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According to the book I'm studying, the cutoff frequency for this circuit Collector network is given by $$f_c=\frac{1}{2\pi(R_C+R_L)C_3}.$$

I tried to derive it by solving for the frequency at which \$V_{out}\$ is \$V_{collector}/\sqrt{2}\$: $$\frac{V_{collector}R_L}{\sqrt{(R_C+R_L)^2+(\frac{1}{2\pi fc})^2}}=\frac{V_{collector}}{\sqrt{2}}$$ but when I simplify I get: $$f=\frac{1}{2\pi C_3\sqrt{R_L^2-R_C^2-2R_CR_L}}$$

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User1726, your start is not correct. The cut-off frequency of the first order highpass is defined as the frequency where the magnitude of Vout is 3dB less (factor 0.7071) than the MAXIMUM of Vout (which is Vout,max=Vcoll*RL/(RL+Rc).

There is a simpler method for finding the cut-off frequency. You have nothing to do than to find the time constant of the circuit (simple visual inspection) which is T=C3*(Rc+RL). The inverse of T is the (angular) frequency for cut-off.

And don´t overlook that the capacitive impedance is 1/2*Pi*f*C (and not 1/2*Pi*f)

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  • \$\begingroup\$ Missing parentheses: Indeed 1/(2*PifC). \$\endgroup\$ – Dirceu Rodrigues Jr Jul 16 '14 at 14:55
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Remember that

$$(A - B)^2 = A^2 - 2AB + B^2$$

This means that your last equation can be simplified to

$$f = \frac{1}{2 \pi C_3 \sqrt{(R_L - R_C)^2}}$$

$$f = \frac{1}{2 \pi C_3 (R_L - R_C)}$$

This means that you got a sign wrong somewhere along the way.

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  • \$\begingroup\$ I found my mistake in the sign, but it's still not right. \$\endgroup\$ – Phil S Jul 16 '14 at 12:19
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It turns out I understood the concept of cutoff frequencies wrong. The cutoff frequency should be \$1/\sqrt{2}\$ of the mid-band voltage (at which \$X_c\$ is zero). So the derivation should start with:

$$\frac{V_{collector}R_L}{\sqrt{(R_C+R_L)^2+(\frac{1}{2\pi fc})^2}}=\frac{V_{collector}R_L}{\sqrt{2}(R_C+R_L)^2}$$

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