Re: Matching source and input impedances in power amplifiers
- From: theduder2005@xxxxxxxxx
- Date: 21 Jun 2006 08:28:55 -0700
So the biggest reason to match is to prevent reflections?
It makes sense why you'd want to match Zl to Zo (assuming everything is
unilateral) to transfer the most power FROM the amplifier to the load.
But why would you want to transfer the most power TO the amplifier?
Aren't most amplifiers inherently voltage or current sensing?
If the amplifier had say a common gate structure, won't there be more
output power (with a matched load) if you maximized the input voltage
(vgs) by making Zin >> Zs (given the frequency is low enough to
disregard reflections)?
rick H wrote:
theduder2005@xxxxxxxxx wrote:
Hi.
I have a question concerning impedance matching in RF power amplifiers.
Given a power amplifier with input impedance Zin and output impedance
Zo, a source with impedance Zs and a load Zl - does conjugate matching
Zin and Zs give the highest output power?
Yes, if the amplifier is unilateral (S12=0).
In the more general case of a bilateral network (S12 is not 0)
changing the output's load will change the impedance looking into
the input of the amplifier, and changing the input's load will change
the impedance looking into the output of the amplifier.
In terms of reflection coefficients and S-parameters, you end up
with a set of two simultaneous equations:
conjugate(Gamma_source) = S11 + (S12*S21*Gamma_load)/(1-S22*Gamma_load)
conjugate(Gamma_load) = S22 + (S12*S21*Gamma_source)/(1-S11*Gamma_source)
which can be solved to give Gamma_source and Gamma_load required for
maximum transducer power gain.
I know that matching Zo to Zl will transfer the most power to the load,
but shouldn't Zin be much greater than Zs so the amplifier "sees" the
whole input signal? (ignoring reflections and noise performance)
You can't ignore reflections. If you make Zin very high, you will indeed
maximise the voltage incident on the Zin terminal, but the magnitude of
the reflection coefficient between the transmission line and Zin will
be close to unity - i.e. almost all of the power incident on the PA's
input will be reflected back to the source. If little of the power
available from the source is absorbed by the amplifier, then whatever
the power-gain of the amplifier, little will be available at its output
for delivery to the load.
--
Rick
.
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