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To: Larry Benko <>, 'Amps' <>
From: peter chadwick <>
Date: Thu, 05 Dec 2013 00:08:34 +0100
List-post: <">>

the argument used by all the authorities is this:

At an open circuit the forward voltage is Vf. Because it is reflected as VR and 
is the same voltage - necessarily - the two voltages add to give a reflected 
voltage of 2Vf. To get more requires energy to be made.

So on a loss less line, the only voltage reflected back to the tx is 2Vf - 
where Vf is the rms volts that would appear at the load when the line is 

So far, so good. Now let's look at typical pi net values. say Rp 2000 ohms, Xc1 
167 ohms, XL188 ohms and Xc2 31 ohms. At 1500 watts in 50 ohms, Vc2 = 274 rms. 
Ic2 = 8.8 amps, so the cap needs to have substantial current handling - this 
isn't likely to be the gang out of an old BC set. Vplate is 1662 volts rms - 
reasonable for around 2700 volts DC on the plate. So let's assume that we open 
circuit at the antenna end and we just happen to get +j50 at the tx which we 
then tune to resonance with a bit more C2. So we will have -j19 ohms at C2: if 
we are to get twice the 1:1 voltage across it, there must be  now be 28.8 amps 
in C2 - and in Xl. But to get that current in XL the volts across it must be 
5.4kV rms, or 15kV peak to peak. What's going to flash over first - C2 or C1 or 
the tube or the bandswitch?

So to have a requirement for very high voltages for output padders, you have to 
assume not the volts caused by SWR, but the effect of trying to tune that 
impedance - which you likely can't do with any power going to it. What you do 
need to do is to ensure that the padding caps are generously rated for the 
current that they will see under high SWR conditions, although hopefully, that 
won't be for long enough for them to fry unless the op is not keeping an eye on 

OK, this isn't simulated. It's just Kirchoff!

Never forget the famous (and to my mind, profound) comment made at a European 
Telecommunications Standards Institute meeting by Dr. John McKown of Motorola 
who, after some academic guy had made a BS presentation on the virtues of his 
simulation, John said " Simulation is like masturbation: the more of it you do, 
the more you prefer it to the real thing."


Peter G3RZP
 Message Received: Dec 04 2013, 08:43 PM
 From: "Larry Benko" <>
 To:, "'Amps'" <>
 This not a question about who got what right but what is really correct.
 _Please follow this example:_
 1.) Start with a 1W signal generator with a 50 ohm output will generate 
 7.07Vrms into a 50 ohm load resistor.  Disconnecting the output of the 
 signal generator will generate 14.14Vrms into an open circuit and if you 
 short he output of the signal generator you will have 282.8mArms in the 
 2.) Now lets connect the same signal generator directly to a 200 ohm 
 resistors with NO transmission line.  Doing the calculation again the 
 voltage across the 200 ohm resistor is now 11.312Vrms.  The voltage is 
 11.312/7.07 = 1.6 times the matched voltage which is what you think 
 Philip Smith predicted.  Personally I think he really knew exactly how 
 things worked and the error is in your interpretation. An SWR of 4:1 
 (200 ohms) has a reflection coefficient of (SWR-1)/(SWR+1) = 0.6 which 
 when added to 1 exactly matches what the increase in voltage is.  
 However the power in the 200 ohm resistor is 11.312^2/200 = .639W or 
 -1.93dB less than the matched case.  This loss is mismatch loss (look it 
 up on Wikipedia) and can be predicted to be ML = -10*log10(1-p^2) = 
 -1.938dB.  This is why we match for maximum power transfer!
 3.) Now lets include a lossless matching circuit on the generator output 
 before the 200 ohm resistor.   The match type does not matter and I will 
 use a low pass "L" network which is a +j50 inductor in series with the 
 generator output and a -j50 capacitor to ground on the 200 ohm resistor 
 side of the inductor.  Now the voltage across the 200 ohm resistor is 
 14.41Vrms = 1W.  This voltage is 2 times the voltage in the original 50 
 ohm load case which is exactly the square root of the SWR.  Granted I 
 used the worst case 2:1 SWR for voltage but that is what we are talking 
 about.  Had I used 12.5 ohms then that would have been the worst case 
 current case and all other impedances which are SWRs of 2:1 would have 
 been somewhere in between.
 This experiment can be simulated in any Spice program with the exact 
 same results.
 Larry, W0QE
 On 12/4/2013 12:46 PM, peter chadwick wrote:
 > Larry
 > So Phil Smith got it wrong, did he?
 > looking at my books, so did Terman (Radio Engineering), Lamont V. Blake 
 > (Antennas), Glazier and Lamont (The Services Text Book of Radio Vol 5 
 > Antennas and Propagation) AND the ITT Reference Data Book. And in case we 
 > prefer an amateur book, albeit written by a professional antenna designer, 
 > so did Walt Maxwell, W2DU.
 > Six tomes from professionals ALL with the same mistake?
 > 73
 > Peter G3RZP
 > ========================================
 >   Message Received: Dec 04 2013, 07:10 PM
 >   From: "Larry Benko" <>
 >   To:
 >   Cc:
 >   Subject: Re: [Amps] PARALLEL CAPS IN OUTPUT
 >   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" <>
 >   >   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
 >   >   _______________________________________________
 >   >   Amps mailing list
 >   >
 >   >
 >   >
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