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Mythbusters: V/I ratio is forced to Z0
The myth: Measurements with a Bird 43 of the conditions on the
Thruline section are invalid unless it has some minimum length of 50 ohm line on both sides of itself. I have performed a test using components at hand where the Bird 43 has 75 ohm line on both sides of itself, and the test configuration is designed to present a 50+j0 ohm load at the point where the Bird 43 sampling element is located. The question is, how does the Bird respond? The test in detail. Each component is in a list from the source to the load: IC706IIG 1m RG58 with UHF connectors MFJ949E ATU 3m RG6 (Zo=75 ohm) with BNC connectors Bird 43 5.27m Belden 9275 (Zo=75 ohm, vf=0.83) with BNC connectors 50 ohm dummy load Short adapters were used to connect to the Bird's Type N connectors, the dummy load's Type N connectors and the MFJ949 UHF connector. The half wave resonance of the 5.27m length of RG6 was determined by s/c one end and connecting the other end via an adapter to a MFJ259B and finding the impedance dip at 23.05MHz. The calculated vf from this test is 0.81, which reconciles reasonably with the specs. Free space wavelenght at the test frequency is 13m. The transmitter was set to 23.05MHz, and the ATU tuned to develop rated power output. The ATU is only used to present the rated load to the transmitter so as to obtain 100W for the test, to suit the Bird 43 element. It is inconsequential to the DUT (the Bird 43). With this configuration, it is expected that the impedance at the Bird 43 is approximately 50+j0, and that there would be almost zero reflected power. The Bird 43 indicated 100W forward power and a quarter of a needle width detection on reflected power. The Bird 43 would appear to provide valid readings for the conditions on the Bird 43 Thruline section in this case, notwithstanding that there is not any 50 ohm transmission line attached to the Bird 43 + N-BNC adapters. The myth that measurements with a Bird 43 of the conditions on the Thruline section are invalid unless it has some minimum length of 50 ohm line on both sides of itself is BUSTED. Has anyone experimental evidence to the contrary? Owen PS: I hasten to add / apologise, I do not watch much television, but I was forced to endure Mythbusters when visiting a friend recently. It's about as scientific as what goes on here, so I thought the style appropriate! -- |
Owen Duffy wrote: The myth: Measurements with a Bird 43 of the conditions on the Thruline section are invalid unless it has some minimum length of 50 ohm line on both sides of itself. I have performed a test using components at hand where the Bird 43 has 75 ohm line on both sides of itself, and the test configuration is designed to present a 50+j0 ohm load at the point where the Bird 43 sampling element is located. The question is, how does the Bird respond? The test in detail. Each component is in a list from the source to the load: IC706IIG 1m RG58 with UHF connectors MFJ949E ATU 3m RG6 (Zo=75 ohm) with BNC connectors Bird 43 5.27m Belden 9275 (Zo=75 ohm, vf=0.83) with BNC connectors 50 ohm dummy load Short adapters were used to connect to the Bird's Type N connectors, the dummy load's Type N connectors and the MFJ949 UHF connector. The half wave resonance of the 5.27m length of RG6 was determined by s/c one end and connecting the other end via an adapter to a MFJ259B and finding the impedance dip at 23.05MHz. The calculated vf from this test is 0.81, which reconciles reasonably with the specs. Free space wavelenght at the test frequency is 13m. The transmitter was set to 23.05MHz, and the ATU tuned to develop rated power output. The ATU is only used to present the rated load to the transmitter so as to obtain 100W for the test, to suit the Bird 43 element. It is inconsequential to the DUT (the Bird 43). With this configuration, it is expected that the impedance at the Bird 43 is approximately 50+j0, and that there would be almost zero reflected power. The Bird 43 indicated 100W forward power and a quarter of a needle width detection on reflected power. The Bird 43 would appear to provide valid readings for the conditions on the Bird 43 Thruline section in this case, notwithstanding that there is not any 50 ohm transmission line attached to the Bird 43 + N-BNC adapters. The myth that measurements with a Bird 43 of the conditions on the Thruline section are invalid unless it has some minimum length of 50 ohm line on both sides of itself is BUSTED. Has anyone experimental evidence to the contrary? Owen What made you decide to use 23.05 MHz and 5.27 meters of .81 VF feedline? ac6xg |
On Wed, 12 Oct 2005 16:13:24 -0700, Jim Kelley
wrote: Owen Duffy wrote: The myth: Measurements with a Bird 43 of the conditions on the Thruline section are invalid unless it has some minimum length of 50 ohm line on both sides of itself. I have performed a test using components at hand where the Bird 43 has 75 ohm line on both sides of itself, and the test configuration is designed to present a 50+j0 ohm load at the point where the Bird 43 sampling element is located. The question is, how does the Bird respond? The test in detail. Each component is in a list from the source to the load: IC706IIG 1m RG58 with UHF connectors MFJ949E ATU 3m RG6 (Zo=75 ohm) with BNC connectors Bird 43 5.27m Belden 9275 (Zo=75 ohm, vf=0.83) with BNC connectors 50 ohm dummy load Short adapters were used to connect to the Bird's Type N connectors, the dummy load's Type N connectors and the MFJ949 UHF connector. The half wave resonance of the 5.27m length of RG6 was determined by s/c one end and connecting the other end via an adapter to a MFJ259B and finding the impedance dip at 23.05MHz. The calculated vf from this test is 0.81, which reconciles reasonably with the specs. Free space wavelenght at the test frequency is 13m. The transmitter was set to 23.05MHz, and the ATU tuned to develop rated power output. The ATU is only used to present the rated load to the transmitter so as to obtain 100W for the test, to suit the Bird 43 element. It is inconsequential to the DUT (the Bird 43). With this configuration, it is expected that the impedance at the Bird 43 is approximately 50+j0, and that there would be almost zero reflected power. The Bird 43 indicated 100W forward power and a quarter of a needle width detection on reflected power. The Bird 43 would appear to provide valid readings for the conditions on the Bird 43 Thruline section in this case, notwithstanding that there is not any 50 ohm transmission line attached to the Bird 43 + N-BNC adapters. The myth that measurements with a Bird 43 of the conditions on the Thruline section are invalid unless it has some minimum length of 50 ohm line on both sides of itself is BUSTED. Has anyone experimental evidence to the contrary? Owen What made you decide to use 23.05 MHz and 5.27 meters of .81 VF feedline? I have performed a test using components at hand where the Bird 43 has ac6xg -- |
Owen Duffy wrote:
The myth: Measurements with a Bird 43 of the conditions on the Thruline section are invalid unless it has some minimum length of 50 ohm line on both sides of itself. I have performed a test using components at hand where the Bird 43 has 75 ohm line on both sides of itself, and the test configuration is designed to present a 50+j0 ohm load at the point where the Bird 43 sampling element is located. The question is, how does the Bird respond? The test in detail. Each component is in a list from the source to the load: IC706IIG 1m RG58 with UHF connectors MFJ949E ATU 3m RG6 (Zo=75 ohm) with BNC connectors Bird 43 5.27m Belden 9275 (Zo=75 ohm, vf=0.83) with BNC connectors 50 ohm dummy load Short adapters were used to connect to the Bird's Type N connectors, the dummy load's Type N connectors and the MFJ949 UHF connector. The half wave resonance of the 5.27m length of RG6 was determined by s/c one end and connecting the other end via an adapter to a MFJ259B and finding the impedance dip at 23.05MHz. The calculated vf from this test is 0.81, which reconciles reasonably with the specs. Free space wavelenght at the test frequency is 13m. The transmitter was set to 23.05MHz, and the ATU tuned to develop rated power output. The ATU is only used to present the rated load to the transmitter so as to obtain 100W for the test, to suit the Bird 43 element. It is inconsequential to the DUT (the Bird 43). With this configuration, it is expected that the impedance at the Bird 43 is approximately 50+j0, and that there would be almost zero reflected power. The Bird 43 indicated 100W forward power and a quarter of a needle width detection on reflected power. The Bird 43 would appear to provide valid readings for the conditions on the Bird 43 Thruline section in this case, notwithstanding that there is not any 50 ohm transmission line attached to the Bird 43 + N-BNC adapters. The myth that measurements with a Bird 43 of the conditions on the Thruline section are invalid unless it has some minimum length of 50 ohm line on both sides of itself is BUSTED. Actually, the results of your experiment proves the myth to be true and not to be a myth at all. There's 104.17 watts of forward power through the Bird and 4.17 watts of reflected power back through the Bird. Why does the Bird ignore those actual power values? Has anyone experimental evidence to the contrary? Yes, your experiment. Assuming 100 watts delivered to the load, the forward power on the 75 ohm coax is actually about 104.17 watts so the Bird's forward power reading is in error by 4.17 watts. The reflected power on the 75 ohm coax is about 4.17 watts so the Bird's reflected power reading is in error by close to an infinite percentage. The Bird 43 is reading neither of the actual power values correctly. All you have just proven is that the Bird 43 gives invalid readings when it is in a 75 ohm environment. THERE ARE ABOUT 4.17 WATTS OF REFLECTED ENERGY FLOWING BACK THROUGH THE BIRD AND THE BIRD COMPLETELY IGNORES IT. So the Bird is not even yielding valid readings for forward and reflected power through itself. That's exactly what I have been saying all along. If it were calibrated for 75 ohms, it would indicate the correct values. -- 73, Cecil http://www.qsl.net/w5dxp |
"Cecil Moore" wrote in message . .. Owen Duffy wrote: The myth: Measurements with a Bird 43 of the conditions on the Thruline section are invalid unless it has some minimum length of 50 ohm line on both sides of itself. I have performed a test using components at hand where the Bird 43 has 75 ohm line on both sides of itself, and the test configuration is designed to present a 50+j0 ohm load at the point where the Bird 43 sampling element is located. The question is, how does the Bird respond? The test in detail. Each component is in a list from the source to the load: IC706IIG 1m RG58 with UHF connectors MFJ949E ATU 3m RG6 (Zo=75 ohm) with BNC connectors Bird 43 5.27m Belden 9275 (Zo=75 ohm, vf=0.83) with BNC connectors 50 ohm dummy load Short adapters were used to connect to the Bird's Type N connectors, the dummy load's Type N connectors and the MFJ949 UHF connector. The half wave resonance of the 5.27m length of RG6 was determined by s/c one end and connecting the other end via an adapter to a MFJ259B and finding the impedance dip at 23.05MHz. The calculated vf from this test is 0.81, which reconciles reasonably with the specs. Free space wavelenght at the test frequency is 13m. The transmitter was set to 23.05MHz, and the ATU tuned to develop rated power output. The ATU is only used to present the rated load to the transmitter so as to obtain 100W for the test, to suit the Bird 43 element. It is inconsequential to the DUT (the Bird 43). With this configuration, it is expected that the impedance at the Bird 43 is approximately 50+j0, and that there would be almost zero reflected power. The Bird 43 indicated 100W forward power and a quarter of a needle width detection on reflected power. The Bird 43 would appear to provide valid readings for the conditions on the Bird 43 Thruline section in this case, notwithstanding that there is not any 50 ohm transmission line attached to the Bird 43 + N-BNC adapters. The myth that measurements with a Bird 43 of the conditions on the Thruline section are invalid unless it has some minimum length of 50 ohm line on both sides of itself is BUSTED. Actually, the results of your experiment proves the myth to be true and not to be a myth at all. There's 104.17 watts of forward power through the Bird and 4.17 watts of reflected power back through the Bird. Why does the Bird ignore those actual power values? Has anyone experimental evidence to the contrary? Yes, your experiment. Assuming 100 watts delivered to the load, the forward power on the 75 ohm coax is actually about 104.17 watts so the Bird's forward power reading is in error by 4.17 watts. The reflected power on the 75 ohm coax is about 4.17 watts so the Bird's reflected power reading is in error by close to an infinite percentage. The Bird 43 is reading neither of the actual power values correctly. All you have just proven is that the Bird 43 gives invalid readings when it is in a 75 ohm environment. THERE ARE ABOUT 4.17 WATTS OF REFLECTED ENERGY FLOWING BACK THROUGH THE BIRD AND THE BIRD COMPLETELY IGNORES IT. So the Bird is not even yielding valid readings for forward and reflected power through itself. That's exactly what I have been saying all along. If it were calibrated for 75 ohms, it would indicate the correct values. -- 73, Cecil http://www.qsl.net/w5dxp math please??? where do you get 4.17watts?? to me it looks like a 50 ohm load on a 50 ohm meter so zero reflected power. |
Dave wrote:
math please??? where do you get 4.17watts?? to me it looks like a 50 ohm load on a 50 ohm meter so zero reflected power. Here's the setup. 100W--tuner--75 ohm coax--Bird--1/2WL 75 ohm coax--50 ohm load Assumptions: Losses are negligible. The tuner provides a Z0-match so 100 watts is delivered to the load, i.e. all reflected power is re-reflected. I think this was actually the case for the experiment. Also assume that the impedance bump caused by the insertion of the Bird is negligible, i.e. the same net voltage and net current exists whether the Bird is in or out of the circuit. The voltage reflection coefficient at the load is (50-75)/(50+75) equals -0.2. That makes the power reflection coefficient (-0.2)^2 equals 0.04, i.e. 4% of the power incident upon the load is reflected. If 100 watts is delivered to the load while 4% is being reflected then the power incident upon the load is 100w/(1-0.04)=100w/0.96= 104.1667 watts. Incident power minus delivered power = reflected power so the reflected power on the 75 ohm coax is 4.1667 watts. The SWR on the 75 ohm coax is (1+|rho|)/(1-|rho|)= 1.2/0.8 = 1.5 THE SWR ON THE 75 OHM COAX IS 1.5:1. The Bird is in error when it reports the SWR to be 1:1. The SWR is *NOT* 1:1 anywhere on the load side of the system. The Bird reports a forward power of 100w. The actual forward power is 104.1667w. The Bird reports a reflected power of near zero. The actual reflected power is 4.1667w. The Bird reports an SWR of 1:1. The actual SWR is 1.5:1. The Bird is reporting false values because it is embedded in a 75 ohm environment and is calibrated for 50 ohms. If the ratio of voltage to current equals 50 within a 75 ohm system, there exists an SWR of 1.5:1 and an SWR meter calibrated for 75 ohms will verify that fact. -- 73, Cecil http://www.qsl.net/w5dxp |
Cecil Moore wrote:
The SWR is *NOT* 1:1 anywhere on the load side of the system. Sorry, this should have read: The SWR is *NOT* 1:1 anywhere on the load side of the tuner. -- 73, Cecil http://www.qsl.net/w5dxp |
Dave wrote:
math please??? where do you get 4.17watts?? to me it looks like a 50 ohm load on a 50 ohm meter so zero reflected power. Let's compare the following two configurations. 100W--tuner--75 ohm coax--Bird--1/2WL 75 ohm coax--50 ohm load 100W--tuner--75 ohm coax--Bird--1/4WL 75 ohm coax--50 ohm load The actual SWR on the coax is the same in both cases. The actual forward power and reflected power are the same in both cases. The SWR on the coax is 75/50 = 1.5:1. The actual forward power is 104.1667w and the actual reflected power is 4.1667w. In the first case the Bird will report an SWR of 1:1, a forward power of 100w, and a reflected power of zero simply because the ratio of net voltage to net current is 50 ohms at the Bird measurement point. In the second case the Bird will report an SWR of 2.25:1, a forward power of 117.4w and a reflected power of 17.4w simply because the ratio of net voltage to net current is 112.5 ohms at the Bird measurement point. The ratio of net voltage to net current alone is useless for calculating SWR unless the Z0 environment (characteristic impedance) exists and is known. Minus the humor, I believe this is Reg's main point. -- 73, Cecil http://www.qsl.net/w5dxp |
"Cecil Moore" wrote in message .. . Dave wrote: math please??? where do you get 4.17watts?? to me it looks like a 50 ohm load on a 50 ohm meter so zero reflected power. Here's the setup. 100W--tuner--75 ohm coax--Bird--1/2WL 75 ohm coax--50 ohm load Assumptions: Losses are negligible. The tuner provides a Z0-match so 100 watts is delivered to the load, i.e. all reflected power is re-reflected. I think this was actually the case for the experiment. Also assume that the impedance bump caused by the insertion of the Bird is negligible, i.e. the same net voltage and net current exists whether the Bird is in or out of the circuit. STOP! you can not assume this, especially if you consider the actual results of the experiment. the bird is a 50 ohm transmission line segment and it is seeing a 50 ohm load. the 50 ohm load plus 1/2 wave of anything is 50 ohms. so as far as the bird knows there is no reflected power and the real world reading is correct. |
"Cecil Moore" wrote in message
.. . Dave wrote: math please??? where do ---- ----alone is useless for calculating SWR unless the Z0 environment (characteristic impedance) exists and is known. Minus the humor, I believe this is Reg's main point. -- 73, Cecil http://www.qsl.net/w5dxp Did I miss something? As far as the Bird is concerned, all it sees is 50 ohms, indistinguishable from a 50 ohm termination, on its connector. 73, Frank |
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