Linear RF Power amplifier

Chinku_420 said:

Rload will be 9/(2* 3.162m) = 1.43 kOhm

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Where does the "9" and "2" come from? I don´t understand the 1.43 kOhms.

Usually the input and output power in dBm is referenced to the characteristic impedance. Usually 50 Ohms.

Klaus

KlausST said:

Where does the "9" and "2" come from? I don´t understand the 1.43 kOhms.

Usually the input and output power in dBm is referenced to the characteristic impedance. Usually 50 Ohms.

Klaus

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Pout= VDD^2/(2 Rl).
Is that right formula?

You are apparently assuming an ideal class A output stage with output inductor, achieving 2*Vdd swing. I think that's completely unrealistic and no suitable way to design microwave amplifiers. Also 1.4k:50 ohm impedance matching is no good design idea.

FvM said:

You are apparently assuming an ideal class A output stage with output inductor, achieving 2*Vdd swing. I think that's completely unrealistic and no suitable way to design microwave amplifiers. Also 1.4k:50 ohm impedance matching is no good design idea.

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Thanks for the reply.

At this point, I am only doing calculations while assuming everything ideal. I know that the the drain inductor has very low Q factor. But, my question is how can I chose the load resistance with the same specs.

I know about the load-pull analysis and conjugate match concept. But at this point, I want to do the simple calculations.
Thanks

Your theoretical load resistance equation is fine. Rload=(Vcc^2)/(2*Pout). The reason that you get such high load resistance is because the output power is low (3mW).
For the given Vcc, if you increase the output power to about 100mW you get a load resistance close to 50 ohms.
The high load resistance in very low power amplifiers should not be an issue.
Problems in PA design starts to appear in high power amplifiers, when the load resistance is very low (under 1 ohm).