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

billcalley · 2005-07-17T01:36:15+00:00

Hi All! I have to admit I am really confused: When dealing with the RFpower transfer in a 50 ohm system, I have been told repeatedly that wewould 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, letssay, the output of a radio receiver into its demodulator*. But thenhow come the power does not decrease when a 50 ohm IF signal isinserted directly into a very high impedance (buffer) op-amp? Whyisn't most of that energy reflected right back to the 50 ohm stagesand wasted? I just don't get this! And if we can safely mix a lowimpedance stage with a high impedance stage and still be just fine,then how would I calculate the actual system gain and final outputpower of such a setup?Signed;Bill* In other words, if the stages are not all matched to the samereference impedance, then the output power would be less than what itshould be.

Jul 17, 2005 41 Replies

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Fred Abse 21 years ago

T'was ever thus. The classical maximum power transfer theorem is all about loading a generator which has a finite internal impedance, not the other way round.

Fixing the load and varying the source impedance isn't the same thing.

It's only a potential divider, FFS, yet people get hung up on it.

"Electricity is of two kinds, positive and negative. The difference is, I presume, that one comes a little more expensive, but is more durable; the other is a cheaper thing, but the moths get into it." (Stephen Leacock)

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