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

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

1. ### billcalleyGuest

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?

Signed;

Bill

* 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 EldredGuest

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