Re: Using HV COAX without stripping the shield

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: "Bob (R.A.) Jones" <a1accounting@xxxxxxxxxxxxx>


Hi


> Original poster: "Gerry  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>
>
 the reflection coefficient (Kr) at the SG end is -1 and at the pig
> end is +1.  If the intial charged voltage at the time of SG breakdown
> is 15KV , then a pulse (change of, or delta waveform) of -15KV will
> start propagating toward the pig end.  When the pulse arrives at the
> pig end (incident voltage Vi),  Vi will get totally reflected  (Vr =
> Kr*Vi),  and will result in Vr of -15KV.  The voltage at the pig end
> is the sum of Vi and Vr or -30KV.  Now the reflected wave will arrive
> back at the SG end as Vi = -15KV and will reflect as +15KV (K= -1).
> When the 15KV reaches the pig end again. it will double to +30KV and
> so on. The voltage at the pig end will oscillate between -30KV and
> +30KV (note this waveform is superimposed on the initial charge of 15KV).
>
 I think the above is incorrect.
A -15kV(ie a voltage of 0v) step does propagates from the SG
and at the OC end the step does double to -30kV.
But thats relative to the original +15kV so its -15kv wrt ground.

The over all effect (at the OC) is alternating polarity pulses about 0V that
exponentially decay to 0V which it must do with a short at one end.
Note that the transformer end reverses polarity but does not increases in
amplitude.

Perhaps the rapid voltage reversal causes the bushing to track because
surface charge of the original polarity is stored on its surfaces.


Robert (R. A.) Jones
A1 Accounting, Inc., Fl
407 649 6400