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Re: [softrock40] Local oscillator: don't understand the circuit

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  • G8SQH
    As already said, it s a Colpitts Looking at C12 - remember the LO runs at 4* band freq. Hence for 40m the reactance is 1/(2*pi*F*C) = 254 ohms at 28.4 MHz.
    Message 1 of 3 , Apr 29 4:57 PM
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      As already said, it's a Colpitts

      Looking at C12 - remember the LO runs at 4* band freq. Hence for 40m the reactance is 1/(2*pi*F*C) = 254 ohms at 28.4 MHz. This gives enough coupling to the common emitter amplifer, so no reason to load the oscillator any more

      (I can assume you that 22 pF in my 30/40 6.1 tx/rx works fine - now up to 57 countries on one Watt into a 1/4 vertical.. Seriously - I can recommend 30m for PSK31 qrp - I haven't switched to 40m for ages)

      73

      --David G8SQH

      On Thu, 29 Apr 2010 18:03:25 -0000
      "cam0r0" <estolavi@...> wrote:

      > Hi,
      >
      > I just received my 40m softrock. I started to assemble it, but I also wanted to understand exactly how the circuit works. The more I learn, the more I see that my problem are always related to capacitors.
      > Basically, in these schema: http://www.wb5rvz.com/sdr/sr_lite_ii/02_lo.htm, what are doing C10 & C11 ? Why are they connected to the Emitter of the transistor (It looks like a Colpitt oscillator, but...) ? C12 has the role of coupling capacitor I believe, how one can come up with the value of 22pF ?
      >
      > Thanks a lot for your help !
      >
      > c0
      >


      --
      <djch-yahoo@...>
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