Re: Soft transformer turn on without a variac

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Robert Jones" <alwynj48-at-earthlink-dot-net> 

 >  >          dwp
 >
 > My reading of the graphs shown at the website pointed to by the url
 > Bob Jones posted suggests that the real problem with a zero voltage
 > crossing turnon is that the flux in the core is forced to climb in a
 > unipolar fashion for half a cycle instead of just a quarter. There is
 > no instantaneous turn-on transient when switching at this particular
 > time, nor should there be since the applied voltage at turn-on is
 > zero with zero resultant current flow at that instant. Does this
 > sound like the correct story Bob?
 >
 > Malcolm
 >

Hi Malcolm. Ignoring the probagation time in the core as I believe thats
going to be sub ms to us or less. I am not certain about the quarter half
thing and its not just the first cycle. I think the profile is more
cerrectly described as an offset cosine(without saturation effects) if thats
your meaning I believe its correct

Here is a bit of my first post with a bit more detail on the end that may
help.

"The standard text book explanation for the turn on surge goes something
like
this. At turn on the response consists of a steady state response and a
transient response. The steady state response is the current that flows
after the transient has died away. The transient response is an
exponentially decay with a time constant of L/R. At the instant of or just
before turn on the input current is zero hence the two responses must be
equal and opposite in sign.

If the turn on occurs at a point when the steady state input current is zero
(approximately at the peak of the input voltage) then the transient current
must also be zero i.e. no surge.

If the turn on occurs at a point when the steady state input current is at a
maximum the transient current must be equal and opposite in direction. If
the transformer is efficient (usually this mean kVA and up) the time
constant (L<<R) of the transient response decay time is larger than the
cycle time of the input. Hence the transient response is still large
compared to the steady state current when the steady state current has the
same sign as the transient. In the no loss limit the peak of the surge is
twice the steady state current (ignoring saturation) and repeats cycle by
cycle."

So just after turn on at zero input voltage the current input is the
difference between the steadystate current (90 deg lag sine current) and the
exponential decaying transient.which are equal but have opposite signs.

If the time constant of the transient is several cycles of the power
frequency (L<<R) it will be almost constant for the first few cycles.  Hence
as the steady state current goes from minus to plus it sums with the
transient ineffect the first few cycles of current are an offset cosine
starting at zero and hence the peak current would be x2.( If durring the
peak the core saturates the current can rise to vaules more determined by
just the primary resistance hence x10 or more.

I did a quick calc on the probagation thing even if you assume a perm L  of
10,000 and perm C of 10 than add x10 for saftey that still produces a
velocity of 300,000m/s (squareroot in the velcoity equ) thats still only
about 3us to penetrate a core with a radius of 1m. Its difficult for me to
to see how this effect can contribute significantly to the rms value of the
turn on surge,

Bob