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(It's really a continuation of a previous question but it is old so I decided to ask a new one with more detail)

I have a colpitts oscillator designed for 27 MHz going into a short monopole antenna of 1/20 of the wavelength, or 55 cm. The radiation resistance is 2 Ohms, with capacitance of 5 pF (calculated with this calculator). This means that to bring it to the resonance I am to use 6.9 uH inductor, according to the same calculator. My colpitts oscillator has output impedance of 300 ohms, which I matched to 50 ohm load. If I understand correctly now I have to make my 2 ohm real resistance load look like it is 50 ohm load. So here's what I came up with:

schematic

simulate this circuit – Schematic created using CircuitLab

I am not sure if this is really done this way. Is the impedance of the antenna (after I include tuning inductor of 6.9uH) really 2 ohm real resistance, or is it calculated in some other way? The other question is if matching both the source and the load to 50 ohms is done this way, or the calculation for when both of them should be matched is different (I am not sure if the L and C elements from different L networks would influence each other and the outcome?

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  • \$\begingroup\$ LC ccts change impedance of input and output (f) but does not boost power, only minimize reflected power if matched. Consider Google images for better 27MHz power solutions with antenna \$\endgroup\$
    – Tony Stewart EE75
    Aug 24, 2017 at 14:58
  • \$\begingroup\$ Congrats on building that oscillator. Short whip antennas do require some series inductance to make them appear resistive (that resistance tends to be small). But you're forgetting about ground impedance: an excellent RF ground is required to drive low-Z antennas efficiently. Otherwise, your whip is like the tail wagging the dog. \$\endgroup\$
    – glen_geek
    Aug 24, 2017 at 17:00
  • \$\begingroup\$ How did you get to the matching circuit shown in your question? \$\endgroup\$
    – Andy aka
    Aug 24, 2017 at 17:09
  • \$\begingroup\$ @glen_geek Thanks! Hm.. Yeah I haven't gotten to that part yet. I've seen monopoles with several conductors facing downwards as a ground plane, not sure if I will be able to implement that though. \$\endgroup\$
    – Anthropomorphous Dodecahedron
    Aug 24, 2017 at 17:49
  • \$\begingroup\$ @Andyaka I've calculated the first LC pair as if I was matching the 300 ohm source to 50 ohm load. Calculating that Q = 2.23 and from that calculating the impedances of capacitor and inductor. The same thought process for the second pair just with different Q = 4.9. I may have not done it correctly, I am not too experienced in the whole thing. \$\endgroup\$
    – Anthropomorphous Dodecahedron
    Aug 24, 2017 at 17:53

1 Answer 1

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I suggest you to Download AppCAD DOS version from http://www.hp.woodshot.com/appcad/appcadcl.exe

Run the APPCAD.BAT and choose impedance matching L type. This can easily solve this circuit by only 2 components basically.

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