[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

Re: Using HV COAX without stripping the shield

Original poster: "Gerry  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>

Hi Bob,

You're right, they dont oscillate between -30 and +30 KV, they oscillate with -30 and +30 KV deltas. The descriptions below were meant to be delta waveforms but I didn't follow thru :o(( Converting to ground reference, the voltage at the pig goes from 15KV to -15V (a 30KV delta) and two prop delays later will go back to +15KV and so on (a 30KV pk to pk swing). This continues to oscillate during the SG conduction time cause no damping is provided for (assuming the SG is zero ohms and the pig doesn't load the line).

Anyway, going to the intent of the original post, I'm trying to understand the mechanism of voltage growth when using a shielded coax (not because I want to use shielded coaxes, but because I want to understand). Is the phenominum transmission line effects being underdamped while energy is feeding the line, is it a resonant charging effect (series LC) due to the capacitance of the coax, or is it some other mechanism??? Also in the TC context, is the blumlein effect the same as the transmission line effect I described above or something else???

Thankyou for your response and any further enlightenment.

Gerry R


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

> 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