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[SOLVED] RF Input Power of Power Amplifier ADS

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darwinxie

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Hi all,

I am designing a RF power amplifier for my final project. The PA adapted Doherty architecture.
I intend to design it for input power of 25 dBm (which is 0.316 W). If the input signal is a sinusoidal signal, how much is the peak voltage?
The 25 dBm, can I consider it as a RMS value and using P = Vrms^2 / R formula?
I look forward for your help.

Thank you..
 

Yes, if your amplifier input impedance is properly matched with you input port.
Since you already use ADS for design assistant, you could just use a voltage probe at the node you are interesting with to see what happen. That's why we need a simulation software, to measure what is not possible to see in practice. :smile:
 

Hi all,

I am designing a RF power amplifier for my final project. The PA adapted Doherty architecture.
I intend to design it for input power of 25 dBm (which is 0.316 W). If the input signal is a sinusoidal signal, how much is the peak voltage?
The 25 dBm, can I consider it as a RMS value and using P = Vrms^2 / R formula?
I look forward for your help.

Thank you..

Your equation is close, all you need to is convert Vrms to Vpeak... so \[{V}_{Peak} = {V}_{RMS}*\sqrt{2}\].

Another handy equation to remember is this one (moves the sqrt(2) to the denominator).
\[{P}_{AVG} = \frac{{{V}_{PK}}^{2}}{2*R}\]

Remember, this works when you know the characteristic impedance at a particular point in a circuit. For a nice 50 ohm transmission line (like a coax cable or SMA connector, Z = R = 50 ohms). However, when you start matching your transistor for the amplifier, you probably won't be matching to 50 ohms, so be careful when and how you apply that voltage conversion, especially in/after your matching circuit. Simulation will tell you what you're really getting after all of the L's and C's are taken into account.
 
Ok, I get it. However, I do the matching impedance using microstrip line, not lumped elements of C and L. Is there any different?
 

Ok, I get it. However, I do the matching impedance using microstrip line, not lumped elements of C and L. Is there any different?

By using lumped elements or microstrip lines, you are still adjusting the circuit impedance seen by the device or load. It's the same effect, just two different paths to the same destination.
 

All right then,, big thank you for your help guys..
 

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