bjtcircuit analysiscutoff frequencyfilter
According to the book I'm studying, the cutoff frequency for this circuit
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:
$$f=\frac{1}{2\pi C_3\sqrt{R_L^2-R_C^2-2R_CR_L}}$$
The cut-off frequency is where the gain is 3dB below the passband gain, \$=K\$, say. Assuming your equation relates to voltage gain, -3dB equates to \$\dfrac{1}{\sqrt2}K\$
The equation represents a 2nd order low-pass filter and the passband gain is found when \$\omega=0\$. This gives a passband gain of \$K=10\$, hence the cut-off frequency can be determined by setting \$|H(j\omega)|=\dfrac{10}{\sqrt2}\$
simulate this circuit – Schematic created using CircuitLab
Applying KVL,
$$ V_{in} = V_{R} + V_{C} = I\cdot Z_C + I\cdot Z_R \\ \implies I = \frac{V_{in}}{Z_R+Z_C} $$
Setting \$f=\frac{1}{2\pi RC}\$, we have
$$ Z_C=\frac{1}{j2\pi fC}=\frac{1}{j2\pi \frac{1}{2\pi RC} C} = \frac{R}{j} = -jR $$
In the low-pass filter, the output voltage is taken across the capacitor:
$$ V_{out,lpf}=I\cdot Z_C = V_{in} \cdot \frac{Z_C}{Z_R+Z_C} = V_{in} \cdot \frac{-jR}{R-jR} = V_{in} \cdot \frac{-j}{1-j} \\ \implies \left|\frac{V_{out,lpf}}{V_{in}}\right| = \left|\frac{-j}{1-j}\right| = \frac{1}{\sqrt{1+1}} = \frac{1}{\sqrt{2}} $$
In the high-pass filter, the output voltage is taken across the resistor:
$$ V_{out,hpf}=I\cdot Z_R = V_{in} \cdot \frac{Z_R}{Z_R+Z_C} = V_{in} \cdot \frac{R}{R-jR} = V_{in} \cdot \frac{1}{1-j} \\ \implies \left|\frac{V_{out,hpf}}{V_{in}}\right| = \left|\frac{1}{1-j}\right| = \frac{1}{\sqrt{1+1}} = \frac{1}{\sqrt{2}} $$
Best Answer
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)