To: |
g8on@fsmail.net, 'Amps' <Amps@contesting.com> |
---|---|

Subject: |
Re: [Amps] PARALLEL CAPS IN OUTPUT |

From: |
Larry Benko <xxw0qe@comcast.net> |

Date: |
Wed, 04 Dec 2013 13:43:19 -0700 |

List-post: |
<amps@contesting.com">mailto:amps@contesting.com> |

Peter, 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.
73, 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" <xxw0qe@comcast.net> To: amps@contesting.com Cc: Subject: Re: [Amps] PARALLEL CAPS IN OUTPUT _______________________________________________ Amps mailing list Amps@contesting.com http://lists.contesting.com/mailman/listinfo/amps |

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