status than BS7671, so we should always refer to the COP and follow its recommendations rather trying to decide how to do it ourselves based on our own interpretation of the requirements in BS7671.

What is the issue you have? I have stated mine. Did not need the reg’s or a COP. As an engineer you should make sure the work you do os faultless.

I can help ya if you are unsure but you have a responsibility to be electrically safe.

my round next

As an engineer you should make sure the work you do os faultless.

Unfortunately the real world gets in the way of that admiral sentiment. There are many things - in BS 7671 and elsewhere - that are far from ideal, perfect or faultless. Any amount of RCD testing today isn't going to guarantee it'll work as required tomorrow. Shock currents and disconnection times are based on assumptions that probably only hold safe for around 95% of the population - the remaining 5% have trust to luck. Earth faults from distribution and larger final circuits are permitted disconnection times that don't align with shock protection at all. Faults between conductors of different circuits aren't generally considered. Small overloads of long duration are another know problem. L-N voltage operated open-PEN detection devices are known to be fooled by a variety of conditions that can arise from polyphase distribution systems. Errors and failures are to some extent natural and unavoidable.Really the best we can do is to recognise and do our best to control, mitigate or work around the worst of the problems, at an acceptable cost. Generally things are acceptable if they are safer overall than all the alternatives - rather than being free of all possible risk.

   - Andy. 

Hi Andy. 
impedance. We rely on that. I agree with you but live electrical resistance is how fault protection is designed.

impedance. We rely on that. I agree with you but live electrical resistance is how fault protection is designed.

Sorry? I'm not quite following. Yes I know how the calculations are performed, my point is that we only consider impedances related to a single circuit at a time - faults between conductors of different circuits may well have a different range of impedances, and indeed different cross-sectional-areas of conductors to be protected (and possibly different insulation an other factors that provide limits to acceptable temperature rise) but we don't consider them - in many cases it would be impractical to do so - yet the possibility of such faults does exist.

    - Andy.

Give an example 

Give an example 

Give an example 

Sorry, of what?

   - Andy.

The multiple circuit  fault issue. My experience using design software was that it calculated this. You may know something I dont

Say a couple of T&E cables go though a common hole in a stud in a partition wall - plasterboard screw or picture nail or DIY's drill bit then hits them - clipping say L on a 50A/10mm² shower circuit and N on a 6A/1mm² lighting circuit. The only thing "protecting" the 1.0mm² lighting N conductor is the shower circuit's 50A OPD and also the fault loop impedance is increased as half of the circuit is now in 1.0mm² rather than 10mm² so disconnection times may well be extended.

In commercial and industrial environments it's common to have outgoing circuits from DBs wired in singles in common trunking - so any damage to insulation there is just as likely to result in a short between two different circuits as two wires of the same circuit.

Such faults are thankfully pretty rare, but not impossible.

AFAIK BS 7671 doesn't ask for such situations to be considered at all (e.g. see note below reg 434) - and I'm not aware of any software that does (but I have very limited experience of modern circuit design software, so willing to be presently surprised if it does).

   - Andy.

Fault current rating and disconnection time,  time tripping curve required of the protective device prevents this. 

Fault current rating and disconnection time,  time tripping curve required of the protective device prevents this. 

...or directly using energy let-though for MCBs. The point though is that normally such things are calculated for the 50A device and 10mm² conductors on one hand, and the 6A device and 1mm² conductors on the other - but not other combinations if inter-circuit faults had to be considered. (FWTW a 6kA 50A B-type MCB has an energy let-though of anything up to 54,000 A²s - so can't easily be shown to protect conductors below about 2mm²).

   - Andy.

I don’t know what you are talking about . Use an RCD/ RCBO  if you are not confident in the MCB. You have to select the correct  CPD for your circuit. 

You have to select the correct  CPD for your circuit. 

Indeed - but that doesn't cover faults between conductors of different circuits - which was what I was trying to illustrate about BS 7671 (and engineering in general) not covering every eventuality and so nothing can really be considered totally faultless however hard we try.

Use an RCD/ RCBO  if you are not confident in the MCB

Er no. If the RCD is common to both circuits it won't even see the fault. If each circuit had its own RCBO then the L-N fault between circuits would be seen, but even then the guaranteed opening time (e.g. 40ms) would likely be far too slow to protect conductors against fault currents (in many conditions, k=115), fault currents above 575A would exceed the withstand of a 1mm² conductor (which is why RCDs are almost always backed up by OCPDs).

   - Andy.

Tell that to a TT earthing arrangement. I am unsure what your issue is, OCPD is not the same as fault protection.

and it is 400 ms for a TNC system before going to 5 seconds. 200 ms on a TT unless you have MPb and a max operation time of 300ms via RCD  will satisfy a TNC.

if you connect two circuits on the same phase via a ‘screw’ the OCPD will have to be relied on. Other than that your fault protection becomes what youd rely on. 
if you are worried about arcing faults fit an AFDD

Hi Mark. Just picking up on energy let-through of a device. For example, suppose a MCB is installed where the fault level is 10KA and the MCB has a energy let-through of 0.52 x10E6 A2s. And you want to know if a 6mm2 CPC with thermoplastic insulation be enough for this device? According to table 43.1 of BS 7671, the value of K is 115. So, we need to find the minimum cross-sectional area (S) of the CPC that satisfies this equation: S squared x 115 squared is greater than or equal to 0.52 x 10E6. If we solve for S, we get: S is greater than or equal to the square root of (0.52 x 10E6) divided by 115, which is about 6.27mm2. Therefore, a 6mm2 CPC would not be sufficient for this device, and we would need to use a 10mm2 CPC instead.

Thank ypu for the lesson in Adiabatic calculation  Of we are designing imaginary circuits… What if the fault current was 5kA and the CPD was rated at 6kA ?