This session wasn’t necessarily poor but there was little to stand out as exciting aside from the trans-Atlantic reports in the North eastern US. It seems the geoeffective coronal hole that was supposed to begin impacting the Earth a few days ago is only just now impacting radio propagation. The Kp is elevated and solar wind velocities are in excess of 500 km/s.
Roelof, PA0RDT, reported very easy copy of Joe, VO1NA, who was transmitting CW at 477.7 kHz.
David, G0MRF, reports that his portable operation has ceased for the season but not before a nice QRPP report from WG2XJM:
WSPR activity was very good once again, with MF WSPR station numbers in the 80’s during the North American evening as reported on the WSPRnet activity page.
Regional and continental WSPR breakdowns follow:
There were no reports from the trans-African path. UA0SNV was present from Asiatic Russia but had no reports.
Trans-Atlantic reports were the bright spot in the session for four North American stations that either heard or were heard in Europe.
Eden, ZF1EJ, and Roger, ZF1RC, provide reports from WG2XXM and WG2XIQ.
Laurence, KL7L / WE2XPQ, was another bright spot in the session, with a few nice reports from his portable outpost in VE6.
Laurence’s receiver in Alaska also reported many of the typical stations in the West during the session:
In Hawaii, Merv, K9FD/KH6 / WH2XCR, experienced a typical session, with an open two-way path to Australia and reports from WG2XJM in western Pennsylvania.
In Australia, Phil, VK3ELV, and Roger, VK4YB, receive reports from WH2XCR.
Jim, W5EST, continues his discussion on RF current in lossy systems entitled, “PART 2: RF CURRENTS ON 630/2200M WITH LOSSY ATU”:
“Lossy ATUs get airtime today. I’ve patterned this post for side-by-side comparison with the lossless ATU case discussed yesterday. Let’s talk about 630m/2200m ATUs when their losses exceed 5% of the sum of antenna ohmic losses and grounding system losses for the whole antenna system. You know what ATU loss is. It’s what makes the ATU feel warm when you shut off the TX and open the ATU box. It’s what melts snow off the ATU box if it snows in winter where you live. The question is whether that warmth represents significant ATU losses compared to TPO.
Whatever the SWR, the net power into the ATU is forward power minus the reflected power minus a little transmission line loss. Because reflected power gets back to the transmitter (TX) the output stage heats up because dissipation is where that reflected power goes.
Now let’s assume you do match the SWR 1:1 with the ATU, but ATU is lossy. What happens to the net power into ATU (forward power minus reflected power)? What does all this mean for current, voltage and impedance transformations by a lossy ATU at 1:1 SWR ?
First, the lossy ATU matches its 50Ω input not to the total ohmic system resistance Rsystem but to some combined impedance involving Rsystem and an equivalent ATU dissipative resistance RATU.
Second: Not surprisingly, loss in the ATU diminishes power-out because power-out equals power-in minus the ATU’s power loss PATU:
Iout2 Rsystem = Iin2 50Ω – Iin2 RATU.
RATU = PATU/ Iin2 where PATU is the power dissipation in the ATU. Recall that Iin2 50Ω is the TPO when coax losses are negligible.
The output current Iout of the ATU is related to the TPO and the ATU losses by:
Iout = sqrt [(TPO- PATU) / Rsystem]
So the output current of the ATU is related to the input current to the ATU by
Iout/Iin = sqrt [(50Ω- RATU) / Rsystem].
Same as for lossless ATU, if TPO is 100 watts, RF coax current into the ATU is:
Iin = sqrt(TPO/50Ω) = sqrt (100 watts/50Ω) = 1.41A rms.
Suppose system resistance were 25Ω and the ATU dissipates 20 watts. Then the antenna base current delivered by the lossy ATU is:
Iout= sqrt (100w -20w / 25Ω) = 1.8A rms. (Reduced from 2A a lossless ATU delivers).
In a real system, the ATU loss is probably the unknown that you would like to find. One more good reason to have an RF ammeter (besides using RF amps to guesstimate EIRP) is to help you determine whether the ATU loss is significant. With antenna base disconnected from the ATU, suppose an antenna analyzer accurately measures the antenna base impedance Rsystem+jX to be 25 –j3000 ohms (capacitive). Then Rsystem = 25Ω, and you’re ready to use this next equation for ATU loss:
PATU = Iin2 50Ω – Iout2 Rsystem = (1.41A)250Ω – (1.8A)225Ω = 20w.
Third, the voltage output of the ATU is the voltage developed by the current Iout flowing through the system resistance Rsystem and reactance X. Then the ATU output voltage Vout at input of antenna base is
Vout = Iout (Rsystem + j Xsystem) = 1.8A rms x (25 – j3000)Ω = (45 – j5400)V, or about 5.4KV rms. That means about 7.6 KV peak RF voltage since peak voltage is 1.41 x rms voltage.
Fourth, the ATU’s lossy power transfer means that VsysΩ Iout =Vin Iin – PATU Consequently, the voltage VsysΩ developed in the antenna system resistance is given by:
VsysΩ / Vin = Iin / Iout – PATU/(Vin Iout)
The ATU input voltage is Vin = sqrt(TPO x 50Ω) = sqrt (100w x 50Ω) = 70.7V rms in the coax or 100V peak RF voltage on the coax input to ATU, same as for a lossless ATU case. A “current step-up ratio” for the lossy ATU is:
Iout / Iin = sqrt [(TPO- PATU) / (Iin250Ω)) (50Ω /Rsystem)]
Iin250Ω is just the TPO. ATU power transfer efficiency eta η is the ratio of the power that reaches the antenna system divided by the TPO. Efficiency then is (TPO- PATU) /TPO. Also, the ratio 50Ω /Rsystem is the square of the current step-up ratio for a lossless ATU. So the current step-up ratio for a lossy ATU is that ratio for a lossless ATU discounted by the square root of the efficiency, or:
Iout / Iin = sqrt(50Ω /Rsystem) sqrt(η) = sqrt(50Ω /25Ω)sqrt(0.80) = 1.26.
A further blog post could discuss losses and behavior of particular ATU circuits. We invite your words of wisdom from your own experience, and let’s blog them. TU & GL!”
Additions, corrections, clarifications, etc? Send me a message on the Contact page or directly to KB5NJD <at> gmail dot (com)!