Complex power. V and I of lumped elements

[kavea]

Hi guys,
Can some one explain me how is it possible to get these voltages and currents on this circuit ? If P=1kW, f=2MHz

And/or these amplitudes of voltage/current at the power supply in ADS:


It doesn't match my calculus, because i find \[U_{rms}=264.5V\] and \[I_{rms}=4.52A(-33.3^{\circ})\]

 

[chuckey]

You do not say exactly where Urms and I rms are in the circuit. If this is meant to be a resonant circuit then X2 (output L) can be replaced by two Ls in parallel, one whose reactance is the same as the aerial but with the opposite sign ( i.e. +j519) and another to bring the total reactance back to +j 417. By a back of a fag packet calculation, I make this about +j2000. So this L and the series C and the remaining L, is a 1:1 matching circuit, so the 3 ohms is reflected through to the generator, this is not a good matching unit!! . I suspect your figures are correct.
Frank

 

[FvM]

Do you own a text book about AC circuits?

Not particularly dedicated to RF applications, but with a good introduction to RLC circuit calculations: https://www.ibiblio.org/kuphaldt/electricCircuits/AC/index.html

 

[kavea]

You do not say exactly where Urms and I rms are in the circuit. If this is meant to be a resonant circuit then X2 (output L) can be replaced by two Ls in parallel, one whose reactance is the same as the aerial but with the opposite sign ( i.e. +j519) and another to bring the total reactance back to +j 417. By a back of a fag packet calculation, I make this about +j2000. So this L and the series C and the remaining L, is a 1:1 matching circuit, so the 3 ohms is reflected through to the generator, this is not a good matching unit!! . I suspect your figures are correct.
Frank

Urms and Irms are calculated at the power source, as well as U_SRC and I_SRC on 3rd figure (but max values). No, it isn't a good matching circuit, the equivalent impedance of all circuit is about \[Zin=48.85+j32.15\]. But it's ok, i want just understand the values given by ADS and (especially) by another calculator at the 1st screenshot. I also suspect my figures correct, but if i could find these values by calculus it would be great. There is all to be be calculated by hand [F=2Mhz, values of components, P=1kW, Z0=50 Ohms]. The goal is to replace this matching circuit by an other (5 or more elements) and calculate V and I at every single component, because voltages of several kV are too high in 3 element matching circuit.

Do you own a text book about AC circuits?

Not particularly dedicated to RF applications, but with a good introduction to RLC circuit calculations: **broken link removed**

Normally yes, but i will take a look

 

[chuckey]

I know that the best way to deal with these short aerials is to put a L in series with them, at resonance, the load presented will be 3 ohms. There will be a high voltage across the L, so wind it with insulated wire on a large former perhaps using basket weave. Now you have to match 50 ohm to 3 ohm, no ultra high voltages, just the output current ( 1 KW across 3 ohms = ( P=I^2 R) = 1000/3 ~ 17 A.
I worked on a 400 KHZ 80W transmitter that had to be matched to 3 ohms+ 100 PF, it had a huge coil in series with the dummy aerial. Coil former was about 100 mm in diameter. To adjust it to resonance, it had a smaller coil former inside the main one with windings on it that were in series with the main coil. One rotated the inner coil ( + / - mutual inductance) to resonate the dummy aerial.
Frank