# What is the difference between an RF amplifier's DC or quiescent operating point and the RF operating point? I chose the resistor values in the following schematic to set the DC bias point of VCE=6V and IC=5mA because the transistor data sheet gave the S-parameters at 162MHz for these values. I learned how to do this by studying the book "Rf Circuit Design" by Chris Bowick. Early on in the book he mentions that stray capacitance inherent in the DC bias network resistors can effect the "RF operating point". I know what the DC operating point is ,obviously, but what is the RF operating point and how is it effected by the stray parallel capacitance of the bias network resistors? In this case I'm referring to the stray capacitance of the 3.240K Ohm resistor only because the other three resistors are RF bypassed by the .1 micro Farad capacitors. Bowick does explain how the impedance of a resistor can lessen with an increase in frequency but I just don't know what the RF operating point of the bias network is much less how the reduced impedance of the resistor effects it.

• In typical circuit analysis, it's customary to consider the DC model (short all inductors, open all caps, resistors all ideal) and AC model separately. The AC model is simplified by assuming that resistors are ideal and that the AC component simply "rides along" on the DC op point. However, at RF frequencies, the inherent parasitic capacitance and inductance of resistors changes the operating point (imagine a capacitor in parallel and inductor in series with the 3.24k base biasing resistor). I think this is what he means by RF operating point. Dec 27 '20 at 10:13
• @Buck8pe that sounds like an answer to me. Dec 28 '20 at 4:51
• So exactly what happens to the amplifier if the DC and AC operating points don't match? Does it change the AC, or in this case, the RF load line like a class A audio amplifier? And if it changes the load line, how is the collector saturating current and the emitter to collector cutoff voltage calculated? Dec 28 '20 at 5:33