Gerald,
You are correct except you are unable to generate a 1,000,000:1 SWR. A
line loss of .001dB will produce a return loss of never less than .002dB
which is a reflection coefficient of .999769 which is an SWR of 8685:1
and no where near the 1,000,000:1 SWR. I am 100% positive the equations
are correct. Would you rather take this offline? It is freezing cold
and snowing here today and I can give you a call also if you wish since
I'm not going anywhere.
73,
Larry, W0QE
On 12/4/2013 12:31 PM, TexasRF@aol.com wrote:
Larry, that doesn't seem intuitive at all. For example, if there was a
short or open circuit load producing a swr of 1,000,000 then the
square root of 1000 times 274v (1500w & 50R) = 274,000 volts.
There ain't that much voltage anywhere, is there?
The unmatched example seems ok, with p=very close to 1 added to 1 =
2X matched voltage.
73,
Gerald K5GW
In a message dated 12/4/2013 11:10:44 A.M. Pacific Standard Time,
xxw0qe@comcast.net writes:
Not true Peter,
Unmatched (assuming 50 ohm output Z in 50 ohm circuit) the max
possible
voltage is (1+p) times the 1:1 SWR voltage. The reflection
coefficient
p = (SWR-1)/(SWR+1).
In a matched circuit (if the matching has no loss) the maximum
voltage
possible is square root of the SWR times the 1:1 SWR voltage.
Undergrad EE classes cover these topics and programs such as
LTSpice can
show it as well. Obviously the maximum posible voltage may not be
what
you see depending on electrical distance to the load and the loss
in the
transmission but the above formulas bound the upper limit.
73,
Larry, W0QE
On 12/4/2013 11:49 AM, peter chadwick wrote:
> If one is to believe Philip H. Smith in 'Electronic Applications
of the Smith Chart', McGraw-Hill 1969, page 6, Fig 1.3, the
maximum voltage appearing on a lossless transmission line with an
SWR of infinity is twice the voltage when matched.
>
> So a 1kV rating is adequate.
>
> It makes sense when you think about it, too.
>
> But of course, Smith might have got it wrong.......
>
> 73
>
> Peter G3RZP
>
>
> ========================================
> Message Received: Dec 04 2013, 06:40 PM
> From: "Bill Turner" <dezrat1242@wildblue.net>
> To: "Amps" <amps@contesting.com>
> Cc:
> Subject: Re: [Amps] PARALLEL CAPS IN OUTPUT
>
> ORIGINAL MESSAGE: (may be snipped)
>
> On Wed, 4 Dec 2013 09:16:06 -0500 (EST), K5GW wrote:
>
> >
> >The voltage rating is not the problem; after all there is
less than 300v
> >rms across a 50 ohm load with 1500 watts power.
>
> REPLY:
>
> Capacitors don't arc at the RMS voltage. They arc at the peak
of the RF
> cycle. For 1500 watts into 50 ohms, the peak is about 387 VAC.
And that's
> with a 1:1 SWR.
>
> A high SWR can cause voltage nodes many times the normal
voltage to appear
> on the coax, and if the coax is just the wrong length, one of
those nodes
> may appear right at your load cap. Have you ever transmitted
into the wrong
> antenna?
>
> IMO, padder caps rated at 5 or 6 kV are NOT overkill.
>
> Once a capacitor arcs, even if it survives, little blisters
form at the
> point of the arc and, due to corona effect, are prone to arc
again but at
> even lower voltage. It is always best to prevent the arc in
the first place.
> High voltage caps are your friend.
>
> 73, Bill W6WRT
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