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How come its OK to mix impedances in a radio system??

Discussion in 'Electronic Basics' started by billcalley, Jul 17, 2005.

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  1. billcalley

    billcalley Guest

    Hi All!
    I have to admit I am really confused: When dealing with the RF
    power transfer in a 50 ohm system, I have been told repeatedly that we
    would want ALL of the amplifier (and filter) stages matched to 50 ohms.
    That way, I have been told, we would get the most power from, lets
    say, the output of a radio receiver into its demodulator*. But then
    how come the power does not decrease when a 50 ohm IF signal is
    inserted directly into a very high impedance (buffer) op-amp? Why
    isn't most of that energy reflected right back to the 50 ohm stages
    and wasted? I just don't get this! And if we can safely mix a low
    impedance stage with a high impedance stage and still be just fine,
    then how would I calculate the actual system gain and final output
    power of such a setup?



    * In other words, if the stages are not all matched to the same
    reference impedance, then the output power would be less than what it
    should be.
  2. Bob Eldred

    Bob Eldred Guest

    You are confusing power transfer with voltage transfer. Maximum power is
    transfered when the impedances are matched (really complex conjugates).
    However, maximum voltage occurs when the source is not loaded, but then the
    power transfer is nill. A high input impedance amplifier does this. It
    "samples" the voltage without loading the source. But, by ohms law the power
    transferred is very small P = V^2/Z. If Z is big, P is small. By not
    loading sources, lower distorion and higher voltage gain is achieved but not
    power gain. Furthermore, the "Q" of tuned circuits is not degraded by
    loading when the impedances are high.
  3. billcalley

    billcalley Guest

    Thanks for the great info Bob -- I'll have to mull it over a bit to get
    it straight in my thick head!

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