To: |
Larry Benko <xxw0qe@comcast.net>, 'Amps' <Amps@contesting.com> |
---|---|

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

From: |
peter chadwick <g8on@fsmail.net> |

Reply-to: |
g8on@fsmail.net |

Date: |
Thu, 05 Dec 2013 00:08:34 +0100 |

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

Larry, 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 matched. 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 things. 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." 73 Peter G3RZP ======================================== Message Received: Dec 04 2013, 08:43 PM From: "Larry Benko" <xxw0qe@comcast.net> To: g8on@fsmail.net, "'Amps'" <Amps@contesting.com> Cc: Subject: Re: [Amps] PARALLEL CAPS IN OUTPUT 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: > |

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