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.

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 ?

Depends on the manufacturer energy let through table. different energy let-through values for various prospective fault current 

Exactly. Imaginary circuits can be designed to fail as an example

If the fault current is 5KA and your MCB rated short circuit capacity is 6 KA this means it’s adequately rated to interrupt fault currents safely. Icn - max can interrupt safely but may no longer be usable,  Ics max without loss of performance 

It’s kA the capital would be Amperes shortened name. The one Coulomb a second person. Cheers for the education. It is good to see there are people in the industry still interested. Enjoy ya night mate 

Wonderful thread drift going on!

It can be difficult to get energy let through data, but in my humble domestic environment, 1kA is much more likely than 10 kA.

Cheers Mark, you too !

Have a look at TIS.  They have very good testers at a cheaper price than Megger. 

Have a look at TIS.  They have very good testers at a cheaper price than Megger.