[Top] [All Lists]


To:, 'Amps' <>
From: Larry Benko <>
Date: Wed, 04 Dec 2013 13:43:19 -0700
List-post: <">>

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 short.

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:

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?


Peter G3RZP

  Message Received: Dec 04 2013, 07:10 PM
  From: "Larry Benko" <>
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.
  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 
  > 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 
  >   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 
  >   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 
  >   High voltage caps are your friend.
  >   73, Bill W6WRT
  >   _______________________________________________
  >   Amps mailing list
  > _______________________________________________
  > Amps mailing list
  Amps mailing list

Amps mailing list

<Prev in Thread] Current Thread [Next in Thread>