skatefast08
Full Member level 4
I have been reading this article on Practical RF Amplifier Design Method from Keysight, as shown on a PDF here: https://www.google.com/url?sa=t&sou...FjAAegQIBhAB&usg=AOvVaw1p-gaSExXoWSgc0aBLURwJ
If you look at page 20 at the end of the first paragraph they say, "The battery is connected to the 5 Ω resistor
and 0.1 μF capacitor junction in the final circuit." If they connect the capacitor in series with the battery, then they are going to block the DC power to the drain, unless the capacitor is in parallel with the 5 ohm and 27nH inductor, right?
Another question, if you look at page 49, why do they use the ideal feed inductor and 5 ohm resistor that connect to the drain? Why couldn't they just have connected the 27nH (RF choke) coilcraft inductor where the ideal inductor is? Instead they place the same series circuit that suppose to go to the drain and place it off to the side near node 3a. I understand that they are trying to represent the actual component response, but when they place the circuit near the battery, then the 0.1uF capacitor will block the DC component, which kind of leads to my first question. I hope you can understand my question, thanks.
If you look at page 20 at the end of the first paragraph they say, "The battery is connected to the 5 Ω resistor
and 0.1 μF capacitor junction in the final circuit." If they connect the capacitor in series with the battery, then they are going to block the DC power to the drain, unless the capacitor is in parallel with the 5 ohm and 27nH inductor, right?
Another question, if you look at page 49, why do they use the ideal feed inductor and 5 ohm resistor that connect to the drain? Why couldn't they just have connected the 27nH (RF choke) coilcraft inductor where the ideal inductor is? Instead they place the same series circuit that suppose to go to the drain and place it off to the side near node 3a. I understand that they are trying to represent the actual component response, but when they place the circuit near the battery, then the 0.1uF capacitor will block the DC component, which kind of leads to my first question. I hope you can understand my question, thanks.