How do I calculate the voltage rating I'll need for capacitors used to
make traps?

I am making a portable field antenna for 160 through 40 meters, for use
with 100 watts absolute maximum (normal use will be from 5 to 50 watts).

On Apr 5, 4:29 pm, Rick wrote:
How do I calculate the voltage rating I'll need for capacitors used to
make traps?

I am making a portable field antenna for 160 through 40 meters, for use
with 100 watts absolute maximum (normal use will be from 5 to 50 watts).

One way to put an absolute maximum rating on the trap voltage and
current is to assume that ALL your power is dissipated in the trap.
Obviously, that would be a real bummer, but it's a limit. In the case
of a symmetric dipole, the power would divide, probably fairly close
to equally, between the two sides. Remember that you want to be sure
the cap doesn't flash over, so take into account peak voltages: peaks
of modulation, etc.

So if you have a trap that, at resonance, has Xc=Xl, and has some Q,
then at power P dissipated in the trap, the voltage will be
sqrt(Xl*Q*P). Circulating current will be sqrt(P*Q/Xl). You
mentioned in another posting 100 watts absolute max, and 200pF
resonant at 80 meters, so Xc=Xl=200 ohms, approximately. You want to
make them small, so I'd guess Q300...let's say Q=200. So for the
case of 100W in one trap (that coil's gonna get really hot if it
really is required to dissipate that for long...), that's
Vrms=sqrt(200*200*100)=2000V. Irms=sqrt(100*200/200)=10A. Please
note that the CURRENT in the cap may be a bigger problem than the
voltage across it!

That's absolute worst-case. I'd guess you'll see more like 5 watts
max per side dissipated with 100 watts input, the rest being radiated,
and if that is indeed the case, Vrms=450V and Irms=2.23A. If there's
only 1 watt dissipated, it's still 200V and 1A.

At about 30pF/foot, RG-58 could do the job, but 200pF would be almost
7 feet of coax. That would be economical, to be sure, and should be
up to the job with no problems. Paralleling some high voltage ceramic
discs would probably work...use NPO (C0G) type, maybe 5 or 10 in
parallel to avoid current problems.

Cheers,
Tom

Rick wrote:
How do I calculate the voltage rating I'll need for capacitors used to
make traps?

I am making a portable field antenna for 160 through 40 meters, for use
with 100 watts absolute maximum (normal use will be from 5 to 50 watts).

I don't know of any general way to calculate it, but you can easily get
the data from a model. In EZNEC, clicking Load Dat will show you the
voltage across and current through a "load" (in this case, trap), as
well as its power loss. Other programs have an equivalent feature.

Roy Lewallen, W7EL

Rick wrote in newsan.2007.04.05.23.29.53.793507
@reply.in.gp:


How do I calculate the voltage rating I'll need for capacitors used to
make traps?

I am making a portable field antenna for 160 through 40 meters, for use
with 100 watts absolute maximum (normal use will be from 5 to 50
watts).


The antenna can be designed so that traps are not resonant at the
operating frequency, and that tends to reduce the voltage impressed
across the traps.

Modelling might give you the best indication of the operating voltage on
the traps. Don't forget to allow for peak and a safety margin.

Another approach is the so called coax trap. I have seen many
descriptions of these things, and they are all different, so I thought I
would write my own which, in keeping with the others, is also different.
My approach was to describe them using a transmission line explanation.
Such an explanation suggests they are a lot better than at first might
meet the eye. My notes are at http://www.vk1od.net/coaxtrap/index.htm .

Owen