Low frequency response of a BJT amplifier.How is the cutoff frequency equal to \$ \frac{1}{2\pi (R_S+R_{in})C_{in}}\$ in fig 2

Question

In a low frequency region of the single stage BJT amplifier, it is the RC combinations formed by the capacitors \$ C_{in},C_E,C_{out} \$ -Electronic Devices and Circuit Theory-Boylestad

Considering the equivalent circuit formed as shown in fig.1

The output voltage and input voltage are related by:

$$\Bbb V_{out}=\frac{\Bbb R}{\Bbb R-j\Bbb X_C}\Bbb V_{in}$$ The magnitude is given by $$V_{out}=\frac{R}{\left[ R^2+X^2_C \right]^{1/2}}V_{in}$$ $$\text{When } X_C=R$$ $$V_{out}=0.707V_{in}$$ The frequency at which this occur, is given by the equation, $$R=X_C=\frac{1}{2\pi f_LC}$$ or, 3dB cutoff frequency $$f_L=\frac{1}{2\pi RC}$$

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In case the equivalent circuit formed by the BJT circuit is something like fig 2, which is the case when we are considering the input portion of the BJT circuit, the analysis is something like this:

$$\Bbb V_{in}=\frac{\Bbb R_{in}}{\Bbb R_{in}+\Bbb R_S-j\Bbb X_C}\Bbb V_{in}$$ The magnitude is given by $$V_{in}=\frac{R_{in}}{\left[ (R_{in}+R_S)^2+X^2_C \right]^{1/2}}V_{in}$$ $$\text{When } X_C=R_S+R_{in}$$ $$=>\frac{1}{2\pi f C_{in}}=R_{in}+R_S$$ $$=>f=\frac{1}{2\pi(R_{in}+R_S)C_{in}}$$

But for \$ X_C=R_S+R_{in} \$ $$V_{in}=0.707 \frac{R_{in}}{(R_S+R_{in})}V_{S}$$

Which is not $$V_{in}=0.707V_S$$

How is then the cutoff frequency equal to \$ \frac{1}{2\pi (R_S+R_{in})C_{in}}\$ in case of Figure 2 ?

Furthur,

For

$$V_{in}=0.707V_S$$

$$\frac{R_{in}}{\left[ (R_{in}+R_S)^2+X^2_C \right]^{1/2}}V_{in}=\frac{1}{2^{1/2}}$$ $$=>\frac{R_{in}}{(R_{in}+R_S)^2+X_C^2}=\frac{1}{2}$$ $$=>2R_{in}^2=(R_{in}+R_S)^2+X_C^2$$ $$=>X_C^2=(R_{in}-R_S)^2$$ $$=>X_c=R_{in}-R_S$$

Answer 1

The answer is simple: The 3dB cut-off (fc) is NOT defined for the frequency fc that gives Vin=0.707*Vs. Instead, it is defined for the frequency fc where the filters output is 3dB BELOW the maximum output.

For a simple R-C lowpass resp. C-R highpass this maximum is unity (for very low resp. very high frequencies). But in Fig.2 this maximum (for infinite frequencies) is Vin=Vs*Rin/(Rin+Rs) .