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From: Larry Benko <>
Date: Wed, 04 Dec 2013 13:14:12 -0700
List-post: <">>

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.

Larry, W0QE
On 12/4/2013 12:31 PM, 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.
Gerald K5GW
In a message dated 12/4/2013 11:10:44 A.M. Pacific Standard Time, writes:

    Not true Peter,

    Unmatched (assuming 50 ohm output Z in 50 ohm circuit) the max
    voltage is (1+p) times the 1:1 SWR voltage.  The reflection
    p = (SWR-1)/(SWR+1).

    In a matched circuit (if the matching has no loss) the maximum
    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
    you see depending on electrical distance to the load and the loss
    in the
    transmission but the above formulas bound the upper limit.

    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" <>
    >   To: "Amps" <>
    >   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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