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RF Power Amplifier Output question?

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Y.C Park

Member level 2
Dec 20, 2004
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Hi all

I am a newbie. In fact I am an analog Eng'r.
In the designing PA, I have a big problem.
if Pout=30dBm, load impedance is 50ohm. VCC=3.5V
So output voltage swing is much larger than VCC.
Is it reasonable?


What about the current your PA can source?

You have to operate the transistors in a low impedance circuit and use a matching network to transform the output to drive a 50ohm load.

An LC network that also supresses harmonic output is often used. Sometimes transmission line matching is used.


I've still confused ....
sometimes power amplifier output voltage swing is smaller than input. we just consider power. But fixed 50ohm driving load... increase current means increase voltage...
which point is wrong for me?

for example) if target output power is 30dBm, 50ohm driving load
power supply is 3.5V. I wonder output voltage swing...
my thought is 50ohm load , output voltage should be 10V.
output current is 0.2A... Something wrong ?
I think this question looks so stupid but I can not understand....TT

Thanks in advance


Your impedance network "takes" care of the transformation of the voltage and currents. The way we insert the voltage at either drain/collector through fx. inductor. The voltage swing at the drain is larger than your supply voltage: Vcc+Vd. This can happen because of your LC network on the output, which transforms the voltage and currents.


For a class AB PA that would be a reasonable assumption.

Make sure you "de-couple" your measurement device, so your voltmeter is not influenced by RF. If you want me to elaborate please post here.

Let us do a simple calculation. Then you will know why output swing can be much higher than VCC.

Suppose the impedance seen from the drain or collector of the transistor is 2 Ohm, voltage swing at the drain or collector is 3V. Also, suppose the efficiency of matching network is 66% and the impedance of terminal load is 50 Ohm.

Then the power output from the transistor is 3^2/(2*2)=2.25W. And the final output power on 50 Ohm load is 2.25*66%=1.5W. So, the final output swing is about 12V.

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