Hello
I have a quick question about deriving the lower critical frequency for the output circuit of an amplifier. Thevenizing the output circuit we get this:
The output voltage can be written as
\(V_{out} = \frac{R_L}{R_L + R_C +\frac{1}{j \omega C}}V_{coll}\)
so just concentrating on the magnitudes we get
\(V_{out} = \frac{R_L}{\sqrt{(R_L + R_C)^2 +X_C^2}}V_{coll}\)
I think I am right in saying that the lower critical frequency is defined as the frequency at which the voltage has decreased by a factor of 1/√2 (i.e. 3dB)? So should I not solve the above equation for \(X_C\) in terms of \(R_L\) and \(R_C\) when \(V_{out} = \frac{1}{\sqrt2}V_{coll}\)?
My book states the answer as
\(f_{cl(output)} = \frac{1}{2 \pi (R_C + R_L)C_3}\)
i.e. \(X_C = R_L + R_C\), but this does not give an attenuation of -3dB does it?! The answer I get is
\(X_C = \sqrt{R_L^2 - R_C^2 - 2R_LR_C}\)
Thanks for your help!
I have a quick question about deriving the lower critical frequency for the output circuit of an amplifier. Thevenizing the output circuit we get this:
The output voltage can be written as
\(V_{out} = \frac{R_L}{R_L + R_C +\frac{1}{j \omega C}}V_{coll}\)
so just concentrating on the magnitudes we get
\(V_{out} = \frac{R_L}{\sqrt{(R_L + R_C)^2 +X_C^2}}V_{coll}\)
I think I am right in saying that the lower critical frequency is defined as the frequency at which the voltage has decreased by a factor of 1/√2 (i.e. 3dB)? So should I not solve the above equation for \(X_C\) in terms of \(R_L\) and \(R_C\) when \(V_{out} = \frac{1}{\sqrt2}V_{coll}\)?
My book states the answer as
\(f_{cl(output)} = \frac{1}{2 \pi (R_C + R_L)C_3}\)
i.e. \(X_C = R_L + R_C\), but this does not give an attenuation of -3dB does it?! The answer I get is
\(X_C = \sqrt{R_L^2 - R_C^2 - 2R_LR_C}\)
Thanks for your help!
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