SPD - cable selection

am I right in thinking it’s only there to provide short circuit protection rather than overload? 

The 2nd half of 534.4.10 would seem to agree!

   - Andy.

One hopes that there is almost no current in the MCB or the wiring to it, most of the time. The only times there is would be 

1) by design for some microseconds during a transient over-voltage when a few kA might flow and the SPD dissipates at peak perhaps hundreds of kW but only for a few microseconds,  and  thinking adiabatically, a wiring copper area of a few mm2 would probably do.

2) When the SPD finally dies, if it does so in a short circuit sort of way, when the large current will only flow at most for the time to operate the breaker, if the SPD does not fuse first, and to be honest a thinner cable would probably be OK then as well...
Mike
PS

In terms of responding to fast edges for lower inductance,  and best clamping a pair of flat palm braids would be better than twins of  round wire ,but such things are not compatible with normal wiring terminals and methods.

or 3). when a fault (short) occurs on the wiring between the MCB and SPD...

"or 3)."

Very true. A case where again, a thinner cable would probably suffice. There is no safe condition where 63 amps flows for very long at all, given the volume of the SPD itself and the absence of cooling air, water or rotating shafts to remove the energy.
Mike

mapj1 said:

In terms of responding to fast edges for lower inductance,  and best clamping a pair of flat palm braids would be better than twins of  round wire ,but such things are not compatible with normal wiring terminals and methods.

It may not explain the relatively high rating of the MCB, but are the short and fat cables not required to minimize inductance, else the surge might, as it were, not be able to reach the SPD.

it is indeed all about the surface area of the copper - the outer 'skin' if you like. but if it was not so impractical, the centre of the wire could be replaced with empty space, or even craft cheese for all the effect it has on the high speed waveforms. Actually inductance per unit length is a reasonably gentle function of diameter, compared to resistance, and far more impedance reduction is to be gained by bringing the equal and opposite 'return' current as close as possible.

This is because then the magnetic fields cancel in the region "far away", observed from which, the 2 currents appear more or less centred on the same line. The closer you can get the two currents to each other, the nearer that "we are far enough away so the rest of the magnetic field from here on out is cancelled " boundary gets, and the less voltage drop is needed to induce whatever close-in magnetic field remains having been not quite cancelled.

There is quite a nice free calculator here 
https://k7mem.com/Ind_Wires_Strip.html
that allows one to see the effect of bringing the ground return path close to a wire. Be aware that the defaults are inches and American wire gauges but with a bit of effort wires a sensible few mm in diameter and a metre long can be calculated as they are far away (many diameters) and close (perhaps a few mm)  from the return current path.

regards Mike.

mapj1 said:

because then the magnetic fields cancel i the region "far away",

And if you completely box it in, you get a microwave waveguide .  (but that's taking it too far!)

Thanks all, it seems to be, as per regulation 534.4.10. It is sufficient as it states “referring to regulation 433.3.1(ii)….. 6mm for type 1 or 2.5mm for type 2”. 

as we know 433.3.1(ii) refers to not using a device for protection of overload due to the nature of the load is not likely to carry overload as long as 434 is met (fault current) in which the mccb does provide 

feel free to correct me if I’m incorrect 

Coaxial cable with the thinnest possible dielectric  is perhaps more practical in such cases where very low impedance and no external fields at all are the requirements-)  

Mike

Coaxial cable with the thinnest possible dielectric  is perhaps more practical in such cases where very low impedance and no external fields at all are the requirements-)  

Mike

Especially as no electrons move in almost all these [AC] cases. It's the polarisation of free space  .

(at 50 Hz, at the fusing current of copper, the 'electrons' move a mere fraction of an atomic spacing - it's just 'polarisation')