# 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

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 $$f_L=\frac{1}{2\pi RC}$$

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

$$V_{in}=0.707V_S$$

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

• You forget the "static" gain $\frac{R_\text{in}}{R_\text{in}+R_\text{s}}$. That's your zero dB line which you shall use as a reference to find the -3dB frequency Commented Mar 2, 2017 at 7:35

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) .

• Sir, how can I model the transistor as a simple RC circuit looking from the bypass capacitor? I understand that there will be an equivalent resistance seen through $C_E$, but from where should I take the output in the equivalent RC circuit if the output is taken from the collector terminal in the original transistor circuit? imgur.com/EajgWO8 Commented Mar 2, 2017 at 13:12
• Soumee, do you speak about a transistor in common-base configuration? In this case, emitter E is input and collector C is output.
– LvW
Commented Mar 2, 2017 at 13:33
• No sir, I am talking about common emitter configuration(the one which is in the question figure, where $R_E$ has a bypass capacitor connected across it and the output is taken from the collector). Commented Mar 2, 2017 at 13:38
• But when you write "looking from the bypass capacitor" I suppose you speak about Cout in your circuit. And in this case, I am looking into the emitter - which means: I consider the emitter as input. Or what is the meaning of your question?
– LvW
Commented Mar 2, 2017 at 15:15
• Sir My question is if we want to model figure 3(I have edited the question figure), in terms of resistances and capacitance $C_E$ how should we do that(as we have done in figure 2 for fig -A). I am confused about this as I cannot understand from where should I take the output terminal in the equivalent RC circuit(For figure 3). Commented Mar 3, 2017 at 17:29